Selected Stacks Project statements and proofs for the algebraic Zariski Main Theorem, including the conductor-radical coefficient lemma and the exclusion of quasi-finite points by strong transcendence. The selection also includes dimension at a point under arbitrary field extension and finite module presentation across a finite, finitely presented algebra.
By the Stacks Project Authors, from the AI Integrated Stacks Project. Attribution and edition notice · GNU FDL 1.2 · Complete selected LaTeX source.
Stacks definition · LaTeX source
Definition
Let \(R\) be a ring. Let \(M\) be an \(R\)-module.
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 \(n \in \mathbf{N}\).
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 \[R^{\oplus m} \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0\]
Stacks tag 00F3 · LaTeX source
Definition
Let \(R \to S\) be a ring map.
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 \(R[x_1, \ldots, x_n] \to S\).
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 \(R[x_1, \ldots, x_n]/(f_1, \ldots, f_m) \cong S\).
Stacks definition · LaTeX source
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.
Stacks tag 0564 · LaTeX source
Lemma
Let \(R \to S\) be a finite and finitely presented ring map. Let \(M\) be an \(S\)-module. Then \(M\) is finitely presented as an \(R\)-module if and only if \(M\) is finitely presented as an \(S\)-module.
Proof
One of the implications follows from Lemma 0561. To see the other assume that \(M\) is finitely presented as an \(S\)-module. Pick a presentation \[S^{\oplus m} \longrightarrow S^{\oplus n} \longrightarrow M \longrightarrow 0\] As \(S\) is finite as an \(R\)-module, the kernel of \(S^{\oplus n} \to M\) is a finite \(R\)-module. Thus from Lemma 0519 we see that it suffices to prove that \(S\) is finitely presented as an \(R\)-module.
Pick \(y_1, \ldots, y_n \in S\) such that \(y_1, \ldots, y_n\) generate \(S\) as an \(R\)-module. By Lemma 052I each \(y_i\) is integral over \(R\). Choose monic polynomials \(P_i(x) \in R[x]\) with \(P_i(y_i) = 0\). Consider the ring \[S' = R[x_1, \ldots, x_n]/(P_1(x_1), \ldots, P_n(x_n))\] Then we see that \(S\) is of finite presentation as an \(S'\)-algebra by Lemma 00F4. Since \(S' \to S\) is surjective, the kernel \(J = \Ker(S' \to S)\) is finitely generated as an ideal by Lemma 00R2. Hence \(J\) is a finite \(S'\)-module (immediate from the definitions). Thus \(S = \Coker(J \to S')\) is of finite presentation as an \(S'\)-module by Lemma 0519. Hence, arguing as in the first paragraph, it suffices to show that \(S'\) is of finite presentation as an \(R\)-module. Actually, \(S'\) is free as an \(R\)-module with basis the monomials \(x_1^{e_1} \ldots x_n^{e_n}\) for \(0 \leq e_i < \deg(P_i)\). Namely, write \(R \to S'\) as the composition \[R \to R[x_1]/(P_1(x_1)) \to R[x_1, x_2]/(P_1(x_1), P_2(x_2)) \to \ldots \to S'\] This shows that the \(i\)th ring in this sequence is free as a module over the \((i - 1)\)st one with basis \(1, x_i, \ldots, x_i^{\deg(P_i) - 1}\). The result follows easily from this by induction. Some details omitted.
Stacks tag 00P2 · LaTeX source
Lemma
Let \(k\) be a field. Let \(S' \to S\) be a surjection of finite type \(k\) algebras. Let \(\mathfrak p \subset S\) be a prime ideal, and let \(\mathfrak p'\) be the corresponding prime ideal of \(S'\). Let \(X = \Spec(S)\), resp. \(X' = \Spec(S')\), and let \(x \in X\), resp. \(x'\in X'\) be the point corresponding to \(\mathfrak p\), resp. \(\mathfrak p'\). Then \[\dim_{x'} X' - \dim_x X = \text{height}(\mathfrak p') - \text{height}(\mathfrak p).\]
Proof
Immediate from Lemma 00P1.
Stacks tag 00P4 · LaTeX source
Lemma
Let \(k\) be a field. Let \(S\) be a finite type \(k\)-algebra. Set \(X = \Spec(S)\). Let \(K/k\) be a field extension. Set \(S_K = K \otimes_k S\), and \(X_K = \Spec(S_K)\). Let \(\mathfrak q \subset S\) be a prime corresponding to \(x \in X\) and let \(\mathfrak q_K \subset S_K\) be a prime corresponding to \(x_K \in X_K\) lying over \(\mathfrak q\). Then \(\dim_x X = \dim_{x_K} X_K\).
Proof
Choose a presentation \(S = k[x_1, \ldots, x_n]/I\). This gives a presentation \(K \otimes_k S = K[x_1, \ldots, x_n]/(K \otimes_k I)\). Let \(\mathfrak q_K' \subset K[x_1, \ldots, x_n]\), resp. \(\mathfrak q' \subset k[x_1, \ldots, x_n]\) be the corresponding primes. Consider the following commutative diagram of Noetherian local rings \[\begin{array}{ccc} K[x_1, \ldots, x_n]_{\mathfrak q_K'} & \longrightarrow & (K \otimes_k S)_{\mathfrak q_K} \\ \big\uparrow & & \big\uparrow \\ k[x_1, \ldots, x_n]_{\mathfrak q'} & \longrightarrow & S_{\mathfrak q} \end{array}\] Both vertical arrows are flat because they are localizations of the flat ring maps \(S \to S_K\) and \(k[x_1, \ldots, x_n] \to K[x_1, \ldots, x_n]\). Moreover, the vertical arrows have the same fibre rings. Hence, we see from Lemma 00ON that \(\text{height}(\mathfrak q') - \text{height}(\mathfrak q) = \text{height}(\mathfrak q_K') - \text{height}(\mathfrak q_K)\). Denote \(x' \in X' = \Spec(k[x_1, \ldots, x_n])\) and \(x'_K \in X'_K = \Spec(K[x_1, \ldots, x_n])\) the points corresponding to \(\mathfrak q'\) and \(\mathfrak q_K'\). By Lemma 00P2 and what we showed above we have \[\begin{eqnarray*} n - \dim_x X & = & \dim_{x'} X' - \dim_x X \\ & = & \text{height}(\mathfrak q') - \text{height}(\mathfrak q) \\ & = & \text{height}(\mathfrak q_K') - \text{height}(\mathfrak q_K) \\ & = & \dim_{x'_K} X'_K - \dim_{x_K} X_K \\ & = & n - \dim_{x_K} X_K \end{eqnarray*}\] and the lemma follows.
Stacks tag 0307 · LaTeX source
Lemma
Integral closure commutes with localization: If \(A \to B\) is a ring map, and \(S \subset A\) is a multiplicative subset, then the integral closure of \(S^{-1}A\) in \(S^{-1}B\) is \(S^{-1}B'\), where \(B' \subset B\) is the integral closure of \(A\) in \(B\).
Proof
Since localization is exact we see that \(S^{-1}B' \subset S^{-1}B\). Suppose \(x \in B'\) and \(f \in S\). Then \(x^d + \sum_{i = 1, \ldots, d} a_i x^{d - i} = 0\) in \(B\) for some \(a_i \in A\). Hence also \[(x/f)^d + \sum\nolimits_{i = 1, \ldots, d} a_i/f^i (x/f)^{d - i} = 0\] in \(S^{-1}B\). In this way we see that \(S^{-1}B'\) is contained in the integral closure of \(S^{-1}A\) in \(S^{-1}B\). Conversely, suppose that \(x/f \in S^{-1}B\) is integral over \(S^{-1}A\). Then we have \[(x/f)^d + \sum\nolimits_{i = 1, \ldots, d} (a_i/f_i) (x/f)^{d - i} = 0\] in \(S^{-1}B\) for some \(a_i \in A\) and \(f_i \in S\). This means that \[(f'f_1 \ldots f_d x)^d + \sum\nolimits_{i = 1, \ldots, d} f^i(f')^if_1^i \ldots f_i^{i - 1} \ldots f_d^i a_i (f'f_1 \ldots f_dx)^{d - i} = 0\] for a suitable \(f' \in S\). Hence \(f'f_1\ldots f_dx \in B'\) and thus \(x/f \in S^{-1}B'\) as desired.
Stacks tag 00PN · LaTeX source
Lemma
Let \[\begin{array}{ccc@{\qquad}ccc} S & \longrightarrow & S' & \mathfrak q & \rule{1.4em}{.04em} & \mathfrak q' \\ \big\uparrow & & \big\uparrow & \big\vert & & \big\vert \\ R & \longrightarrow & R' & \mathfrak p & \rule{1.4em}{.04em} & \mathfrak p' \end{array}\] be a commutative diagram of rings with primes as indicated. Assume \(R \to S\) of finite type, and \(S \otimes_R R' \to S'\) surjective. If \(R \to S\) is quasi-finite at \(\mathfrak q\), then \(R' \to S'\) is quasi-finite at \(\mathfrak q'\).
Proof
Write \(S \otimes_R \kappa(\mathfrak p) = S_1 \times S_2\) with \(S_1\) finite over \(\kappa(\mathfrak p)\) and such that \(\mathfrak q\) corresponds to a point of \(S_1\) as in Lemma 00PJ. This product decomposition induces a corresponding product decomposition for any \(S \otimes_R \kappa(\mathfrak p)\)-algebra. In particular, we obtain \(S' \otimes_{R'} \kappa(\mathfrak p') = S'_1 \times S'_2\). Because \(S \otimes_R R' \to S'\) is surjective the canonical map \((S \otimes_R \kappa(\mathfrak p)) \otimes_{\kappa(\mathfrak p)} \kappa(\mathfrak p') \to S' \otimes_{R'} \kappa(\mathfrak p')\) is surjective and hence \(S_i \otimes_{\kappa(\mathfrak p)} \kappa(\mathfrak p') \to S'_i\) is surjective. It follows that \(S'_1\) is finite over \(\kappa(\mathfrak p')\). The map \(S' \otimes_{R'} \kappa(\mathfrak p') \to \kappa(\mathfrak q')\) factors through \(S_1'\) (i.e. it annihilates the factor \(S_2'\)) because the map \(S \otimes_R \kappa(\mathfrak p) \to \kappa(\mathfrak q)\) factors through \(S_1\) (i.e. it annihilates the factor \(S_2\)). Thus \(\mathfrak q'\) corresponds to a point of \(\Spec(S_1')\) in the disjoint union decomposition of the fibre: \(\Spec(S' \otimes_{R'} \kappa(\mathfrak p')) = \Spec(S_1') \amalg \Spec(S_2')\), see Lemma 00ED. Since \(S_1'\) is finite over a field, it is an Artinian ring, and hence \(\Spec(S_1')\) is a finite discrete set. (See Proposition 00KJ.) We conclude \(\mathfrak q'\) is isolated in its fibre as desired.
Stacks tag 00PT · LaTeX source
Lemma
Let \(R\) be a ring. Let \(\varphi : R[x] \to S\) be a ring map. Let \(t \in S\). Assume that (a) \(t\) is integral over \(R[x]\), and (b) there exists a monic \(p \in R[x]\) such that \(t \varphi(p) \in \Im(\varphi)\). Then there exists a \(q \in R[x]\) such that \(t - \varphi(q)\) is integral over \(R\).
Proof
Write \(t \varphi(p) = \varphi(r)\) for some \(r \in R[x]\). Using euclidean division, write \(r = qp + r'\) with \(q, r' \in R[x]\) and \(\deg(r') < \deg(p)\). We may replace \(t\) by \(t - \varphi(q)\) which is still integral over \(R[x]\), so that we obtain \(t \varphi(p) = \varphi(r')\). In the ring \(S_t\) we may write this as \(\varphi(p) - (1/t) \varphi(r') = 0\). This implies that \(\varphi(x)\) gives an element of the localization \(S_t\) which is integral over \(\varphi(R)[1/t] \subset S_t\). On the other hand, \(t\) is integral over the subring \(\varphi(R)[\varphi(x)] \subset S\). Combined we conclude that \(t\) is integral over the subring \(\varphi(R)[1/t] \subset S_t\), see Lemma 00GN. In other words there exists an equation of the form \[t^d + \sum\nolimits_{i < d} \left(\sum\nolimits_{j = 0, \ldots, n_i} \varphi(r_{i, j})/t^j\right) t^i = 0\] in \(S_t\) with \(r_{i, j} \in R\). This means that \(t^{d + N} + \sum_{i < d} \sum_{j = 0, \ldots, n_i} \varphi(r_{i, j}) t^{i + N - j} = 0\) in \(S\) for some \(N\) large enough. In other words \(t\) is integral over \(R\).
Stacks tag 00PV · LaTeX source
Lemma
Let \(R\) be a ring. Let \(\varphi : R[x] \to S\) be a ring map. Let \(t \in S\). Assume \(t\) is integral over \(R[x]\). Let \(p \in R[x]\), \(p = a_0 + a_1x + \ldots + a_k x^k\) such that \(t \varphi(p) \in \Im(\varphi)\). Then there exists a \(q \in R[x]\) and \(n \geq 0\) such that \(\varphi(a_k)^n t - \varphi(q)\) is integral over \(R\).
Proof
Let \(R'\) and \(S'\) be the localization of \(R\) and \(S\) at the element \(a_k\). Let \(\varphi' : R'[x] \to S'\) be the localization of \(\varphi\). Let \(t' \in S'\) be the image of \(t\). Set \(p' = p/a_k \in R'[x]\). Then \(t' \varphi'(p') \in \Im(\varphi')\) since \(t \varphi(p) \in \Im(\varphi)\). As \(p'\) is monic, by Lemma 00PT there exists a \(q' \in R'[x]\) such that \(t' - \varphi'(q')\) is integral over \(R'\). We may choose an \(n \geq 0\) and an element \(q \in R[x]\) such that \(a_k^n q'\) is the image of \(q\). Then \(\varphi(a_k)^n t - \varphi(q)\) is an element of \(S\) whose image in \(S'\) is integral over \(R'\). By Lemma 0307 there exists an \(m \geq 0\) such that \(\varphi(a_k)^m(\varphi(a_k)^n t - \varphi(q))\) is integral over \(R\). Thus \(\varphi(a_k)^{m + n}t - \varphi(a_k^m q)\) is integral over \(R\) as desired.
Stacks tag 00PW · LaTeX source
Situation
Let \(R\) be a ring. Let \(\varphi : R[x] \to S\) be finite. Let \[J = \{ g \in S \mid gS \subset \Im(\varphi)\}\] be the “conductor ideal” of \(\varphi\). Assume that \(\varphi(R)\) is integrally closed in \(S\).
Stacks tag 00PX · LaTeX source
Lemma
In Situation 00PW. Suppose \(u \in S\), \(a_0, \ldots, a_k \in R\), \(u \varphi(a_0 + a_1x + \ldots + a_k x^k) \in J\). Then there exists an \(m \geq 0\) such that \(u \varphi(a_k)^m \in J\).
Proof
Assume that \(S\) is generated by \(t_1, \ldots, t_n\) as an \(R[x]\)-module. In this case \(J = \{ g \in S \mid gt_i \in \Im(\varphi)\text{ for all }i\}\). Note that each element \(u t_i\) is integral over \(R[x]\), see Lemma 00GK. We have \(\varphi(a_0 + a_1x + \ldots + a_k x^k) u t_i \in \Im(\varphi)\). By Lemma 00PV, for each \(i\) there exists an integer \(n_i\) and an element \(q_i \in R[x]\) such that \(\varphi(a_k^{n_i}) u t_i - \varphi(q_i)\) is integral over \(R\). By assumption this element is in \(\varphi(R)\) and hence \(\varphi(a_k^{n_i}) u t_i \in \Im(\varphi)\). It follows that \(m = \max\{n_1, \ldots, n_n\}\) works.
Stacks tag 00PY · LaTeX source
Lemma
In Situation 00PW. Suppose \(u \in S\), \(a_0, \ldots, a_k \in R\), \(u \varphi(a_0 + a_1x + \ldots + a_k x^k) \in \sqrt{J}\). Then \(u \varphi(a_i) \in \sqrt{J}\) for all \(i\).
Proof
Under the assumptions of the lemma we have \(u^n \varphi(a_0 + a_1x + \ldots + a_k x^k)^n \in J\) for some \(n \geq 1\). By Lemma 00PX we deduce \(u^n \varphi(a_k^{nm}) \in J\) for some \(m \geq 1\). Thus \(u \varphi(a_k) \in \sqrt{J}\), and so \(u \varphi(a_0 + a_1x + \ldots + a_k x^k) - u \varphi(a_k x^k) = u \varphi(a_0 + a_1x + \ldots + a_{k-1} x^{k-1}) \in \sqrt{J}\). We win by induction on \(k\).
Stacks definition · LaTeX source
Definition
Given an inclusion of rings \(R \subset S\) and an element \(x \in S\) we say that \(x\) is strongly transcendental over \(R\) if whenever \(u(a_0 + a_1 x + \ldots + a_k x^k) = 0\) with \(u \in S\) and \(a_i \in R\), then we have \(ua_i = 0\) for all \(i\).
Stacks tag 00Q0 · LaTeX source
Lemma
Suppose \(R \subset S\) is an inclusion of reduced rings and suppose that \(x \in S\) is strongly transcendental over \(R\). Let \(\mathfrak q \subset S\) be a minimal prime and let \(\mathfrak p = R \cap \mathfrak q\). Then the image of \(x\) in \(S/\mathfrak q\) is strongly transcendental over the subring \(R/\mathfrak p\).
Proof
Suppose \(u(a_0 + a_1x + \ldots + a_k x^k) \in \mathfrak q\). By Lemma 00EU the local ring \(S_{\mathfrak q}\) is a field, and hence \(u(a_0 + a_1x + \ldots + a_k x^k)\) is zero in \(S_{\mathfrak q}\). Thus \(uu'(a_0 + a_1x + \ldots + a_k x^k) = 0\) for some \(u' \in S\), \(u' \not\in \mathfrak q\). Since \(x\) is strongly transcendental over \(R\) we get \(uu'a_i = 0\) for all \(i\). This in turn implies that \(ua_i \in \mathfrak q\).
Stacks tag 00Q1 · LaTeX source
Lemma
Suppose \(R\subset S\) is an inclusion of domains and let \(x \in S\). Assume \(x\) is (strongly) transcendental over \(R\) and that \(S\) is finite over \(R[x]\). Then \(R \to S\) is not quasi-finite at any prime of \(S\).
Proof
As a first case, assume that \(R\) is normal, see Definition 00GV. By Lemma 00H1 we see that \(R[x]\) is normal. Take a prime \(\mathfrak q \subset S\), and set \(\mathfrak p = R \cap \mathfrak q\). Assume that the extension \(\kappa(\mathfrak p) \subset \kappa(\mathfrak q)\) is finite. This would be the case if \(R \to S\) is quasi-finite at \(\mathfrak q\). Let \(\mathfrak r = R[x] \cap \mathfrak q\). Then since \(\kappa(\mathfrak p) \subset \kappa(\mathfrak r) \subset \kappa(\mathfrak q)\) we see that the extension \(\kappa(\mathfrak p) \subset \kappa(\mathfrak r)\) is finite too. Thus the inclusion \(\mathfrak r \supset \mathfrak p R[x]\) is strict. By going down for \(R[x] \subset S\), see Proposition 00H8, we find a prime \(\mathfrak q' \subset \mathfrak q\), lying over the prime \(\mathfrak pR[x]\). Hence the fibre \(\Spec(S \otimes_R \kappa(\mathfrak p))\) contains a point not equal to \(\mathfrak q\), namely \(\mathfrak q'\), whose closure contains \(\mathfrak q\) and hence \(\mathfrak q\) is not isolated in its fibre.
If \(R\) is not normal, let \(R \subset R' \subset K\) be the integral closure \(R'\) of \(R\) in its field of fractions \(K\). Let \(S \subset S' \subset L\) be the subring \(S'\) of the field of fractions \(L\) of \(S\) generated by \(R'\) and \(S\). Note that by construction the map \(S \otimes_R R' \to S'\) is surjective. This implies that \(R'[x] \subset S'\) is finite. Also, the map \(S \subset S'\) induces a surjection on \(\Spec\), see Lemma 00GQ. We conclude by Lemma 00PN and the normal case we just discussed.
Stacks tag 00Q2 · LaTeX source
Lemma
Suppose \(R \subset S\) is an inclusion of reduced rings. Assume \(x \in S\) is strongly transcendental over \(R\), and \(S\) finite over \(R[x]\). Then \(R \to S\) is not quasi-finite at any prime of \(S\).
Proof
Let \(\mathfrak q \subset S\) be any prime. Choose a minimal prime \(\mathfrak q' \subset \mathfrak q\). According to Lemmas 00Q0 and 00Q1 the extension \(R/(R \cap \mathfrak q') \subset S/\mathfrak q'\) is not quasi-finite at the prime corresponding to \(\mathfrak q\). By Lemma 00PN the extension \(R \to S\) is not quasi-finite at \(\mathfrak q\).