Commutative Algebra

Introduction

Basic commutative algebra will be explained in this document. A reference is [MatCA].

Conventions

A ring is commutative with \(1\). The zero ring is a ring. In fact it is the only ring that does not have a prime ideal. The Kronecker symbol \(\delta_{ij}\) will be used. If \(R \to S\) is a ring map and \(\mathfrak q\) a prime of \(S\), then we use the notation “\(\mathfrak p = R \cap \mathfrak q\)” to indicate the prime which is the inverse image of \(\mathfrak q\) under \(R \to S\) even if \(R\) is not a subring of \(S\) and even if \(R \to S\) is not injective.

Basic notions

The following is a list of basic notions in commutative algebra. Some of these notions are discussed in more detail in the text that follows and some are defined in the list, but others are considered basic and will not be defined. If you are not familiar with most of the italicized concepts, then we suggest looking at an introductory text on algebra before continuing.

1. \(R\) is a ring, [00AS]

2. \(x\in R\) is nilpotent, [00AT]

3. \(x\in R\) is a zerodivisor, [00AU]

4. \(x\in R\) is a unit, [00AV]

5. \(e \in R\) is an idempotent, [00AW]

6. an idempotent \(e \in R\) is called trivial if \(e = 1\) or \(e = 0\), [00AX]

7. \(\varphi : R_1 \to R_2\) is a ring homomorphism, [00AY]

8. \(\varphi : R_1 \to R_2\) is of finite presentation, or \(R_2\) is a finitely presented \(R_1\)-algebra, see Definition 00F3,

9. \(\varphi : R_1 \to R_2\) is of finite type, or \(R_2\) is a finite type \(R_1\)-algebra, see Definition 00F3,

10. \(\varphi : R_1 \to R_2\) is finite, or \(R_2\) is a finite \(R_1\)-algebra,

11. \(R\) is a (integral) domain, [00B2]

12. \(R\) is reduced, [00B3]

13. \(R\) is Noetherian, [00B4]

14. \(R\) is a principal ideal domain or a PID, [00B5]

15. \(R\) is a Euclidean domain, [00B6]

16. \(R\) is a unique factorization domain or a UFD, [00B7]

17. \(R\) is a discrete valuation ring or a dvr, [00B8]

18. \(K\) is a field, [00B9]

19. \(L/K\) is a field extension, [00BA]

20. \(L/K\) is an algebraic field extension, [00BB]

21. \(\{t_i\}_{i\in I}\) is a transcendence basis for \(L\) over \(K\), [00BC]

22. the transcendence degree \(\text{trdeg}(L/K)\) of \(L\) over \(K\), [00BD]

23. the field \(k\) is algebraically closed, [00BE]

24. if \(L/K\) is algebraic, and \(\Omega/K\) an extension with \(\Omega\) algebraically closed, then there exists a ring map \(L \to \Omega\) extending the map on \(K\),

25. \(I \subset R\) is an ideal, [00BG]

26. \(I \subset R\) is radical, [00BH]

27. if \(I\) is an ideal then we have its radical \(\sqrt{I}\), [00BI]

28. \(I \subset R\) is nilpotent means that \(I^n = 0\) for some \(n \in \mathbf{N}\),

29. \(I \subset R\) is locally nilpotent means that every element of \(I\) is nilpotent,

30. \(\mathfrak p \subset R\) is a prime ideal, [00BK]

31. if \(\mathfrak p \subset R\) is prime and if \(I, J \subset R\) are ideal, and if \(IJ\subset \mathfrak p\), then \(I \subset \mathfrak p\) or \(J \subset \mathfrak p\).

32. \(\mathfrak m \subset R\) is a maximal ideal, [00BM]

33. any nonzero ring has a maximal ideal, [00BN]

34. the Jacobson radical of \(R\) is \(\text{rad}(R) = \bigcap_{\mathfrak m \subset R} \mathfrak m\) the intersection of all the maximal ideals of \(R\),

35. the ideal \((T)\) generated by a subset \(T \subset R\), [00BP]

36. the quotient ring \(R/I\), [00BQ]

37. an ideal \(I\) in the ring \(R\) is prime if and only if \(R/I\) is a domain, [00BR]

38. an ideal \(I\) in the ring \(R\) is maximal if and only if the ring \(R/I\) is a field,

39. if \(\varphi : R_1 \to R_2\) is a ring homomorphism, and if \(I \subset R_2\) is an ideal, then \(\varphi^{-1}(I)\) is an ideal of \(R_1\),

40. if \(\varphi : R_1 \to R_2\) is a ring homomorphism, and if \(I \subset R_1\) is an ideal, then \(\varphi(I) \cdot R_2\) (sometimes denoted \(I \cdot R_2\), or \(IR_2\)) is the ideal of \(R_2\) generated by \(\varphi(I)\),

41. if \(\varphi : R_1 \to R_2\) is a ring homomorphism, and if \(\mathfrak p \subset R_2\) is a prime ideal, then \(\varphi^{-1}(\mathfrak p)\) is a prime ideal of \(R_1\),

42. \(M\) is an \(R\)-module, [00BW]

43. for \(m \in M\) the annihilator \(I = \{f \in R \mid fm = 0\}\) of \(m\) in \(R\),

44. \(N \subset M\) is an \(R\)-submodule, [00BX]

45. \(M\) is a Noetherian \(R\)-module, [00BY]

46. \(M\) is a finite \(R\)-module, [00BZ]

47. \(M\) is a finitely generated \(R\)-module, [00C0]

48. \(M\) is a finitely presented \(R\)-module, [00C1]

49. \(M\) is a free \(R\)-module, [00C2]

50. if \(0 \to K \to L \to M \to 0\) is a short exact sequence of \(R\)-modules and \(K\), \(M\) are free, then \(L\) is free,

51. if \(N \subset M \subset L\) are \(R\)-modules, then \(L/M = (L/N)/(M/N)\), [00C3]

52. \(S\) is a multiplicative subset of \(R\), [00C4]

53. the localization \(R \to S^{-1}R\) of \(R\), [00C5]

54. if \(R\) is a ring and \(S\) is a multiplicative subset of \(R\) then \(S^{-1}R\) is the zero ring if and only if \(S\) contains \(0\),

55. if \(R\) is a ring and if the multiplicative subset \(S\) consists completely of nonzerodivisors, then \(R \to S^{-1}R\) is injective,

56. if \(\varphi : R_1 \to R_2\) is a ring homomorphism, and \(S\) is a multiplicative subset of \(R_1\), then \(\varphi(S)\) is a multiplicative subset of \(R_2\),

57. if \(S\), \(S'\) are multiplicative subsets of \(R\), and if \(SS'\) denotes the set of products \(SS' = \{r \in R \mid \exists s\in S, \exists s' \in S', r = ss'\}\) then \(SS'\) is a multiplicative subset of \(R\),

58. if \(S\), \(S'\) are multiplicative subsets of \(R\), and if \(\overline{S}\) denotes the image of \(S\) in \((S')^{-1}R\), then \((SS')^{-1}R = \overline{S}^{-1}((S')^{-1}R)\),

59. the localization \(S^{-1}M\) of the \(R\)-module \(M\), [00CA]

60. the functor \(M \mapsto S^{-1}M\) preserves injective maps, surjective maps, and exactness,

61. if \(S\), \(S'\) are multiplicative subsets of \(R\), and if \(M\) is an \(R\)-module, then \((SS')^{-1}M = S^{-1}((S')^{-1}M)\),

62. if \(R\) is a ring, \(I\) an ideal of \(R\), and \(S\) a multiplicative subset of \(R\), then \(S^{-1}I\) is an ideal of \(S^{-1}R\), and we have \(S^{-1}R/S^{-1}I = \overline{S}^{-1}(R/I)\), where \(\overline{S}\) is the image of \(S\) in \(R/I\),

63. if \(R\) is a ring, and \(S\) a multiplicative subset of \(R\), then any ideal \(I'\) of \(S^{-1}R\) is of the form \(S^{-1}I\), where one can take \(I\) to be the inverse image of \(I'\) in \(R\),

64. if \(R\) is a ring, \(M\) an \(R\)-module, and \(S\) a multiplicative subset of \(R\), then any submodule \(N'\) of \(S^{-1}M\) is of the form \(S^{-1}N\) for some submodule \(N \subset M\), where one can take \(N\) to be the inverse image of \(N'\) in \(M\),

65. if \(S = \{1, f, f^2, \ldots\}\) then \(R_f = S^{-1}R\) and \(M_f = S^{-1}M\), [00CG]

66. if \(S = R \setminus \mathfrak p = \{x\in R \mid x\not\in \mathfrak p\}\) for some prime ideal \(\mathfrak p\), then it is customary to denote \(R_{\mathfrak p} = S^{-1}R\) and \(M_{\mathfrak p} = S^{-1}M\),

67. a local ring is a ring with exactly one maximal ideal, [00CI]

68. a semi-local ring is a ring with finitely many maximal ideals, [03C0]

69. if \(\mathfrak p\) is a prime in \(R\), then \(R_{\mathfrak p}\) is a local ring with maximal ideal \(\mathfrak p R_{\mathfrak p}\),

70. the residue field, denoted \(\kappa(\mathfrak p)\), of the prime \(\mathfrak p\) in the ring \(R\) is the field of fractions of the domain \(R/\mathfrak p\); it is equal to \(R_\mathfrak p/\mathfrak pR_\mathfrak p = (R \setminus \mathfrak p)^{-1}R/\mathfrak p\),

71. given \(R\) and \(M_1\), \(M_2\) the tensor product \(M_1 \otimes_R M_2\), [00CL]

72. given matrices \(A\) and \(B\) in a ring \(R\) of sizes \(m \times n\) and \(n \times m\) we have \(\det(AB) = \sum \det(A_S)\det({}_SB)\) in \(R\) where the sum is over subsets \(S \subset \{1, \ldots, n\}\) of size \(m\) and \(A_S\) is the \(m \times m\) submatrix of \(A\) with columns corresponding to \(S\) and \({}_SB\) is the \(m \times m\) submatrix of \(B\) with rows corresponding to \(S\),

73. etc.

Snake lemma

The snake lemma and its variants are discussed in the setting of abelian categories in Homology, Section 00ZX.

Lemma

Given a commutative diagram \[\xymatrix{ & X \ar[r] \ar[d]^\alpha & Y \ar[r] \ar[d]^\beta & Z \ar[r] \ar[d]^\gamma & 0 \\ 0 \ar[r] & U \ar[r] & V \ar[r] & W }\] of abelian groups with exact rows, there is a canonical exact sequence \[\Ker(\alpha) \to \Ker(\beta) \to \Ker(\gamma) \to \Coker(\alpha) \to \Coker(\beta) \to \Coker(\gamma)\] Moreover: if \(X \to Y\) is injective, then the first map is injective; if \(V \to W\) is surjective, then the last map is surjective.

Proof

The map \(\partial : \Ker(\gamma) \to \Coker(\alpha)\) is defined as follows. Take \(z \in \Ker(\gamma)\). Choose \(y \in Y\) mapping to \(z\). Then \(\beta(y) \in V\) maps to zero in \(W\). Hence \(\beta(y)\) is the image of some \(u \in U\). Set \(\partial z = \overline{u}\), the class of \(u\) in the cokernel of \(\alpha\). Proof of exactness is omitted.

Finite modules and finitely presented modules

Just some basic notation and lemmas.

Definition

Let \(R\) be a ring. Let \(M\) be an \(R\)-module.

1. We say \(M\) is a finite \(R\)-module, or a finitely generated \(R\)-module if there exist \(n \in \mathbf{N}\) and \(x_1, \ldots, x_n \in M\) such that every element of \(M\) is an \(R\)-linear combination of the \(x_i\). Equivalently, this means there exists a surjection \(R^{\oplus n} \to M\) for some \(n \in \mathbf{N}\).

2. We say \(M\) is a finitely presented \(R\)-module or an \(R\)-module of finite presentation if there exist integers \(n, m \in \mathbf{N}\) and an exact sequence \[R^{\oplus m} \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0\]

Informally, \(M\) is a finitely presented \(R\)-module if and only if it is finitely generated and the module of relations among these generators is finitely generated as well. A choice of an exact sequence as in the definition is called a presentation of \(M\).

Lemma

Let \(R\) be a ring. Let \(\alpha : R^{\oplus n} \to M\) and \(\beta : N \to M\) be module maps. If \(\Im(\alpha) \subset \Im(\beta)\), then there exists an \(R\)-module map \(\gamma : R^{\oplus n} \to N\) such that \(\alpha = \beta \circ \gamma\).

Proof

Let \(e_i = (0, \ldots, 0, 1, 0, \ldots, 0)\) be the \(i\)th basis vector of \(R^{\oplus n}\). Let \(x_i \in N\) be an element with \(\alpha(e_i) = \beta(x_i)\) which exists by assumption. Set \(\gamma(a_1, \ldots, a_n) = \sum a_i x_i\). By construction \(\alpha = \beta \circ \gamma\).

Lemma

Let \(R\) be a ring. Let \[0 \to M_1 \to M_2 \to M_3 \to 0\] be a short exact sequence of \(R\)-modules.

1. If \(M_1\) and \(M_3\) are finite \(R\)-modules, then \(M_2\) is a finite \(R\)-module.

2. If \(M_1\) and \(M_3\) are finitely presented \(R\)-modules, then \(M_2\) is a finitely presented \(R\)-module.

3. If \(M_2\) is a finite \(R\)-module, then \(M_3\) is a finite \(R\)-module.

4. If \(M_2\) is a finitely presented \(R\)-module and \(M_1\) is a finite \(R\)-module, then \(M_3\) is a finitely presented \(R\)-module.

5. If \(M_3\) is a finitely presented \(R\)-module and \(M_2\) is a finite \(R\)-module, then \(M_1\) is a finite \(R\)-module.

Proof

Proof of (1). If \(x_1, \ldots, x_n\) are generators of \(M_1\) and \(y_1, \ldots, y_m \in M_2\) are elements whose images in \(M_3\) are generators of \(M_3\), then \(x_1, \ldots, x_n, y_1, \ldots, y_m\) generate \(M_2\).

Part (3) is immediate from the definition.

Proof of (5). Assume \(M_3\) is finitely presented and \(M_2\) finite. Choose a presentation \[R^{\oplus m} \to R^{\oplus n} \to M_3 \to 0\] By Lemma 07JX there exists a map \(R^{\oplus n} \to M_2\) such that the solid diagram \[\xymatrix{ & R^{\oplus m} \ar[r] \ar@{..>}[d] & R^{\oplus n} \ar[r] \ar[d] & M_3 \ar[r] \ar[d]^{\text{id}} & 0 \\ 0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0 }\] commutes. This produces the dotted arrow. By the snake lemma (Lemma 07JW) we see that we get an isomorphism \[\Coker(R^{\oplus m} \to M_1) \cong \Coker(R^{\oplus n} \to M_2)\] In particular we conclude that \(\Coker(R^{\oplus m} \to M_1)\) is a finite \(R\)-module. Since \(\Im(R^{\oplus m} \to M_1)\) is finite by (3), we see that \(M_1\) is finite by part (1).

Proof of (4). Assume \(M_2\) is finitely presented and \(M_1\) is finite. Choose a presentation \(R^{\oplus m} \to R^{\oplus n} \to M_2 \to 0\). Choose a surjection \(R^{\oplus k} \to M_1\). By Lemma 07JX there exists a factorization \(R^{\oplus k} \to R^{\oplus n} \to M_2\) of the composition \(R^{\oplus k} \to M_1 \to M_2\). Then \(R^{\oplus k + m} \to R^{\oplus n} \to M_3 \to 0\) is a presentation.

Proof of (2). Assume that \(M_1\) and \(M_3\) are finitely presented. The argument in the proof of part (1) produces a commutative diagram \[\xymatrix{ 0 \ar[r] & R^{\oplus n} \ar[d] \ar[r] & R^{\oplus n + m} \ar[d] \ar[r] & R^{\oplus m} \ar[d] \ar[r] & 0 \\ 0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0 }\] with surjective vertical arrows. By the snake lemma we obtain a short exact sequence \[0 \to \Ker(R^{\oplus n} \to M_1) \to \Ker(R^{\oplus n + m} \to M_2) \to \Ker(R^{\oplus m} \to M_3) \to 0\] By part (5) we see that the outer two modules are finite. Hence the middle one is finite too. By (4) we see that \(M_2\) is of finite presentation.

Lemma

Let \(R\) be a ring, and let \(M\) be a finite \(R\)-module. There exists a filtration by finite \(R\)-submodules \[0 = M_0 \subset M_1 \subset \ldots \subset M_n = M\] such that each quotient \(M_i/M_{i - 1}\) is isomorphic to \(R/I_i\) for some ideal \(I_i\) of \(R\).

Proof

By induction on the number of generators of \(M\). Let \(x_1, \ldots, x_r \in M\) be generators. Let \(M' = Rx_1 \subset M\). Then \(M/M'\) has \(r - 1\) generators and the induction hypothesis applies. And clearly \(M' \cong R/I_1\) with \(I_1 = \{f \in R \mid fx_1 = 0\}\).

Lemma

Let \(R \to S\) be a ring map. Let \(M\) be an \(S\)-module. If \(M\) is finite as an \(R\)-module, then \(M\) is finite as an \(S\)-module.

Proof

In fact, any \(R\)-generating set of \(M\) is also an \(S\)-generating set of \(M\), since the \(R\)-module structure is induced by the image of \(R\) in \(S\).

Ring maps of finite type and of finite presentation

Definition

Let \(R \to S\) be a ring map.

1. We say \(R \to S\) is of finite type, or that \(S\) is a finite type \(R\)-algebra if there exist an \(n \in \mathbf{N}\) and a surjection of \(R\)-algebras \(R[x_1, \ldots, x_n] \to S\).

2. We say \(R \to S\) is of finite presentation if there exist integers \(n, m \in \mathbf{N}\) and polynomials \(f_1, \ldots, f_m \in R[x_1, \ldots, x_n]\) and an isomorphism of \(R\)-algebras \(R[x_1, \ldots, x_n]/(f_1, \ldots, f_m) \cong S\).

Informally, \(R \to S\) is of finite presentation if and only if \(S\) is finitely generated as an \(R\)-algebra and the ideal of relations among the generators is finitely generated. A choice of a surjection \(R[x_1, \ldots, x_n] \to S\) as in the definition is sometimes called a presentation of \(S\).

Lemma

The notions finite type and finite presentation have the following permanence properties.

1. A composition of ring maps of finite type is of finite type.

2. A composition of ring maps of finite presentation is of finite presentation.

3. Given \(R \to S' \to S\) with \(R \to S\) of finite type, then \(S' \to S\) is of finite type.

4. Given \(R \to S' \to S\), with \(R \to S\) of finite presentation, and \(R \to S'\) of finite type, then \(S' \to S\) is of finite presentation.

Proof

We only prove the last assertion. Write \(S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)\) and \(S' = R[y_1, \ldots, y_a]/I\). Say that the class \(\bar y_i\) of \(y_i\) maps to \(h_i \bmod (f_1, \ldots, f_m)\) in \(S\). Then it is clear that \(S = S'[x_1, \ldots, x_n]/(f_1, \ldots, f_m, h_1 - \bar y_1, \ldots, h_a - \bar y_a)\).

Lemma

Let \(R \to S\) be a ring map of finite presentation. For any surjection \(\alpha : R[x_1, \ldots, x_n] \to S\) the kernel of \(\alpha\) is a finitely generated ideal in \(R[x_1, \ldots, x_n]\).

Proof

Write \(S = R[y_1, \ldots, y_m]/(f_1, \ldots, f_k)\). Choose \(g_i \in R[y_1, \ldots, y_m]\) which are lifts of \(\alpha(x_i)\). Then we see that \(S = R[x_i, y_j]/(f_l, x_i - g_i)\). Choose \(h_j \in R[x_1, \ldots, x_n]\) such that \(\alpha(h_j)\) corresponds to \(y_j \bmod (f_1, \ldots, f_k)\). Consider the map \(\psi : R[x_i, y_j] \to R[x_i]\), \(x_i \mapsto x_i\), \(y_j \mapsto h_j\). Then the kernel of \(\alpha\) is the image of \((f_l, x_i - g_i)\) under \(\psi\) and we win.

Lemma

Let \(R \to S\) be a ring map. Let \(M\) be an \(S\)-module. Assume \(R \to S\) is of finite type and \(M\) is finitely presented as an \(R\)-module. Then \(M\) is finitely presented as an \(S\)-module.

Proof

This is similar to the proof of part (4) of Lemma 00F4. We may assume \(S = R[x_1, \ldots, x_n]/J\). Choose \(y_1, \ldots, y_m \in M\) which generate \(M\) as an \(R\)-module and choose relations \(\sum a_{ij} y_j = 0\), \(i = 1, \ldots, t\) which generate the kernel of \(R^{\oplus m} \to M\). For any \(i = 1, \ldots, n\) and \(j = 1, \ldots, m\) write \[x_i y_j = \sum a_{ijk} y_k\] for some \(a_{ijk} \in R\). Consider the \(S\)-module \(N\) generated by \(y_1, \ldots, y_m\) subject to the relations \(\sum a_{ij} y_j = 0\), \(i = 1, \ldots, t\) and \(x_i y_j = \sum a_{ijk} y_k\), \(i = 1, \ldots, n\) and \(j = 1, \ldots, m\). Then \(N\) has a presentation \[S^{\oplus nm + t} \longrightarrow S^{\oplus m} \longrightarrow N \longrightarrow 0\] By construction there is a surjective map \(\varphi : N \to M\). To finish the proof we show \(\varphi\) is injective. Suppose \(z = \sum b_j y_j \in N\) for some \(b_j \in S\). We may think of \(b_j\) as a polynomial in \(x_1, \ldots, x_n\) with coefficients in \(R\). By applying the relations of the form \(x_i y_j = \sum a_{ijk} y_k\) we can inductively lower the degree of the polynomials. Hence we see that \(z = \sum c_j y_j\) for some \(c_j \in R\). Hence if \(\varphi(z) = 0\) then the vector \((c_1, \ldots, c_m)\) is an \(R\)-linear combination of the vectors \((a_{i1}, \ldots, a_{im})\) and we conclude that \(z = 0\) as desired.

Finite ring maps

Here is the definition.

Definition

Let \(\varphi : R \to S\) be a ring map. We say \(\varphi : R \to S\) is finite if \(S\) is finite as an \(R\)-module.

Lemma

Let \(R \to S\) be a finite ring map. Let \(M\) be an \(S\)-module. Then \(M\) is finite as an \(R\)-module if and only if \(M\) is finite as an \(S\)-module.

Proof

One of the implications follows from Lemma 0560. To see the other assume that \(M\) is finite as an \(S\)-module. Pick \(x_1, \ldots, x_n \in S\) which generate \(S\) as an \(R\)-module. Pick \(y_1, \ldots, y_m \in M\) which generate \(M\) as an \(S\)-module. Then \(x_i y_j\) generate \(M\) as an \(R\)-module.

Lemma

Suppose that \(R \to S\) and \(S \to T\) are finite ring maps. Then \(R \to T\) is finite.

Proof

If \(t_i\) generate \(T\) as an \(S\)-module and \(s_j\) generate \(S\) as an \(R\)-module, then \(t_i s_j\) generate \(T\) as an \(R\)-module. (Also follows from Lemma 00GJ.)

Lemma

Let \(\varphi : R \to S\) be a ring map.

1. If \(\varphi\) is finite, then \(\varphi\) is of finite type.

2. If \(S\) is of finite presentation as an \(R\)-module, then \(\varphi\) is of finite presentation.

Proof

For (1) if \(x_1, \ldots, x_n \in S\) generate \(S\) as an \(R\)-module, then \(x_1, \ldots, x_n\) generate \(S\) as an \(R\)-algebra. For (2), suppose that \(\sum r_j^ix_i = 0\), \(j = 1, \ldots, m\) is a set of generators of the relations among the \(x_i\) when viewed as \(R\)-module generators of \(S\). Furthermore, write \(1 = \sum r_ix_i\) for some \(r_i \in R\) and \(x_ix_j = \sum r_{ij}^k x_k\) for some \(r_{ij}^k \in R\). Then \[S = R[t_1, \ldots, t_n]/ (\sum r_j^it_i,\ 1 - \sum r_it_i,\ t_it_j - \sum r_{ij}^k t_k)\] as an \(R\)-algebra which proves (2).

For more information on finite ring maps, please see Section 00GH.

Colimits

Some of the material in this section overlaps with the general discussion on colimits in Categories, Sections 002D – 002Z. The notion of a preordered set is defined in Categories, Definition 00D3. It is a slightly weaker notion than a partially ordered set.

Definition

Let \((I, \leq)\) be a preordered set. A system \((M_i, \mu_{ij})\) of \(R\)-modules over \(I\) consists of a family of \(R\)-modules \(\{M_i\}_{i\in I}\) indexed by \(I\) and a family of \(R\)-module maps \(\{\mu_{ij} : M_i \to M_j\}_{i \leq j}\) such that for all \(i \leq j \leq k\) \[\mu_{ii} = \text{id}_{M_i}\quad \mu_{ik} = \mu_{jk}\circ \mu_{ij}\] We say \((M_i, \mu_{ij})\) is a directed system if \(I\) is a directed set.

This is the same as the notion defined in Categories, Definition 0030 and Section 002Z. We refer to Categories, Definition 002F for the definition of a colimit of a diagram/system in any category.

Lemma

Let \((M_i, \mu_{ij})\) be a system of \(R\)-modules over the preordered set \(I\). The colimit of the system \((M_i, \mu_{ij})\) is the quotient \(R\)-module \((\bigoplus_{i\in I} M_i) /Q\) where \(Q\) is the \(R\)-submodule generated by all elements \[\iota_i(x_i) - \iota_j(\mu_{ij}(x_i))\] where \(\iota_i : M_i \to \bigoplus_{i\in I} M_i\) is the natural inclusion. We denote the colimit \(M = \colim_i M_i\). We denote \(\pi : \bigoplus_{i\in I} M_i \to M\) the projection map and \(\phi_i = \pi \circ \iota_i : M_i \to M\).

Proof

This lemma is a special case of Categories, Lemma 002P but we will also prove it directly in this case. Namely, note that \(\phi_i = \phi_j\circ \mu_{ij}\) in the above construction. To show the pair \((M, \phi_i)\) is the colimit we have to show it satisfies the universal property: for any other such pair \((Y, \psi_i)\) with \(\psi_i : M_i \to Y\), \(\psi_i = \psi_j\circ \mu_{ij}\), there is a unique \(R\)-module homomorphism \(g : M \to Y\) such that the following diagram commutes: \[\xymatrix{ M_i \ar[rr]^{\mu_{ij}} \ar[dr]^{\phi_i} \ar[ddr]_{\psi_i} & & M_j\ar[dl]_{\phi_j} \ar[ddl]^{\psi_j} \\ & M \ar[d]^{g}\\ & Y }\] And this is clear because we can define \(g\) by taking the map \(\psi_i\) on the summand \(M_i\) in the direct sum \(\bigoplus M_i\).

Lemma

Let \((M_i, \mu_{ij})\) be a system of \(R\)-modules over the preordered set \(I\). Assume that \(I\) is directed. The colimit of the system \((M_i, \mu_{ij})\) is canonically isomorphic to the module \(M\) defined as follows:

1. as a set let \[M = \left(\coprod\nolimits_{i \in I} M_i\right)/\sim\] where for \(m \in M_i\) and \(m' \in M_{i'}\) we have \[m \sim m' \Leftrightarrow \mu_{ij}(m) = \mu_{i'j}(m')\text{ for some }j \geq i, i'\]

2. as an abelian group for \(m \in M_i\) and \(m' \in M_{i'}\) we define the sum of the classes of \(m\) and \(m'\) in \(M\) to be the class of \(\mu_{ij}(m) + \mu_{i'j}(m')\) where \(j \in I\) is any index with \(i \leq j\) and \(i' \leq j\), and

3. as an \(R\)-module define for \(m \in M_i\) and \(x \in R\) the product of \(x\) and the class of \(m\) in \(M\) to be the class of \(xm\) in \(M\).

The canonical maps \(\phi_i : M_i \to M\) are induced by the canonical maps \(M_i \to \coprod_{i \in I} M_i\).

Proof

Omitted. Compare with Categories, Section 04AX.

Lemma

Let \((M_i, \mu_{ij})\) be a directed system. Let \(M = \colim M_i\) with \(\mu_i : M_i \to M\). Then, \(\mu_i(x_i) = 0\) for \(x_i \in M_i\) if and only if there exists \(j \geq i\) such that \(\mu_{ij}(x_i) = 0\).

Proof

This is clear from the description of the directed colimit in Lemma 00D6.

Example

Consider the partially ordered set \(I = \{a, b, c\}\) with \(a < b\) and \(a < c\) and no other strict inequalities. A system \((M_a, M_b, M_c, \mu_{ab}, \mu_{ac})\) over \(I\) consists of three \(R\)-modules \(M_a, M_b, M_c\) and two \(R\)-module homomorphisms \(\mu_{ab} : M_a \to M_b\) and \(\mu_{ac} : M_a \to M_c\). The colimit of the system is just \[M := \colim_{i \in I} M_i = \Coker(M_a \to M_b \oplus M_c)\] where the map is \(\mu_{ab} \oplus -\mu_{ac}\). Thus the kernel of the canonical map \(M_a \to M\) is \(\Ker(\mu_{ab}) + \Ker(\mu_{ac})\). And the kernel of the canonical map \(M_b \to M\) is the image of \(\Ker(\mu_{ac})\) under the map \(\mu_{ab}\). Hence clearly the result of Lemma 00D7 is false for general systems.

Definition

Let \((M_i, \mu_{ij})\), \((N_i, \nu_{ij})\) be systems of \(R\)-modules over the same preordered set \(I\). A homomorphism of systems \(\Phi\) from \((M_i, \mu_{ij})\) to \((N_i, \nu_{ij})\) is by definition a family of \(R\)-module homomorphisms \(\phi_i : M_i \to N_i\) such that \(\phi_j \circ \mu_{ij} = \nu_{ij} \circ \phi_i\) for all \(i \leq j\).

This is the same notion as a transformation of functors between the associated diagrams \(M : I \to \text{Mod}_R\) and \(N : I \to \text{Mod}_R\), in the language of categories. The following lemma is a special case of Categories, Lemma 002K.

Lemma

Let \((M_i, \mu_{ij})\), \((N_i, \nu_{ij})\) be systems of \(R\)-modules over the same preordered set. A morphism of systems \(\Phi = (\phi_i)\) from \((M_i, \mu_{ij})\) to \((N_i, \nu_{ij})\) induces a unique homomorphism \[\colim \phi_i : \colim M_i \longrightarrow \colim N_i\] such that \[\xymatrix{ M_i \ar[r] \ar[d]_{\phi_i} & \colim M_i \ar[d]^{\colim \phi_i} \\ N_i \ar[r] & \colim N_i }\] commutes for all \(i \in I\).

Proof

Write \(M = \colim M_i\) and \(N = \colim N_i\) and \(\phi = \colim \phi_i\) (as yet to be constructed). We will use the explicit description of \(M\) and \(N\) in Lemma 00D5 without further mention. The condition of the lemma is equivalent to the condition that \[\xymatrix{ \bigoplus_{i\in I} M_i \ar[r] \ar[d]_{\bigoplus\phi_i} & M \ar[d]^\phi \\ \bigoplus_{i\in I} N_i \ar[r] & N }\] commutes. Hence it is clear that if \(\phi\) exists, then it is unique. To see that \(\phi\) exists, it suffices to show that the kernel of the upper horizontal arrow is mapped by \(\bigoplus \phi_i\) to the kernel of the lower horizontal arrow. To see this, let \(j \leq k\) and \(x_j \in M_j\). Then \[(\bigoplus \phi_i)(x_j - \mu_{jk}(x_j)) = \phi_j(x_j) - \phi_k(\mu_{jk}(x_j)) = \phi_j(x_j) - \nu_{jk}(\phi_j(x_j))\] which is in the kernel of the lower horizontal arrow as required.

Lemma

Let \(I\) be a directed set. Let \((L_i, \lambda_{ij})\), \((M_i, \mu_{ij})\), and \((N_i, \nu_{ij})\) be systems of \(R\)-modules over \(I\). Let \(\varphi_i : L_i \to M_i\) and \(\psi_i : M_i \to N_i\) be morphisms of systems over \(I\). Assume that for all \(i \in I\) the sequence of \(R\)-modules \[\xymatrix{ L_i \ar[r]^{\varphi_i} & M_i \ar[r]^{\psi_i} & N_i }\] is a complex with homology \(H_i\). Then the \(R\)-modules \(H_i\) form a system over \(I\), the sequence of \(R\)-modules \[\xymatrix{ \colim_i L_i \ar[r]^\varphi & \colim_i M_i \ar[r]^\psi & \colim_i N_i }\] is a complex as well, and denoting \(H\) its homology we have \[H = \colim_i H_i.\]

Proof

It is clear that \(\xymatrix{ \colim_i L_i \ar[r]^\varphi & \colim_i M_i \ar[r]^\psi & \colim_i N_i }\) is a complex. For each \(i \in I\), there is a canonical \(R\)-module morphism \(H_i \to H\) (sending each \([m] \in H_i = \Ker(\psi_i) / \Im(\varphi_i)\) to the residue class in \(H = \Ker(\psi) / \Im(\varphi)\) of the image of \(m\) in \(\colim_i M_i\)). These give rise to a morphism \(\colim_i H_i \to H\). It remains to show that this morphism is surjective and injective.

We are going to repeatedly use the description of colimits over \(I\) as in Lemma 00D6 without further mention. Let \(h \in H\). Since \(H = \Ker(\psi)/\Im(\varphi)\) we see that \(h\) is the class mod \(\Im(\varphi)\) of an element \([m]\) in \(\Ker(\psi) \subset \colim_i M_i\). Choose an \(i\) such that \([m]\) comes from an element \(m \in M_i\). Choose a \(j \geq i\) such that \(\nu_{ij}(\psi_i(m)) = 0\) which is possible since \([m] \in \Ker(\psi)\). After replacing \(i\) by \(j\) and \(m\) by \(\mu_{ij}(m)\) we see that we may assume \(m \in \Ker(\psi_i)\). This shows that the map \(\colim_i H_i \to H\) is surjective.

Suppose that \(h_i \in H_i\) has image zero in \(H\). Since \(H_i = \Ker(\psi_i)/\Im(\varphi_i)\) we may represent \(h_i\) by an element \(m \in \Ker(\psi_i) \subset M_i\). The assumption on the vanishing of \(h_i\) in \(H\) means that the class of \(m\) in \(\colim_i M_i\) lies in the image of \(\varphi\). Hence there exists a \(j \geq i\) and an \(l \in L_j\) such that \(\varphi_j(l) = \mu_{ij}(m)\). Clearly this shows that the image of \(h_i\) in \(H_j\) is zero. This proves the injectivity of \(\colim_i H_i \to H\).

Example

Taking colimits is not exact in general. Consider the partially ordered set \(I = \{a, b, c\}\) with \(a < b\) and \(a < c\) and no other strict inequalities, as in Example 00D8. Consider the map of systems \((0, \mathbf{Z}, \mathbf{Z}, 0, 0) \to (\mathbf{Z}, \mathbf{Z}, \mathbf{Z}, 1, 1)\). From the description of the colimit in Example 00D8 we see that the associated map of colimits is not injective, even though the map of systems is injective on each object. Hence the result of Lemma 00DB is false for general systems.

Lemma

Let \(\mathcal{I}\) be an index category satisfying the assumptions of Categories, Lemma 002X. Then taking colimits of diagrams of abelian groups over \(\mathcal{I}\) is exact (i.e., the analogue of Lemma 00DB holds in this situation).

Proof

By Categories, Lemma 002X we may write \(\mathcal{I} = \coprod_{j \in J} \mathcal{I}_j\) with each \(\mathcal{I}_j\) a filtered category, and \(J\) possibly empty. By Categories, Lemma 0032 taking colimits over the index categories \(\mathcal{I}_j\) is the same as taking the colimit over some directed set. Hence Lemma 00DB applies to these colimits. This reduces the problem to showing that coproducts in the category of \(R\)-modules over the set \(J\) are exact. In other words, exact sequences \(L_j \to M_j \to N_j\) of \(R\) modules we have to show that \[\bigoplus\nolimits_{j \in J} L_j \longrightarrow \bigoplus\nolimits_{j \in J} M_j \longrightarrow \bigoplus\nolimits_{j \in J} N_j\] is exact. This can be verified by hand, and holds even if \(J\) is empty.

Localization

Definition

Let \(R\) be a ring, \(S\) a subset of \(R\). We say \(S\) is a multiplicative subset of \(R\) if \(1\in S\) and \(S\) is closed under multiplication, i.e., \(s, s' \in S \Rightarrow ss' \in S\).

Given a ring \(A\) and a multiplicative subset \(S\), we define a relation on \(A \times S\) as follows: \[(x, s) \sim (y, t) \Leftrightarrow \exists u \in S \text{ such that } (xt-ys)u = 0\] It is easily checked that this is an equivalence relation. Let \(x/s\) (or \(\frac{x}{s}\)) be the equivalence class of \((x, s)\) and \(S^{-1}A\) be the set of all equivalence classes. Define addition and multiplication in \(S^{-1}A\) as follows: \[x/s + y/t = (xt + ys)/st, \quad x/s \cdot y/t = xy/st\] One can check that \(S^{-1}A\) becomes a ring under these operations.

Definition

This ring is called the localization of \(A\) with respect to \(S\).

We have a natural ring map from \(A\) to its localization \(S^{-1}A\), \[A \longrightarrow S^{-1}A, \quad x \longmapsto x/1\] which is sometimes called the localization map. In general the localization map is not injective, unless \(S\) contains no zerodivisors. For, if \(x/1 = 0\), then there is a \(u\in S\) such that \(xu = 0\) in \(A\) and hence \(x = 0\) since there are no zerodivisors in \(S\). The localization of a ring has the following universal property.

Proposition

Let \(f : A \to B\) be a ring map that sends every element in \(S\) to a unit of \(B\). Then there is a unique homomorphism \(g : S^{-1}A \to B\) such that the following diagram commutes. \[\xymatrix{ A \ar[rr]^{f} \ar[dr] & & B \\ & S^{-1}A \ar[ur]_g }\]

Proof

Existence. We define a map \(g\) as follows. For \(x/s\in S^{-1}A\), let \(g(x/s) = f(x)f(s)^{-1}\in B\). It is easily checked from the definition that this is a well-defined ring map. And it is also clear that this makes the diagram commutative.

Uniqueness. We now show that if \(g' : S^{-1}A \to B\) satisfies \(g'(x/1) = f(x)\), then \(g = g'\). Hence \(f(s) = g'(s/1)\) for \(s \in S\) by the commutativity of the diagram. But then \(g'(1/s)f(s) = 1\) in \(B\), which implies that \(g'(1/s) = f(s)^{-1}\) and hence \(g'(x/s) = g'(x/1)g'(1/s) = f(x)f(s)^{-1} = g(x/s)\).

Lemma

The localization \(S^{-1}A\) is the zero ring if and only if \(0\in S\).

Proof

If \(0\in S\), any pair \((a, s)\sim (0, 1)\) by definition. If \(0\not \in S\), then clearly \(1/1 \neq 0/1\) in \(S^{-1}A\).

Lemma

Let \(R\) be a ring. Let \(S \subset R\) be a multiplicative subset. The category of \(S^{-1}R\)-modules is equivalent to the category of \(R\)-modules \(N\) with the property that every \(s \in S\) acts as an automorphism on \(N\).

Proof

The functor which defines the equivalence associates to an \(S^{-1}R\)-module \(M\) the same module but now viewed as an \(R\)-module via the localization map \(R \to S^{-1}R\). Conversely, if \(N\) is an \(R\)-module, such that every \(s \in S\) acts via an automorphism \(s_N\), then we can think of \(N\) as an \(S^{-1}R\)-module by letting \(x/s\) act via \(x_N \circ s_N^{-1}\). We omit the verification that these two functors are quasi-inverse to each other.

The notion of localization of a ring can be generalized to the localization of a module. Let \(A\) be a ring, \(S\) a multiplicative subset of \(A\) and \(M\) an \(A\)-module. We define a relation on \(M \times S\) as follows \[(m, s) \sim (n, t) \Leftrightarrow \exists u\in S \text{ such that } (mt-ns)u = 0\] This is clearly an equivalence relation. Denote by \(m/s\) (or \(\frac{m}{s}\)) the equivalence class of \((m, s)\) and by \(S^{-1}M\) the set of all equivalence classes. For \(a \in A\), \(m, n \in M\), and \(s, t \in S\), define addition and scalar multiplication by \[m/s + n/t = (mt + ns)/st,\quad (a/s)\cdot(m/t) = am/st\] It is clear that this makes \(S^{-1}M\) an \(S^{-1}A\)-module.

Definition

The \(S^{-1}A\)-module \(S^{-1}M\) is called the localization of \(M\) with respect to \(S\).

Note that there is an \(A\)-module map \(M \to S^{-1}M\), \(m \mapsto m/1\) which is sometimes called the localization map. It satisfies the following universal property.

Lemma

Let \(R\) be a ring. Let \(S \subset R\) be a multiplicative subset. Let \(M\), \(N\) be \(R\)-modules. Assume all the elements of \(S\) act as automorphisms on \(N\). Then the canonical map \[\Hom_R(S^{-1}M, N) \longrightarrow \Hom_R(M, N)\] induced by the localization map, is an isomorphism.

Proof

It is clear that the map is well-defined and \(R\)-linear. Injectivity: Let \(\alpha \in \Hom_R(S^{-1}M, N)\) and take an arbitrary element \(m/s \in S^{-1}M\). Then, since \(s \cdot \alpha(m/s) = \alpha(m/1)\), we have \(\alpha(m/s) =s^{-1}(\alpha (m/1))\), so \(\alpha\) is completely determined by what it does on the image of \(M\) in \(S^{-1}M\). Surjectivity: Let \(\beta : M \rightarrow N\) be a given \(R\)-linear map. We need to show that it can be "extended" to \(S^{-1}M\). Define a map of sets \[M \times S \rightarrow N,\quad (m,s) \mapsto s^{-1}\beta(m)\] Clearly, this map respects the equivalence relation from above, so it descends to a well-defined map \(\alpha : S^{-1}M \rightarrow N\). It remains to show that this map is \(R\)-linear, so take \(r, r' \in R\) as well as \(s, s' \in S\) and \(m, m' \in M\). Then \[\begin{align*} \alpha(r \cdot m/s + r' \cdot m' /s') & = \alpha((r \cdot s' \cdot m + r' \cdot s \cdot m') /(ss')) \\ & = (ss')^{-1}\beta(r \cdot s' \cdot m + r' \cdot s \cdot m') \\ & = (ss')^{-1} (r \cdot s' \beta (m) + r' \cdot s \beta (m')) \\ & = r \alpha (m/s) + r' \alpha (m' /s') \end{align*}\] and we win.

Example

Let \(A\) be a ring and let \(M\) be an \(A\)-module. Here are some important examples of localizations.

1. Given \(\mathfrak p\) a prime ideal of \(A\) consider \(S = A\setminus\mathfrak p\). It is immediately checked that \(S\) is a multiplicative set. In this case we denote \(A_\mathfrak p\) and \(M_\mathfrak p\) the localization of \(A\) and \(M\) with respect to \(S\) respectively. These are called the localization of \(A\), resp. \(M\) at \(\mathfrak p\).

2. Let \(f\in A\). Consider \(S = \{1, f, f^2, \ldots\}\). This is clearly a multiplicative subset of \(A\). In this case we denote \(A_f\) (resp. \(M_f\)) the localization \(S^{-1}A\) (resp. \(S^{-1}M\)). This is called the localization of \(A\), resp. \(M\) with respect to \(f\). Note that \(A_f = 0\) if and only if \(f\) is nilpotent in \(A\).

3. Let \(S = \{f \in A \mid f \text{ is not a zerodivisor in }A\}\). This is a multiplicative subset of \(A\). In this case the ring \(Q(A) = S^{-1}A\) is called either the total quotient ring, or the total ring of fractions of \(A\).

4. If \(A\) is a domain, then the total quotient ring \(Q(A)\) is the field of fractions of \(A\). Please see Fields, Example 09FJ.

Lemma

Let \(R\) be a ring. Let \(S \subset R\) be a multiplicative subset. Let \(M\) be an \(R\)-module. Then \[S^{-1}M = \colim_{f \in S} M_f\] where the preorder on \(S\) is given by \(f \geq f' \Leftrightarrow f = f'f''\) for some \(f'' \in R\) in which case the map \(M_{f'} \to M_f\) is given by \(m/(f')^e \mapsto m(f'')^e/f^e\).

Proof

Omitted. Hint: Use the universal property of Lemma 07K0.

In the following paragraph, let \(A\) denote a ring, and \(M, N\) denote modules over \(A\).

If \(S\) and \(S'\) are multiplicative sets of \(A\), then it is clear that \[SS' = \{ss' : s\in S, \ s'\in S'\}\] is also a multiplicative set of \(A\). Then the following holds.

Proposition

Let \(\overline{S}\) be the image of \(S\) in \(S'^{-1}A\), then \((SS')^{-1}A\) is isomorphic to \(\overline{S}^{-1}(S'^{-1}A)\).

Proof

The map sending \(x\in A\) to \(x/1\in (SS')^{-1}A\) induces a map sending \(x/s\in S'^{-1}A\) to \(x/s \in (SS')^{-1}A\), by universal property. The image of the elements in \(\overline{S}\) are invertible in \((SS')^{-1}A\). By the universal property we get a map \(f : \overline{S}^{-1}(S'^{-1}A) \to (SS')^{-1}A\) which maps \((x/s')/(s/1)\) to \(x/ss'\).

On the other hand, the map from \(A\) to \(\overline{S}^{-1}(S'^{-1}A)\) sending \(x\in A\) to \((x/1)/(1/1)\) also induces a map \(g : (SS')^{-1}A \to \overline{S}^{-1}(S'^{-1}A)\) which sends \(x/ss'\) to \((x/s')/(s/1)\), by the universal property again. It is immediately checked that \(f\) and \(g\) are inverse to each other, hence they are both isomorphisms.

For the module \(M\) we have

Proposition

View \(S'^{-1}M\) as an \(A\)-module, then \(S^{-1}(S'^{-1}M)\) is isomorphic to \((SS')^{-1}M\).

Proof

Note that given an \(A\)-module \(M\), we have not proved any universal property for \(S^{-1}M\). Hence we cannot reason as in the preceding proof; we have to construct the isomorphism explicitly.

We define the maps as follows \[\begin{align*} & f : S^{-1}(S'^{-1}M) \longrightarrow (SS')^{-1}M, \quad \frac{x/s'}{s}\mapsto x/ss'\\ & g : (SS')^{-1}M \longrightarrow S^{-1}(S'^{-1}M), \quad x/t\mapsto \frac{x/s'}{s}\ \text{for some }s\in S, s'\in S', \text{ and } t = ss' \end{align*}\] We have to check that these homomorphisms are well-defined, that is, independent of the choice of the fraction. This is easily checked and it is also straightforward to show that they are inverse to each other.

If \(u : M \to N\) is an \(A\)-module homomorphism, then the localization indeed induces a well-defined \(S^{-1}A\) homomorphism \(S^{-1}u : S^{-1}M \to S^{-1}N\) which sends \(x/s\) to \(u(x)/s\). It is immediately checked that this construction is functorial, so that \(S^{-1}\) is actually a functor from the category of \(A\)-modules to the category of \(S^{-1}A\)-modules. Moreover this functor is exact, as we show in the following proposition.

Proposition

Let \(L\xrightarrow{u} M\xrightarrow{v} N\) be an exact sequence of \(R\)-modules. Then \(S^{-1}L \to S^{-1}M \to S^{-1}N\) is also exact.

Proof

First it is clear that \(S^{-1}L \to S^{-1}M \to S^{-1}N\) is a complex since localization is a functor. Next suppose that \(x/s\) maps to zero in \(S^{-1}N\) for some \(x/s \in S^{-1}M\). Then by definition there is a \(t\in S\) such that \(v(xt) = v(x)t = 0\) in \(N\), which means \(xt \in \Ker(v)\). By the exactness of \(L \to M \to N\) we have \(xt = u(y)\) for some \(y\) in \(L\). Then \(x/s\) is the image of \(y/st\). This proves the exactness.

Lemma

Localization respects quotients, i.e. if \(N\) is a submodule of \(M\), then \(S^{-1}(M/N)\simeq (S^{-1}M)/(S^{-1}N)\).

Proof

From the exact sequence \[0 \longrightarrow N \longrightarrow M \longrightarrow M/N \longrightarrow 0\] we have \[0 \longrightarrow S^{-1}N \longrightarrow S^{-1}M \longrightarrow S^{-1}(M/N) \longrightarrow 0\] The corollary then follows.

If, in the preceding Lemma, we take \(N = I\) and \(M = A\) for an ideal \(I\) of \(A\), we see that \(S^{-1}A/S^{-1}I \simeq S^{-1}(A/I)\) as \(A\)-modules. The next proposition shows that they are isomorphic as rings.

Proposition

Let \(I\) be an ideal of \(A\), \(S\) a multiplicative set of \(A\). Then \(S^{-1}I\) is an ideal of \(S^{-1}A\) and \(\overline{S}^{-1}(A/I)\) is isomorphic to \(S^{-1}A/S^{-1}I\), where \(\overline{S}\) is the image of \(S\) in \(A/I\).

Proof

The fact that \(S^{-1}I\) is an ideal is clear since \(I\) itself is an ideal. Define \[f : S^{-1}A\longrightarrow \overline{S}^{-1}(A/I), \quad x/s\mapsto \overline{x}/\overline{s}\] where \(\overline{x}\) and \(\overline{s}\) are the images of \(x\) and \(s\) in \(A/I\). We shall keep similar notations in this proof. This map is well-defined by the universal property of \(S^{-1}A\), and \(S^{-1}I\) is contained in the kernel of it, therefore it induces a map \[\overline{f} : S^{-1}A/S^{-1}I \longrightarrow \overline{S}^{-1}(A/I), \quad \overline{x/s}\mapsto \overline{x}/\overline{s}\]

On the other hand, the map \(A \to S^{-1}A/S^{-1}I\) sending \(x\) to \(\overline{x/1}\) induces a map \(A/I \to S^{-1}A/S^{-1}I\) sending \(\overline{x}\) to \(\overline{x/1}\). The image of \(\overline{S}\) is invertible in \(S^{-1}A/S^{-1}I\), thus induces a map \[g : \overline{S}^{-1}(A/I) \longrightarrow S^{-1}A/S^{-1}I, \quad \frac{\overline{x}}{\overline{s}}\mapsto \overline{x/s}\] by the universal property. It is then clear that \(\overline{f}\) and \(g\) are inverse to each other, hence are both isomorphisms.

We now consider how submodules behave in localization.

Lemma

Any submodule \(N'\) of \(S^{-1}M\) is of the form \(S^{-1}N\) for some \(N\subset M\). Indeed one can take \(N\) to be the inverse image of \(N'\) in \(M\).

Proof

Let \(N\) be the inverse image of \(N'\) in \(M\). Then one can see that \(S^{-1}N\supset N'\). To show they are equal, take \(x/s\) in \(S^{-1}N\), where \(s\in S\) and \(x\in N\). This yields that \(x/1\in N'\). Since \(N'\) is an \(S^{-1}R\)-submodule we have \(x/s = x/1\cdot 1/s\in N'\). This finishes the proof.

Taking \(M = A\) and \(N = I\) an ideal of \(A\), we have the following corollary, which can be viewed as a converse of the first part of Proposition 00CT.

Lemma

Each ideal \(I'\) of \(S^{-1}A\) takes the form \(S^{-1}I\), where one can take \(I\) to be the inverse image of \(I'\) in \(A\).

Proof

Immediate from Lemma 00CU.

Internal Hom

If \(R\) is a ring, and \(M\), \(N\) are \(R\)-modules, then \[\Hom_R(M, N) = \{ \varphi : M \to N\}\] is the set of \(R\)-linear maps from \(M\) to \(N\). This set comes with the structure of an abelian group by setting \((\varphi + \psi)(m) = \varphi(m) + \psi(m)\), as usual. In fact, \(\Hom_R(M, N)\) is also an \(R\)-module via the rule \((x \varphi)(m) = x \varphi(m) = \varphi(xm)\).

Given maps \(a : M \to M'\) and \(b : N \to N'\) of \(R\)-modules, we can pre-compose and post-compose homomorphisms by \(a\) and \(b\). This leads to the following commutative diagram \[\xymatrix{ \Hom_R(M', N) \ar[d]_{- \circ a} \ar[r]_{b \circ -} & \Hom_R(M', N') \ar[d]^{- \circ a} \\ \Hom_R(M, N) \ar[r]^{b \circ -} & \Hom_R(M, N') }\] In fact, the maps in this diagram are \(R\)-module maps. Thus \(\Hom_R\) defines an additive functor \[\text{Mod}_R^{opp} \times \text{Mod}_R \longrightarrow \text{Mod}_R, \quad (M, N) \longmapsto \Hom_R(M, N)\]

Lemma

Exactness and \(\Hom_R\). Let \(R\) be a ring. Let \(M_1\), \(M_2\), \(M_3\) be \(R\)-modules. Let \(M_1 \to M_2\) and \(M_2 \to M_3\) be \(R\)-module maps.

1. \(M_1 \to M_2 \to M_3 \to 0\) is exact if and only if \(0 \to \Hom_R(M_3, N) \to \Hom_R(M_2, N) \to \Hom_R(M_1, N)\) is exact for all \(R\)-modules \(N\).

2. \(0 \to M_1 \to M_2 \to M_3\) is exact if and only if \(0 \to \Hom_R(N, M_1) \to \Hom_R(N, M_2) \to \Hom_R(N, M_3)\) is exact for all \(R\)-modules \(N\).

Proof

Omitted.

Lemma

Let \(R\) be a ring. Let \(M\) be a finitely presented \(R\)-module. Let \(N\) be an \(R\)-module.

1. For \(f \in R\) we have \(\Hom_R(M, N)_f = \Hom_{R_f}(M_f, N_f) = \Hom_R(M_f, N_f)\),

2. for a multiplicative subset \(S\) of \(R\) we have \[S^{-1}\Hom_R(M, N) = \Hom_{S^{-1}R}(S^{-1}M, S^{-1}N) = \Hom_R(S^{-1}M, S^{-1}N).\]

Proof

Part (1) is a special case of part (2). The second equality in (2) follows from Lemma 07JY. Choose a presentation \[\bigoplus\nolimits_{j = 1, \ldots, m} R \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, n} R \to M \to 0.\] By Lemma 0582 this gives an exact sequence \[0 \to \Hom_R(M, N) \to \bigoplus\nolimits_{i = 1, \ldots, n} N \longrightarrow \bigoplus\nolimits_{j = 1, \ldots, m} N.\] Inverting \(S\) and using Proposition 00CS we get an exact sequence \[0 \to S^{-1}\Hom_R(M, N) \to \bigoplus\nolimits_{i = 1, \ldots, n} S^{-1}N \longrightarrow \bigoplus\nolimits_{j = 1, \ldots, m} S^{-1}N\] and the result follows since \(S^{-1}M\) sits in an exact sequence \[\bigoplus\nolimits_{j = 1, \ldots, m} S^{-1}R \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, n} S^{-1}R \to S^{-1}M \to 0\] which induces (by Lemma 0582) the exact sequence \[0 \to \Hom_{S^{-1}R}(S^{-1}M, S^{-1}N) \to \bigoplus\nolimits_{i = 1, \ldots, n} S^{-1}N \longrightarrow \bigoplus\nolimits_{j = 1, \ldots, m} S^{-1}N\] which is the same as the one above.

Characterizing finite and finitely presented modules

Given a module \(N\) over a ring \(R\), you can characterize whether or not \(N\) is a finite module or a finitely presented module in terms of the functor \(\Hom_R(N, -)\).

Lemma

Let \(R\) be a ring. Let \(N\) be an \(R\)-module. The following are equivalent

1. \(N\) is a finite \(R\)-module,

2. for any filtered colimit \(M = \colim M_i\) of \(R\)-modules the map \(\colim \Hom_R(N, M_i) \to \Hom_R(N, M)\) is injective.

Proof

Assume (1) and choose generators \(x_1, \ldots, x_m\) for \(N\). If \(N \to M_i\) is a module map and the composition \(N \to M_i \to M\) is zero, then because \(M = \colim_{i' \geq i} M_{i'}\) for each \(j \in \{1, \ldots, m\}\) we can find a \(i' \geq i\) such that \(x_j\) maps to zero in \(M_{i'}\). Since there are finitely many \(x_j\) we can find a single \(i'\) which works for all of them. Then the composition \(N \to M_i \to M_{i'}\) is zero and we conclude the map is injective, i.e., part (2) holds.

Assume (2). For a finite subset \(E \subset N\) denote \(N_E \subset N\) the \(R\)-submodule generated by the elements of \(E\). Then \(0 = \colim N/N_E\) is a filtered colimit. Hence we see that \(\text{id} : N \to N\) maps into \(N_E\) for some \(E\), i.e., \(N\) is finitely generated.

For purposes of reference, we define what it means to have a relation between elements of a module.

Definition

Let \(R\) be a ring. Let \(M\) be an \(R\)-module. Let \(n \geq 0\) and \(x_i \in M\) for \(i = 1, \ldots, n\). A relation between \(x_1, \ldots, x_n\) in \(M\) is a sequence of elements \(f_1, \ldots, f_n \in R\) such that \(\sum_{i = 1, \ldots, n} f_i x_i = 0\).

Lemma

Let \(R\) be a ring and let \(M\) be an \(R\)-module. Then \(M\) is the colimit of a directed system \((M_i, \mu_{ij})\) of \(R\)-modules with all \(M_i\) finitely presented \(R\)-modules.

Proof

Consider any finite subset \(S \subset M\) and any finite collection of relations \(E\) among the elements of \(S\). So each \(s \in S\) corresponds to \(x_s \in M\) and each \(e \in E\) consists of a vector of elements \(f_{e, s} \in R\) such that \(\sum f_{e, s} x_s = 0\). Let \(M_{S, E}\) be the cokernel of the map \[R^{\# E} \longrightarrow R^{\# S}, \quad (g_e)_{e\in E} \longmapsto (\sum g_e f_{e, s})_{s\in S}.\] There are canonical maps \(M_{S, E} \to M\). If \(S \subset S'\) and if the elements of \(E\) correspond, via this map, to relations in \(E'\), then there is an obvious map \(M_{S, E} \to M_{S', E'}\) commuting with the maps to \(M\). Let \(I\) be the set of pairs \((S, E)\) with ordering by inclusion as above. It is clear that the colimit of this directed system is \(M\).

Lemma

Let \(R\) be a ring. Let \(N\) be an \(R\)-module. The following are equivalent

1. \(N\) is a finitely presented \(R\)-module,

2. for any filtered colimit \(M = \colim M_i\) of \(R\)-modules the map \(\colim \Hom_R(N, M_i) \to \Hom_R(N, M)\) is bijective.

Proof

Assume (1) and choose an exact sequence \(F_{-1} \to F_0 \to N \to 0\) with \(F_i\) finite free. Then we have an exact sequence \[0 \to \Hom_R(N, M) \to \Hom_R(F_0, M) \to \Hom_R(F_{-1}, M)\] functorial in the \(R\)-module \(M\). The functors \(\Hom_R(F_i, M)\) commute with filtered colimits as \(\Hom_R(R^{\oplus n}, M) = M^{\oplus n}\). Since filtered colimits are exact (Lemma 00DB) we see that (2) holds.

Assume (2). By Lemma 00HA we can write \(N = \colim N_i\) as a filtered colimit such that \(N_i\) is of finite presentation for all \(i\). Thus \(\text{id}_N\) factors through \(N_i\) for some \(i\). This means that \(N\) is a direct summand of a finitely presented \(R\)-module (namely \(N_i\)) and hence finitely presented.

Tensor products

Definition

Let \(R\) be a ring, \(M, N, P\) be three \(R\)-modules. A mapping \(f : M \times N \to P\) (where \(M \times N\) is viewed only as Cartesian product of two \(R\)-modules) is said to be \(R\)-bilinear if for each \(x \in M\) the mapping \(y\mapsto f(x, y)\) of \(N\) into \(P\) is \(R\)-linear, and for each \(y\in N\) the mapping \(x\mapsto f(x, y)\) is also \(R\)-linear.

Lemma

Let \(M, N\) be \(R\)-modules. Then there exists a pair \((T, g)\) where \(T\) is an \(R\)-module, and \(g : M \times N \to T\) an \(R\)-bilinear mapping, with the following universal property: For any \(R\)-module \(P\) and any \(R\)-bilinear mapping \(f : M \times N \to P\), there exists a unique \(R\)-linear mapping \(\tilde{f} : T \to P\) such that \(f = \tilde{f} \circ g\). In other words, the following diagram commutes: \[\xymatrix{ M \times N \ar[rr]^f \ar[dr]_g & & P\\ & T \ar[ur]_{\tilde f} }\] Moreover, if \((T, g)\) and \((T', g')\) are two pairs with this property, then there exists a unique isomorphism \(j : T \to T'\) such that \(j\circ g = g'\).

The \(R\)-module \(T\) which satisfies the above universal property is called the tensor product of \(R\)-modules \(M\) and \(N\), denoted as \(M \otimes_R N\).

Proof

We first prove the existence of such \(R\)-module \(T\). Let \(M, N\) be \(R\)-modules. Let \(T\) be the quotient module \(P/Q\), where \(P\) is the free \(R\)-module \(R^{(M \times N)}\) and \(Q\) is the \(R\)-module generated by all elements of the following types: (\(x\in M, y\in N\)) \[\begin{align*} (x + x', y) - (x, y) - (x', y), \\ (x, y + y') - (x, y) - (x, y'), \\ (ax, y) - a(x, y), \\ (x, ay) - a(x, y) \end{align*}\] Let \(\pi : M \times N \to T\) denote the natural map. This map is \(R\)-bilinear, as implied by the above relations when we check the bilinearity conditions. Denote the image \(\pi(x, y) = x \otimes y\), then these elements generate \(T\). Now let \(f : M \times N \to P\) be an \(R\)-bilinear map, then we can define \(f' : T \to P\) by extending the mapping \(f'(x \otimes y) = f(x, y)\). Clearly \(f = f'\circ \pi\). Moreover, \(f'\) is uniquely determined by the value on the generating sets \(\{x \otimes y : x\in M, y\in N\}\). Suppose there is another pair \((T', g')\) satisfying the same properties. Then there is a unique \(j : T \to T'\) and also \(j' : T' \to T\) such that \(g' = j\circ g\), \(g = j'\circ g'\). But then both the maps \((j'\circ j) \circ g\) and \(g\) satisfy the universal properties, so by uniqueness they are equal, and hence \(j'\circ j\) is identity on \(T\). Similarly \((j\circ j') \circ g' = g'\) and \(j\circ j'\) is identity on \(T'\). So \(j\) is an isomorphism.

Lemma

Let \(M, N, P\) be \(R\)-modules, then the bilinear maps \[\begin{align*} (x, y) & \mapsto y \otimes x\\ (x + y, z) & \mapsto x \otimes z + y \otimes z\\ (r, x) & \mapsto rx \end{align*}\] induce unique isomorphisms \[\begin{align*} M \otimes_R N & \to N \otimes_R M, \\ (M\oplus N)\otimes_R P & \to (M \otimes_R P)\oplus(N \otimes_R P), \\ R \otimes_R M & \to M \end{align*}\]

Proof

Omitted.

We may generalize the tensor product of two \(R\)-modules to finitely many \(R\)-modules, and set up a correspondence between the multi-tensor product with multilinear mappings. Using almost the same construction one can prove that:

Lemma

Let \(M_1, \ldots, M_r\) be \(R\)-modules. Then there exists a pair \((T, g)\) consisting of an \(R\)-module \(T\) and an \(R\)-multilinear mapping \(g : M_1\times \ldots \times M_r \to T\) with the universal property: For any \(R\)-multilinear mapping \(f : M_1\times \ldots \times M_r \to P\) there exists a unique \(R\)-module homomorphism \(f' : T \to P\) such that \(f'\circ g = f\). Such a module \(T\) is unique up to unique isomorphism. We denote it \(M_1\otimes_R \ldots \otimes_R M_r\) and we denote the universal multilinear map \((m_1, \ldots, m_r) \mapsto m_1 \otimes \ldots \otimes m_r\).

Proof

Omitted.

Lemma

The homomorphisms \[(M \otimes_R N)\otimes_R P \to M \otimes_R N \otimes_R P \to M \otimes_R (N \otimes_R P)\] such that \(f((x \otimes y)\otimes z) = x \otimes y \otimes z\) and \(g(x \otimes y \otimes z) = x \otimes (y \otimes z)\), \(x\in M, y\in N, z\in P\) are well-defined and are isomorphisms.

Proof

We shall prove \(f\) is well-defined and is an isomorphism, and this proof carries analogously to \(g\). Fix any \(z\in P\), then the mapping \((x, y)\mapsto x \otimes y \otimes z\), \(x\in M, y\in N\), is \(R\)-bilinear in \(x\) and \(y\), and hence induces homomorphism \(f_z : M \otimes N \to M \otimes N \otimes P\) which sends \(f_z(x \otimes y) = x \otimes y \otimes z\). Then consider \((M \otimes N)\times P \to M \otimes N \otimes P\) given by \((w, z)\mapsto f_z(w)\). The map is \(R\)-bilinear and thus induces \(f : (M \otimes_R N)\otimes_R P \to M \otimes_R N \otimes_R P\) and \(f((x \otimes y)\otimes z) = x \otimes y \otimes z\). To construct the inverse, we note that the map \(\pi : M \times N \times P \to (M \otimes N)\otimes P\) is \(R\)-trilinear. Therefore, it induces an \(R\)-linear map \(h : M \otimes N \otimes P \to (M \otimes N)\otimes P\) which agrees with the universal property. Here we see that \(h(x \otimes y \otimes z) = (x \otimes y)\otimes z\). From the explicit expression of \(f\) and \(h\), \(f\circ h\) and \(h\circ f\) are identity maps of \(M \otimes N \otimes P\) and \((M \otimes N)\otimes P\) respectively, hence \(f\) is our desired isomorphism.

Doing induction we see that this extends to multi-tensor products. Combined with Lemma 00CY we see that the tensor product operation on the category of \(R\)-modules is associative, commutative and distributive.

Definition

An abelian group \(N\) is called an \((A, B)\)-bimodule if it is both an \(A\)-module and a \(B\)-module and for all \(a \in A\) and \(b \in B\) the multiplication by \(a\) and \(b\) commute, so \(b(an) = a(bn)\) for all \(n \in N\). In this situation we usually write the \(B\)-action on the right: so for \(b \in B\) and \(n \in N\) the result of multiplying \(n\) by \(b\) is denoted \(nb\). With this convention the compatibility above is that \((ax)b = a(xb)\) for all \(a\in A, b\in B, x\in N\). The shorthand \(_AN_B\) is used to denote an \((A, B)\)-bimodule \(N\).

Lemma

For \(A\)-module \(M\), \(B\)-module \(P\) and \((A, B)\)-bimodule \(N\), the modules \((M \otimes_A N)\otimes_B P\) and \(M \otimes_A(N \otimes_B P)\) can both be given \((A, B)\)-bimodule structure, and moreover \[(M \otimes_A N)\otimes_B P \cong M \otimes_A(N \otimes_B P).\]

Proof

A priori \(M \otimes_A N\) is an \(A\)-module, but we can give it a \(B\)-module structure by letting \[(x \otimes y)b = x \otimes yb, \quad x\in M, y\in N, b\in B\] Thus \(M \otimes_A N\) becomes an \((A, B)\)-bimodule. Similarly for \(N \otimes_B P\), and thus for \((M \otimes_A N)\otimes_B P\) and \(M \otimes_A(N \otimes_B P)\). By Lemma 00D0, these two modules are isomorphic both as \(A\)-modules and as \(B\)-modules via the same mapping.

Lemma

For any three \(R\)-modules \(M, N, P\), \[\Hom_R(M \otimes_R N, P) \cong \Hom_R(M, \Hom_R(N, P))\]

Proof

An \(R\)-linear map \(\hat{f}\in \Hom_R(M \otimes_R N, P)\) corresponds to an \(R\)-bilinear map \(f : M \times N \to P\). For each \(x\in M\) the mapping \(y\mapsto f(x, y)\) is \(R\)-linear by the universal property. Thus \(f\) corresponds to a map \(\phi_f : M \to \Hom_R(N, P)\). This map is \(R\)-linear since \[\phi_f(ax + y)(z) = f(ax + y, z) = af(x, z)+f(y, z) = (a\phi_f(x)+\phi_f(y))(z),\] for all \(a \in R\), \(x \in M\), \(y \in M\) and \(z \in N\). Conversely, any \(f \in \Hom_R(M, \Hom_R(N, P))\) defines an \(R\)-bilinear map \(M \times N \to P\), namely \((x, y)\mapsto f(x)(y)\). So this is a natural one-to-one correspondence between the two modules \(\Hom_R(M \otimes_R N, P)\) and \(\Hom_R(M, \Hom_R(N, P))\).

Lemma

Let \((M_i, \mu_{ij})\) be a system over the preordered set \(I\). Let \(N\) be an \(R\)-module. Then \[\colim (M_i \otimes N) \cong (\colim M_i)\otimes N.\] Moreover, the isomorphism is induced by the homomorphisms \(\mu_i \otimes 1: M_i \otimes N \to M \otimes N\) where \(M = \colim_i M_i\) with natural maps \(\mu_i : M_i \to M\).

Proof

First proof. The functor \(M' \mapsto M' \otimes_R N\) is left adjoint to the functor \(N' \mapsto \Hom_R(N, N')\) by Lemma 00DE. Thus \(M' \mapsto M' \otimes_R N\) commutes with all colimits, see Categories, Lemma 0038.

Second direct proof. Let \(P = \colim (M_i \otimes N)\) with coprojections \(\lambda_i : M_i \otimes N \to P\). Let \(M = \colim M_i\) with coprojections \(\mu_i : M_i \to M\). Then for all \(i\leq j\), the following diagram commutes: \[\xymatrix{ M_i \otimes N \ar[r]_{\mu_i \otimes 1} \ar[d]_{\mu_{ij} \otimes 1} & M \otimes N \ar[d]^{\text{id}} \\ M_j \otimes N \ar[r]^{\mu_j \otimes 1} & M \otimes N }\] By Lemma 00DA these maps induce a unique homomorphism \(\psi : P \to M \otimes N\) such that \(\mu_i \otimes 1 = \psi \circ \lambda_i\).

To construct the inverse map, for each \(i\in I\), there is the canonical \(R\)-bilinear mapping \(g_i : M_i \times N \to M_i \otimes N\). This induces a unique mapping \(\widehat{\phi} : M \times N \to P\) such that \(\widehat{\phi} \circ (\mu_i \times 1) = \lambda_i \circ g_i\). It is \(R\)-bilinear. Thus it induces an \(R\)-linear mapping \(\phi : M \otimes N \to P\). From the commutative diagram below: \[\xymatrix{ M_i \times N \ar[r]^{g_i} \ar[d]^{\mu_i \times \text{id}} & M_i \otimes N\ar[r]_{\text{id}} \ar[d]_{\lambda_i} & M_i \otimes N \ar[d]_{\mu_i \otimes \text{id}} \ar[rd]^{\lambda_i} \\ M \times N \ar[r]^{\widehat{\phi}} & P \ar[r]^{\psi} & M \otimes N \ar[r]^{\phi} & P }\] we see that \(\psi\circ\widehat{\phi} = g\), the canonical \(R\)-bilinear mapping \(g : M \times N \to M \otimes N\). So \(\psi\circ\phi\) is identity on \(M \otimes N\). From the right-hand square and triangle, \(\phi\circ\psi\) is also identity on \(P\).

Lemma

Let \[\begin{align*} M_1\xrightarrow{f} M_2\xrightarrow{g} M_3 \to 0 \end{align*}\] be an exact sequence of \(R\)-modules and homomorphisms, and let \(N\) be any \(R\)-module. Then the sequence [00DG]\[\begin{equation} M_1\otimes N\xrightarrow{f \otimes 1} M_2\otimes N \xrightarrow{g \otimes 1} M_3\otimes N \to 0 \end{equation}\] is exact. In other words, the functor \(- \otimes_R N\) is right exact, in the sense that tensoring each term in the original right exact sequence preserves the exactness.

Proof

For every \(R\)-module \(P\) we apply the functor \(\Hom(-, \Hom(N, P))\) to the first exact sequence. We obtain \[0 \to \Hom(M_3, \Hom(N, P)) \to \Hom(M_2, \Hom(N, P)) \to \Hom(M_1, \Hom(N, P))\] which is exact by Lemma 0582 (1). By Lemma 00DE this becomes the sequence \[0 \to \Hom(M_3 \otimes N, P) \to \Hom(M_2 \otimes N, P) \to \Hom(M_1 \otimes N, P)\] which is therefore also exact. Then using Lemma 0582 (1) again, we arrive at the desired exact sequence.

Remark

However, tensor product does NOT preserve exact sequences in general. In other words, if \(M_1 \to M_2 \to M_3\) is exact, then it is not necessarily true that \(M_1 \otimes N \to M_2 \otimes N \to M_3 \otimes N\) is exact for arbitrary \(R\)-module \(N\).

Example

Consider the injective map \(2 : \mathbf{Z}\to \mathbf{Z}\) viewed as a map of \(\mathbf{Z}\)-modules. Let \(N = \mathbf{Z}/2\). Then the induced map \(\mathbf{Z} \otimes \mathbf{Z}/2 \to \mathbf{Z} \otimes \mathbf{Z}/2\) is NOT injective. This is because for \(x \otimes y\in \mathbf{Z} \otimes \mathbf{Z}/2\), \[(2 \otimes 1)(x \otimes y) = 2x \otimes y = x \otimes 2y = x \otimes 0 = 0\] Therefore the induced map is the zero map while \(\mathbf{Z} \otimes N\neq 0\).

Remark

For \(R\)-modules \(N\), if the functor \(-\otimes_R N\) is exact, i.e. tensoring with \(N\) preserves all exact sequences, then \(N\) is said to be a flat \(R\)-module. We will discuss this later in Section 00H9.

Lemma

Let \(R\) be a ring. Let \(M\) and \(N\) be \(R\)-modules.

1. If \(N\) and \(M\) are finite, then so is \(M \otimes_R N\).

2. If \(N\) and \(M\) are finitely presented, then so is \(M \otimes_R N\).

Proof

Suppose \(M\) is finite. Then choose a presentation \(0 \to K \to R^{\oplus n} \to M \to 0\). This gives an exact sequence \(K \otimes_R N \to N^{\oplus n} \to M \otimes_R N \to 0\) by Lemma 00DF. We conclude that if \(N\) is finite too then \(M \otimes_R N\) is a quotient of a finite module, hence finite, see Lemma 0519. Similarly, if both \(N\) and \(M\) are finitely presented, then we see that \(K\) is finite and that \(M \otimes_R N\) is a quotient of the finitely presented module \(N^{\oplus n}\) by a finite module, namely \(K \otimes_R N\), and hence finitely presented, see Lemma 0519.

Lemma

Let \(M\) be an \(R\)-module. Then the \(S^{-1}R\)-modules \(S^{-1}M\) and \(S^{-1}R \otimes_R M\) are canonically isomorphic, and the canonical isomorphism \(f : S^{-1}R \otimes_R M \to S^{-1}M\) is given by \[f((a/s) \otimes m) = am/s, \forall a \in R, m \in M, s \in S\]

Proof

Obviously, the map \(f' : S^{-1}R \times M \to S^{-1}M\) given by \(f'(a/s, m) = am/s\) is bilinear, and thus by the universal property, this map induces a unique \(S^{-1}R\)-module homomorphism \(f : S^{-1}R \otimes_R M \to S^{-1}M\) as in the statement of the lemma. Actually every element in \(S^{-1}M\) is of the form \(m/s\), \(m\in M, s\in S\) and every element in \(S^{-1}R \otimes_R M\) is of the form \(1/s \otimes m\). To see the latter fact, write an element in \(S^{-1}R \otimes_R M\) as \[\sum_k \frac{a_k}{s_k} \otimes m_k = \sum_k \frac{a_k t_k}{s} \otimes m_k = \frac{1}{s} \otimes \sum_k {a_k t_k}m_k = \frac{1}{s} \otimes m\] Where \(m = \sum_k {a_k t_k}m_k\). Then it is obvious that \(f\) is surjective, and if \(f(\frac{1}{s} \otimes m) = m/s = 0\) then there exists \(t \in S\) with \(tm = 0\) in \(M\). Then we have \[\frac{1}{s} \otimes m = \frac{1}{st} \otimes tm = \frac{1}{st} \otimes 0 = 0\] Therefore \(f\) is injective.

Lemma

Let \(M, N\) be \(R\)-modules, then there is a canonical \(S^{-1}R\)-module isomorphism \(f : S^{-1}M \otimes_{S^{-1}R}S^{-1}N \to S^{-1}(M \otimes_R N)\), given by \[f((m/s)\otimes(n/t)) = (m \otimes n)/st\]

Proof

We may use Lemma 00D2 and Lemma 00DK repeatedly to see that these two \(S^{-1}R\)-modules are isomorphic, noting that \(S^{-1}R\) is an \((R, S^{-1}R)\)-bimodule: \[\begin{align*} S^{-1}(M \otimes_R N) & \cong S^{-1}R \otimes_R (M \otimes_R N)\\ & \cong S^{-1}M \otimes_R N\\ & \cong (S^{-1}M \otimes_{S^{-1}R}S^{-1}R)\otimes_R N\\ & \cong S^{-1}M \otimes_{S^{-1}R}(S^{-1}R \otimes_R N)\\ & \cong S^{-1}M \otimes_{S^{-1}R}S^{-1}N \end{align*}\] This isomorphism is easily seen to be the one stated in the lemma.

Tensor algebra

Let \(R\) be a ring. Let \(M\) be an \(R\)-module. We define the tensor algebra of \(M\) over \(R\) to be the noncommutative \(R\)-algebra \[\text{T}(M) = \text{T}_R(M) = \bigoplus\nolimits_{n \geq 0} \text{T}^n(M)\] with \(\text{T}^0(M) = R\), \(\text{T}^1(M) = M\), \(\text{T}^2(M) = M \otimes_R M\), \(\text{T}^3(M) = M \otimes_R M \otimes_R M\), and so on. Multiplication is defined by the rule that on pure tensors we have \[(x_1 \otimes x_2 \otimes \ldots \otimes x_n) \cdot (y_1 \otimes y_2 \otimes \ldots \otimes y_m) = x_1 \otimes x_2 \otimes \ldots \otimes x_n \otimes y_1 \otimes y_2 \otimes \ldots \otimes y_m\] and we extend this by linearity.

We define the exterior algebra \(\wedge(M)\) of \(M\) over \(R\) to be the quotient of \(\text{T}(M)\) by the two sided ideal generated by the elements \(x \otimes x \in \text{T}^2(M)\). The image of a pure tensor \(x_1 \otimes \ldots \otimes x_n\) in \(\wedge^n(M)\) is denoted \(x_1 \wedge \ldots \wedge x_n\). These elements generate \(\wedge^n(M)\), they are \(R\)-linear in each \(x_i\) and they are zero when two of the \(x_i\) are equal (i.e., they are alternating as functions of \(x_1, x_2, \ldots, x_n\)). The multiplication on \(\wedge(M)\) is graded commutative, i.e., every \(x \in M\) and \(y \in M\) satisfy \(x \wedge y = - y \wedge x\).

An example of this is when \(M = Rx_1 \oplus \ldots \oplus Rx_n\) is a finite free module. In this case \(\wedge(M)\) is free over \(R\) with basis the elements \[x_{i_1} \wedge \ldots \wedge x_{i_r}\] with \(0 \leq r \leq n\) and \(1 \leq i_1 < i_2 < \ldots < i_r \leq n\).

We define the symmetric algebra \(\text{Sym}(M)\) of \(M\) over \(R\) to be the quotient of \(\text{T}(M)\) by the two sided ideal generated by the elements \(x \otimes y - y \otimes x \in \text{T}^2(M)\). The image of a pure tensor \(x_1 \otimes \ldots \otimes x_n\) in \(\text{Sym}^n(M)\) is denoted just \(x_1 \ldots x_n\). These elements generate \(\text{Sym}^n(M)\), these are \(R\)-linear in each \(x_i\) and \(x_1 \ldots x_n = x_1' \ldots x_n'\) if the sequence of elements \(x_1, \ldots, x_n\) is a permutation of the sequence \(x_1', \ldots, x_n'\). Thus we see that \(\text{Sym}(M)\) is commutative.

An example of this is when \(M = Rx_1 \oplus \ldots \oplus Rx_n\) is a finite free module. In this case \(\text{Sym}(M) = R[x_1, \ldots, x_n]\) is a polynomial algebra.

Lemma

Let \(R\) be a ring. Let \(M\) be an \(R\)-module. If \(M\) is a free \(R\)-module, so is each symmetric and exterior power.

Proof

Omitted, but see above for the finite free case.

Lemma

Let \(R\) be a ring. Let \(M_2 \to M_1 \to M \to 0\) be an exact sequence of \(R\)-modules. There are exact sequences \[M_2 \otimes_R \text{Sym}^{n - 1}(M_1) \to \text{Sym}^n(M_1) \to \text{Sym}^n(M) \to 0\] and similarly \[M_2 \otimes_R \wedge^{n - 1}(M_1) \to \wedge^n(M_1) \to \wedge^n(M) \to 0\]

Proof

Omitted.

Lemma

Let \(R\) be a ring. Let \(M\) be an \(R\)-module. Let \(x_i\), \(i \in I\) be a given system of generators of \(M\) as an \(R\)-module. Let \(n \geq 2\). There exists a canonical exact sequence \[\bigoplus_{1 \leq j_1 < j_2 \leq n} \bigoplus_{i_1, i_2 \in I} \text{T}^{n - 2}(M) \oplus \bigoplus_{1 \leq j_1 < j_2 \leq n} \bigoplus_{i \in I} \text{T}^{n - 2}(M) \to \text{T}^n(M) \to \wedge^n(M) \to 0\] where the pure tensor \(m_1 \otimes \ldots \otimes m_{n - 2}\) in the first summand maps to \[\begin{align*} \underbrace{ m_1 \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_1} \text{ and } x_{i_2} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \\ + \underbrace{ m_1 \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_2} \text{ and } x_{i_1} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \end{align*}\] and \(m_1 \otimes \ldots \otimes m_{n - 2}\) in the second summand maps to \[\underbrace{ m_1 \otimes \ldots \otimes x_i \otimes \ldots \otimes x_i \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i} \text{ and } x_{i} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}}\] There is also a canonical exact sequence \[\bigoplus_{1 \leq j_1 < j_2 \leq n} \bigoplus_{i_1, i_2 \in I} \text{T}^{n - 2}(M) \to \text{T}^n(M) \to \text{Sym}^n(M) \to 0\] where the pure tensor \(m_1 \otimes \ldots \otimes m_{n - 2}\) maps to \[\begin{align*} \underbrace{ m_1 \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_1} \text{ and } x_{i_2} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \\ - \underbrace{ m_1 \otimes \ldots \otimes x_{i_2} \otimes \ldots \otimes x_{i_1} \otimes \ldots \otimes m_{n - 2} }_{\text{with } x_{i_2} \text{ and } x_{i_1} \text{ occupying slots } j_1 \text{ and } j_2 \text{ in the tensor}} \end{align*}\]

Proof

Omitted.

Lemma

Let \(A \to B\) be a ring map. Let \(M\) be a \(B\)-module. Let \(n > 1\). The kernel of the \(A\)-linear map \(M \otimes_A \ldots \otimes_A M \to \wedge^n_B(M)\) is generated as an \(A\)-module by the elements \(m_1 \otimes \ldots \otimes m_n\) with \(m_i = m_j\) for \(i \not = j\), \(m_1, \ldots, m_n \in M\) and the elements \(m_1 \otimes \ldots \otimes bm_i \otimes \ldots \otimes m_n - m_1 \otimes \ldots \otimes bm_j \otimes \ldots \otimes m_n\) for \(i \not = j\), \(m_1, \ldots, m_n \in M\), and \(b \in B\).

Proof

Omitted.

Lemma

Let \(R\) be a ring. Let \(M_i\) be a directed system of \(R\)-modules. Then \(\colim_i \text{T}(M_i) = \text{T}(\colim_i M_i)\) and similarly for the symmetric and exterior algebras.

Proof

Omitted. Hint: Apply Lemma 00DD.

Lemma

Let \(R\) be a ring and let \(S \subset R\) be a multiplicative subset. Then \(S^{-1}T_R(M) = T_{S^{-1}R}(S^{-1}M)\) for any \(R\)-module \(M\). Similar for symmetric and exterior algebras.

Proof

Omitted. Hint: Apply Lemma 00DL.

Base change

We formally introduce base change in algebra as follows.

Definition

Let \(\varphi : R \to S\) be a ring map. Let \(M\) be an \(S\)-module. Let \(R \to R'\) be any ring map. The base change of \(\varphi\) by \(R \to R'\) is the ring map \(R' \to S \otimes_R R'\). In this situation we often write \(S' = S \otimes_R R'\). The base change of the \(S\)-module \(M\) is the \(S'\)-module \(M \otimes_R R'\).

If \(S = R[x_i]/(f_j)\) for some collection of variables \(x_i\), \(i \in I\) and some collection of polynomials \(f_j \in R[x_i]\), \(j \in J\), then \(S \otimes_R R' = R'[x_i]/(f'_j)\), where \(f'_j \in R'[x_i]\) is the image of \(f_j\) under the map \(R[x_i] \to R'[x_i]\) induced by \(R \to R'\). This simple remark is the key to understanding base change.

Lemma

Let \(R \to S\) be a ring map. Let \(M\) be an \(S\)-module. Let \(R \to R'\) be a ring map and let \(S' = S \otimes_R R'\) and \(M' = M \otimes_R R'\) be the base changes.

1. If \(M\) is a finite \(S\)-module, then the base change \(M'\) is a finite \(S'\)-module.

2. If \(M\) is an \(S\)-module of finite presentation, then the base change \(M'\) is an \(S'\)-module of finite presentation.

3. If \(R \to S\) is of finite type, then the base change \(R' \to S'\) is of finite type.

4. If \(R \to S\) is of finite presentation, then the base change \(R' \to S'\) is of finite presentation.

Proof

Proof of (1). Take a surjective, \(S\)-linear map \(S^{\oplus n} \to M \to 0\). By Lemma 00CY and 00DF the result after tensoring with \(R^\prime\) is a surjection \({S^\prime}^{\oplus n} \to M^\prime \rightarrow 0\), so \(M^\prime\) is a finitely generated \(S^\prime\)-module. Proof of (2). Take a presentation \(S^{\oplus m} \to S^{\oplus n} \to M \to 0\). By Lemma 00CY and 00DF the result after tensoring with \(R^\prime\) gives a finite presentation \({S^\prime}^{\oplus m} \to {S^\prime}^{\oplus n} \to M^\prime \to 0\), of the \(S^\prime\)-module \(M^\prime\). Proof of (3). This follows by the remark preceding the lemma as we can take \(I\) to be finite by assumption. Proof of (4). This follows by the remark preceding the lemma as we can take \(I\) and \(J\) to be finite by assumption.

Let \(\varphi : R \to S\) be a ring map. Given an \(S\)-module \(N\) we obtain an \(R\)-module \(N_R\) by the rule \(r \cdot n = \varphi(r)n\). This is sometimes called the restriction of \(N\) to \(R\).

Lemma

Let \(R \to S\) be a ring map. The functors \(\text{Mod}_S \to \text{Mod}_R\), \(N \mapsto N_R\) (restriction) and \(\text{Mod}_R \to \text{Mod}_S\), \(M \mapsto M \otimes_R S\) (base change) are adjoint functors. In a formula \[\Hom_R(M, N_R) = \Hom_S(M \otimes_R S, N)\]

Proof

If \(\alpha : M \to N_R\) is an \(R\)-module map, then we define \(\alpha' : M \otimes_R S \to N\) by the rule \(\alpha'(m \otimes s) = s\alpha(m)\). If \(\beta : M \otimes_R S \to N\) is an \(S\)-module map, we define \(\beta' : M \to N_R\) by the rule \(\beta'(m) = \beta(m \otimes 1)\). We omit the verification that these constructions are mutually inverse.

The lemma above tells us that restriction has a left adjoint, namely base change. It also has a right adjoint.

Lemma

Let \(R \to S\) be a ring map. The functors \(\text{Mod}_S \to \text{Mod}_R\), \(N \mapsto N_R\) (restriction) and \(\text{Mod}_R \to \text{Mod}_S\), \(M \mapsto \Hom_R(S, M)\) are adjoint functors. In a formula \[\Hom_R(N_R, M) = \Hom_S(N, \Hom_R(S, M))\]

Proof

If \(\alpha : N_R \to M\) is an \(R\)-module map, then we define \(\alpha' : N \to \Hom_R(S, M)\) by the rule \(\alpha'(n) = (s \mapsto \alpha(sn))\). If \(\beta : N \to \Hom_R(S, M)\) is an \(S\)-module map, we define \(\beta' : N_R \to M\) by the rule \(\beta'(n) = \beta(n)(1)\). We omit the verification that these constructions are mutually inverse.

Lemma

Let \(R \to S\) be a ring map. Given \(S\)-modules \(M, N\) and an \(R\)-module \(P\) we have \[\Hom_R(M \otimes_S N, P) = \Hom_S(M, \Hom_R(N, P))\]

Proof

This can be proved directly, but it is also a consequence of Lemmas 08YP and 00DE. Namely, we have \[\begin{align*} \Hom_R(M \otimes_S N, P) & = \Hom_S(M \otimes_S N, \Hom_R(S, P)) \\ & = \Hom_S(M, \Hom_S(N, \Hom_R(S, P))) \\ & = \Hom_S(M, \Hom_R(N, P)) \end{align*}\] as desired.

Miscellany

The proofs in this section should not refer to any results except those from the section on basic notions, Section 00AR.

Lemma

Let \(R\) be a ring, \(I\) and \(J\) two ideals and \(\mathfrak p\) a prime ideal containing the product \(IJ\). Then \(\mathfrak{p}\) contains \(I\) or \(J\).

Proof

Assume the contrary and take \(x \in I \setminus \mathfrak p\) and \(y \in J \setminus \mathfrak p\). Their product is an element of \(IJ \subset \mathfrak p\), which contradicts the assumption that \(\mathfrak p\) was prime.

Lemma

Let \(R\) be a ring. Let \(I_i \subset R\), \(i = 1, \ldots, r\), and \(J \subset R\) be ideals. Assume

1. \(J \not\subset I_i\) for \(i = 1, \ldots, r\), and

2. all but two of \(I_i\) are prime ideals.

Then there exists an \(x \in J\), \(x\not\in I_i\) for all \(i\).

Proof

The result is true for \(r = 1\). If \(r = 2\), then let \(x, y \in J\) with \(x \not \in I_1\) and \(y \not \in I_2\). We are done unless \(x \in I_2\) and \(y \in I_1\). Then the element \(x + y\) cannot be in \(I_1\) (since that would mean \(x + y - y \in I_1\)) and it also cannot be in \(I_2\).

For \(r \geq 3\), assume the result holds for \(r - 1\). After renumbering we may assume that \(I_r\) is prime. We may also assume there are no inclusions among the \(I_i\). Pick \(x \in J\), \(x \not \in I_i\) for all \(i = 1, \ldots, r - 1\). If \(x \not\in I_r\) we are done. So assume \(x \in I_r\). If \(J I_1 \ldots I_{r - 1} \subset I_r\) then \(J \subset I_r\) (by Lemma 07K1) a contradiction. Pick \(y \in J I_1 \ldots I_{r - 1}\), \(y \not \in I_r\). Then \(x + y\) works.

Lemma

Let \(R\) be a ring. Let \(x \in R\), \(I \subset R\) an ideal, and \(\mathfrak p_i\), \(i = 1, \ldots, r\) be prime ideals. Suppose that \(x + I \not \subset \mathfrak p_i\) for \(i = 1, \ldots, r\). Then there exists a \(y \in I\) such that \(x + y \not \in \mathfrak p_i\) for all \(i\).

Proof

We may assume there are no inclusions among the \(\mathfrak p_i\). After reordering we may assume \(x \not \in \mathfrak p_i\) for \(i < s\) and \(x \in \mathfrak p_i\) for \(i \geq s\). If \(s = r + 1\) then we are done. If not, then we can find \(y \in I\) with \(y \not \in \mathfrak p_s\). Choose \(f \in \bigcap_{i < s} \mathfrak p_i\) with \(f \not \in \mathfrak p_s\). Then \(x + fy\) does not belong to any of \(\mathfrak p_1, \ldots, \mathfrak p_s\). Thus we win by induction on \(s\).

Lemma

Let \(R\) be a ring.

1. If \(I_1, \ldots, I_r\) are ideals such that \(I_a + I_b = R\) when \(a \not = b\), then \(I_1 \cap \ldots \cap I_r = I_1I_2\ldots I_r\) and \(R/(I_1I_2\ldots I_r) \cong R/I_1 \times \ldots \times R/I_r\).

2. If \(\mathfrak m_1, \ldots, \mathfrak m_r\) are pairwise distinct maximal ideals then \(\mathfrak m_a + \mathfrak m_b = R\) for \(a \not = b\) and the above applies.

Proof

Let us first prove \(I_1 \cap \ldots \cap I_r = I_1 \ldots I_r\) as this will also imply the injectivity of the induced ring homomorphism \(R/(I_1 \ldots I_r) \rightarrow R/I_1 \times \ldots \times R/I_r\). The inclusion \(I_1 \cap \ldots \cap I_r \supset I_1 \ldots I_r\) is always fulfilled since ideals are closed under multiplication with arbitrary ring elements. To prove the other inclusion, we claim that the ideals \[I_1 \ldots \hat I_i \ldots I_r,\quad i = 1, \ldots, r\] generate the ring \(R\). We prove this by induction on \(r\). It holds when \(r = 2\). If \(r > 2\), then we see that \(R\) is the sum of the ideals \(I_1 \ldots \hat I_i \ldots I_{r - 1}\), \(i = 1, \ldots, r - 1\). Hence \(I_r\) is the sum of the ideals \(I_1 \ldots \hat I_i \ldots I_r\), \(i = 1, \ldots, r - 1\). Applying the same argument with the reverse ordering on the ideals we see that \(I_1\) is the sum of the ideals \(I_1 \ldots \hat I_i \ldots I_r\), \(i = 2, \ldots, r\). Since \(R = I_1 + I_r\) by assumption we see that \(R\) is the sum of the ideals displayed above. Therefore we can find elements \(a_i \in I_1 \ldots \hat I_i \ldots I_r\) such that their sum is one. Multiplying this equation by an element of \(I_1 \cap \ldots \cap I_r\) gives the other inclusion. It remains to show that the canonical map \(R/(I_1 \ldots I_r) \rightarrow R/I_1 \times \ldots \times R/I_r\) is surjective. For this, consider its action on the equation \(1 = \sum_{i=1}^r a_i\) we derived above. On the one hand, a ring morphism sends 1 to 1 and on the other hand, the image of any \(a_i\) is zero in \(R/I_j\) for \(j \neq i\). Therefore, the image of \(a_i\) in \(R/I_i\) is the identity. So given any element \((\bar{b_1}, \ldots, \bar{b_r}) \in R/I_1 \times \ldots \times R/I_r\), the element \(\sum_{i=1}^r a_i \cdot b_i\) is an inverse image in \(R\).

To see (2), by the very definition of being distinct maximal ideals, we have \(\mathfrak{m}_a + \mathfrak{m}_b = R\) for \(a \neq b\) and so the above applies.

Lemma

Let \(R\) be a ring. Let \(n \geq m\). Let \(A\) be an \(n \times m\) matrix with coefficients in \(R\). Let \(J \subset R\) be the ideal generated by the \(m \times m\) minors of \(A\).

1. For any \(f \in J\) there exists a \(m \times n\) matrix \(B\) such that \(BA = f 1_{m \times m}\).

2. If \(f \in R\) and \(BA = f 1_{m \times m}\) for some \(m \times n\) matrix \(B\), then \(f^m \in J\).

Proof

For \(I \subset \{1, \ldots, n\}\) with \(|I| = m\), we denote by \(E_I\) the \(m \times n\) matrix of the projection \[R^{\oplus n} = \bigoplus\nolimits_{i \in \{1, \ldots, n\}} R \longrightarrow \bigoplus\nolimits_{i \in I} R\] and set \(A_I = E_I A\), i.e., \(A_I\) is the \(m \times m\) matrix whose rows are the rows of \(A\) with indices in \(I\). Let \(B_I\) be the adjugate (transpose of cofactor) matrix to \(A_I\), i.e., such that \(A_I B_I = B_I A_I = \det(A_I) 1_{m \times m}\). The \(m \times m\) minors of \(A\) are the determinants \(\det A_I\) for all the \(I \subset \{1, \ldots, n\}\) with \(|I| = m\). If \(f \in J\) then we can write \(f = \sum c_I \det(A_I)\) for some \(c_I \in R\). Set \(B = \sum c_I B_I E_I\) to see that (1) holds.

If \(f 1_{m \times m} = BA\) then by the Cauchy-Binet formula (0F0K) we have \(f^m = \sum b_I \det(A_I)\) where \(b_I\) is the determinant of the \(m \times m\) matrix whose columns are the columns of \(B\) with indices in \(I\).

Lemma

Let \(R\) be a ring. Let \(n \geq m\). Let \(A = (a_{ij})\) be an \(n \times m\) matrix with coefficients in \(R\), written in block form as \[A = \left( \begin{matrix} A_1 \\ A_2 \end{matrix} \right)\] where \(A_1\) has size \(m \times m\). Let \(B\) be the adjugate (transpose of cofactor) matrix to \(A_1\). Then \[AB = \left( \begin{matrix} f 1_{m \times m} \\ C \end{matrix} \right)\] where \(f = \det(A_1)\) and \(c_{ij}\) is (up to sign) the determinant of the \(m \times m\) minor of \(A\) corresponding to the rows \(1, \ldots, \hat j, \ldots, m, i\).

Proof

Since the adjugate has the property \(A_1B = B A_1 = fI_m\) the first block of the expression for \(AB\) is correct. Note that \[c_{ij} = \sum\nolimits_k a_{ik}b_{kj} = \sum (-1)^{j + k}a_{ik} \det(A_1^{jk})\] where \(A_1^{jk}\) means \(A_1\) with the \(j\)th row and \(k\)th column removed. This last expression is the row expansion of the determinant of the matrix in the statement of the lemma.

Lemma

Let \(R\) be a nonzero ring. Let \(n \geq 1\). Let \(M\) be an \(R\)-module generated by \(< n\) elements. Then any \(R\)-module map \(f : R^{\oplus n} \to M\) has a nonzero kernel.

Proof

Choose a surjection \(R^{\oplus n - 1} \to M\). We may lift the map \(f\) to a map \(f' : R^{\oplus n} \to R^{\oplus n - 1}\) (Lemma 07JX). It suffices to prove \(f'\) has a nonzero kernel. The map \(f' : R^{\oplus n} \to R^{\oplus n - 1}\) is given by a matrix \(A = (a_{ij})\). If one of the \(a_{ij}\) is not nilpotent, say \(a = a_{ij}\) is not, then we can replace \(R\) by the localization \(R_a\) and we may assume \(a_{ij}\) is a unit. Since if we find a nonzero kernel after localization then there was a nonzero kernel to start with as localization is exact, see Proposition 00CS. In this case we can do a base change on both \(R^{\oplus n}\) and \(R^{\oplus n - 1}\) and reduce to the case where \[A = \left( \begin{matrix} 1 & 0 & 0 & \ldots \\ 0 & a_{22} & a_{23} & \ldots \\ 0 & a_{32} & \ldots \\ \ldots & \ldots \end{matrix} \right)\] Hence in this case we win by induction on \(n\). If not then each \(a_{ij}\) is nilpotent. Set \(I = (a_{ij}) \subset R\). Note that \(I^{m + 1} = 0\) for some \(m \geq 0\). Let \(m\) be the largest integer such that \(I^m \not = 0\). Then we see that \((I^m)^{\oplus n}\) is contained in the kernel of the map and we win.

Lemma

Let \(R\) be a nonzero ring. Let \(n, m \geq 0\) be integers. If \(R^{\oplus n}\) is isomorphic to \(R^{\oplus m}\) as \(R\)-modules, then \(n = m\).

Proof

Immediate from Lemma 05WI.

Cayley-Hamilton

Lemma

Let \(R\) be a ring. Let \(A = (a_{ij})\) be an \(n \times n\) matrix with coefficients in \(R\). Let \(P(x) \in R[x]\) be the characteristic polynomial of \(A\) (defined as \(\det(x\text{id}_{n \times n} - A)\)). Then \(P(A) = 0\) in \(\text{Mat}(n \times n, R)\).

Proof

We reduce the question to the well-known Cayley-Hamilton theorem from linear algebra in several steps:

1. If \(\phi :S \rightarrow R\) is a ring morphism and \(b_{ij}\) are inverse images of the \(a_{ij}\) under this map, then it suffices to show the statement for \(S\) and \((b_{ij})\) since \(\phi\) is a ring morphism.

2. If \(\psi :R \hookrightarrow S\) is an injective ring morphism, it clearly suffices to show the result for \(S\) and the \(a_{ij}\) considered as elements of \(S\).

3. Thus we may first reduce to the case \(R = \mathbf{Z}[X_{ij}]\), \(a_{ij} = X_{ij}\) of a polynomial ring and then further to the case \(R = \mathbf{Q}(X_{ij})\) where we may finally apply Cayley-Hamilton.

Lemma

Let \(R\) be a ring. Let \(M\) be a finite \(R\)-module. Let \(\varphi : M \to M\) be an endomorphism. Then there exists a monic polynomial \(P \in R[T]\) such that \(P(\varphi) = 0\) as an endomorphism of \(M\).

Proof

Choose a surjective \(R\)-module map \(R^{\oplus n} \to M\), given by \((a_1, \ldots, a_n) \mapsto \sum a_ix_i\) for some generators \(x_i \in M\). Choose \((a_{i1}, \ldots, a_{in}) \in R^{\oplus n}\) such that \(\varphi(x_i) = \sum a_{ij} x_j\). In other words the diagram \[\xymatrix{ R^{\oplus n} \ar[d]_A \ar[r] & M \ar[d]^\varphi \\ R^{\oplus n} \ar[r] & M }\] is commutative where \(A = (a_{ij})\). By Lemma 00DX there exists a monic polynomial \(P\) such that \(P(A) = 0\). Then it follows that \(P(\varphi) = 0\).

Lemma

Let \(R\) be a ring. Let \(I \subset R\) be an ideal. Let \(M\) be a finite \(R\)-module. Let \(\varphi : M \to M\) be an endomorphism such that \(\varphi(M) \subset IM\). Then there exists a monic polynomial \(P = T^n + a_1 T^{n - 1} + \ldots + a_n \in R[T]\) such that \(a_j \in I^j\) and \(P(\varphi) = 0\) as an endomorphism of \(M\).

Proof

Choose a surjective \(R\)-module map \(R^{\oplus n} \to M\), given by \((a_1, \ldots, a_n) \mapsto \sum a_ix_i\) for some generators \(x_i \in M\). Choose \((a_{i1}, \ldots, a_{in}) \in I^{\oplus n}\) such that \(\varphi(x_i) = \sum a_{ij} x_j\). In other words the diagram \[\xymatrix{ R^{\oplus n} \ar[d]_A \ar[r] & M \ar[d]^\varphi \\ I^{\oplus n} \ar[r] & M }\] is commutative where \(A = (a_{ij})\). By Lemma 00DX the polynomial \(P(t) = \det(t\text{id}_{n \times n} - A)\) has all the desired properties.

As a fun example application we prove the following surprising lemma.

Lemma

Let \(R\) be a ring. Let \(M\) be a finite \(R\)-module. Let \(\varphi : M \to M\) be a surjective \(R\)-module map. Then \(\varphi\) is an isomorphism.

Proof

Write \(R' = R[x]\) and think of \(M\) as a finite \(R'\)-module with \(x\) acting via \(\varphi\). Set \(I = (x) \subset R'\). By our assumption that \(\varphi\) is surjective we have \(IM = M\). Hence we may apply Lemma 05G7 to \(M\) as an \(R'\)-module, the ideal \(I\) and the endomorphism \(\text{id}_M\). We conclude that \((1 + a_1 + \ldots + a_n)\text{id}_M = 0\) with \(a_j \in I\). Write \(a_j = b_j(x)x\) for some \(b_j(x) \in R[x]\). Translating back into \(\varphi\) we see that \(\text{id}_M = -(\sum_{j = 1}^{n} b_j(\varphi)) \varphi\), and hence \(\varphi\) is invertible.

Proof

We perform induction on the number of generators of \(M\) over \(R\). If \(M\) is generated by one element, then \(M \cong R/I\) for some ideal \(I \subset R\). In this case we may replace \(R\) by \(R/I\) so that \(M = R\). In this case \(\varphi : R \to R\) is given by multiplication on \(M\) by an element \(r \in R\). The surjectivity of \(\varphi\) forces \(r\) invertible, since \(\varphi\) must hit \(1\), which implies that \(\varphi\) is invertible.

Now assume that we have proven the lemma in the case of modules generated by \(n - 1\) elements, and are examining a module \(M\) generated by \(n\) elements. Let \(A\) mean the ring \(R[t]\), and regard the module \(M\) as an \(A\)-module by letting \(t\) act via \(\varphi\); since \(M\) is finite over \(R\), it is finite over \(R[t]\) as well, and since we’re trying to prove \(\varphi\) injective, a set-theoretic property, we might as well prove the endomorphism \(t : M \to M\) over \(A\) injective. We have reduced our problem to the case our endomorphism is multiplication by an element of the ground ring. Let \(M' \subset M\) denote the sub-\(A\)-module generated by the first \(n - 1\) of the generators of \(M\), and consider the diagram \[\xymatrix{ 0 \ar[r] & M' \ar[r]\ar[d]^{\varphi\mid_{M'}} & M\ar[d]^\varphi \ar[r] & M/M' \ar[d]^{\varphi \bmod M'} \ar[r] & 0 \\ 0 \ar[r] & M' \ar[r] & M \ar[r] & M/M' \ar[r] & 0, }\] where the restriction of \(\varphi\) to \(M'\) and the map induced by \(\varphi\) on the quotient \(M/M'\) are well-defined since \(\varphi\) is multiplication by an element in the base, and \(M'\) and \(M/M'\) are \(A\)-modules in their own right. By the case \(n = 1\) the map \(M/M' \to M/M'\) is an isomorphism. A diagram chase implies that \(\varphi|_{M'}\) is surjective hence by induction \(\varphi|_{M'}\) is an isomorphism. This forces the middle column to be an isomorphism by the snake lemma.

The spectrum of a ring

We arbitrarily decide that the spectrum of a ring as a topological space is part of the algebra chapter, whereas an affine scheme is part of the chapter on schemes.

Definition

Let \(R\) be a ring.

1. The spectrum of \(R\) is the set of prime ideals of \(R\). It is usually denoted \(\Spec(R)\).

2. Given a subset \(T \subset R\) we let \(V(T) \subset \Spec(R)\) be the set of primes containing \(T\), i.e., \(V(T) = \{ \mathfrak p \in \Spec(R) \mid \forall f\in T, f\in \mathfrak p\}\).

3. Given an element \(f \in R\) we let \(D(f) \subset \Spec(R)\) be the set of primes not containing \(f\).

Lemma

Let \(R\) be a ring.

1. The spectrum of a ring \(R\) is empty if and only if \(R\) is the zero ring.

2. Every nonzero ring has a maximal ideal.

3. Every nonzero ring has a minimal prime ideal.

4. Given an ideal \(I \subset R\) and a prime ideal \(I \subset \mathfrak p\) there exists a prime \(I \subset \mathfrak q \subset \mathfrak p\) such that \(\mathfrak q\) is minimal over \(I\).

5. If \(T \subset R\), and if \((T)\) is the ideal generated by \(T\) in \(R\), then \(V((T)) = V(T)\).

6. If \(I\) is an ideal and \(\sqrt{I}\) is its radical, see basic notion (00BI), then \(V(I) = V(\sqrt{I})\).

7. Given an ideal \(I\) of \(R\) we have \(\sqrt{I} = \bigcap_{I \subset \mathfrak p} \mathfrak p\).

8. If \(I\) is an ideal then \(V(I) = \emptyset\) if and only if \(I\) is the unit ideal.

9. If \(I\), \(J\) are ideals of \(R\) then \(V(I) \cup V(J) = V(I \cap J)\).

10. If \((I_a)_{a\in A}\) is a set of ideals of \(R\) then \(\bigcap_{a\in A} V(I_a) = V(\bigcup_{a\in A} I_a)\).

11. If \(f \in R\), then \(D(f) \amalg V(f) = \Spec(R)\).

12. If \(f \in R\) then \(D(f) = \emptyset\) if and only if \(f\) is nilpotent.

13. If \(f = u f'\) for some unit \(u \in R\), then \(D(f) = D(f')\).

14. If \(I \subset R\) is an ideal, and \(\mathfrak p\) is a prime of \(R\) with \(\mathfrak p \not\in V(I)\), then there exists an \(f \in R\) such that \(\mathfrak p \in D(f)\), and \(D(f) \cap V(I) = \emptyset\).

15. If \(f, g \in R\), then \(D(fg) = D(f) \cap D(g)\).

16. If \(f_i \in R\) for \(i \in I\), then \(\bigcup_{i\in I} D(f_i)\) is the complement of \(V(\{f_i \}_{i\in I})\) in \(\Spec(R)\).

17. If \(f \in R\) and \(D(f) = \Spec(R)\), then \(f\) is a unit.

Proof

We address each part in the corresponding item below.

1. This is a direct consequence of (2) or (3).

2. Let \(\mathfrak{A}\) be the set of all proper ideals of \(R\). This set is ordered by inclusion and is non-empty, since \((0) \in \mathfrak{A}\) is a proper ideal. Let \(A\) be a totally ordered subset of \(\mathfrak A\). Then \(\bigcup_{I \in A} I\) is in fact an ideal. Since \(1 \notin I\) for all \(I \in A\), the union does not contain \(1\) and thus is proper. Hence \(\bigcup_{I \in A} I\) is in \(\mathfrak{A}\) and is an upper bound for the set \(A\). Thus by Zorn’s lemma \(\mathfrak{A}\) has a maximal element, which is the sought-after maximal ideal.

3. Since \(R\) is nonzero, it contains a maximal ideal which is a prime ideal. Thus the set \(\mathfrak{A}\) of all prime ideals of \(R\) is nonempty. \(\mathfrak{A}\) is ordered by reverse-inclusion. Let \(A\) be a totally ordered subset of \(\mathfrak{A}\). It’s pretty clear that \(J = \bigcap_{I \in A} I\) is in fact an ideal. Not so clear, however, is that it is prime. Let \(xy \in J\). Then \(xy \in I\) for all \(I \in A\). Now let \(B = \{I \in A | y \in I\}\). Let \(K = \bigcap_{I \in B} I\). Since \(A\) is totally ordered, either \(K = J\) (and we’re done, since then \(y \in J\)) or \(K \supset J\) and for all \(I \in A\) such that \(I\) is properly contained in \(K\), we have \(y \notin I\). But that means that for all those \(I, x \in I\), since they are prime. Hence \(x \in J\). In either case, \(J\) is prime as desired. Hence by Zorn’s lemma we get a maximal element which in this case is a minimal prime ideal.

4. This is the same exact argument as (3) except you only consider prime ideals contained in \(\mathfrak{p}\) and containing \(I\).

5. \((T)\) is the smallest ideal containing \(T\). Hence if \(T \subset I\), some ideal, then \((T) \subset I\) as well. Hence if \(I \in V(T)\), then \(I \in V((T))\) as well. The other inclusion is obvious.

6. Since \(I \subset \sqrt{I}, V(\sqrt{I}) \subset V(I)\). Now let \(\mathfrak{p} \in V(I)\). Let \(x \in \sqrt{I}\). Then \(x^n \in I\) for some \(n\). Hence \(x^n \in \mathfrak{p}\). But since \(\mathfrak{p}\) is prime, a boring induction argument gets you that \(x \in \mathfrak{p}\). Hence \(\sqrt{I} \subset \mathfrak{p}\) and \(\mathfrak{p} \in V(\sqrt{I})\).

7. Let \(f \in R \setminus \sqrt{I}\). Then \(f^n \notin I\) for all \(n\). Hence \(S = \{1, f, f^2, \ldots\}\) is a multiplicative subset, not containing \(0\). Take a prime ideal \(\bar{\mathfrak{p}} \subset S^{-1}R\) containing \(S^{-1}I\). Then the pull-back \(\mathfrak{p}\) in \(R\) of \(\bar{\mathfrak{p}}\) is a prime ideal containing \(I\) that does not intersect \(S\). This shows that \(\bigcap_{I \subset \mathfrak p} \mathfrak p \subset \sqrt{I}\). Now if \(a \in \sqrt{I}\), then \(a^n \in I\) for some \(n\). Hence if \(I \subset \mathfrak{p}\), then \(a^n \in \mathfrak{p}\). But since \(\mathfrak{p}\) is prime, we have \(a \in \mathfrak{p}\). Thus the equality is shown.

8. \(I\) is not the unit ideal if and only if \(I\) is contained in some maximal ideal (to see this, apply (2) to the ring \(R/I\)) which is therefore prime.

9. If \(\mathfrak{p} \in V(I) \cup V(J)\), then \(I \subset \mathfrak{p}\) or \(J \subset \mathfrak{p}\) which means that \(I \cap J \subset \mathfrak{p}\). Now if \(I \cap J \subset \mathfrak{p}\), then \(IJ \subset \mathfrak{p}\) and hence either \(I \subset \mathfrak{p}\) or \(J \subset \mathfrak{p}\), since \(\mathfrak{p}\) is prime.

10. \(\mathfrak{p} \in \bigcap_{a \in A} V(I_a) \Leftrightarrow I_a \subset \mathfrak{p}, \forall a \in A \Leftrightarrow \mathfrak{p} \in V(\bigcup_{a\in A} I_a)\)

11. If \(\mathfrak{p}\) is a prime ideal and \(f \in R\), then either \(f \in \mathfrak{p}\) or \(f \notin \mathfrak{p}\) (strictly) which is what the disjoint union says.

12. If \(a \in R\) is nilpotent, then \(a^n = 0\) for some \(n\). Hence \(a^n \in \mathfrak{p}\) for any prime ideal. Thus \(a \in \mathfrak{p}\) as can be shown by induction and \(D(a) = \emptyset\). Now, as shown in (7), if \(a \in R\) is not nilpotent, then there is a prime ideal that does not contain it.

13. \(f \in \mathfrak{p} \Leftrightarrow uf \in \mathfrak{p}\), since \(u\) is invertible.

14. If \(\mathfrak{p} \notin V(I)\), then \(\exists f \in I \setminus \mathfrak{p}\). Then \(f \notin \mathfrak{p}\) so \(\mathfrak{p} \in D(f)\). Also if \(\mathfrak{q} \in D(f)\), then \(f \notin \mathfrak{q}\) and thus \(I\) is not contained in \(\mathfrak{q}\). Thus \(D(f) \cap V(I) = \emptyset\).

15. If \(fg \in \mathfrak{p}\), then \(f \in \mathfrak{p}\) or \(g \in \mathfrak{p}\). Hence if \(f \notin \mathfrak{p}\) and \(g \notin \mathfrak{p}\), then \(fg \notin \mathfrak{p}\). Since \(\mathfrak{p}\) is an ideal, if \(fg \notin \mathfrak{p}\), then \(f \notin \mathfrak{p}\) and \(g \notin \mathfrak{p}\).

16. \(\mathfrak{p} \in \bigcup_{i \in I} D(f_i) \Leftrightarrow \exists i \in I, f_i \notin \mathfrak{p} \Leftrightarrow \mathfrak{p} \in \Spec(R) \setminus V(\{f_i\}_{i \in I})\)

17. If \(D(f) = \Spec(R)\), then \(V(f) = \emptyset\) and hence \(fR = R\), so \(f\) is a unit.

The lemma implies that the subsets \(V(T)\) from Definition 00DZ form the closed subsets of a topology on \(\Spec(R)\). And it also shows that the sets \(D(f)\) are open and form a basis for this topology.

Definition

Let \(R\) be a ring. The topology on \(\Spec(R)\) whose closed sets are the sets \(V(T)\) is called the Zariski topology. The open subsets \(D(f)\) are called the standard opens of \(\Spec(R)\).

It should be clear from context whether we consider \(\Spec(R)\) just as a set or as a topological space.

Lemma

Suppose that \(\varphi : R \to R'\) is a ring homomorphism. The induced map \[\Spec(\varphi) : \Spec(R') \longrightarrow \Spec(R), \quad \mathfrak p' \longmapsto \varphi^{-1}(\mathfrak p')\] is continuous for the Zariski topologies. In fact, for any element \(f \in R\) we have \(\Spec(\varphi)^{-1}(D(f)) = D(\varphi(f))\).

Proof

It is basic notion (00BV) that \(\mathfrak p := \varphi^{-1}(\mathfrak p')\) is indeed a prime ideal of \(R\). The last assertion of the lemma follows directly from the definitions, and implies the first.

If \(\varphi' : R' \to R''\) is a second ring homomorphism then the composition \[\Spec(R'') \longrightarrow \Spec(R') \longrightarrow \Spec(R)\] equals \(\Spec(\varphi' \circ \varphi)\). In other words, \(\Spec\) is a contravariant functor from the category of rings to the category of topological spaces.

Lemma

Let \(\varphi : R \to S\) be a ring map. Assume that every \(g \in S\) can be written as \(g = u\varphi(f)\) for some \(f \in R\) and some unit \(u \in S\). Then \[\Spec(S) \longrightarrow \Spec(R)\] is a homeomorphism onto its image.

Proof

The map is continuous by Lemma 00E2. If \(\mathfrak q\) and \(\mathfrak q'\) have the same inverse image in \(R\), then the assumption shows that every \(g \in S\) is in \(\mathfrak q\) if and only if it is in \(\mathfrak q'\). Thus the map is injective. Finally, if \(g = u\varphi(f)\) as in the statement, then \[D(g) = \Spec(\varphi)^{-1}(D(f)).\] Since the standard opens form a basis, the map is a homeomorphism onto its image.

Lemma

Let \(R\) be a ring. Let \(S \subset R\) be a multiplicative subset. The map \(R \to S^{-1}R\) induces via the functoriality of \(\Spec\) a homeomorphism \[\Spec(S^{-1}R) \longrightarrow \{\mathfrak p \in \Spec(R) \mid S \cap \mathfrak p = \emptyset \}\] where the topology on the right hand side is that induced from the Zariski topology on \(\Spec(R)\). The inverse map is given by \(\mathfrak p \mapsto S^{-1}\mathfrak p = \mathfrak p(S^{-1}R)\).

Proof

Denote the right hand side of the arrow of the lemma by \(D\). Choose a prime \(\mathfrak p' \subset S^{-1}R\) and let \(\mathfrak p\) be the inverse image of \(\mathfrak p'\) in \(R\). Since \(\mathfrak p'\) does not contain \(1\) we see that \(\mathfrak p\) does not contain any element of \(S\). Hence \(\mathfrak p \in D\) and we see that the image is contained in \(D\). Let \(\mathfrak p \in D\). By assumption the image \(\overline{S}\) does not contain \(0\). By basic notion (00C6) \(\overline{S}^{-1}(R/\mathfrak p)\) is not the zero ring. By basic notion (00CD) we see \(S^{-1}R / S^{-1}\mathfrak p = \overline{S}^{-1}(R/\mathfrak p)\) is a domain, and hence \(S^{-1}\mathfrak p\) is a prime. The equality of rings also shows that the inverse image of \(S^{-1}\mathfrak p\) in \(R\) is equal to \(\mathfrak p\), because \(R/\mathfrak p \to \overline{S}^{-1}(R/\mathfrak p)\) is injective by basic notion (00C7). This proves that the map \(\Spec(S^{-1}R) \to \Spec(R)\) is bijective onto \(D\) with inverse as given. Every element of \(S^{-1}R\) is a unit times the image of an element of \(R\). Hence the map is a homeomorphism onto its image by Lemma algebra-lemma-spec-homeomorphism-onto-image-units.

Lemma

Let \(R\) be a ring. Let \(f \in R\). The map \(R \to R_f\) induces via the functoriality of \(\Spec\) a homeomorphism \[\Spec(R_f) \longrightarrow D(f) \subset \Spec(R).\] The inverse is given by \(\mathfrak p \mapsto \mathfrak p \cdot R_f\).

Proof

This is a special case of Lemma 00E3.

It is not the case that every “affine open” of a spectrum is a standard open. See Example 00F1.

Lemma

Let \(R\) be a ring. Let \(I \subset R\) be an ideal. The map \(R \to R/I\) induces via the functoriality of \(\Spec\) a homeomorphism \[\Spec(R/I) \longrightarrow V(I) \subset \Spec(R).\] The inverse is given by \(\mathfrak p \mapsto \mathfrak p / I\).

Proof

It is immediate that the image is contained in \(V(I)\). On the other hand, if \(\mathfrak p \in V(I)\) then \(\mathfrak p \supset I\) and we may consider the ideal \(\mathfrak p /I \subset R/I\). Using basic notion (00C3) we see that \((R/I)/(\mathfrak p/I) = R/\mathfrak p\) is a domain and hence \(\mathfrak p/I\) is a prime ideal. From this and Lemma algebra-lemma-spec-homeomorphism-onto-image-units, applied to the surjection \(R \to R/I\), the result follows.

Lemma

Let \(R\) be a ring. The space \(\Spec(R)\) is quasi-compact.

Proof

It suffices to prove that any covering of \(\Spec(R)\) by standard opens can be refined by a finite covering. Thus suppose that \(\Spec(R) = \cup D(f_i)\) for a set of elements \(\{f_i\}_{i\in I}\) of \(R\). This means that \(\cap V(f_i) = \emptyset\). According to Lemma 00E0 this means that \(V(\{f_i \}) = \emptyset\). According to the same lemma this means that the ideal generated by the \(f_i\) is the unit ideal of \(R\). This means that we can write \(1\) as a finite sum: \(1 = \sum_{i \in J} r_i f_i\) with \(J \subset I\) finite. And then it follows that \(\Spec(R) = \cup_{i \in J} D(f_i)\).

Lemma

Let \(R\) be a ring. The topology on \(X = \Spec(R)\) has the following properties:

1. \(X\) is quasi-compact,

2. \(X\) has a basis for the topology consisting of quasi-compact opens, and

3. the intersection of any two quasi-compact opens is quasi-compact.

Proof

The spectrum of a ring is quasi-compact, see Lemma 00E8. It has a basis for the topology consisting of the standard opens \(D(f) = \Spec(R_f)\) (Lemma 00E4) which are quasi-compact by the first remark. The intersection of two standard opens is quasi-compact as \(D(f) \cap D(g) = D(fg)\). Given any two quasi-compact opens \(U, V \subset X\) we may write \(U = D(f_1) \cup \ldots \cup D(f_n)\) and \(V = D(g_1) \cup \ldots \cup D(g_m)\). Then \(U \cap V = \bigcup_{1 \leq i \leq n,\ 1 \leq j \leq m} D(f_i g_j)\) which is quasi-compact.

Local rings

Local rings are the bread and butter of algebraic geometry.

Definition

A local ring is a ring with exactly one maximal ideal. If \(R\) is a local ring, then the maximal ideal is often denoted \(\mathfrak m_R\) and the field \(R/\mathfrak m_R\) is called the residue field of the local ring \(R\). We often say “let \((R, \mathfrak m)\) be a local ring” or “let \((R, \mathfrak m, \kappa)\) be a local ring” to indicate that \(R\) is local, \(\mathfrak m\) is its unique maximal ideal and \(\kappa = R/\mathfrak m\) is its residue field. A local homomorphism of local rings is a ring map \(\varphi : R \to S\) such that \(R\) and \(S\) are local rings and such that \(\varphi(\mathfrak m_R) \subset \mathfrak m_S\). If it is given that \(R\) and \(S\) are local rings, then the phrase “local ring map \(\varphi : R \to S\)” means that \(\varphi\) is a local homomorphism of local rings.

A field is a local ring. Any ring map between fields is a local homomorphism of local rings.

The localization \(R_\mathfrak p\) of a ring \(R\) at a prime \(\mathfrak p\) is a local ring with maximal ideal \(\mathfrak p R_\mathfrak p\). Namely, by Lemma 00E3 every prime ideal of \(R_\mathfrak p\) is contained in the prime ideal \(\mathfrak p R_\mathfrak p\) (hence this is a maximal ideal and the only maximal ideal of \(R_\mathfrak p\)). The residue field of \(R_\mathfrak p\) is denoted \(\kappa(\mathfrak p)\); we call it the residue field of \(\mathfrak p\); by Proposition 00CT we may identify \(\kappa(\mathfrak p)\) with the field of fractions of the domain \(R/\mathfrak p\). Via the composition \[\Spec(\kappa(\mathfrak p)) \to \Spec(R_\mathfrak p) \to \Spec(R)\] the unique point of the source maps to the point \(\mathfrak p\) of the target.

Let \(\varphi : R \to S\) be a ring map. Let \(\mathfrak q \subset S\) be a prime and consider the prime \(\mathfrak p = \varphi^{-1}(\mathfrak q)\) of \(R\). Since \(\varphi(\mathfrak p) \subset \mathfrak q\) the induced ring map \[R_\mathfrak p \to S_\mathfrak q,\quad r/g \mapsto \varphi(r)/\varphi(g)\] is a local ring map and we obtain an induced map of residue fields \(\kappa(\mathfrak p) \to \kappa(\mathfrak q)\).

Example

If \(R\) is a local ring and \(\mathfrak p \subset R\) is a non-maximal prime ideal, then \(R \to R_\mathfrak p\) is not a local homomorphism.

Lemma

Let \(R\) be a ring. The following are equivalent:

1. \(R\) is a local ring,

2. \(\Spec(R)\) has exactly one closed point,

3. \(R\) has a maximal ideal \(\mathfrak m\) and every element of \(R \setminus \mathfrak m\) is a unit, and

4. \(R\) is not the zero ring and for every \(x \in R\) either \(x\) or \(1 - x\) is invertible or both.

Proof

Let \(R\) be a ring, and \(\mathfrak m\) a maximal ideal. If \(x \in R \setminus \mathfrak m\), and \(x\) is not a unit then there is a maximal ideal \(\mathfrak m'\) containing \(x\). Hence \(R\) has at least two maximal ideals. Conversely, if \(\mathfrak m'\) is another maximal ideal, then choose \(x \in \mathfrak m'\), \(x \not \in \mathfrak m\). Clearly \(x\) is not a unit. This proves the equivalence of (1) and (3). The equivalence of (1) and (2) is tautological. If \(R\) is local then (4) holds since \(x\) is either in \(\mathfrak m\) or not. If (4) holds, and \(\mathfrak m\), \(\mathfrak m'\) are distinct maximal ideals then we may choose \(x \in R\) such that \(x \bmod \mathfrak m' = 0\) and \(x \bmod \mathfrak m = 1\) by the Chinese remainder theorem (Lemma 00DT). This element \(x\) is not invertible and neither is \(1 - x\) which is a contradiction. Thus (4) and (1) are equivalent.

Lemma

Let \(\varphi : R \to S\) be a ring map. Assume \(R\) and \(S\) are local rings. The following are equivalent:

1. \(\varphi\) is a local ring map,

2. \(\varphi(\mathfrak m_R) \subset \mathfrak m_S\),

3. \(\varphi^{-1}(\mathfrak m_S) = \mathfrak m_R\), and

4. for any \(x \in R\), if \(\varphi(x)\) is invertible in \(S\), then \(x\) is invertible in \(R\).

Proof

Conditions (1) and (2) are equivalent by definition. If (3) holds then (2) holds. Conversely, if (2) holds, then \(\varphi^{-1}(\mathfrak m_S)\) is a prime ideal containing the maximal ideal \(\mathfrak m_R\), hence \(\varphi^{-1}(\mathfrak m_S) = \mathfrak m_R\). Finally, (4) is the contrapositive of (2) by Lemma 00E9.

Remark

A fundamental commutative diagram associated to a ring map \(\varphi : R \to S\) and a prime \(\mathfrak p \subset R\) is the following \[\xymatrix{ \kappa(\mathfrak p) \otimes_R S = S_{\mathfrak p}/{\mathfrak p}S_{\mathfrak p} & S_{\mathfrak p} \ar[l] & S \ar[r] \ar[l] & S/\mathfrak pS \ar[r] & (R \setminus \mathfrak p)^{-1}S/\mathfrak pS \\ \kappa(\mathfrak p) = R_{\mathfrak p}/{\mathfrak p}R_{\mathfrak p} \ar[u] & R_{\mathfrak p} \ar[u] \ar[l] & R \ar[u] \ar[r] \ar[l] & R/\mathfrak p \ar[u] \ar[r] & \kappa(\mathfrak p) \ar[u] }\] In this diagram the outer left and outer right columns are identical. On spectra the horizontal maps induce homeomorphisms onto their images and the squares induce fibre squares of topological spaces (see Lemmas 00E3 and 00E5). This shows that \(\mathfrak p\) is in the image of the map on Spec if and only if \(S \otimes_R \kappa(\mathfrak p)\) is not the zero ring. If there does exist a prime \(\mathfrak q \subset S\) lying over \(\mathfrak p\), i.e., with \(\mathfrak p = \varphi^{-1}(\mathfrak q)\) then we can extend the diagram to the following diagram \[\xymatrix{ \kappa(\mathfrak q) = S_{\mathfrak q}/{\mathfrak q}S_{\mathfrak q} & S_{\mathfrak q} \ar[l] & S \ar[r] \ar[l] & S/\mathfrak q \ar[r] & \kappa(\mathfrak q) \\ \kappa(\mathfrak p) \otimes_R S = S_{\mathfrak p}/{\mathfrak p}S_{\mathfrak p} \ar[u] & S_{\mathfrak p} \ar[u] \ar[l] & S \ar[u] \ar[r] \ar[l] & S/\mathfrak pS \ar[u] \ar[r] & (R \setminus \mathfrak p)^{-1}S/\mathfrak pS \ar[u] \\ \kappa(\mathfrak p) = R_{\mathfrak p}/{\mathfrak p}R_{\mathfrak p} \ar[u] & R_{\mathfrak p} \ar[u] \ar[l] & R \ar[u] \ar[r] \ar[l] & R/\mathfrak p \ar[u] \ar[r] & \kappa(\mathfrak p) \ar[u] }\] In this diagram it is still the case that the outer left and outer right columns are identical and that on spectra the horizontal maps induce homeomorphisms onto their image.

Lemma

Let \(\varphi : R \to S\) be a ring map. Let \(\mathfrak p\) be a prime of \(R\). The following are equivalent

1. \(\mathfrak p\) is in the image of \(\Spec(S) \to \Spec(R)\),

2. \(S \otimes_R \kappa(\mathfrak p) \not = 0\),

3. \(S_{\mathfrak p}/\mathfrak p S_{\mathfrak p} \not = 0\),

4. \((S/\mathfrak pS)_{\mathfrak p} \not = 0\), and

5. \(\mathfrak p = \varphi^{-1}(\mathfrak pS)\).

Proof

We have already seen the equivalence of the first two in Remark 00E6. The others are just reformulations of this.

Remark

Let \(R \to S\) be a ring map. Let \(\mathfrak q\) be a prime ideal of \(S\) lying over the prime ideal \(\mathfrak p\) of \(R\). According to Remark 00E6 the prime \(\mathfrak q\) corresponds to a unique prime \(\overline{\mathfrak q}\) of the fibre ring \(F = S \otimes_R \kappa(\mathfrak p)\). Then we have \[F_{\overline{\mathfrak q}} \cong S_\mathfrak q \otimes_{R_\mathfrak p} \kappa(\mathfrak p) \cong S_\mathfrak q/\mathfrak p S_\mathfrak q\] Namely, there is an obvious ring map \(F \to S_\mathfrak q \otimes_{R_\mathfrak p} \kappa(\mathfrak p)\) which, under the displayed isomorphism, identifies with \(F \to F_{\overline{\mathfrak q}}\). The second isomorphism follows from the fact that \(\kappa(\mathfrak p)\) is the quotient of \(R_\mathfrak p\) by \(\mathfrak pR_\mathfrak p\).

The Jacobson radical of a ring

We recall that the Jacobson radical \(\text{rad}(R)\) of a ring \(R\) is the intersection of all maximal ideals of \(R\). If \(R\) is local then \(\text{rad}(R)\) is the maximal ideal of \(R\).

Lemma

Let \(R\) be a ring with Jacobson radical \(\text{rad}(R)\). Let \(I \subset R\) be an ideal. The following are equivalent

1. \(I \subset \text{rad}(R)\), and

2. every element of \(1 + I\) is a unit in \(R\).

In this case every element of \(R\) which maps to a unit of \(R/I\) is a unit.

Proof

If \(f \in \text{rad}(R)\), then \(f \in \mathfrak m\) for all maximal ideals \(\mathfrak m\) of \(R\). Hence \(1 + f \not \in \mathfrak m\) for all maximal ideals \(\mathfrak m\) of \(R\). Thus the closed subset \(V(1 + f)\) of \(\Spec(R)\) is empty. This implies that \(1 + f\) is a unit, see Lemma 00E0.

Conversely, assume that \(1 + f\) is a unit for all \(f \in I\). If \(\mathfrak m\) is a maximal ideal and \(I \not \subset \mathfrak m\), then \(I + \mathfrak m = R\). Hence \(1 = f + g\) for some \(g \in \mathfrak m\) and \(f \in I\). Then \(g = 1 + (-f)\) is not a unit, contradiction.

For the final statement let \(f \in R\) map to a unit in \(R/I\). Then we can find \(g \in R\) mapping to the multiplicative inverse of \(f \bmod I\). Then \(fg = 1 \bmod I\). Hence \(fg\) is a unit of \(R\) by (2) which implies that \(f\) is a unit.

Lemma

Let \(R\) be a ring and let \(I \subset R\) be an ideal contained in the Jacobson radical of \(R\). If \(U \subset \Spec(R)\) is an open subset and \(V(I) \subset U\), then \(U = \Spec(R)\).

Proof

Suppose that \(Z = \Spec(R) \setminus U\) is nonempty. Write \(Z = V(J)\) for an ideal \(J \subset R\). Then \(J\) is a proper ideal, so it is contained in a maximal ideal \(\mathfrak m\). Since \(I\) is contained in every maximal ideal, we have \(\mathfrak m \in V(I) \cap V(J)\). This contradicts \(V(I) \subset U\).

Lemma

Let \(\varphi : R \to S\) be a ring map such that the induced map \(\Spec(S) \to \Spec(R)\) is surjective. Then an element \(x \in R\) is a unit if and only if \(\varphi(x) \in S\) is a unit.

Proof

If \(x\) is a unit, then so is \(\varphi(x)\). Conversely, if \(\varphi(x)\) is a unit, then \(\varphi(x) \not \in \mathfrak q\) for all \(\mathfrak q \in \Spec(S)\). Hence \(x \not \in \varphi^{-1}(\mathfrak q) = \Spec(\varphi)(\mathfrak q)\) for all \(\mathfrak q \in \Spec(S)\). Since \(\Spec(\varphi)\) is surjective we conclude that \(x\) is a unit by part (17) of Lemma 00E0.

Nakayama’s lemma

We quote from [MatCA]: “This simple but important lemma is due to T. Nakayama, G. Azumaya and W. Krull. Priority is obscure, and although it is usually called the Lemma of Nakayama, late Prof. Nakayama did not like the name.”

Lemma

Let \(R\) be a ring with Jacobson radical \(\text{rad}(R)\). Let \(M\) be an \(R\)-module. Let \(I \subset R\) be an ideal.

1. If \(IM = M\) and \(M\) is finite, then there exists an \(f \in 1 + I\) such that \(fM = 0\).

2. If \(IM = M\), \(M\) is finite, and \(I \subset \text{rad}(R)\), then \(M = 0\).

3. If \(N, N' \subset M\), \(M = N + IN'\), and \(N'\) is finite, then there exists an \(f \in 1 + I\) such that \(fM \subset N\) and \(M_f = N_f\).

4. If \(N, N' \subset M\), \(M = N + IN'\), \(N'\) is finite, and \(I \subset \text{rad}(R)\), then \(M = N\).

5. If \(N \to M\) is a module map, \(N/IN \to M/IM\) is surjective, and \(M\) is finite, then there exists an \(f \in 1 + I\) such that \(N_f \to M_f\) is surjective.

6. If \(N \to M\) is a module map, \(N/IN \to M/IM\) is surjective, \(M\) is finite, and \(I \subset \text{rad}(R)\), then \(N \to M\) is surjective.

7. If \(x_1, \ldots, x_n \in M\) generate \(M/IM\) and \(M\) is finite, then there exists an \(f \in 1 + I\) such that \(x_1, \ldots, x_n\) generate \(M_f\) over \(R_f\).

8. If \(x_1, \ldots, x_n \in M\) generate \(M/IM\), \(M\) is finite, and \(I \subset \text{rad}(R)\), then \(M\) is generated by \(x_1, \ldots, x_n\).

9. If \(IM = M\), \(I\) is nilpotent, then \(M = 0\).

10. If \(N, N' \subset M\), \(M = N + IN'\), and \(I\) is nilpotent then \(M = N\).

11. If \(N \to M\) is a module map, \(I\) is nilpotent, and \(N/IN \to M/IM\) is surjective, then \(N \to M\) is surjective.

12. If \(\{x_\alpha\}_{\alpha \in A}\) is a set of elements of \(M\) which generate \(M/IM\) and \(I\) is nilpotent, then \(M\) is generated by the \(x_\alpha\).

Proof

Proof of (00DW). Choose generators \(y_1, \ldots, y_m\) of \(M\) over \(R\). For each \(i\) we can write \(y_i = \sum z_{ij} y_j\) with \(z_{ij} \in I\) (since \(M = IM\)). In other words \(\sum_j (\delta_{ij} - z_{ij})y_j = 0\). Let \(f\) be the determinant of the \(m \times m\) matrix \(A = (\delta_{ij} - z_{ij})\). Note that \(f \in 1 + I\) (since the matrix \(A\) is entrywise congruent to the \(m \times m\) identity matrix modulo \(I\)). By Lemma 07DQ (1), there exists an \(m \times m\) matrix \(B\) such that \(BA = f 1_{m \times m}\). Writing out we see that \(\sum_{i} b_{hi} a_{ij} = f \delta_{hj}\) for all \(h\) and \(j\); hence, \(\sum_{i, j} b_{hi} a_{ij} y_j = \sum_{j} f \delta_{hj} y_j = f y_h\) for every \(h\). In other words, \(0 = f y_h\) for every \(h\) (since each \(i\) satisfies \(\sum_j a_{ij} y_j = 0\)). This implies that \(f\) annihilates \(M\).

By Lemma 0AME an element of \(1 + \text{rad}(R)\) is an invertible element of \(R\). Hence we see that (00DW) implies (2). We obtain (3) by applying (1) to \(M/N\) which is finite as \(N'\) is finite. We obtain (4) by applying (2) to \(M/N\) which is finite as \(N'\) is finite. We obtain (5) by applying (3) to \(M\) and the submodules \(\Im(N \to M)\) and \(M\). We obtain (6) by applying (4) to \(M\) and the submodules \(\Im(N \to M)\) and \(M\). We obtain (7) by applying (5) to the map \(R^{\oplus n} \to M\), \((a_1, \ldots, a_n) \mapsto a_1x_1 + \ldots + a_nx_n\). We obtain (8) by applying (6) to the map \(R^{\oplus n} \to M\), \((a_1, \ldots, a_n) \mapsto a_1x_1 + \ldots + a_nx_n\).

Part (9) holds because if \(M = IM\) then \(M = I^nM\) for all \(n \geq 0\) and \(I\) being nilpotent means \(I^n = 0\) for some \(n \gg 0\). Parts (10), (11), and (12) follow from (9) by the arguments used above.

Lemma

Let \(R\) be a ring, let \(S \subset R\) be a multiplicative subset, let \(I \subset R\) be an ideal, and let \(M\) be a finite \(R\)-module. If \(x_1, \ldots, x_r \in M\) generate \(S^{-1}(M/IM)\) as an \(S^{-1}(R/I)\)-module, then there exists an \(f \in S + I\) such that \(x_1, \ldots, x_r\) generate \(M_f\) as an \(R_f\)-module.¹

Proof

Special case \(I = 0\). Let \(y_1, \ldots, y_s\) be generators for \(M\) over \(R\). Since \(S^{-1}M\) is generated by \(x_1, \ldots, x_r\), for each \(i\) we can write \(y_i = \sum (a_{ij}/s_{ij})x_j\) in \(S^{-1}M\) for some \(a_{ij} \in R\) and \(s_{ij} \in S\). Multiplying by the product \(s \in S\) of the \(s_{ij}\) we see that \(sy_i = \sum a'_{ij}x_j\) in \(S^{-1}M\) for some \(a'_{ij} \in R\). This in turn means there exist \(t_i \in S\) such that \(t_isy_i = \sum t_ia'_{ij}x_j\) in \(M\). Thus if \(t \in S\) is the product of the \(t_i\), then we see that \(y_i\) is in the \(R_{st}\)-submodule generated by \(x_1, \ldots, x_r\) of \(M_{st}\). Hence \(x_1, \ldots, x_r\) generate \(M_{st}\).

General case. By the special case, we can find an \(s \in S\) such that \(x_1, \ldots, x_r\) generate \((M/IM)_s\) over \((R/I)_s\). By Lemma 00DV we can find a \(g \in 1 + I_s \subset R_s\) such that \(x_1, \ldots, x_r\) generate \((M_s)_g\) over \((R_s)_g\). Write \(g = 1 + i/s'\). Then \(f = ss' + is\) works; details omitted.

Lemma

Let \(A \to B\) be a local homomorphism of local rings. Assume

1. \(B\) is finite as an \(A\)-module,

2. \(\mathfrak m_B\) is a finitely generated ideal,

3. \(A \to B\) induces an isomorphism on residue fields, and

4. \(\mathfrak m_A/\mathfrak m_A^2 \to \mathfrak m_B/\mathfrak m_B^2\) is surjective.

Then \(A \to B\) is surjective.

Proof

To show that \(A \to B\) is surjective, we view it as a map of \(A\)-modules and apply Lemma 00DV (6). We conclude it suffices to show that \(A/\mathfrak m_A \to B/\mathfrak m_AB\) is surjective. As \(A/\mathfrak m_A = B/\mathfrak m_B\) it suffices to show that \(\mathfrak m_AB \to \mathfrak m_B\) is surjective. View \(\mathfrak m_AB \to \mathfrak m_B\) as a map of \(B\)-modules and apply Lemma 00DV (6). We conclude it suffices to see that \(\mathfrak m_AB/\mathfrak m_A\mathfrak m_B \to \mathfrak m_B/\mathfrak m_B^2\) is surjective. This follows from assumption (4).

Open and closed subsets of spectra

It turns out that open and closed subsets of a spectrum correspond to idempotents of the ring.

Lemma

Let \(R\) be a ring. Let \(e \in R\) be an idempotent. In this case \[\Spec(R) = D(e) \amalg D(1-e).\]

Proof

Note that an idempotent \(e\) of a domain is either \(1\) or \(0\). Hence we see that \[\begin{eqnarray*} D(e) & = & \{ \mathfrak p \in \Spec(R) \mid e \not\in \mathfrak p \} \\ & = & \{ \mathfrak p \in \Spec(R) \mid e \not = 0\text{ in }\kappa(\mathfrak p) \} \\ & = & \{ \mathfrak p \in \Spec(R) \mid e = 1\text{ in }\kappa(\mathfrak p) \} \end{eqnarray*}\] Similarly we have \[\begin{eqnarray*} D(1-e) & = & \{ \mathfrak p \in \Spec(R) \mid 1 - e \not\in \mathfrak p \} \\ & = & \{ \mathfrak p \in \Spec(R) \mid e \not = 1\text{ in }\kappa(\mathfrak p) \} \\ & = & \{ \mathfrak p \in \Spec(R) \mid e = 0\text{ in }\kappa(\mathfrak p) \} \end{eqnarray*}\] Since the image of \(e\) in any residue field is either \(1\) or \(0\) we deduce that \(D(e)\) and \(D(1-e)\) cover all of \(\Spec(R)\).

Lemma

Let \(R_1\) and \(R_2\) be rings. Let \(R = R_1 \times R_2\). The maps \(R \to R_1\), \((x, y) \mapsto x\) and \(R \to R_2\), \((x, y) \mapsto y\) induce continuous maps \(\Spec(R_1) \to \Spec(R)\) and \(\Spec(R_2) \to \Spec(R)\). The induced map \[\Spec(R_1) \amalg \Spec(R_2) \longrightarrow \Spec(R)\] is a homeomorphism. In other words, the spectrum of \(R = R_1\times R_2\) is the disjoint union of the spectrum of \(R_1\) and the spectrum of \(R_2\).

Proof

Write \(1 = e_1 + e_2\) with \(e_1 = (1, 0)\) and \(e_2 = (0, 1)\). Note that \(e_1\) and \(e_2 = 1 - e_1\) are idempotents. We leave it to the reader to show that \(R_1 = R_{e_1}\) is the localization of \(R\) at \(e_1\). Similarly for \(e_2\). Thus the statement of the lemma follows from Lemma 00EC combined with Lemma 00E4.

We reprove the following lemma later after introducing a glueing lemma for functions. See Section 00EI.

Lemma

Let \(R\) be a ring. For each \(U \subset \Spec(R)\) which is open and closed there exists a unique idempotent \(e \in R\) such that \(U = D(e)\). This induces a 1-1 correspondence between open and closed subsets \(U \subset \Spec(R)\) and idempotents \(e \in R\).

Proof

Let \(U \subset \Spec(R)\) be open and closed. Since \(U\) is closed it is quasi-compact by Lemma 00E8, and similarly for its complement. Write \(U = \bigcup_{i = 1}^n D(f_i)\) as a finite union of standard opens. Similarly, write \(\Spec(R) \setminus U = \bigcup_{j = 1}^m D(g_j)\) as a finite union of standard opens. Since \(\emptyset = D(f_i) \cap D(g_j) = D(f_i g_j)\) we see that \(f_i g_j\) is nilpotent by Lemma 00E0. Let \(I = (f_1, \ldots, f_n) \subset R\) and let \(J = (g_1, \ldots, g_m) \subset R\). Note that \(V(J)\) equals \(U\), that \(V(I)\) equals the complement of \(U\), so \(\Spec(R) = V(I) \amalg V(J)\). By the remark on nilpotency above, we see that \((IJ)^N = (0)\) for some sufficiently large integer \(N\). Since \(\bigcup D(f_i) \cup \bigcup D(g_j) = \Spec(R)\) we see that \(I + J = R\), see Lemma 00E0. By raising this equation to the \(2N\)th power we conclude that \(I^N + J^N = R\). Write \(1 = x + y\) with \(x \in I^N\) and \(y \in J^N\). Then \(0 = xy = x(1 - x)\) as \(I^N J^N = (0)\). Thus \(x = x^2\) is idempotent and contained in \(I^N \subset I\). The idempotent \(y = 1 - x\) is contained in \(J^N \subset J\). This shows that the idempotent \(x\) maps to \(1\) in every residue field \(\kappa(\mathfrak p)\) for \(\mathfrak p \in V(J)\) and that \(x\) maps to \(0\) in \(\kappa(\mathfrak p)\) for every \(\mathfrak p \in V(I)\).

To see uniqueness suppose that \(e_1, e_2\) are distinct idempotents in \(R\). We have to show there exists a prime \(\mathfrak p\) such that \(e_1 \in \mathfrak p\) and \(e_2 \not \in \mathfrak p\), or conversely. Write \(e_i' = 1 - e_i\). If \(e_1 \not = e_2\), then \(0 \not = e_1 - e_2 = e_1(e_2 + e_2') - (e_1 + e_1')e_2 = e_1 e_2' - e_1' e_2\). Hence either the idempotent \(e_1 e_2' \not = 0\) or \(e_1' e_2 \not = 0\). A nonzero idempotent is not nilpotent, and hence we find a prime \(\mathfrak p\) such that either \(e_1e_2' \not \in \mathfrak p\) or \(e_1'e_2 \not \in \mathfrak p\), by Lemma 00E0. It is easy to see this gives the desired prime.

Lemma

Let \(R\) be a nonzero ring. Then \(\Spec(R)\) is connected if and only if \(R\) has no nontrivial idempotents.

Proof

Obvious from Lemma 00EE and the definition of a connected topological space.

Lemma

Let \(I \subset R\) be a finitely generated ideal of a ring \(R\) such that \(I = I^2\). Then

1. there exists an idempotent \(e \in R\) such that \(I = (e)\),

2. \(R/I \cong R_{e'}\) for the idempotent \(e' = 1 - e \in R\), and

3. \(V(I)\) is open and closed in \(\Spec(R)\).

Proof

By Nakayama’s Lemma 00DV there exists an element \(f = 1 + i\), \(i \in I\) such that \(fI = 0\). Then \(f^2 = f + fi = f\) is an idempotent. Consider the idempotent \(e = 1 - f = -i \in I\). For \(j \in I\) we have \(ej = j - fj = j\) hence \(I = (e)\). This proves (1).

Parts (2) and (3) follow from (1). Namely, we have \(V(I) = V(e) = \Spec(R) \setminus D(e)\) which is open and closed by either Lemma 00EC or Lemma 00EE. This proves (3). For (2) observe that the map \(R \to R_{e'}\) is surjective since \(x/(e')^n = x/e' = xe'/(e')^2 = xe'/e' = x/1\) in \(R_{e'}\). The kernel of the map \(R \to R_{e'}\) is the set of elements of \(R\) annihilated by a positive power of \(e'\). Since \(e'\) is idempotent this is the ideal of elements annihilated by \(e'\) which is the ideal \(I = (e)\) as \(e + e' = 1\) is a pair of orthogonal idempotents. This proves (2).

Connected components of spectra

Connected components of spectra are not as easy to understand as one may think at first. This is because we are used to the topology of locally connected spaces, but the spectrum of a ring is in general not locally connected.

Lemma

Let \(R\) be a ring. Let \(T \subset \Spec(R)\) be a subset of the spectrum. The following are equivalent

1. \(T\) is closed and is a union of connected components of \(\Spec(R)\),

2. \(T\) is an intersection of open and closed subsets of \(\Spec(R)\), and

3. \(T = V(I)\) where \(I \subset R\) is an ideal generated by idempotents.

Moreover, the ideal in (3) if it exists is unique.

Proof

By Lemma 04PM and Topology, Lemma 04PL we see that (1) and (2) are equivalent. Assume (2) and write \(T = \bigcap U_\alpha\) with \(U_\alpha \subset \Spec(R)\) open and closed. Then \(U_\alpha = D(e_\alpha)\) for some idempotent \(e_\alpha \in R\) by Lemma 00EE. Then setting \(I = (1 - e_\alpha)\) we see that \(T = V(I)\), i.e., (3) holds. Finally, assume (3). Write \(T = V(I)\) and \(I = (e_\alpha)\) for some collection of idempotents \(e_\alpha\). Then it is clear that \(T = \bigcap V(e_\alpha) = \bigcap D(1 - e_\alpha)\).

Suppose that \(I\) is an ideal generated by idempotents. Let \(e \in R\) be an idempotent such that \(V(I) \subset V(e)\). Then by Lemma 00E0 we see that \(e^n \in I\) for some \(n \geq 1\). As \(e\) is an idempotent this means that \(e \in I\). Hence we see that \(I\) is generated by exactly those idempotents \(e\) such that \(T \subset V(e)\). In other words, the ideal \(I\) is completely determined by the closed subset \(T\) which proves uniqueness.

Lemma

Let \(R\) be a ring. A connected component of \(\Spec(R)\) is of the form \(V(I)\), where \(I\) is an ideal generated by idempotents such that every idempotent of \(R\) either maps to \(0\) or \(1\) in \(R/I\).

Proof

Let \(\mathfrak p\) be a prime of \(R\). By Lemma 04PM we see that the hypotheses of Topology, Lemma 005F are satisfied for the topological space \(\Spec(R)\). Hence the connected component of \(\mathfrak p\) in \(\Spec(R)\) is the intersection of open and closed subsets of \(\Spec(R)\) containing \(\mathfrak p\). Hence it equals \(V(I)\) where \(I\) is generated by the idempotents \(e \in R\) such that \(e\) maps to \(0\) in \(\kappa(\mathfrak p)\), see Lemma 00EE. Any idempotent \(e\) which is not in this collection clearly maps to \(1\) in \(R/I\).

Glueing properties

In this section we put a number of standard results of the form: if something is true for all members of a standard open covering then it is true. In fact, it often suffices to check things on the level of local rings as in the following lemma.

Lemma

Let \(R\) be a ring.

1. For an element \(x\) of an \(R\)-module \(M\) the following are equivalent

   1. \(x = 0\),

   2. \(x\) maps to zero in \(M_\mathfrak p\) for all \(\mathfrak p \in \Spec(R)\),

   3. \(x\) maps to zero in \(M_{\mathfrak m}\) for all maximal ideals \(\mathfrak m\) of \(R\).

   In other words, the map \(M \to \prod_{\mathfrak m} M_{\mathfrak m}\) is injective.

2. Given an \(R\)-module \(M\) the following are equivalent

   1. \(M\) is zero,

   2. \(M_{\mathfrak p}\) is zero for all \(\mathfrak p \in \Spec(R)\),

   3. \(M_{\mathfrak m}\) is zero for all maximal ideals \(\mathfrak m\) of \(R\).

3. Given a complex \(M_1 \to M_2 \to M_3\) of \(R\)-modules the following are equivalent

   1. \(M_1 \to M_2 \to M_3\) is exact,

   2. for every prime \(\mathfrak p\) of \(R\) the localization \(M_{1, \mathfrak p} \to M_{2, \mathfrak p} \to M_{3, \mathfrak p}\) is exact,

   3. for every maximal ideal \(\mathfrak m\) of \(R\) the localization \(M_{1, \mathfrak m} \to M_{2, \mathfrak m} \to M_{3, \mathfrak m}\) is exact.

4. Given a map \(f : M \to M'\) of \(R\)-modules the following are equivalent

   1. \(f\) is injective,

   2. \(f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p\) is injective for all primes \(\mathfrak p\) of \(R\),

   3. \(f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m\) is injective for all maximal ideals \(\mathfrak m\) of \(R\).

5. Given a map \(f : M \to M'\) of \(R\)-modules the following are equivalent

   1. \(f\) is surjective,

   2. \(f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p\) is surjective for all primes \(\mathfrak p\) of \(R\),

   3. \(f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m\) is surjective for all maximal ideals \(\mathfrak m\) of \(R\).

6. Given a map \(f : M \to M'\) of \(R\)-modules the following are equivalent

   1. \(f\) is bijective,

   2. \(f_{\mathfrak p} : M_\mathfrak p \to M'_\mathfrak p\) is bijective for all primes \(\mathfrak p\) of \(R\),

   3. \(f_{\mathfrak m} : M_\mathfrak m \to M'_\mathfrak m\) is bijective for all maximal ideals \(\mathfrak m\) of \(R\).

Proof

Let \(x \in M\) as in (1). Let \(I = \{f \in R \mid fx = 0\}\). It is easy to see that \(I\) is an ideal (it is the annihilator of \(x\)). Condition (1)(c) means that for all maximal ideals \(\mathfrak m\) there exists an \(f \in R \setminus \mathfrak m\) such that \(fx =0\). In other words, \(V(I)\) does not contain a closed point. By Lemma 00E0 we see \(I\) is the unit ideal. Hence \(x\) is zero, i.e., (1)(a) holds. This proves (1).

Part (2) follows by applying (1) to all elements of \(M\) simultaneously.

Proof of (3). Let \(H\) be the homology of the sequence, i.e., \(H = \Ker(M_2 \to M_3)/\Im(M_1 \to M_2)\). By Proposition 00CS we have that \(H_\mathfrak p\) is the homology of the sequence \(M_{1, \mathfrak p} \to M_{2, \mathfrak p} \to M_{3, \mathfrak p}\). Hence (3) is a consequence of (2).

Parts (4) and (5) are special cases of (3). Part (6) follows formally on combining (4) and (5).

Lemma

Let \(R\) be a ring. Let \(M\) be an \(R\)-module. Let \(S\) be an \(R\)-algebra. Suppose that \(f_1, \ldots, f_n\) is a finite list of elements of \(R\) such that \(\bigcup D(f_i) = \Spec(R)\), in other words \((f_1, \ldots, f_n) = R\).

1. If each \(M_{f_i} = 0\) then \(M = 0\).

2. If each \(M_{f_i}\) is a finite \(R_{f_i}\)-module, then \(M\) is a finite \(R\)-module.

3. If each \(M_{f_i}\) is a finitely presented \(R_{f_i}\)-module, then \(M\) is a finitely presented \(R\)-module.

4. Let \(M \to N\) be a map of \(R\)-modules. If \(M_{f_i} \to N_{f_i}\) is an isomorphism for each \(i\) then \(M \to N\) is an isomorphism.

5. Let \(0 \to M'' \to M \to M' \to 0\) be a complex of \(R\)-modules. If \(0 \to M''_{f_i} \to M_{f_i} \to M'_{f_i} \to 0\) is exact for each \(i\), then \(0 \to M'' \to M \to M' \to 0\) is exact.

6. If each \(R_{f_i}\) is Noetherian, then \(R\) is Noetherian.

7. If each \(S_{f_i}\) is a finite type \(R_{f_i}\)-algebra, then \(S\) is a finite type \(R\)-algebra.

8. If each \(S_{f_i}\) is of finite presentation over \(R_{f_i}\), then \(S\) is a finitely presented \(R\)-algebra.

Proof

We prove each of the parts in turn.

1. By Proposition 02C6 this implies \(M_\mathfrak p = 0\) for all \(\mathfrak p \in \Spec(R)\), so we conclude by Lemma 00HN.

2. For each \(i\) take a finite generating set \(X_i\) of \(M_{f_i}\). Without loss of generality, we may assume that the elements of \(X_i\) are in the image of the localization map \(M \rightarrow M_{f_i}\), so we take a finite set \(Y_i\) of preimages of the elements of \(X_i\) in \(M\). Let \(Y\) be the union of these sets. This is still a finite set. Consider the obvious \(R\)-linear map \(R^Y \rightarrow M\) sending the basis element \(e_y\) to \(y\). By assumption this map is surjective after localizing at an arbitrary prime ideal \(\mathfrak p\) of \(R\), so it is surjective by Lemma 00HN and \(M\) is finitely generated.

3. By (2) we have a short exact sequence \[0 \rightarrow K \rightarrow R^n \rightarrow M \rightarrow 0\] Since localization is an exact functor and \(M_{f_i}\) is finitely presented we see that \(K_{f_i}\) is finitely generated for all \(1 \leq i \leq n\) by Lemma 0519. By (2) this implies that \(K\) is a finite \(R\)-module and therefore \(M\) is finitely presented.

4. By Proposition 02C6 the assumption implies that the induced morphism on localizations at all prime ideals is an isomorphism, so we conclude by Lemma 00HN.

5. By Proposition 02C6 the assumption implies that the induced sequence of localizations at all prime ideals is short exact, so we conclude by Lemma 00HN.

6. We will show that every ideal of \(R\) has a finite generating set: For this, let \(I \subset R\) be an arbitrary ideal. By Proposition 00CS each \(I_{f_i} \subset R_{f_i}\) is an ideal. These are all finitely generated by assumption, so we conclude by (2).

7. For each \(i\) take a finite generating set \(X_i\) of \(S_{f_i}\). Without loss of generality, we may assume that the elements of \(X_i\) are in the image of the localization map \(S \rightarrow S_{f_i}\), so we take a finite set \(Y_i\) of preimages of the elements of \(X_i\) in \(S\). Let \(Y\) be the union of these sets. This is still a finite set. Consider the algebra homomorphism \(R[X_y]_{y \in Y} \rightarrow S\) induced by \(Y\). Since it is an algebra homomorphism, the image \(T\) is an \(R\)-submodule of the \(R\)-module \(S\), so we can consider the quotient module \(S/T\). By assumption, this is zero if we localize at the \(f_i\), so it is zero by (1) and therefore \(S\) is an \(R\)-algebra of finite type.

8. By the previous item, there exists a surjective \(R\)-algebra homomorphism \(R[X_1, \ldots, X_n] \rightarrow S\). Let \(K\) be the kernel of this map. This is an ideal in \(R[X_1, \ldots, X_n]\), finitely generated in each localization at \(f_i\). Since the \(f_i\) generate the unit ideal in \(R\), they also generate the unit ideal in \(R[X_1, \ldots, X_n]\), so an application of (2) finishes the proof.

Lemma

Let \(R \to S\) be a ring map. Suppose that \(g_1, \ldots, g_n\) is a finite list of elements of \(S\) such that \(\bigcup D(g_i) = \Spec(S)\) in other words \((g_1, \ldots, g_n) = S\).

1. If each \(S_{g_i}\) is of finite type over \(R\), then \(S\) is of finite type over \(R\).

2. If each \(S_{g_i}\) is of finite presentation over \(R\), then \(S\) is of finite presentation over \(R\).

Proof

Choose \(h_1, \ldots, h_n \in S\) such that \(\sum h_i g_i = 1\).

Proof of (1). For each \(i\) choose a finite list of elements \(x_{i, j} \in S_{g_i}\), \(j = 1, \ldots, m_i\) which generate \(S_{g_i}\) as an \(R\)-algebra. Write \(x_{i, j} = y_{i, j}/g_i^{n_{i, j}}\) for some \(y_{i, j} \in S\) and some \(n_{i, j} \ge 0\). Consider the \(R\)-subalgebra \(S' \subset S\) generated by \(g_1, \ldots, g_n\), \(h_1, \ldots, h_n\) and \(y_{i, j}\), \(i = 1, \ldots, n\), \(j = 1, \ldots, m_i\). Since localization is exact (Proposition 00CS), we see that \(S'_{g_i} \to S_{g_i}\) is injective. On the other hand, it is surjective by our choice of \(y_{i, j}\). The elements \(g_1, \ldots, g_n\) generate the unit ideal in \(S'\) as \(h_1, \ldots, h_n \in S'\). Thus \(S' \to S\) viewed as an \(S'\)-module map is an isomorphism by Lemma 00EO.

Proof of (2). We already know that \(S\) is of finite type. Write \(S = R[x_1, \ldots, x_m]/J\) for some ideal \(J\). For each \(i\) choose a lift \(g'_i \in R[x_1, \ldots, x_m]\) of \(g_i\) and we choose a lift \(h'_i \in R[x_1, \ldots, x_m]\) of \(h_i\). Then we see that \[S_{g_i} = R[x_1, \ldots, x_m, y_i]/(J_i + (1 - y_ig'_i))\] where \(J_i\) is the ideal of \(R[x_1, \ldots, x_m, y_i]\) generated by \(J\). Small detail omitted. By Lemma 00R2 we may choose a finite list of elements \(f_{i, j} \in J\), \(j = 1, \ldots, m_i\) such that the images of \(f_{i, j}\) in \(J_i\) and \(1 - y_ig'_i\) generate the ideal \(J_i + (1 - y_ig'_i)\). Set \[S' = R[x_1, \ldots, x_m]/\left(\sum h'_ig'_i - 1, f_{i, j}; i = 1, \ldots, n, j = 1, \ldots, m_i\right)\] There is a surjective \(R\)-algebra map \(S' \to S\). The classes of the elements \(g'_1, \ldots, g'_n\) in \(S'\) generate the unit ideal and by construction the maps \(S'_{g'_i} \to S_{g_i}\) are injective. Thus we conclude as in part (1).

Glueing functions

In this section we show that given an open covering \[\Spec(R) = \bigcup\nolimits_{i = 1}^n D(f_i)\] by standard opens, and given an element \(h_i \in R_{f_i}\) for each \(i\) such that \(h_i = h_j\) as elements of \(R_{f_i f_j}\) then there exists a unique \(h \in R\) such that the image of \(h\) in \(R_{f_i}\) is \(h_i\). This result can be interpreted in two ways:

1. The rule \(D(f) \mapsto R_f\) is a sheaf of rings on the standard opens, see Sheaves, Section 009H.

2. If we think of elements of \(R_f\) as the “algebraic” or “regular” functions on \(D(f)\), then these glue as would continuous, resp. differentiable functions on a topological, resp. differentiable manifold.

Lemma

Let \(R\) be a ring. Let \(f_1, \ldots, f_n\) be elements of \(R\) generating the unit ideal. Let \(M\) be an \(R\)-module. The sequence \[0 \to M \xrightarrow{\alpha} \bigoplus\nolimits_{i = 1}^n M_{f_i} \xrightarrow{\beta} \bigoplus\nolimits_{i, j = 1}^n M_{f_i f_j}\] is exact, where \(\alpha(m) = (m/1, \ldots, m/1)\) and \(\beta(m_1/f_1^{e_1}, \ldots, m_n/f_n^{e_n}) = (m_i/f_i^{e_i} - m_j/f_j^{e_j})_{(i, j)}\).

Proof

It suffices to show that the localization of the sequence at any maximal ideal \(\mathfrak m\) is exact, see Lemma 00HN. Since \(f_1, \ldots, f_n\) generate the unit ideal, there is an \(i\) such that \(f_i \not \in \mathfrak m\). After renumbering we may assume \(i = 1\). Note that \((M_{f_i})_\mathfrak m = (M_\mathfrak m)_{f_i}\) and \((M_{f_if_j})_\mathfrak m = (M_\mathfrak m)_{f_if_j}\), see Proposition 02C7. In particular \((M_{f_1})_\mathfrak m = M_\mathfrak m\) and \((M_{f_1 f_i})_\mathfrak m = (M_\mathfrak m)_{f_i}\), because \(f_1\) is a unit. Note that the maps in the sequence are the canonical ones coming from Lemma 07K0 and the identity map on \(M\). Having said all of this, after replacing \(R\) by \(R_\mathfrak m\), \(M\) by \(M_\mathfrak m\), and \(f_i\) by their image in \(R_\mathfrak m\), and \(f_1\) by \(1 \in R_\mathfrak m\), we reduce to the case where \(f_1 = 1\).

Assume \(f_1 = 1\). Injectivity of \(\alpha\) is now trivial. Let \(m = (m_i) \in \bigoplus_{i = 1}^n M_{f_i}\) be in the kernel of \(\beta\). Then \(m_1 \in M_{f_1} = M\). Moreover, \(\beta(m) = 0\) implies that \(m_1\) and \(m_i\) map to the same element of \(M_{f_1f_i} = M_{f_i}\). Thus \(\alpha(m_1) = m\) and the proof is complete.

Lemma

Let \(R\) be a ring, and let \(f_1, f_2, \ldots, f_n \in R\) generate the unit ideal in \(R\). Then the following sequence is exact: \[0 \longrightarrow R \longrightarrow \bigoplus\nolimits_i R_{f_i} \longrightarrow \bigoplus\nolimits_{i, j}R_{f_if_j}\] where the maps \(\alpha : R \longrightarrow \bigoplus_i R_{f_i}\) and \(\beta : \bigoplus_i R_{f_i} \longrightarrow \bigoplus_{i, j} R_{f_if_j}\) are defined as \[\alpha(x) = \left(\frac{x}{1}, \ldots, \frac{x}{1}\right) \text{ and } \beta\left(\frac{x_1}{f_1^{r_1}}, \ldots, \frac{x_n}{f_n^{r_n}}\right) = \left(\frac{x_i}{f_i^{r_i}}-\frac{x_j}{f_j^{r_j}}~\text{in}~R_{f_if_j}\right).\]

Proof

Special case of Lemma 00EK.

The following we have already seen above, but we state it explicitly here for convenience.

Lemma

Let \(R\) be a ring. If \(\Spec(R) = U \amalg V\) with both \(U\) and \(V\) open then \(R \cong R_1 \times R_2\) with \(U \cong \Spec(R_1)\) and \(V \cong \Spec(R_2)\) via the maps in Lemma 00ED. Moreover, both \(R_1\) and \(R_2\) are localizations as well as quotients of the ring \(R\).

Proof

By Lemma 00EE we have \(U = D(e)\) and \(V = D(1-e)\) for some idempotent \(e\). By Lemma 00EJ we see that \(R \cong R_e \times R_{1 - e}\) (since clearly \(R_{e(1-e)} = 0\) so the glueing condition is trivial; of course it is trivial to prove the product decomposition directly in this case). The lemma follows.

Lemma

Let \(R\) be a ring. Let \(f_1, \ldots, f_n \in R\). Let \(M\) be an \(R\)-module. Then \(M \to \bigoplus M_{f_i}\) is injective if and only if \[M \longrightarrow \bigoplus\nolimits_{i = 1, \ldots, n} M, \quad m \longmapsto (f_1m, \ldots, f_nm)\] is injective.

Proof

The map \(M \to \bigoplus M_{f_i}\) is injective if and only if for all \(m \in M\) and \(e_1, \ldots, e_n \geq 1\) such that \(f_i^{e_i}m = 0\), \(i = 1, \ldots, n\) we have \(m = 0\). This clearly implies the displayed map is injective. Conversely, suppose the displayed map is injective and \(m \in M\) and \(e_1, \ldots, e_n \geq 1\) are such that \(f_i^{e_i}m = 0\), \(i = 1, \ldots, n\). If \(e_i = 1\) for all \(i\), then we immediately conclude that \(m = 0\) from the injectivity of the displayed map. Next, we prove this holds for any such data by induction on \(e = \sum e_i\). The base case is \(e = n\), and we have just dealt with this. If some \(e_i > 1\), then set \(m' = f_im\). By induction we see that \(m' = 0\). Hence we see that \(f_i m = 0\), i.e., we may take \(e_i = 1\) which decreases \(e\) and we win.

The following lemma is better stated and proved in the more general context of flat descent. However, it makes sense to state it here since it fits well with the above.

Lemma

Let \(R\) be a ring. Let \(f_1, \ldots, f_n \in R\). Suppose we are given the following data:

1. For each \(i\) an \(R_{f_i}\)-module \(M_i\).

2. For each pair \(i, j\) an \(R_{f_if_j}\)-module isomorphism \(\psi_{ij} : (M_i)_{f_j} \to (M_j)_{f_i}\).

which satisfy the “cocycle condition” that all the diagrams \[\xymatrix{ (M_i)_{f_jf_k} \ar[rd]_{\psi_{ij}} \ar[rr]^{\psi_{ik}} & & (M_k)_{f_if_j} \\ & (M_j)_{f_if_k} \ar[ru]_{\psi_{jk}} }\] commute (for all triples \(i, j, k\)). Given this data define \[M = \Ker\left( \bigoplus\nolimits_{1 \leq i \leq n} M_i \longrightarrow \bigoplus\nolimits_{1 \leq i, j \leq n} (M_i)_{f_j} \right)\] where \((m_1, \ldots, m_n)\) maps to the element whose \((i, j)\)th entry is \(m_i/1 - \psi_{ji}(m_j/1)\). Then the natural map \(M \to M_i\) induces an isomorphism \(M_{f_i} \to M_i\). Moreover \(\psi_{ij}(m/1) = m/1\) for all \(m \in M\) (with obvious notation).

Proof

To show that \(M_{f_1} \to M_1\) is an isomorphism, it suffices to show that its localization at every prime \(\mathfrak p'\) of \(R_{f_1}\) is an isomorphism, see Lemma 00HN. Write \(\mathfrak p' = \mathfrak p R_{f_1}\) for some prime \(\mathfrak p \subset R\), \(f_1 \not \in \mathfrak p\), see Lemma 00E4. Since localization is exact (Proposition 00CS), we see that \[\begin{align*} (M_{f_1})_{\mathfrak p'} & = M_\mathfrak p \\ & = \Ker\left( \bigoplus\nolimits_{1 \leq i \leq n} M_{i, \mathfrak p} \longrightarrow \bigoplus\nolimits_{1 \leq i, j \leq n} ((M_i)_{f_j})_\mathfrak p \right) \\ & = \Ker\left( \bigoplus\nolimits_{1 \leq i \leq n} M_{i, \mathfrak p} \longrightarrow \bigoplus\nolimits_{1 \leq i, j \leq n} (M_{i, \mathfrak p})_{f_j} \right) \end{align*}\] Here we also used Proposition 02C7. Since \(f_1\) is a unit in \(R_\mathfrak p\), this reduces us to the case where \(f_1 = 1\) by replacing \(R\) by \(R_\mathfrak p\), \(f_i\) by the image of \(f_i\) in \(R_\mathfrak p\), \(M\) by \(M_\mathfrak p\), and \(f_1\) by \(1\).

Assume \(f_1 = 1\). Then \(\psi_{1j} : (M_1)_{f_j} \to M_j\) is an isomorphism for \(j = 2, \ldots, n\). If we use these isomorphisms to identify \(M_j = (M_1)_{f_j}\), then we see that \(\psi_{ij} : (M_1)_{f_if_j} \to (M_1)_{f_if_j}\) is the canonical identification. Thus the complex \[0 \to M_1 \to \bigoplus\nolimits_{1 \leq i \leq n} (M_1)_{f_i} \longrightarrow \bigoplus\nolimits_{1 \leq i, j \leq n} (M_1)_{f_if_j}\] is exact by Lemma 00EK. Thus the first map identifies \(M_1\) with \(M\) in this case and everything is clear.

Zerodivisors and total rings of fractions

The local ring at a minimal prime has the following properties.

Lemma

Let \(\mathfrak p\) be a minimal prime of a ring \(R\). Every element of the maximal ideal of \(R_{\mathfrak p}\) is nilpotent. If \(R\) is reduced then \(R_{\mathfrak p}\) is a field.

Proof

If some element \(x\) of \({\mathfrak p}R_{\mathfrak p}\) is not nilpotent, then \(D(x) \not = \emptyset\), see Lemma 00E0. This contradicts the minimality of \(\mathfrak p\). If \(R\) is reduced, then \({\mathfrak p}R_{\mathfrak p} = 0\) and hence \(R_{\mathfrak p}\) is a field.

Lemma

Let \(R\) be a reduced ring. Then

1. \(R\) is a subring of a product of fields,

2. \(R \to \prod_{\mathfrak p\text{ minimal}} R_{\mathfrak p}\) is an embedding into a product of fields,

3. \(\bigcup_{\mathfrak p\text{ minimal}} \mathfrak p\) is the set of zerodivisors of \(R\).

Proof

By Lemma 00EU each of the rings \(R_\mathfrak p\) is a field. In particular, the kernel of the ring map \(R \to R_\mathfrak p\) is \(\mathfrak p\). By Lemma 00E0 we have \(\bigcap_{\mathfrak p} \mathfrak p = (0)\). Hence (2) and (1) are true. If \(x y = 0\) and \(y \not = 0\), then \(y \not \in \mathfrak p\) for some minimal prime \(\mathfrak p\). Hence \(x \in \mathfrak p\). Thus every zerodivisor of \(R\) is contained in \(\bigcup_{\mathfrak p\text{ minimal}} \mathfrak p\). Conversely, suppose that \(x \in \mathfrak p\) for some minimal prime \(\mathfrak p\). Then \(x\) maps to zero in \(R_\mathfrak p\), hence there exists \(y \in R\), \(y \not \in \mathfrak p\) such that \(xy = 0\). In other words, \(x\) is a zerodivisor. This finishes the proof of (3) and the lemma.

The total ring of fractions \(Q(R)\) of a ring \(R\) was introduced in Example 02C5.

Lemma

Let \(R\) be a ring. Let \(S \subset R\) be a multiplicative subset consisting of nonzerodivisors. Then \(Q(R) \cong Q(S^{-1}R)\). In particular \(Q(R) \cong Q(Q(R))\).

Proof

If \(x \in S^{-1}R\) is a nonzerodivisor, and \(x = r/f\) for some \(r \in R\), \(f \in S\), then \(r\) is a nonzerodivisor in \(R\). Whence the lemma.

We can apply glueing results to prove something about total rings of fractions \(Q(R)\) which we introduced in Example 02C5.

Lemma

Let \(R\) be a ring. Assume that \(R\) has finitely many minimal primes \(\mathfrak q_1, \ldots, \mathfrak q_t\), and that \(\mathfrak q_1 \cup \ldots \cup \mathfrak q_t\) is the set of zerodivisors of \(R\). Then the total ring of fractions \(Q(R)\) is equal to \(R_{\mathfrak q_1} \times \ldots \times R_{\mathfrak q_t}\).

Proof

There are natural maps \(Q(R) \to R_{\mathfrak q_i}\) since any nonzerodivisor lies in \(R \setminus \mathfrak q_i\). Hence a natural map \(Q(R) \to R_{\mathfrak q_1} \times \ldots \times R_{\mathfrak q_t}\). For any nonminimal prime \(\mathfrak p \subset R\) we see that \(\mathfrak p \not \subset \mathfrak q_1 \cup \ldots \cup \mathfrak q_t\) by Lemma 00DS. Hence \(\Spec(Q(R)) = \{\mathfrak q_1, \ldots, \mathfrak q_t\}\) (as subsets of \(\Spec(R)\), see Lemma 00E3). Therefore \(\Spec(Q(R))\) is a finite discrete set and it follows that \(Q(R) = A_1 \times \ldots \times A_t\) with \(\Spec(A_i) = \{\mathfrak{q}_i\}\), see Lemma 00EM. Moreover \(A_i\) is a local ring, which is a localization of \(R\). Hence \(A_i \cong R_{\mathfrak q_i}\).

Irreducible components of spectra

We show that irreducible components of the spectrum of a ring correspond to the minimal primes in the ring.

Lemma

Let \(R\) be a ring.

1. For a prime \(\mathfrak p \subset R\) the closure of \(\{\mathfrak p\}\) in the Zariski topology is \(V(\mathfrak p)\). In a formula \(\overline{\{\mathfrak p\}} = V(\mathfrak p)\).

2. The irreducible closed subsets of \(\Spec(R)\) are exactly the subsets \(V(\mathfrak p)\), with \(\mathfrak p \subset R\) a prime.

3. The irreducible components (see Topology, Definition 004V) of \(\Spec(R)\) are exactly the subsets \(V(\mathfrak p)\), with \(\mathfrak p \subset R\) a minimal prime.

Proof

Note that if \(\mathfrak p \in V(I)\), then \(I \subset \mathfrak p\). Hence, clearly \(\overline{\{\mathfrak p\}} = V(\mathfrak p)\). In particular \(V(\mathfrak p)\) is the closure of a singleton and hence irreducible. The second assertion implies the third. To show the second, let \(V(I) \subset \Spec(R)\) with \(I\) a radical ideal. If \(I\) is not prime, then choose \(a, b\in R\), \(a, b\not \in I\) with \(ab\in I\). In this case \(V(I, a) \cup V(I, b) = V(I)\), but neither \(V(I, b) = V(I)\) nor \(V(I, a) = V(I)\), by Lemma 00E0. Hence \(V(I)\) is not irreducible.

In other words, this lemma shows that every irreducible closed subset of \(\Spec(R)\) is of the form \(V(\mathfrak p)\) for some prime \(\mathfrak p\). Since \(V(\mathfrak p) = \overline{\{\mathfrak p\}}\) we see that each irreducible closed subset has a unique generic point, see Topology, Definition 004X. In particular, \(\Spec(R)\) is a sober topological space. We record this fact in the following lemma.

Lemma

The spectrum of a ring is a spectral space, see Topology, Definition 08YG.

Proof

Formally this follows from Lemma 00ES and Lemma 04PM. See also discussion above.

Lemma

Let \(R\) be a ring. Let \(\mathfrak p \subset R\) be a prime.

1. the set of irreducible closed subsets of \(\Spec(R)\) passing through \(\mathfrak p\) is in one-to-one correspondence with primes \(\mathfrak q \subset R_{\mathfrak p}\).

2. The set of irreducible components of \(\Spec(R)\) passing through \(\mathfrak p\) is in one-to-one correspondence with minimal primes \(\mathfrak q \subset R_{\mathfrak p}\).

Proof

Follows from Lemma 00ES and the description of \(\Spec(R_\mathfrak p)\) in Lemma 00E3 which shows that \(\Spec(R_\mathfrak p)\) corresponds to primes \(\mathfrak q\) in \(R\) with \(\mathfrak q \subset \mathfrak p\).

Lemma

Let \(R\) be a ring. Let \(\mathfrak p\) be a minimal prime of \(R\). Let \(W \subset \Spec(R)\) be a quasi-compact open not containing the point \(\mathfrak p\). Then there exists an \(f \in R\), \(f \not \in \mathfrak p\) such that \(D(f) \cap W = \emptyset\).

Proof

Since \(W\) is quasi-compact we may write it as a finite union of standard affine opens \(D(g_i)\), \(i = 1, \ldots, n\). Since \(\mathfrak p \not \in W\) we have \(g_i \in \mathfrak p\) for all \(i\). By Lemma 00EU each \(g_i\) is nilpotent in \(R_{\mathfrak p}\). Hence we can find an \(f \in R\), \(f \not \in \mathfrak p\) such that for all \(i\) we have \(f g_i^{n_i} = 0\) for some \(n_i > 0\). Then \(D(f)\) works.

Lemma

Let \(R\) be a ring. Let \(X = \Spec(R)\) as a topological space. The following are equivalent

1. \(X\) is profinite,

2. \(X\) is Hausdorff,

3. \(X\) is totally disconnected.

4. every quasi-compact open of \(X\) is closed,

5. there are no nontrivial inclusions between its prime ideals,

6. every prime ideal is a maximal ideal,

7. every prime ideal is minimal,

8. every standard open \(D(f) \subset X\) is closed.

Proof

First proof. It is clear that (5), (6), and (7) are equivalent. It is clear that (4) and (8) are equivalent as every quasi-compact open is a finite union of standard opens. The implication (7) \(\Rightarrow\) (4) follows from Lemma 00EV. Assume (4) holds. Let \(\mathfrak p, \mathfrak p'\) be distinct primes of \(R\). Choose an \(f \in \mathfrak p'\), \(f \not \in \mathfrak p\) (if needed switch \(\mathfrak p\) with \(\mathfrak p'\)). Then \(\mathfrak p' \not \in D(f)\) and \(\mathfrak p \in D(f)\). By (4) the open \(D(f)\) is also closed. Hence \(\mathfrak p\) and \(\mathfrak p'\) are in disjoint open neighbourhoods whose union is \(X\). Thus \(X\) is Hausdorff and totally disconnected. Thus (4) \(\Rightarrow\) (2) and (3). If (3) holds then there cannot be any specializations between points of \(\Spec(R)\) and we see that (5) holds. If \(X\) is Hausdorff then every point is closed, so (2) implies (6). Thus (2), (3), (4), (5), (6), (7) and (8) are equivalent. Any profinite space is Hausdorff, so (1) implies (2). If \(X\) satisfies (2) and (3), then \(X\) (being quasi-compact by Lemma 00E8) is profinite by Topology, Lemma 08ZY.

Second proof. Besides the equivalence of (4) and (8) this follows from Lemma 090M and purely topological facts, see Topology, Lemma 0905.

Examples of spectra of rings

In this section we put some examples of spectra.

Example

In this example we describe \(X = \Spec(\mathbf{Z}[x]/(x^2 - 4))\). Let \(\mathfrak{p}\) be an arbitrary prime in \(X\). Let \(\phi : \mathbf{Z} \to \mathbf{Z}[x]/(x^2 - 4)\) be the natural ring map. Then, \(\phi^{-1}(\mathfrak p)\) is a prime in \(\mathbf{Z}\). If \(\phi^{-1}(\mathfrak p) = (2)\), then since \(\mathfrak p\) contains \(2\), it corresponds to a prime ideal in \(\mathbf{Z}[x]/(x^2 - 4, 2) \cong (\mathbf{Z}/2\mathbf{Z})[x]/(x^2)\) via the map \(\mathbf{Z}[x]/(x^2 - 4) \to \mathbf{Z}[x]/(x^2 - 4, 2)\). Any prime in \((\mathbf{Z}/2\mathbf{Z})[x]/(x^2)\) corresponds to a prime in \((\mathbf{Z}/2\mathbf{Z})[x]\) containing \((x^2)\). Such primes will then contain \(x\). Since \((\mathbf{Z}/2\mathbf{Z}) \cong (\mathbf{Z}/2\mathbf{Z})[x]/(x)\) is a field, \((x)\) is a maximal ideal. Since any prime contains \((x)\) and \((x)\) is maximal, the ring contains only one prime \((x)\). Thus, in this case, \(\mathfrak p = (2, x)\). Now, if \(\phi^{-1}(\mathfrak p) = (q)\) for \(q > 2\), then since \(\mathfrak p\) contains \(q\), it corresponds to a prime ideal in \(\mathbf{Z}[x]/(x^2 - 4, q) \cong (\mathbf{Z}/q\mathbf{Z})[x]/(x^2 - 4)\) via the map \(\mathbf{Z}[x]/(x^2 - 4) \to \mathbf{Z}[x]/(x^2 - 4, q)\). Any prime in \((\mathbf{Z}/q\mathbf{Z})[x]/(x^2 - 4)\) corresponds to a prime in \((\mathbf{Z}/q\mathbf{Z})[x]\) containing \((x^2 - 4) = (x -2)(x + 2)\). Hence, these primes must contain either \(x -2\) or \(x + 2\). Since \((\mathbf{Z}/q\mathbf{Z})[x]\) is a PID, all nonzero primes are maximal, and so there are precisely 2 primes in \((\mathbf{Z}/q\mathbf{Z})[x]\) containing \((x-2)(x + 2)\), namely \((x-2)\) and \((x + 2)\). In conclusion, there exist two primes \((q, x-2)\) and \((q, x + 2)\) since \(2 \neq -2 \in \mathbf{Z}/(q)\). Finally, we treat the case where \(\phi^{-1}(\mathfrak p) = (0)\). Notice that \(\mathfrak p\) corresponds to a prime ideal in \(\mathbf{Z}[x]\) that contains \((x^2 - 4) = (x -2)(x + 2)\). Hence, \(\mathfrak p\) contains either \((x-2)\) or \((x + 2)\). Hence, \(\mathfrak p\) corresponds to a prime in \(\mathbf{Z}[x]/(x - 2)\) or one in \(\mathbf{Z}[x]/(x + 2)\) that intersects \(\mathbf{Z}\) only at \(0\), by assumption. Since \(\mathbf{Z}[x]/(x - 2) \cong \mathbf{Z}\) and \(\mathbf{Z}[x]/(x + 2) \cong \mathbf{Z}\), this means that \(\mathfrak p\) must correspond to \(0\) in one of these rings. Thus, \(\mathfrak p = (x - 2)\) or \(\mathfrak p = (x + 2)\) in the original ring.

Example

In this example we describe \(X = \Spec(\mathbf{Z}[x])\). Fix \(\mathfrak p \in X\). Let \(\phi : \mathbf{Z} \to \mathbf{Z}[x]\) and notice that \(\phi^{-1}(\mathfrak p) \in \Spec(\mathbf{Z})\). If \(\phi^{-1}(\mathfrak p) = (q)\) for \(q\) a prime number \(q > 0\), then \(\mathfrak p\) corresponds to a prime in \((\mathbf{Z}/(q))[x]\), which must be generated by a polynomial that is irreducible in \((\mathbf{Z}/(q))[x]\). If we choose a representative of this polynomial with minimal degree, then it will also be irreducible in \(\mathbf{Z}[x]\). Hence, in this case \(\mathfrak p = (q, f_q)\) where \(f_q\) is an irreducible polynomial in \(\mathbf{Z}[x]\) that is irreducible when viewed in \((\mathbf{Z}/(q) [x])\). Now, assume that \(\phi^{-1}(\mathfrak p) = (0)\). In this case, if \(\mathfrak p = (0)\) there is nothing more to prove. Otherwise, \(\mathfrak p\) contains nonconstant polynomials which, since \(\mathfrak p\) is prime, may be assumed to be irreducible in \(\mathbf{Z}[x]\). By Gauss’ lemma, these polynomials are also irreducible in \(\mathbf{Q}[x]\). Since \(\mathbf{Q}[x]\) is a Euclidean domain, if there are at least two distinct irreducibles \(f, g\) generating \(\mathfrak p\), then \(1 = af + bg\) for \(a, b \in \mathbf{Q}[x]\). Multiplying through by a common denominator, we see that \(m = \bar{a}f + \bar{b} g\) for \(\bar{a}, \bar{b} \in \mathbf{Z}[x]\) and nonzero \(m \in \mathbf{Z}\). This is a contradiction. Hence, \(\mathfrak p\) is generated by one irreducible polynomial in \(\mathbf{Z}[x]\).

Example

In this example we describe \(X = \Spec(k[x, y])\) when \(k\) is an arbitrary field. Clearly \((0)\) is prime, and any principal ideal generated by an irreducible polynomial will also be a prime since \(k[x, y]\) is a unique factorization domain. Now assume \(\mathfrak p\) is an element of \(X\) that is not principal. Since \(k[x, y]\) is a Noetherian UFD, the prime ideal \(\mathfrak p\) can be generated by a finite number of irreducible polynomials \((f_1, \ldots, f_n)\). Now, I claim that if \(f, g\) are irreducible polynomials in \(k[x, y]\) that are not associates, then \((f, g) \cap k[x] \neq 0\). To do this, it is enough to show that \(f\) and \(g\) are relatively prime when viewed in \(k(x)[y]\). In this case, \(k(x)[y]\) is a Euclidean domain, so by applying the Euclidean algorithm and clearing denominators, we obtain \(p = af + bg\) for \(0 \ne p \in k[x]\) and \(a, b \in k[x, y]\). Thus, assume this is not the case, that is, that some nonunit \(h \in k(x)[y]\) divides both \(f\) and \(g\). Then, by Gauss’s lemma, for some \(a, b \in k(x)\) we have \(ah | f\) and \(bh | g\) for \(ah, bh \in k[x, y]\). By irreducibility, \(ah = f\) and \(bh = g\) (since \(h \notin k(x)\)). So, back in \(k(x)[y]\), \(f, g\) are associates, as \(\frac{a}{b} g = f\). Since \(k(x)\) is the fraction field of \(k[x]\), we can write \(g = \frac{r}{s} f\) for elements \(r , s \in k[x]\) sharing no common factors. This implies that \(sg = rf\) in \(k[x, y]\) and so \(s\) must divide \(f\) since \(k[x, y]\) is a UFD. Hence, \(s = 1\) or \(s = f\). If \(s = f\), then \(r = g\), implying \(f, g \in k[x]\) and thus must be units in \(k(x)\) and relatively prime in \(k(x)[y]\), contradicting our hypothesis. If \(s = 1\), then \(g = rf\), another contradiction. Thus, we must have \(f, g\) relatively prime in \(k(x)[y]\), a Euclidean domain. Thus, we have reduced to the case \(\mathfrak p\) contains some irreducible polynomial \(p \in k[x] \subset k[x, y]\). By the above, \(\mathfrak p\) corresponds to a prime in the ring \(k[x, y]/(p) = k(\alpha)[y]\), where \(\alpha\) is an element algebraic over \(k\) with minimum polynomial \(p\). This is a PID, and so any prime ideal corresponds to \((0)\) or an irreducible polynomial in \(k(\alpha)[y]\). Thus, \(\mathfrak p\) is of the form \((p)\) or \((p, f)\) where \(f\) is a polynomial in \(k[x, y]\) that is irreducible in the quotient \(k[x, y]/(p)\).

Example

Consider the ring \[R = \{ f \in \mathbf{Q}[z]\text{ with }f(0) = f(1) \}.\] Consider the map \[\varphi : \mathbf{Q}[A, B] \to R\] defined by \(\varphi(A) = z^2-z\) and \(\varphi(B) = z^3-z^2\). It is easily checked that \((A^3 - B^2 + AB) \subset \Ker(\varphi)\) and that \(A^3 - B^2 + AB\) is irreducible. Assume that \(\varphi\) is surjective; then since \(R\) is an integral domain (it is a subring of an integral domain), \(\Ker(\varphi)\) must be a prime ideal of \(\mathbf{Q}[A, B]\). The prime ideals which contain \((A^3-B^2 + AB)\) are \((A^3-B^2 + AB)\) itself and any maximal ideal \((f, g)\) with \(f, g\in\mathbf{Q}[A, B]\) such that \(f\) is irreducible mod \(g\). But \(R\) is not a field, so the kernel must be \((A^3-B^2 + AB)\); hence \(\varphi\) gives an isomorphism \(R \to \mathbf{Q}[A, B]/(A^3-B^2 + AB)\).

To see that \(\varphi\) is surjective, we must express any \(f\in R\) as a \(\mathbf{Q}\)-coefficient polynomial in \(A(z) = z^2-z\) and \(B(z) = z^3-z^2\). Note the relation \(zA(z) = B(z)\). Let \(a = f(0) = f(1)\). Then \(z(z-1)\) must divide \(f(z)-a\), so we can write \(f(z) = z(z-1)g(z)+a = A(z)g(z)+a\). If \(\deg(g) < 2\), then \(g(z) = c_1z + c_0\) and \(f(z) = A(z)(c_1z + c_0)+a = c_1B(z)+c_0A(z)+a\), so we are done. If \(\deg(g)\geq 2\), then by the polynomial division algorithm, we can write \(g(z) = A(z)h(z)+b_1z + b_0\) (\(\deg(h)\leq\deg(g)-2\)), so \(f(z) = A(z)^2h(z)+b_1B(z)+b_0A(z)+a\). Applying division to \(h(z)\) and iterating, we obtain an expression for \(f(z)\) as a polynomial in \(A(z)\) and \(B(z)\); hence \(\varphi\) is surjective.

Now let \(a \in \mathbf{Q}\), \(a \neq 0, \frac{1}{2}, 1\) and consider \[R_a = \{ f \in \mathbf{Q}[z, \frac{1}{z-a}]\text{ with }f(0) = f(1) \}.\] This is a finitely generated \(\mathbf{Q}\)-algebra as well: it is easy to check that the functions \(z^2-z\), \(z^3-z\), and \(\frac{a^2-a}{z-a}+z\) generate \(R_a\) as an \(\mathbf{Q}\)-algebra. We have the following inclusions: \[R\subset R_a\subset\mathbf{Q}[z, \frac{1}{z-a}], \quad R\subset\mathbf{Q}[z]\subset\mathbf{Q}[z, \frac{1}{z-a}].\] Recall (Lemma 00E3) that for a ring \(T\) and a multiplicative subset \(S\subset T\), the ring map \(T \to S^{-1}T\) induces a map on spectra \(\Spec(S^{-1}T) \to \Spec(T)\) which is a homeomorphism onto the subset \[\{\mathfrak p \in \Spec(T) \mid S \cap \mathfrak p = \emptyset\} \subset \Spec(T).\] When \(S = \{ 1, f, f^2, \ldots\}\) for some \(f\in T\), this is the open set \(D(f)\subset \Spec(T)\). We now verify a corresponding property for the ring map \(R \to R_a\): we will show that the map \(\theta : \Spec(R_a) \to \Spec(R)\) induced by inclusion \(R\subset R_a\) is a homeomorphism onto an open subset of \(\Spec(R)\) by verifying that \(\theta\) is an injective local homeomorphism. We do so with respect to an open cover of \(\Spec(R_a)\) by two distinguished opens, as we now describe. For any \(r\in\mathbf{Q}\), let \(\text{ev}_r : R \to \mathbf{Q}\) be the homomorphism given by evaluation at \(r\). Note that for \(r = 0\) and \(r = 1-a\), this can be extended to a homomorphism \(\text{ev}_r' : R_a \to \mathbf{Q}\) (the latter because \(\frac{1}{z-a}\) is well-defined at \(z = 1-a\), since \(a\neq\frac{1}{2}\)). However, \(\text{ev}_a\) does not extend to \(R_a\). Write \(\mathfrak{m}_r = \Ker(\text{ev}_r)\). We have \[\mathfrak{m}_0 = (z^2-z, z^3-z),\] \[\mathfrak{m}_a = ((z-1 + a)(z-a), (z^2-1 + a)(z-a)), \text{ and}\] \[\mathfrak{m}_{1-a} = ((z-1 + a)(z-a), (z-1 + a)(z^2-a)).\] To verify this, note that the right-hand sides are clearly contained in the left-hand sides. Then check that the right-hand sides are maximal ideals by writing the generators in terms of \(A\) and \(B\), and viewing \(R\) as \(\mathbf{Q}[A, B]/(A^3-B^2 + AB)\). Note that \(\mathfrak{m}_a\) is not in the image of \(\theta\): we have \[(z^2 - z)^2(z - a)\left(\frac{a^2 - a}{z - a} + z\right) = (z^2 - z)^2(a^2 - a) + (z^2 - z)^2(z - a)z\] The left hand side is in \(\mathfrak m_a R_a\) because \((z^2 - z)(z - a)\) is in \(\mathfrak m_a\) and because \((z^2 - z)(\frac{a^2 - a}{z - a} + z)\) is in \(R_a\). Similarly the element \((z^2 - z)^2(z - a)z\) is in \(\mathfrak m_a R_a\) because \((z^2 - z)\) is in \(R_a\) and \((z^2 - z)(z - a)\) is in \(\mathfrak m_a\). As \(a \not \in \{0, 1\}\) we conclude that \((z^2 - z)^2 \in \mathfrak m_a R_a\). Hence no ideal \(I\) of \(R_a\) can satisfy \(I \cap R = \mathfrak m_a\), as such an \(I\) would have to contain \((z^2 - z)^2\), which is in \(R\) but not in \(\mathfrak m_a\). The distinguished open set \(D((z-1 + a)(z-a))\subset\Spec(R)\) is equal to the complement of the closed set \(\{\mathfrak{m}_a, \mathfrak{m}_{1-a}\}\). Then check that \(R_{(z-1 + a)(z-a)} = (R_a)_{(z-1 + a)(z-a)}\); calling this localized ring \(R'\), then, it follows that the map \(R \to R'\) factors as \(R \to R_a \to R'\). By Lemma 00E3, then, these maps express \(\Spec(R') \subset \Spec(R_a)\) and \(\Spec(R') \subset \Spec(R)\) as open subsets; hence \(\theta : \Spec(R_a) \to \Spec(R)\), when restricted to \(D((z-1 + a)(z-a))\), is a homeomorphism onto an open subset. Similarly, \(\theta\) restricted to \(D((z^2 + z + 2a-2)(z-a)) \subset \Spec(R_a)\) is a homeomorphism onto the open subset \(D((z^2 + z + 2a-2)(z-a)) \subset \Spec(R)\). Depending on whether \(z^2 + z + 2a-2\) is irreducible or not over \(\mathbf{Q}\), this former distinguished open set has complement equal to one or two closed points along with the closed point \(\mathfrak{m}_a\). Furthermore, the ideal in \(R_a\) generated by the elements \((z^2 + z + 2a-2)(z-a)\) and \((z-1 + a)(z-a)\) is all of \(R_a\), so these two distinguished open sets cover \(\Spec(R_a)\). Hence in order to show that \(\theta\) is a homeomorphism onto \(\Spec(R)-\{\mathfrak{m}_a\}\), it suffices to show that these one or two points can never equal \(\mathfrak{m}_{1-a}\). And this is indeed the case, since \(1-a\) is a root of \(z^2 + z + 2a-2\) if and only if \(a = 0\) or \(a = 1\), both of which do not occur.

Despite this homeomorphism which mimics the behavior of a localization at an element of \(R\), while \(\mathbf{Q}[z, \frac{1}{z-a}]\) is the localization of \(\mathbf{Q}[z]\) at the element \(z-a\), the ring \(R_a\) is not a localization of \(R\): Any proper localization \(S^{-1}R\) results in more units than the original ring \(R\). The units of \(R\) are \(\mathbf{Q}^\times\), the units of \(\mathbf{Q}\). In fact, it is easy to see that the units of \(R_a\) are \(\mathbf{Q}^*\). Namely, the units of \(\mathbf{Q}[z, \frac{1}{z - a}]\) are \(c (z - a)^n\) for \(c \in \mathbf{Q}^*\) and \(n \in \mathbf{Z}\) and it is clear that these are in \(R_a\) only if \(n = 0\). Hence \(R_a\) has no more units than \(R\) does, and thus cannot be a localization of \(R\).

We used the fact that \(a\neq 0, 1\) to ensure that \(\frac{1}{z-a}\) makes sense at \(z = 0, 1\). We used the fact that \(a\neq 1/2\) in a few places: (1) In order to be able to talk about the kernel of \(\text{ev}_{1-a}\) on \(R_a\), which ensures that \(\mathfrak{m}_{1-a}\) is a point of \(R_a\) (i.e., that \(R_a\) is missing just one point of \(R\)). (2) At the end in order to conclude that \((z-a)^n\) can belong to \(R_a\) only for \(n = 0\); indeed, if \(a = 1/2\), then it belongs to \(R_a\) whenever \(n\) is even. Hence there would indeed be more units in \(R_a\) than in \(R\), and \(R_a\) could possibly be a localization of \(R\).

A meta-observation about prime ideals

This section is taken from the CRing project. Let \(R\) be a ring and let \(S \subset R\) be a multiplicative subset. A consequence of Lemma 00E3 is that an ideal \(I \subset R\) maximal with respect to the property of not intersecting \(S\) is prime. The reason is that \(I = R \cap \mathfrak m\) for some maximal ideal \(\mathfrak m\) of the ring \(S^{-1}R\). It turns out that for many properties of ideals, the maximal ones are prime. A general method of seeing this was developed in [Lam-Reyes]. In this section, we digress to explain this phenomenon.

Let \(R\) be a ring. If \(I\) is an ideal of \(R\) and \(a \in R\), we define \[(I : a) = \left\{ x \in R \mid xa \in I\right\}.\] More generally, if \(J \subset R\) is an ideal, we define \[(I : J) = \left\{ x \in R \mid xJ \subset I\right\}.\]

Lemma

Let \(R\) be a ring. For a principal ideal \(J \subset R\), and for any ideal \(I \subset J\) we have \(I = J (I : J)\).

Proof

Say \(J = (a)\). Then \((I : J) = (I : a)\). Since \(I \subset J\) we see that any \(y \in I\) is of the form \(y = xa\) for some \(x \in (I : a)\). Hence \(I \subset J (I : J)\). Conversely, if \(x \in (I : a)\), then \(xJ = (xa) \subset I\), which proves the other inclusion.

Let \(\mathcal{F}\) be a collection of ideals of \(R\). We are interested in conditions that will guarantee that the maximal elements in the complement of \(\mathcal{F}\) are prime.

Definition

Let \(R\) be a ring. Let \(\mathcal{F}\) be a set of ideals of \(R\). We say \(\mathcal{F}\) is an Oka family if \(R \in \mathcal{F}\) and whenever \(I \subset R\) is an ideal and \((I : a), (I, a) \in \mathcal{F}\) for some \(a \in R\), then \(I \in \mathcal{F}\).

Let us give some examples of Oka families. The first example is the basic example discussed in the introduction to this section.

Example

Let \(R\) be a ring and let \(S\) be a multiplicative subset of \(R\). We claim that \(\mathcal{F} = \{I \subset R \mid I \cap S \not = \emptyset\}\) is an Oka family. Namely, suppose that \((I : a), (I, a) \in \mathcal{F}\) for some \(a \in R\). Then pick \(s \in (I, a) \cap S\) and \(s' \in (I : a) \cap S\). Then \(ss' \in I \cap S\) and hence \(I \in \mathcal{F}\). Thus \(\mathcal{F}\) is an Oka family.

Example

Let \(R\) be a ring, \(I \subset R\) an ideal, and \(a \in R\). If \((I : a)\) is generated by \(a_1, \ldots, a_n\) and \((I, a)\) is generated by \(a, b_1, \ldots, b_m\) with \(b_1, \ldots, b_m \in I\), then \(I\) is generated by \(aa_1, \ldots, aa_n, b_1, \ldots, b_m\). To see this, note that if \(x \in I\), then \(x \in (I, a)\) is a linear combination of \(a, b_1, \ldots, b_m\), but the coefficient of \(a\) must lie in \((I:a)\). As a result, we deduce that the family of finitely generated ideals is an Oka family.

Example

Let us show that the family of principal ideals of a ring \(R\) is an Oka family. Indeed, suppose \(I \subset R\) is an ideal, \(a \in R\), and \((I, a)\) and \((I : a)\) are principal. Note that \((I : a) = (I : (I, a))\). Setting \(J = (I, a)\), we find that \(J\) is principal and \((I : J)\) is too. By Lemma 05K8 we have \(I = J (I : J)\). Thus we find in our situation that since \(J = (I, a)\) and \((I : J)\) are principal, \(I\) is principal.

Example

Let \(R\) be a ring. Let \(\kappa\) be an infinite cardinal. The family of ideals which can be generated by at most \(\kappa\) elements is an Oka family. The argument is analogous to the argument in Example 05KB and is omitted.

Example

Let \(A\) be a ring, \(I \subset A\) an ideal, and \(a \in A\) an element. There is a short exact sequence \(0 \to A/(I : a) \to A/I \to A/(I, a) \to 0\) where the first arrow is given by multiplication by \(a\). Thus if \(P\) is a property of \(A\)-modules that is stable under extensions and holds for \(0\), then the family of ideals \(I\) such that \(A/I\) has \(P\) is an Oka family.

Proposition

If \(\mathcal{F}\) is an Oka family of ideals, then any maximal element of the complement of \(\mathcal{F}\) is prime.

Proof

Suppose \(I \not \in \mathcal{F}\) is maximal with respect to not being in \(\mathcal{F}\) but \(I\) is not prime. Note that \(I \not = R\) because \(R \in \mathcal{F}\). Since \(I\) is not prime we can find \(a, b \in R - I\) with \(ab \in I\). It follows that \((I, a) \neq I\) and \((I : a)\) contains \(b \not \in I\) so also \((I : a) \neq I\). Thus \((I : a), (I, a)\) both strictly contain \(I\), so they must belong to \(\mathcal{F}\). By the Oka condition, we have \(I \in \mathcal{F}\), a contradiction.

At this point we are able to turn most of the examples above into a lemma about prime ideals in a ring.

Lemma

Let \(R\) be a ring. Let \(S\) be a multiplicative subset of \(R\). An ideal \(I \subset R\) which is maximal with respect to the property that \(I \cap S = \emptyset\) is prime.

Proof

This is the example discussed in the introduction to this section. For an alternative proof, combine Example 05KA with Proposition 05KE.

Lemma

Let \(R\) be a ring.

1. An ideal \(I \subset R\) maximal with respect to not being finitely generated is prime.

2. If every prime ideal of \(R\) is finitely generated, then every ideal of \(R\) is finitely generated².

Proof

The first assertion is an immediate consequence of Example 05KB and Proposition 05KE. For the second, suppose that there exists an ideal \(I \subset R\) which is not finitely generated. The union of a totally ordered chain \(\left\{I_\alpha\right\}\) of ideals that are not finitely generated is not finitely generated; indeed, if \(I = \bigcup I_\alpha\) were generated by \(a_1, \ldots, a_n\), then all the generators would belong to some \(I_\alpha\) and would consequently generate it. By Zorn’s lemma, there is an ideal maximal with respect to being not finitely generated. By the first part this ideal is prime.

Lemma

Let \(R\) be a ring.

1. An ideal \(I \subset R\) maximal with respect to not being principal is prime.

2. If every prime ideal of \(R\) is principal, then every ideal of \(R\) is principal.

Proof

The first part follows from Example 05KC and Proposition 05KE. For the second, suppose that there exists an ideal \(I \subset R\) which is not principal. The union of a totally ordered chain \(\left\{I_\alpha\right\}\) of ideals that are not principal is not principal; indeed, if \(I = \bigcup I_\alpha\) were generated by \(a\), then \(a\) would belong to some \(I_\alpha\) and \(a\) would generate it. By Zorn’s lemma, there is an ideal maximal with respect to not being principal. This ideal is necessarily prime by the first part.

Lemma

Let \(R\) be a ring.

1. An ideal maximal among the ideals which do not contain a nonzerodivisor is prime.

2. If \(R\) is nonzero and every nonzero prime ideal in \(R\) contains a nonzerodivisor, then \(R\) is a domain.

Proof

Consider the set \(S\) of nonzerodivisors. It is a multiplicative subset of \(R\). Hence any ideal maximal with respect to not intersecting \(S\) is prime, see Lemma 05KF. Thus, if every nonzero prime ideal contains a nonzerodivisor, then \((0)\) is prime, i.e., \(R\) is a domain.

Remark

Let \(R\) be a ring. Let \(\kappa\) be an infinite cardinal. By applying Example 05KD and Proposition 05KE we see that any ideal maximal with respect to the property of not being generated by \(\kappa\) elements is prime. This result is not so useful because there exists a ring for which every prime ideal of \(R\) can be generated by \(\aleph_0\) elements, but some ideal cannot. Namely, let \(k\) be a field, let \(T\) be a set whose cardinality is greater than \(\aleph_0\) and let \[R = k[\{x_n\}_{n \geq 1}, \{z_{t, n}\}_{t \in T, n \geq 0}]/ (x_n^2, z_{t, n}^2, x_n z_{t, n} - z_{t, n - 1})\] This is a local ring with unique prime ideal \(\mathfrak m = (x_n)\). But the ideal \((z_{t, n})\) cannot be generated by countably many elements.

Example

Let \(R\) be a ring and \(X = \Spec(R)\). Since closed subsets of \(X\) correspond to radical ideals of \(R\) (Lemma 00E0) we see that \(X\) is a Noetherian topological space if and only if we have ACC for radical ideals. This holds if and only if every radical ideal is the radical of a finitely generated ideal (details omitted). Let \[\mathcal{F} = \{I \subset R \mid \sqrt{I} = \sqrt{(f_1, \ldots, f_n)}\text{ for some }n \text{ and }f_1, \ldots, f_n \in R\}.\] The reader can show that \(\mathcal{F}\) is an Oka family by using the identity \[\sqrt{I} = \sqrt{(I, a)(I : a)}\] which holds for any ideal \(I \subset R\) and any element \(a \in R\). On the other hand, if we have a totally ordered chain of ideals \(\{I_\alpha\}\) none of which are in \(\mathcal{F}\), then the union \(I = \bigcup I_\alpha\) cannot be in \(\mathcal{F}\) either. Otherwise \(\sqrt{I} = \sqrt{(f_1, \ldots, f_n)}\), then \(f_i^e \in I\) for some \(e\), then \(f_i^e \in I_\alpha\) for some \(\alpha\) independent of \(i\), then \(\sqrt{I_\alpha} = \sqrt{(f_1, \ldots, f_n)}\), contradiction. Thus if the set of ideals not in \(\mathcal{F}\) is nonempty, then it has maximal elements and exactly as in Lemma 05KG we conclude that \(X\) is a Noetherian topological space if and only if every prime ideal of \(R\) is equal to \(\sqrt{(f_1, \ldots, f_n)}\) for some \(f_1, \ldots, f_n \in R\). If we ever need this result we will carefully state and prove this result here.

Images of ring maps of finite presentation

In this section we prove some results on the topology of maps \(\Spec(S) \to \Spec(R)\) induced by ring maps \(R \to S\), mainly Chevalley’s Theorem. In order to do this we will use the notions of constructible sets, quasi-compact sets, retrocompact sets, and so on which are defined in Topology, Section 04ZC.

Lemma

Let \(U \subset \Spec(R)\) be open. The following are equivalent:

1. \(U\) is retrocompact in \(\Spec(R)\),

2. \(U\) is quasi-compact,

3. \(U\) is a finite union of standard opens, and

4. there exists a finitely generated ideal \(I \subset R\) such that \(X \setminus V(I) = U\).

Proof

We have (1) \(\Rightarrow\) (2) because \(\Spec(R)\) is quasi-compact, see Lemma 00E8. We have (2) \(\Rightarrow\) (3) because standard opens form a basis for the topology. Proof of (3) \(\Rightarrow\) (1). Let \(U = \bigcup_{i = 1\ldots n} D(f_i)\). To show that \(U\) is retrocompact in \(\Spec(R)\) it suffices to show that \(U \cap V\) is quasi-compact for any quasi-compact open \(V\) of \(\Spec(R)\). Write \(V = \bigcup_{j = 1\ldots m} D(g_j)\) which is possible by (2) \(\Rightarrow\) (3). Each standard open is homeomorphic to the spectrum of a ring and hence quasi-compact, see Lemmas 00E4 and 00E8. Thus \(U \cap V = (\bigcup_{i = 1\ldots n} D(f_i)) \cap (\bigcup_{j = 1\ldots m} D(g_j)) = \bigcup_{i, j} D(f_i g_j)\) is a finite union of quasi-compact opens hence quasi-compact. To finish the proof note that (4) is equivalent to (3) by Lemma 00E0.

Lemma

Let \(\varphi : R \to S\) be a ring map. The induced continuous map \(f : \Spec(S) \to \Spec(R)\) is quasi-compact. For any constructible set \(E \subset \Spec(R)\) the inverse image \(f^{-1}(E)\) is constructible in \(\Spec(S)\).

Proof

We first show that the inverse image of any quasi-compact open \(U \subset \Spec(R)\) is quasi-compact. By Lemma 00F6 we may write \(U\) as a finite open of standard opens. Thus by Lemma 00E2 we see that \(f^{-1}(U)\) is a finite union of standard opens. Hence \(f^{-1}(U)\) is quasi-compact by Lemma 00F6 again. The second assertion now follows from Topology, Lemma 005I.

Lemma

Let \(R\) be a ring. A subset of \(\Spec(R)\) is constructible if and only if it can be written as a finite union of subsets of the form \(D(f) \cap V(g_1, \ldots, g_m)\) for \(f, g_1, \ldots, g_m \in R\).

Proof

By Lemma 00F6 the subset \(D(f)\) and the complement of \(V(g_1, \ldots, g_m)\) are retro-compact open. Hence \(D(f) \cap V(g_1, \ldots, g_m)\) is a constructible subset and so is any finite union of such. Conversely, let \(T \subset \Spec(R)\) be constructible. By Topology, Definition 005G, we may assume that \(T = U \cap V^c\), where \(U, V \subset \Spec(R)\) are retrocompact open. By Lemma 00F6 we may write \(U = \bigcup_{i = 1, \ldots, n} D(f_i)\) and \(V = \bigcup_{j = 1, \ldots, m} D(g_j)\). Then \(T = \bigcup_{i = 1, \ldots, n} \big(D(f_i) \cap V(g_1, \ldots, g_m)\big)\).

Lemma

Let \(R\) be a ring and let \(T \subset \Spec(R)\) be constructible. Then there exists a ring map \(R \to S\) of finite presentation such that \(T\) is the image of \(\Spec(S)\) in \(\Spec(R)\).

Proof

The spectrum of a finite product of rings is the disjoint union of the spectra, see Lemma 00ED. Hence if \(T = T_1 \cup T_2\) and the result holds for \(T_1\) and \(T_2\), then the result holds for \(T\). By Lemma 0G1P we may assume that \(T = D(f) \cap V(g_1, \ldots, g_m)\). In this case \(T\) is the image of the map \(\Spec((R/(g_1, \ldots, g_m))_f) \to \Spec(R)\), see Lemmas 00E4 and 00E5.

Lemma

Let \(R\) be a ring. Let \(f\) be an element of \(R\). Let \(S = R_f\). Then the image of a constructible subset of \(\Spec(S)\) is constructible in \(\Spec(R)\).

Proof

We repeatedly use Lemma 00F6 without mention. Let \(U, V\) be quasi-compact open in \(\Spec(S)\). We will show that the image of \(U \cap V^c\) is constructible. Under the identification \(\Spec(S) = D(f)\) of Lemma 00E4 the sets \(U, V\) correspond to quasi-compact opens \(U', V'\) of \(\Spec(R)\). Hence it suffices to show that \(U' \cap (V')^c\) is constructible in \(\Spec(R)\) which is clear.

Lemma

Let \(R\) be a ring. Let \(I\) be a finitely generated ideal of \(R\). Let \(S = R/I\). Then the image of a constructible subset of \(\Spec(S)\) is constructible in \(\Spec(R)\).

Proof

If \(I = (f_1, \ldots, f_m)\), then we see that \(V(I)\) is the complement of \(\bigcup D(f_i)\), see Lemma 00E0. Hence it is constructible, by Lemma 00F6. Denote the map \(R \to S\) by \(f \mapsto \overline{f}\). We have to show that if \(\overline{U}, \overline{V}\) are retrocompact opens of \(\Spec(S)\), then the image of \(\overline{U} \cap \overline{V}^c\) in \(\Spec(R)\) is constructible. By Lemma 00F6 we may write \(\overline{U} = \bigcup D(\overline{g_i})\). Setting \(U = \bigcup D({g_i})\) we see \(\overline{U}\) has image \(U \cap V(I)\) which is constructible in \(\Spec(R)\). Similarly the image of \(\overline{V}\) equals \(V \cap V(I)\) for some retrocompact open \(V\) of \(\Spec(R)\). Hence the image of \(\overline{U} \cap \overline{V}^c\) equals \(U \cap V(I) \cap V^c\) as desired.

Lemma

Let \(R\) be a ring. The map \(\Spec(R[x]) \to \Spec(R)\) is open, and the image of any standard open is a quasi-compact open.

Proof

It suffices to show that the image of a standard open \(D(f)\), \(f\in R[x]\) is quasi-compact open. The image of \(D(f)\) is the image of \(\Spec(R[x]_f) \to \Spec(R)\). Let \(\mathfrak p \subset R\) be a prime ideal. Let \(\overline{f}\) be the image of \(f\) in \(\kappa(\mathfrak p)[x]\). Recall, see Lemma 00E7, that \(\mathfrak p\) is in the image if and only if \(R[x]_f \otimes_R \kappa(\mathfrak p) = \kappa(\mathfrak p)[x]_{\overline{f}}\) is not the zero ring. This is exactly the condition that \(f\) does not map to zero in \(\kappa(\mathfrak p)[x]\), in other words, that some coefficient of \(f\) is not in \(\mathfrak p\). Hence we see: if \(f = a_d x^d + \ldots + a_0\), then the image of \(D(f)\) is \(D(a_d) \cup \ldots \cup D(a_0)\).

We prove a property of characteristic polynomials which will be used below.

Lemma

Let \(R \to A\) be a ring homomorphism. Assume \(A \cong R^{\oplus n}\) as an \(R\)-module. Let \(f \in A\). The multiplication map \(m_f: A \to A\) is \(R\)-linear and hence has a characteristic polynomial \(P(T) = T^n + r_{n-1}T^{n-1} + \ldots + r_0 \in R[T]\). For any prime \(\mathfrak{p} \in \Spec(R)\), \(f\) acts nilpotently on \(A \otimes_R \kappa(\mathfrak{p})\) if and only if \(\mathfrak p \in V(r_0, \ldots, r_{n-1})\).

Proof

This follows quite easily once we prove that the characteristic polynomial \(\bar P(T) \in \kappa(\mathfrak p)[T]\) of the multiplication map \(m_{\bar f}: A \otimes_R \kappa(\mathfrak p) \to A \otimes_R \kappa(\mathfrak p)\) which multiplies elements of \(A \otimes_R \kappa(\mathfrak p)\) by \(\bar f\), the image of \(f\) viewed in \(\kappa(\mathfrak p)\), is just the image of \(P(T)\) in \(\kappa(\mathfrak p)[T]\). Let \((a_{ij})\) be the matrix of the map \(m_f\) with entries in \(R\), using a basis \(e_1, \ldots, e_n\) of \(A\) as an \(R\)-module. Then, \(A \otimes_R \kappa(\mathfrak p) \cong (R \otimes_R \kappa(\mathfrak p))^{\oplus n} = \kappa(\mathfrak p)^n\), which is an \(n\)-dimensional vector space over \(\kappa(\mathfrak p)\) with basis \(e_1 \otimes 1, \ldots, e_n \otimes 1\). The image \(\bar f = f \otimes 1\), and so the multiplication map \(m_{\bar f}\) has matrix \((a_{ij} \otimes 1)\). Thus, the characteristic polynomial is precisely the image of \(P(T)\).

From linear algebra, we know that a linear transformation acts nilpotently on an \(n\)-dimensional vector space if and only if the characteristic polynomial is \(T^n\) (since the characteristic polynomial divides some power of the minimal polynomial). Hence, \(f\) acts nilpotently on \(A \otimes_R \kappa(\mathfrak p)\) if and only if \(\bar P(T) = T^n\). This occurs if and only if \(r_i \in \mathfrak p\) for all \(0 \leq i \leq n - 1\), that is when \(\mathfrak p \in V(r_0, \ldots, r_{n - 1}).\)

Lemma

Let \(R\) be a ring. Let \(f, g \in R[x]\) be polynomials. Assume the leading coefficient of \(g\) is a unit of \(R\). There exists elements \(r_i\in R\), \(i = 1\ldots, n\) such that the image of \(D(f) \cap V(g)\) in \(\Spec(R)\) is \(\bigcup_{i = 1, \ldots, n} D(r_i)\).

Proof

Write \(g = ux^d + a_{d-1}x^{d-1} + \ldots + a_0\), where \(d\) is the degree of \(g\), and hence \(u \in R^*\). Consider the ring \(A = R[x]/(g)\). It is, as an \(R\)-module, finite free with basis the images of \(1, x, \ldots, x^{d-1}\). Consider multiplication by (the image of) \(f\) on \(A\). This is an \(R\)-module map. Hence we can let \(P(T) \in R[T]\) be the characteristic polynomial of this map. Write \(P(T) = T^d + r_{d-1} T^{d-1} + \ldots + r_0\). We claim that \(r_0, \ldots, r_{d-1}\) have the desired property. We will use below the property of characteristic polynomials that \[\mathfrak p \in V(r_0, \ldots, r_{d-1}) \Leftrightarrow \text{multiplication by }f\text{ is nilpotent on } A \otimes_R \kappa(\mathfrak p).\] This was proved in Lemma 00FC.

Suppose \(\mathfrak q\in D(f) \cap V(g)\), and let \(\mathfrak p = \mathfrak q \cap R\). Then there is a nonzero map \(A \otimes_R \kappa(\mathfrak p) \to \kappa(\mathfrak q)\) which is compatible with multiplication by \(f\). And \(f\) acts as a unit on \(\kappa(\mathfrak q)\). Thus we conclude \(\mathfrak p \not \in V(r_0, \ldots, r_{d-1})\).

On the other hand, suppose that \(r_i \not\in \mathfrak p\) for some prime \(\mathfrak p\) of \(R\) and some \(0 \leq i \leq d - 1\). Then multiplication by \(f\) is not nilpotent on the algebra \(A \otimes_R \kappa(\mathfrak p)\). Hence there exists a prime ideal \(\overline{\mathfrak q} \subset A \otimes_R \kappa(\mathfrak p)\) not containing the image of \(f\). The inverse image of \(\overline{\mathfrak q}\) in \(R[x]\) is an element of \(D(f) \cap V(g)\) mapping to \(\mathfrak p\).

Theorem

Suppose that \(R \to S\) is of finite presentation. The image of a constructible subset of \(\Spec(S)\) in \(\Spec(R)\) is constructible.

Proof

Write \(S = R[x_1, \ldots, x_n]/(f_1, \ldots, f_m)\). We may factor \(R \to S\) as \(R \to R[x_1] \to R[x_1, x_2] \to \ldots \to R[x_1, \ldots, x_{n-1}] \to S\). Hence we may assume that \(S = R[x]/(f_1, \ldots, f_m)\). In this case we factor the map as \(R \to R[x] \to S\), and by Lemma 00FA we reduce to the case \(S = R[x]\). By Lemma 00F6 suffices to show that if \(T = (\bigcup_{i = 1\ldots n} D(f_i)) \cap V(g_1, \ldots, g_m)\) for \(f_i , g_j \in R[x]\) then the image in \(\Spec(R)\) is constructible. Since finite unions of constructible sets are constructible, it suffices to deal with the case \(n = 1\), i.e., when \(T = D(f) \cap V(g_1, \ldots, g_m)\).

Note that if \(c \in R\), then we have \[\Spec(R) = V(c) \amalg D(c) = \Spec(R/(c)) \amalg \Spec(R_c),\] and correspondingly \(\Spec(R[x]) = V(c) \amalg D(c) = \Spec(R/(c)[x]) \amalg \Spec(R_c[x])\). The intersection of \(T = D(f) \cap V(g_1, \ldots, g_m)\) with each part still has the same shape, with \(f\), \(g_i\) replaced by their images in \(R/(c)[x]\), respectively \(R_c[x]\). Note that the image of \(T\) in \(\Spec(R)\) is the union of the image of \(T \cap V(c)\) and \(T \cap D(c)\). Using Lemmas 00F9 and 00FA it suffices to prove the images of both parts are constructible in \(\Spec(R/(c))\), respectively \(\Spec(R_c)\).

Let us assume we have \(T = D(f) \cap V(g_1, \ldots, g_m)\) as above, with \(\deg(g_1) \leq \deg(g_2) \leq \ldots \leq \deg(g_m)\). We are going to use induction on \(m\), and on the degrees of the \(g_i\). Let \(d_1 = \deg(g_1)\), i.e., \(g_1 = c x^{d_1} + l.o.t\) with \(c \in R\) not zero. Cutting \(R\) up into the pieces \(R/(c)\) and \(R_c\) we either lower the degree of \(g_1\) (and this is covered by induction) or we reduce to the case where \(c\) is invertible. If \(c\) is invertible, and \(m > 1\), then write \(g_2 = c' x^{d_2} + l.o.t\). In this case consider \(g_2' = g_2 - (c'/c) x^{d_2 - d_1} g_1\). Since the ideals \((g_1, g_2, \ldots, g_m)\) and \((g_1, g_2', g_3, \ldots, g_m)\) are equal we see that \(T = D(f) \cap V(g_1, g_2', g_3\ldots, g_m)\). But here the degree of \(g_2'\) is strictly less than the degree of \(g_2\) and hence this case is covered by induction.

The bases case for the induction above are the cases (a) \(T = D(f) \cap V(g)\) where the leading coefficient of \(g\) is invertible, and (b) \(T = D(f)\). These two cases are dealt with in Lemmas 00FD and 00FB.

More on images

In this section we collect a few additional lemmas concerning the image on \(\Spec\) for ring maps. See also Section 00HU for example.

Lemma

Let \(R \subset S\) be an inclusion of domains. Assume that \(R \to S\) is of finite type. There exists a nonzero \(f \in R\), and a nonzero \(g \in S\) such that \(R_f \to S_{fg}\) is of finite presentation.

Proof

By induction on the number of generators of \(S\) over \(R\). During the proof we may replace \(R\) by \(R_f\) and \(S\) by \(S_f\) for some nonzero \(f \in R\).

Suppose that \(S\) is generated by a single element over \(R\). Then \(S = R[x]/\mathfrak q\) for some prime ideal \(\mathfrak q \subset R[x]\). If \(\mathfrak q = (0)\) there is nothing to prove. If \(\mathfrak q \not = (0)\), then let \(h \in \mathfrak q\) be a nonzero element with minimal degree in \(x\). Write \(h = f x^d + a_{d - 1} x^{d - 1} + \ldots + a_0\) with \(a_i \in R\) and \(f \not = 0\). After inverting \(f\) in \(R\) and \(S\) we may assume that \(h\) is monic. We obtain a surjective \(R\)-algebra map \(R[x]/(h) \to S\). We have \(R[x]/(h) = R \oplus Rx \oplus \ldots \oplus Rx^{d - 1}\) as an \(R\)-module and by minimality of \(d\) we see that \(R[x]/(h)\) maps injectively into \(S\). Thus \(R[x]/(h) \cong S\) is finitely presented over \(R\).

Suppose that \(S\) is generated by \(n > 1\) elements over \(R\). Say \(x_1, \ldots, x_n \in S\) generate \(S\). Denote \(S' \subset S\) the \(R\)-subalgebra generated by \(x_1, \ldots, x_{n-1}\). By induction hypothesis we see that there exist \(f\in R\) and \(g \in S'\) nonzero such that \(R_f \to S'_{fg}\) is of finite presentation. Next we apply the induction hypothesis to \(S'_{fg} \to S_{fg}\) to see that there exist \(f' \in S'_{fg}\) and \(g' \in S_{fg}\) such that \(S'_{fgf'} \to S_{fgf'g'}\) is of finite presentation. We leave it to the reader to conclude.

Lemma

Let \(R \to S\) be a finite type ring map. Denote \(X = \Spec(R)\) and \(Y = \Spec(S)\). Write \(f : Y \to X\) the induced map of spectra. Let \(E \subset Y = \Spec(S)\) be a constructible set. If a point \(\xi \in X\) is in \(f(E)\), then \(\overline{\{\xi\}} \cap f(E)\) contains an open dense subset of \(\overline{\{\xi\}}\).

Proof

Let \(\xi \in X\) be a point of \(f(E)\). Choose a point \(\eta \in E\) mapping to \(\xi\). Let \(\mathfrak p \subset R\) be the prime corresponding to \(\xi\) and let \(\mathfrak q \subset S\) be the prime corresponding to \(\eta\). Consider the diagram \[\xymatrix{ \eta \ar[r] \ar@{|->}[d] & E \cap Y' \ar[r] \ar[d] & Y' = \Spec(S/\mathfrak q) \ar[r] \ar[d] & Y \ar[d] \\ \xi \ar[r] & f(E) \cap X' \ar[r] & X' = \Spec(R/\mathfrak p) \ar[r] & X }\] By Lemma 00F7 the set \(E \cap Y'\) is constructible in \(Y'\). It follows that we may replace \(X\) by \(X'\) and \(Y\) by \(Y'\). Hence we may assume that \(R \subset S\) is an inclusion of domains, \(\xi\) is the generic point of \(X\), and \(\eta\) is the generic point of \(Y\). By Lemma 00FG combined with Chevalley’s theorem (Theorem 00FE) we see that there exist dense opens \(U \subset X\), \(V \subset Y\) such that \(f(V) \subset U\) and such that \(f : V \to U\) maps constructible sets to constructible sets. Note that \(E \cap V\) is constructible in \(V\), see Topology, Lemma 005J. Hence \(f(E \cap V)\) is constructible in \(U\) and contains \(\xi\). By Topology, Lemma 005K we see that \(f(E \cap V)\) contains a dense open \(U' \subset U\).

At the end of this section we present a few more results on images of maps on Spectra that have nothing to do with constructible sets.

Lemma

Let \(\varphi : R \to S\) be a ring map. The following are equivalent:

1. The map \(\Spec(S) \to \Spec(R)\) is surjective.

2. For any ideal \(I \subset R\) the inverse image of \(\sqrt{IS}\) in \(R\) is equal to \(\sqrt{I}\).

3. For any radical ideal \(I \subset R\) the inverse image of \(IS\) in \(R\) is equal to \(I\).

4. For every prime \(\mathfrak p\) of \(R\) the inverse image of \(\mathfrak p S\) in \(R\) is \(\mathfrak p\).

In this case the same is true after any base change: Given a ring map \(R \to R'\) the ring map \(R' \to R' \otimes_R S\) has the equivalent properties (1), (2), (3) as well.

Proof

If \(J \subset S\) is an ideal, then \(\sqrt{\varphi^{-1}(J)} = \varphi^{-1}(\sqrt{J})\). This shows that (2) and (3) are equivalent. The implication (3) \(\Rightarrow\) (4) is immediate. If \(I \subset R\) is a radical ideal, then Lemma 00E0 guarantees that \(I = \bigcap_{I \subset \mathfrak p} \mathfrak p\). Hence (4) \(\Rightarrow\) (2). By Lemma 00E7 we have \(\mathfrak p = \varphi^{-1}(\mathfrak p S)\) if and only if \(\mathfrak p\) is in the image. Hence (1) \(\Leftrightarrow\) (4). Thus (1), (2), (3), and (4) are equivalent.

Assume (1) holds. Let \(R \to R'\) be a ring map. Let \(\mathfrak p' \subset R'\) be a prime ideal lying over the prime \(\mathfrak p\) of \(R\). To see that \(\mathfrak p'\) is in the image of \(\Spec(R' \otimes_R S) \to \Spec(R')\) we have to show that \((R' \otimes_R S) \otimes_{R'} \kappa(\mathfrak p')\) is not zero, see Lemma 00E7. But we have \[(R' \otimes_R S) \otimes_{R'} \kappa(\mathfrak p') = S \otimes_R \kappa(\mathfrak p) \otimes_{\kappa(\mathfrak p)} \kappa(\mathfrak p')\] which is not zero as \(S \otimes_R \kappa(\mathfrak p)\) is not zero by assumption and \(\kappa(\mathfrak p) \to \kappa(\mathfrak p')\) is an extension of fields.

Lemma

Let \(R\) be a domain. Let \(\varphi : R \to S\) be a ring map. The following are equivalent:

1. The ring map \(R \to S\) is injective.

2. The image \(\Spec(S) \to \Spec(R)\) contains a dense set of points.

3. There exists a prime ideal \(\mathfrak q \subset S\) whose inverse image in \(R\) is \((0)\).

Proof

Let \(K\) be the field of fractions of the domain \(R\). Assume that \(R \to S\) is injective. Since localization is exact we see that \(K \to S \otimes_R K\) is injective. Hence there is a prime mapping to \((0)\) by Lemma 00E7.

Note that \((0)\) is dense in \(\Spec(R)\), so that the last condition implies the second.

Suppose the second condition holds. Let \(f \in R\), \(f \not = 0\). As \(R\) is a domain we see that \(V(f)\) is a proper closed subset of \(R\). By assumption there exists a prime \(\mathfrak q\) of \(S\) such that \(\varphi(f) \not \in \mathfrak q\). Hence \(\varphi(f) \not = 0\). Hence \(R \to S\) is injective.

Lemma

Let \(R \subset S\) be an injective ring map. Then \(\Spec(S) \to \Spec(R)\) hits all the minimal primes.

Proof

Let \(\mathfrak p \subset R\) be a minimal prime. In this case \(R_{\mathfrak p}\) has a unique prime ideal. Hence it suffices to show that \(S_{\mathfrak p}\) is not zero. And this follows from the fact that localization is exact, see Proposition 00CS.

Lemma

Let \(R \to S\) be a ring map. The following are equivalent:

1. The kernel of \(R \to S\) consists of nilpotent elements.

2. The minimal primes of \(R\) are in the image of \(\Spec(S) \to \Spec(R)\).

3. The image of \(\Spec(S) \to \Spec(R)\) is dense in \(\Spec(R)\).

Proof

Let \(I = \Ker(R \to S)\). Note that \(\sqrt{(0)} = \bigcap_{\mathfrak q \subset S} \mathfrak q\), see Lemma 00E0. Hence \(\sqrt{I} = \bigcap_{\mathfrak q \subset S} R \cap \mathfrak q\). Thus \(V(I) = V(\sqrt{I})\) is the closure of the image of \(\Spec(S) \to \Spec(R)\). This shows that (1) is equivalent to (3). It is clear that (2) implies (3). Finally, assume (1). We may replace \(R\) by \(R/I\) and \(S\) by \(S/IS\) without affecting the topology of the spectra and the map. Hence the implication (1) \(\Rightarrow\) (2) follows from Lemma 00FK.

Lemma

Let \(R \to S\) be a ring map. If a minimal prime \(\mathfrak p \subset R\) is in the image of \(\Spec(S) \to \Spec(R)\), then it is the image of a minimal prime.

Proof

Say \(\mathfrak p = \mathfrak q \cap R\). Then choose a minimal prime \(\mathfrak r \subset S\) with \(\mathfrak r \subset \mathfrak q\), see Lemma 00E0. By minimality of \(\mathfrak p\) we see that \(\mathfrak p = \mathfrak r \cap R\).

Lemma

Let \(A \subset B\) be an inclusion of domains inducing an algebraic extension of fraction fields. If \(J \subset B\) is a nonzero ideal, then \(A \cap J\) is nonzero too. Thus the image of a proper closed subset of \(\Spec(B)\) is not dense in \(\Spec(A)\).

Proof

Let \(x \in J\) be a nonzero element. Since \(x\) is algebraic over the fraction field of \(A\), there exists a \(d \geq 1\) and \(a_0, \ldots, a_d \in A\) with \(a_0, a_d \not = 0\) such that \(a_d x^d + a_{d - 1} x^{d - 1} + \ldots + a_0 = 0\) in \(B\). Then \(a_0 \in A \cap J\).

Noetherian rings

A ring \(R\) is Noetherian if any ideal of \(R\) is finitely generated. This is clearly equivalent to the ascending chain condition for ideals of \(R\). By Lemma 05KG it suffices to check that every prime ideal of \(R\) is finitely generated.

Lemma

Any finitely generated ring over a Noetherian ring is Noetherian. Any localization of a Noetherian ring is Noetherian.

Proof

The statement on localizations follows from the fact that any ideal \(J \subset S^{-1}R\) is of the form \(I \cdot S^{-1}R\). Any quotient \(R/I\) of a Noetherian ring \(R\) is Noetherian because any ideal \(\overline{J} \subset R/I\) is of the form \(J/I\) for some ideal \(I \subset J \subset R\). Thus it suffices to show that if \(R\) is Noetherian so is \(R[X]\). Suppose \(J_1 \subset J_2 \subset \ldots\) is an ascending chain of ideals in \(R[X]\). Consider the ideals \(I_{i, d}\) defined as the ideal of elements of \(R\) which occur as leading coefficients of degree \(d\) polynomials in \(J_i\). Clearly \(I_{i, d} \subset I_{i', d'}\) whenever \(i \leq i'\) and \(d \leq d'\). By the ascending chain condition in \(R\) there are at most finitely many distinct ideals among all of the \(I_{i, d}\). (Hint: Any infinite set of elements of \(\mathbf{N} \times \mathbf{N}\) contains an increasing infinite sequence.) Take \(i_0\) so large that \(I_{i, d} = I_{i_0, d}\) for all \(i \geq i_0\) and all \(d\). Suppose \(f \in J_i\) for some \(i \geq i_0\). By induction on the degree \(d = \deg(f)\) we show that \(f \in J_{i_0}\). Namely, there exists a \(g\in J_{i_0}\) whose degree is \(d\) and which has the same leading coefficient as \(f\). By induction \(f - g \in J_{i_0}\) and we win.

Lemma

If \(R\) is a Noetherian ring, then so is the formal power series ring \(R[[x_1, \ldots, x_n]]\).

Proof

Since \(R[[x_1, \ldots, x_{n + 1}]] \cong R[[x_1, \ldots, x_n]][[x_{n + 1}]]\) it suffices to prove the statement that \(R[[x]]\) is Noetherian if \(R\) is Noetherian. Let \(I \subset R[[x]]\) be an ideal. We have to show that \(I\) is a finitely generated ideal. For each integer \(d\) denote \(I_d = \{a \in R \mid ax^d + \text{h.o.t.} \in I\}\). Then we see that \(I_0 \subset I_1 \subset \ldots\) stabilizes as \(R\) is Noetherian. Choose \(d_0\) such that \(I_{d_0} = I_{d_0 + 1} = \ldots\). For each \(d \leq d_0\) choose elements \(f_{d, j} \in I \cap (x^d)\), \(j = 1, \ldots, n_d\) such that if we write \(f_{d, j} = a_{d, j}x^d + \text{h.o.t}\) then \(I_d = (a_{d, j})\). Denote \(I' = (\{f_{d, j}\}_{d = 0, \ldots, d_0, j = 1, \ldots, n_d})\). Then it is clear that \(I' \subset I\). Pick \(f \in I\). First we may choose \(c_{d, i} \in R\) such that \[f - \sum c_{d, i} f_{d, i} \in (x^{d_0 + 1}) \cap I.\] Next, we can choose \(c_{i, 1} \in R\), \(i = 1, \ldots, n_{d_0}\) such that \[f - \sum c_{d, i} f_{d, i} - \sum c_{i, 1}xf_{d_0, i} \in (x^{d_0 + 2}) \cap I.\] Next, we can choose \(c_{i, 2} \in R\), \(i = 1, \ldots, n_{d_0}\) such that \[f - \sum c_{d, i} f_{d, i} - \sum c_{i, 1}xf_{d_0, i} - \sum c_{i, 2}x^2f_{d_0, i} \in (x^{d_0 + 3}) \cap I.\] And so on. In the end we see that \[f = \sum c_{d, i} f_{d, i} + \sum\nolimits_i (\sum\nolimits_e c_{i, e} x^e)f_{d_0, i}\] is contained in \(I'\) as desired.

The following lemma, although easy, is useful because finite type \(\mathbf{Z}\)-algebras come up quite often in a technique called “absolute Noetherian reduction”.

Lemma

Any finite type algebra over a field is Noetherian. Any finite type algebra over \(\mathbf{Z}\) is Noetherian.

Proof

This is immediate from Lemma 00FN and the fact that fields are Noetherian rings and that \(\mathbf{Z}\) is Noetherian ring (because it is a principal ideal domain).

Lemma

Let \(R\) be a Noetherian ring.

1. Any finite \(R\)-module is of finite presentation.

2. Any submodule of a finite \(R\)-module is finite.

3. Any finite type \(R\)-algebra is of finite presentation over \(R\).

Proof

Let \(M\) be a finite \(R\)-module. By Lemma 00KZ we can find a finite filtration of \(M\) whose successive quotients are of the form \(R/I\). Since any ideal is finitely generated, each of the quotients \(R/I\) is finitely presented. Hence \(M\) is finitely presented by Lemma 0519. This proves (1).

Let \(N \subset M\) be a submodule. As \(M\) is finite, the quotient \(M/N\) is finite. Thus \(M/N\) is of finite presentation by part (1). Thus we see that \(N\) is finite by Lemma 0519 part (5). This proves part (2).

To see (3) note that any ideal of \(R[x_1, \ldots, x_n]\) is finitely generated by Lemma 00FN.

Lemma

If \(R\) is a Noetherian ring then \(\Spec(R)\) is a Noetherian topological space, see Topology, Definition 0051.

Proof

This is because any closed subset of \(\Spec(R)\) is uniquely of the form \(V(I)\) with \(I\) a radical ideal, see Lemma 00E0. And this correspondence is inclusion reversing. Thus the result follows from the definitions.

Lemma

If \(R\) is a Noetherian ring then \(\Spec(R)\) has finitely many irreducible components. In other words \(R\) has finitely many minimal primes.

Proof

By Lemma 00FQ and Topology, Lemma 0052 we see there are finitely many irreducible components. By Lemma 00ES these correspond to minimal primes of \(R\).

Lemma

Let \(R \to S\) be a ring map. Let \(R \to R'\) be of finite type. If \(S\) is Noetherian, then the base change \(S' = R' \otimes_R S\) is Noetherian.

Proof

By Lemma 05G5 finite type is stable under base change. Thus \(S \to S'\) is of finite type. Since \(S\) is Noetherian we can apply Lemma 00FN.

Lemma

Let \(k\) be a field and let \(R\) be a Noetherian \(k\)-algebra. If \(K/k\) is a finitely generated field extension then \(K \otimes_k R\) is Noetherian.

Proof

Since \(K/k\) is a finitely generated field extension, there exists a finitely generated \(k\)-algebra \(B \subset K\) such that \(K\) is the fraction field of \(B\). In other words, \(K = S^{-1}B\) with \(S = B \setminus \{0\}\). Then \(K \otimes_k R = S^{-1}(B \otimes_k R)\). Then \(B \otimes_k R\) is Noetherian by Lemma 0CY6. Finally, \(K \otimes_k R = S^{-1}(B \otimes_k R)\) is Noetherian by Lemma 00FN.

Here are some fun lemmas that are sometimes useful.

Lemma

Let \(R\) be a ring and \(\mathfrak p \subset R\) be a prime. There exists an \(f \in R\), \(f \not \in \mathfrak p\) such that \(R_f \to R_\mathfrak p\) is injective in each of the following cases

1. \(R\) is a domain,

2. \(R\) is Noetherian, or

3. \(R\) is reduced and has finitely many minimal primes.

Proof

If \(R\) is a domain, then \(R \subset R_\mathfrak p\), hence \(f = 1\) works. If \(R\) is Noetherian, then the kernel \(I\) of \(R \to R_\mathfrak p\) is a finitely generated ideal and we can find \(f \in R\), \(f \not \in \mathfrak p\) such that \(IR_f = 0\). For this \(f\) the map \(R_f \to R_\mathfrak p\) is injective and \(f\) works. If \(R\) is reduced with finitely many minimal primes \(\mathfrak p_1, \ldots, \mathfrak p_n\), then we can choose \(f \in \bigcap_{\mathfrak p_i \not \subset \mathfrak p} \mathfrak p_i\), \(f \not \in \mathfrak p\). Indeed, if \(\mathfrak{p}_i\not\subset \mathfrak{p}\) then there exist \(f_i \in \mathfrak{p}_i\), \(f_i \not\in \mathfrak{p}\) and \(f = \prod f_i\) works. For this \(f\) we have \(R_f \subset R_\mathfrak p\) because the minimal primes of \(R_f\) correspond to minimal primes of \(R_\mathfrak p\) and we can apply Lemma 00EW (some details omitted).

Lemma

Any surjective endomorphism of a Noetherian ring is an isomorphism.

Proof

If \(f : R \to R\) were such an endomorphism but not injective, then \[\Ker(f) \subset \Ker(f \circ f) \subset \Ker(f \circ f \circ f) \subset \ldots\] would be a strictly increasing chain of ideals.

Locally nilpotent ideals

Here is the definition.

Definition

Let \(R\) be a ring. Let \(I \subset R\) be an ideal. We say \(I\) is locally nilpotent if for every \(x \in I\) there exists an \(n \in \mathbf{N}\) such that \(x^n = 0\). We say \(I\) is nilpotent if there exists an \(n \in \mathbf{N}\) such that \(I^n = 0\).

Example

Let \(R = k[x_n | n \in \mathbf{N}]\) be the polynomial ring in infinitely many variables over a field \(k\). Let \(I\) be the ideal generated by the elements \(x_n^n\) for \(n \in \mathbf{N}\) and \(S = R/I\). Then the ideal \(J \subset S\) generated by the images of \(x_n\), \(n \in \mathbf{N}\) is locally nilpotent, but not nilpotent. Indeed, since \(S\)-linear combinations of nilpotents are nilpotent, to prove that \(J\) is locally nilpotent it is enough to observe that all its generators are nilpotent (which they obviously are). On the other hand, for each \(n \in \mathbf{N}\) it holds that \(x_{n + 1}^n \not \in I\), so that \(J^n \not = 0\). It follows that \(J\) is not nilpotent.

Lemma

Let \(R \to R'\) be a ring map and let \(I \subset R\) be a locally nilpotent ideal. Then \(IR'\) is a locally nilpotent ideal of \(R'\).

Proof

This follows from the fact that if \(x, y \in R'\) are nilpotent, then \(x + y\) is nilpotent too. Namely, if \(x^n = 0\) and \(y^m = 0\), then \((x + y)^{n + m - 1} = 0\).

Lemma

Let \(R\) be a ring and let \(I \subset R\) be a locally nilpotent ideal. An element \(x\) of \(R\) is a unit if and only if the image of \(x\) in \(R/I\) is a unit.

Proof

If \(x\) is a unit in \(R\), then its image is clearly a unit in \(R/I\). It remains to prove the converse. Assume the image of \(y \in R\) in \(R/I\) is the inverse of the image of \(x\). Then \(xy = 1 - z\) for some \(z \in I\). This means that \(1\equiv z\) modulo \(xR\). Since \(z\) lies in the locally nilpotent ideal \(I\), we have \(z^N = 0\) for some sufficiently large \(N\). It follows that \(1 = 1^N \equiv z^N = 0\) modulo \(xR\). In other words, \(x\) divides \(1\) and is hence a unit.

Lemma

Let \(R\) be a Noetherian ring. Let \(I, J\) be ideals of \(R\). Suppose \(J \subset \sqrt{I}\). Then \(J^n \subset I\) for some \(n\). In particular, in a Noetherian ring the notions of “locally nilpotent ideal” and “nilpotent ideal” coincide.

Proof

Say \(J = (f_1, \ldots, f_s)\). By assumption \(f_i^{d_i} \in I\). Take \(n = d_1 + d_2 + \ldots + d_s + 1\).

Lemma

Let \(R\) be a ring. Let \(I \subset R\) be a locally nilpotent ideal. Then \(R \to R/I\) induces a bijection on idempotents.

Proof

As \(I\) is locally nilpotent it is contained in every prime ideal. Hence \(\Spec(R/I) = V(I) = \Spec(R)\). Hence the lemma follows from Lemma 00EE.

Proof

Suppose \(\overline{e} \in R/I\) is an idempotent. We have to lift \(\overline{e}\) to an idempotent of \(R\).

First, choose any lift \(f \in R\) of \(\overline{e}\), and set \(x = f^2 - f\). Then, \(x \in I\), so \(x\) is nilpotent (since \(I\) is locally nilpotent). Let now \(J\) be the ideal of \(R\) generated by \(x\). Then, \(J\) is nilpotent (not just locally nilpotent), since it is generated by the nilpotent \(x\).

Now, assume that we have found a lift \(e \in R\) of \(\overline{e}\) such that \(e^2 - e \in J^k\) for some \(k \geq 1\). Let \(e' = e - (2e - 1)(e^2 - e) = 3e^2 - 2e^3\), which is another lift of \(\overline{e}\) (since the idempotency of \(\overline{e}\) yields \(e^2 - e \in I\)). Then \[(e')^2 - e' = (4e^2 - 4e - 3)(e^2 - e)^2 \in J^{2k}\] by a simple computation.

We thus have started with a lift \(e\) of \(\overline{e}\) such that \(e^2 - e \in J^k\), and obtained a lift \(e'\) of \(\overline{e}\) such that \((e')^2 - e' \in J^{2k}\). This way we can successively improve the approximation (starting with \(e = f\), which fits the bill for \(k = 1\)). Eventually, we reach a stage where \(J^k = 0\), and at that stage we have a lift \(e\) of \(\overline{e}\) such that \(e^2 - e \in J^k = 0\), that is, this \(e\) is idempotent.

We thus have seen that if \(\overline{e} \in R/I\) is any idempotent, then there exists a lift of \(\overline{e}\) which is an idempotent of \(R\). It remains to prove that this lift is unique. Indeed, let \(e_1\) and \(e_2\) be two such lifts. We need to show that \(e_1 = e_2\).

By definition of \(e_1\) and \(e_2\), we have \(e_1 \equiv e_2 \mod I\), and both \(e_1\) and \(e_2\) are idempotent. From \(e_1 \equiv e_2 \mod I\), we see that \(e_1 - e_2 \in I\), so that \(e_1 - e_2\) is nilpotent (since \(I\) is locally nilpotent). A straightforward computation (using the idempotency of \(e_1\) and \(e_2\)) reveals that \((e_1 - e_2)^3 = e_1 - e_2\). Using this and induction, we obtain \((e_1 - e_2)^k = e_1 - e_2\) for any positive odd integer \(k\). Since all high enough \(k\) satisfy \((e_1 - e_2)^k = 0\) (since \(e_1 - e_2\) is nilpotent), this shows \(e_1 - e_2 = 0\), so that \(e_1 = e_2\), which completes our proof.

Lemma

Let \(A\) be a possibly noncommutative algebra. Let \(e \in A\) be an element such that \(x = e^2 - e\) is nilpotent. Then there exists an idempotent of the form \(e' = e + x(\sum a_{i, j}e^ix^j) \in A\) with \(a_{i, j} \in \mathbf{Z}\).

Proof

Consider the ring \(R_n = \mathbf{Z}[e]/((e^2 - e)^n)\). It is clear that if we can prove the result for each \(R_n\) then the lemma follows. In \(R_n\) consider the ideal \(I = (e^2 - e)\) and apply Lemma 00J9.

Lemma

Let \(R\) be a ring. Let \(I \subset R\) be a locally nilpotent ideal. Let \(n \geq 1\) be an integer which is invertible in \(R/I\). Then

1. the \(n\)th power map \(1 + I \to 1 + I\), \(1 + x \mapsto (1 + x)^n\) is a bijection,

2. a unit of \(R\) is a \(n\)th power if and only if its image in \(R/I\) is an \(n\)th power.

Proof

Let \(a \in R\) be a unit whose image in \(R/I\) is the same as the image of \(b^n\) with \(b \in R\). Then \(b\) is a unit (Lemma 0AMG) and \(ab^{-n} = 1 + x\) for some \(x \in I\). Hence \(ab^{-n} = c^n\) by part (1). Thus (2) follows from (1).

Proof of (1). This is true because there is an inverse to the map \(1 + x \mapsto (1 + x)^n\). Namely, we can consider the map which sends \(1 + x\) to \[\begin{align*} (1 + x)^{1/n} & = 1 + {1/n \choose 1}x + {1/n \choose 2}x^2 + {1/n \choose 3}x^3 + \ldots \\ & = 1 + \frac{1}{n} x + \frac{1 - n}{2n^2}x^2 + \frac{(1 - n)(1 - 2n)}{6n^3}x^3 + \ldots \end{align*}\] as in elementary calculus. This makes sense because the series is finite as \(x^k = 0\) for all \(k \gg 0\) and each coefficient \({1/n \choose k} \in \mathbf{Z}[1/n]\) (details omitted; observe that \(n\) is invertible in \(R\) by Lemma 0AMG).

Curiosity

Lemma 00EM explains what happens if \(V(I)\) is open for some ideal \(I \subset R\). But what if \(\Spec(S^{-1}R)\) is closed in \(\Spec(R)\)? The next two lemmas give a partial answer. For more information see Section 04PQ.

Lemma

Let \(R\) be a ring. Let \(S \subset R\) be a multiplicative subset. Assume the image of the map \(\Spec(S^{-1}R) \to \Spec(R)\) is closed. Then \(S^{-1}R \cong R/I\) for some ideal \(I \subset R\).

Proof

Let \(I = \Ker(R \to S^{-1}R)\) so that \(V(I)\) contains the image. Say the image is the closed subset \(V(I') \subset \Spec(R)\) for some ideal \(I' \subset R\). So \(V(I') \subset V(I)\). For \(f \in I'\) we see that \(f/1 \in S^{-1}R\) is contained in every prime ideal. Hence \(f^n\) maps to zero in \(S^{-1}R\) for some \(n \geq 1\) (Lemma 00E0). Hence \(V(I') = V(I)\). Then this implies every \(g \in S\) is invertible mod \(I\). Hence we get ring maps \(R/I \to S^{-1}R\) and \(S^{-1}R \to R/I\). The first map is injective by choice of \(I\). The second is the map \(S^{-1}R \to S^{-1}(R/I) = R/I\) which has kernel \(S^{-1}I\) because localization is exact. Since \(S^{-1}I = 0\) we see also the second map is injective. Hence \(S^{-1}R \cong R/I\).

Lemma

Let \(R\) be a ring. Let \(S \subset R\) be a multiplicative subset. Assume the image of the map \(\Spec(S^{-1}R) \to \Spec(R)\) is closed. If \(R\) is Noetherian, or \(\Spec(R)\) is a Noetherian topological space, or \(S\) is finitely generated as a monoid, then \(R \cong S^{-1}R \times R'\) for some ring \(R'\).

Proof

By Lemma 02JH we have \(S^{-1}R \cong R/I\) for some ideal \(I \subset R\). By Lemma 00EM it suffices to show that \(V(I)\) is open. If \(R\) is Noetherian then \(\Spec(R)\) is a Noetherian topological space, see Lemma 00FQ. If \(\Spec(R)\) is a Noetherian topological space, then the complement \(\Spec(R) \setminus V(I)\) is quasi-compact, see Topology, Lemma 04ZA. Hence there exist finitely many \(f_1, \ldots, f_n \in I\) such that \(V(I) = V(f_1, \ldots, f_n)\). Since each \(f_i\) maps to zero in \(S^{-1}R\) there exists a \(g \in S\) such that \(gf_i = 0\) for \(i = 1, \ldots, n\). Hence \(D(g) = V(I)\) as desired. In case \(S\) is finitely generated as a monoid, say \(S\) is generated by \(g_1, \ldots, g_m\), then \(S^{-1}R \cong R_{g_1 \ldots g_m}\) and we conclude that \(V(I) = D(g_1 \ldots g_m)\).