Regular local rings

Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Public domain (CC0).

At a regular point, the number of independent linear coordinates equals the local dimension. The consequences reach much further than a count of tangent directions: every finite module has a bounded free resolution, regularity persists at more general points, and every height-one prime ideal is generated by one equation. The last assertion is the algebra behind comparing codimension-one subvarieties with locally principal equations.

Rings are commutative with identity, and finite means finitely generated. Every local ring in a regularity assertion is explicitly Noetherian. We write \((R,\mathfrak m,\kappa)\) for a nonzero Noetherian local ring. The dimension lesson supplies dimension, height and Hilbert–Samuel theory. The depth lesson, Theorem 6.1, already proves the graded-ring, domain and Cohen–Macaulay foundations of regular local rings. The projective-dimension lesson supplies minimal resolutions, the local global-dimension test and Auslander–Buchsbaum. We use those proved results and establish the remaining implications and factoriality here.

1. Coordinates, graded rings and regular quotients

The embedding dimension is

\[ e(R)=\dim_\kappa(\mathfrak m/\mathfrak m^2). \tag{1} \]

Nakayama identifies it with the smallest number of generators of \(\mathfrak m\). The dimension lesson, Proposition 4.2, proves \(\dim R\leq e(R)\). We call \(R\) regular local when equality holds. If \(D=\dim R\), a minimal list \(x_1,\ldots,x_D\) generating \(\mathfrak m\) is a regular system of parameters. An arbitrary system of parameters need only generate an \(\mathfrak m\)-primary ideal; it need not generate \(\mathfrak m\) itself. [Stacks, Tag 00KU.]

Theorem 1.1. For a Noetherian local ring of dimension \(D\), regularity is equivalent to a graded \(\kappa\)-algebra isomorphism

\[ \operatorname{gr}_{\mathfrak m}R =\bigoplus_{n\geq0}\mathfrak m^n/\mathfrak m^{n+1} \simeq\kappa[T_1,\ldots,T_D],\qquad \deg T_i=1. \tag{2} \]

A regular local ring is a domain and is Cohen–Macaulay. Every minimal generating list \(x_1,\ldots,x_D\) is a regular sequence, and

\[ R/(x_1,\ldots,x_c) \text{ is regular local of dimension }D-c \quad(0\leq c\leq D). \tag{3} \]

Proof. The forward implication, the domain assertion, the regular sequence and all successive quotients are proved in Theorem 6.1 of the depth lesson. Its proof constructs (2) with \(T_i\) mapping to the initial form of \(x_i\); Krull separation detects nonzero elements, and the polynomial initial forms detect injectivity on successive quotients. Conversely (2), in degree one, gives \(e(R)=D\). This is precisely regularity. A regular sequence of length \(D\) gives depth \(D\), by the depth bound, hence Cohen–Macaulayness. \(\square\)

This proof dependency is useful: none of these first properties depends on factoriality or on the homological converse below. Corresponding statements are [Stacks, Tags 00NO, 00NP and 00NQ].

In dimension zero, regularity says \(\mathfrak m/\mathfrak m^2=0\), so Nakayama gives \(\mathfrak m=0\): the regular local rings of dimension zero are exactly fields.

Corollary 1.2. If \(R\) is regular local and \(x\in\mathfrak m\setminus\mathfrak m^2\), then \(R/xR\) is regular of dimension \(D-1\).

Proof. Extend the nonzero class of \(x\) to a basis of \(\mathfrak m/\mathfrak m^2\), lift the basis, and use Nakayama and (3). \(\square\)

Proposition 1.3 (lifting regularity through a regular element). If \(x\in\mathfrak m\) is a nonzerodivisor and \(R/xR\) is regular, then \(R\) is regular. More generally the same holds if a quotient by an \(R\)-regular sequence is regular.

Proof. The one-equation theorem of the dimension lesson, Theorem 3.2, gives

\[ \dim(R/xR)=D-1. \]

Lift \(D-1\) generators of the maximal ideal of \(R/xR\). Together with \(x\) they generate \(\mathfrak m\), since the quotient ideal is generated by their images. Thus \(e(R)\leq D\), and the reverse inequality always holds. For a sequence, apply the one-element assertion successively from the last quotient upward. \(\square\)

The nonzerodivisor hypothesis is essential: \(k[\epsilon]/(\epsilon^2)\), modulo \(\epsilon\), is a field, whereas its embedding dimension is one and its dimension is zero. [Stacks, Tag 00NU.]

Proposition 1.4 (regular quotients). If \(R\) and \(R/I\) are regular local rings, with \(I\) proper, then some regular system of parameters satisfies

\[ I=(x_1,\ldots,x_c),\qquad c=D-\dim(R/I). \tag{4} \]

Proof. The quotient cotangent space is \(\mathfrak m/(I+\mathfrak m^2)\), so the kernel of the map from \(\mathfrak m/\mathfrak m^2\) to it has dimension \(c\). Choose \(x_1,\ldots,x_c\in I\) giving a basis of that kernel and extend to a basis of \(\mathfrak m/\mathfrak m^2\). Put \(Q=R/(x_1,\ldots,x_c)\). By (3), \(Q\) is a regular local domain of dimension \(D-c\), mapping onto \(R/I\) of the same dimension. If the kernel were nonzero, choose a nonzero element \(a\) in it. The element \(a\) is a nonzerodivisor of \(Q\), so \(\dim(Q/aQ)=\dim Q-1\); any further quotient has at most that dimension. This contradicts the dimension of \(R/I\). The kernel is zero. \(\square\)

This is the intrinsic statement that a regular closed subspace inside a regular local space is cut out by part of a coordinate system. [Stacks, Tag 00NR.]

2. The homological converse

The residue field detects a singularity through its free resolution. We first prove the lower bound that makes the detection precise.

Lemma 2.1 (linear coordinates survive in a resolution). For every Noetherian local ring,

\[ \operatorname{pd}_R\kappa\geq e(R). \tag{5} \]

More precisely, its minimal free resolution has Betti numbers

\[ \beta_i^R(\kappa)\geq {e(R)\choose i} \qquad(0\leq i\leq e(R)). \tag{6} \]

Proof. Set \(e=e(R)\), choose minimal generators \(x_1,\ldots,x_e\), and let \(V=\kappa^e\), with basis \(v_j\) corresponding to \(\overline{x_j}\). Let \(K_\bullet\) be their Koszul complex. It has \(K_i=\bigwedge^i R^e\), differential

\[ \partial(e_{j_1}\wedge\cdots\wedge e_{j_i}) =\sum_{a=1}^i(-1)^{a-1}x_{j_a} e_{j_1}\wedge\cdots\widehat{e_{j_a}}\cdots\wedge e_{j_i}. \tag{7} \]

We do not assume this complex exact in positive degrees. Let \(F_\bullet\to\kappa\) be the minimal free resolution from Theorem 2.2 of the projective-dimension lesson. Its \(F_0=R\), and every positive differential has image in \(\mathfrak m F_{i-1}\).

There is a chain map \(\alpha:K_\bullet\to F_\bullet\) inducing the identity of \(\kappa\). Start with \(\alpha_0=1_R\). If \(\alpha_{i-1}\) is already constructed, the image of \(\alpha_{i-1}\partial\) consists of cycles, because \(\partial^2=0\). Exactness of the target resolution lets us lift these cycles through \(F_i\to F_{i-1}\), one basis vector of \(K_i\) at a time. For \(i=1\) use the kernel of the augmentation. This construction uses the freeness of the source terms and exactness of the target; it does not require the source to resolve \(\kappa\).

We prove that \(\overline\alpha_i:K_i/\mathfrak mK_i\to F_i/\mathfrak mF_i\) is injective. The case \(i=0\) is immediate. A differential with entries in \(\mathfrak m\) induces a linear map modulo \(\mathfrak m^2\). For (7) this map is

\[ \sigma_i:\bigwedge^i V\longrightarrow \bigwedge^{i-1}V\otimes_\kappa(\mathfrak m/\mathfrak m^2), \quad v_{j_1}\wedge\cdots\wedge v_{j_i} \longmapsto\sum_a(-1)^{a-1}v_{J\setminus j_a}\otimes\overline{x_{j_a}}. \tag{8} \]

This map is injective in every characteristic. A coordinate labelled by a pair \((J\setminus\{j\},j)\) occurs only in the image of the basis wedge labelled \(J\), with coefficient \(1\) or \(-1\). Distinct basis wedges therefore have disjoint coordinate supports. In particular no division by \(i\) is involved.

Suppose \(w\in K_i\) has image zero under \(\overline\alpha_i\). Then \(\alpha_i(w)\in\mathfrak mF_i\), whence its differential lies in \(\mathfrak m^2F_{i-1}\). The chain-map identity, reduced modulo \(\mathfrak m^2\), says

\[ (\overline\alpha_{i-1}\otimes1)\sigma_i(\overline w)=0. \]

By induction \(\overline\alpha_{i-1}\) is injective, and tensoring vector spaces preserves injectivity. Equation (8) now forces \(\overline w=0\). This completes the induction. The dimensions of the source wedges give (6). If the resolution terminates before degree \(e\), it would have \(F_e=0\), contradicting (6); thus (5) follows, including the case of infinite projective dimension. \(\square\)

This proves the bound of [Stacks, Tag 00OA] with the comparison construction and the characteristic-free linear injectivity made explicit.

Theorem 2.2 (Serre's characterization). For a Noetherian local ring, the following are equivalent:

  1. \(R\) is regular local.
  2. \(R\) has finite global dimension.
  3. \(\operatorname{pd}_R\kappa<\infty\).

When these hold,

\[ \boxed{\operatorname{gl.dim}R =\operatorname{pd}_R\kappa =\operatorname{depth}R =\dim R =e(R).} \tag{9} \]

Every finite module has a finite free resolution of length at most \(D=\dim R\). Global dimension here bounds arbitrary modules, not only finite ones.

Proof. For regular \(R\), the Koszul complex on a regular system of parameters resolves \(\kappa\) in length \(D\). Corollary 4.3 of the projective-dimension lesson proves \(\operatorname{gl.dim}R=\operatorname{pd}_R\kappa=D\), using its local global-dimension test, Theorem 2.3. Hence (1) implies (2), and (2) implies (3).

Assume (3), writing \(p=\operatorname{pd}_R\kappa\). Since \(\operatorname{depth}_R\kappa=0\), Auslander–Buchsbaum, Theorem 3.1 of that lesson, gives \(p=\operatorname{depth}R\). Combine the bounds already proved:

\[ e(R)\leq p=\operatorname{depth}R\leq\dim R\leq e(R). \tag{10} \]

All are equal, proving (1) and (9). Minimal resolutions of finite modules have finite free terms, and their projective dimensions are at most the global dimension, giving the final assertion. \(\square\)

See [Stacks, Tags 00O7 and 00OC]. Our proof of the converse uses Auslander–Buchsbaum rather than an additional exactness criterion for complexes. A singular Noetherian local ring thus has an infinite minimal free resolution of its residue field, with a nonzero term in every degree: a zero term would terminate the minimal construction.

3. Regularity at more general points

Lemma 3.1. For any ring, multiplicative set \(S\), and module \(M\),

\[ \operatorname{pd}_{S^{-1}R}(S^{-1}M)\leq\operatorname{pd}_R M, \qquad \operatorname{gl.dim}(S^{-1}R)\leq\operatorname{gl.dim}R. \tag{11} \]

Proof. Localization is exact. A projective module is a direct summand of a free module, so its localization is projective over \(S^{-1}R\). Localize a projective resolution to obtain the first inequality. For any \(S^{-1}R\)-module \(N\), restrict scalars to \(R\) and observe that \(S^{-1}N\simeq N\). Applying the first inequality to this restricted module and taking the supremum gives the second. Infinite bounds impose no further condition. \(\square\)

Theorem 3.2. If \(R\) is regular local, then \(R_{\mathfrak p}\) is regular local for every prime \(\mathfrak p\), and

\[ \operatorname{gl.dim}R_{\mathfrak p} =\dim R_{\mathfrak p}=\operatorname{ht}_R\mathfrak p. \tag{12} \]

Proof. The localized ring is Noetherian local. By (9) and (11) it has finite global dimension, so Theorem 2.2 makes it regular. The height identity is the prime-chain correspondence for localization. \(\square\)

Thus regularity is preserved when passing from a point to a more general point. [Stacks, Tags 00O8 and 0AFS.]

A Noetherian ring is called regular if all its prime localizations are regular local [Stacks, Tag 00OD]. It suffices to test maximal ideals: every prime lies under a maximal ideal, and its localization is a localization of that maximal localization. This definition makes no assertion that the ring has finite Krull dimension globally.

Proposition 3.3. If a Noetherian ring \(A\) is regular, so is \(A[t]\).

Proof. Take a prime \(\mathfrak q\subset A[t]\), put \(\mathfrak p=\mathfrak q\cap A\), and replace the base by the regular local ring \(A_{\mathfrak p}\), of dimension \(d\). Write its maximal ideal as \((a_1,\ldots,a_d)\), with residue field \(k\). The local map

\[ A_{\mathfrak p}\longrightarrow B=A[t]_{\mathfrak q} \]

is flat, by polynomial freeness and localization. Its closed fibre is \(k[t]_{\overline{\mathfrak q}}\), where \(\overline{\mathfrak q}\) is either zero or generated by one monic irreducible polynomial. These fibres have dimension \(\varepsilon=0\) or \(1\), respectively. The flat local dimension formula, Theorem 5.1 of the dimension lesson, gives \(\dim B=d+\varepsilon\).

If \(\overline{\mathfrak q}=0\), the maximal ideal of \(B\) equals \((a_1,\ldots,a_d)B\). Otherwise lift the coefficients of its monic generator to \(A_{\mathfrak p}\), obtaining \(h\); the maximal ideal is \((a_1,\ldots,a_d,h)B\). Indeed these assertions follow by taking the quotient by \(\mathfrak pB\) and using the stated maximal ideal of the fibre. In both cases \(e(B)\leq d+\varepsilon=\dim B\), so equality holds and \(B\) is regular. \(\square\)

In particular \(k[t_1,\ldots,t_n]_{\mathfrak m}\) is regular for every maximal ideal and every field \(k\). This includes closed points whose residue fields are inseparable over \(k\); no perfection hypothesis has been used. Its dimension is \(n\), by the polynomial height formula and the Nullstellensatz from the earlier lessons.

4. Two tools for unique factorization

A UFD is a domain in which every nonzero nonunit has a factorization into irreducibles, unique up to order and multiplication by units. An element is prime when its principal ideal is a nonzero proper prime ideal.

Lemma 4.1. A Noetherian domain admits factorizations into irreducibles. It is a UFD if and only if every irreducible is prime, and equivalently if and only if every height-one prime ideal is principal.

Proof. If elements without an irreducible factorization existed, choose one, \(a\), whose principal ideal is maximal among such ideals; the ascending chain condition allows this choice. It is reducible, so \(a=bc\) with both factors nonunits. At least one factor also lacks a factorization, but both principal ideals strictly contain \((a)\), a contradiction.

If all irreducibles are prime, compare two factorizations: the first factor in one divides a factor in the other, hence is associated to it. Cancel and continue. This proves uniqueness. Conversely uniqueness forces an irreducible dividing a product to occur in one factor's factorization, so it is prime; a zero factor causes no difficulty.

In a UFD, choose a nonzero element in a height-one prime \(\mathfrak p\) and factor it. Some prime irreducible \(a\) lies in \(\mathfrak p\). Since \((0)\subsetneq(a)\subseteq\mathfrak p\) and \(\mathfrak p\) has height one, \(\mathfrak p=(a)\). Conversely assume every height-one prime principal. For an irreducible \(a\), choose a prime \(\mathfrak p\) minimal over \((a)\). The principal ideal theorem gives height at most one, and the domain and \(a\ne0\) give height exactly one. Write \(\mathfrak p=(b)\). Then \(a=bc\); since \(b\) is a nonunit, irreducibility makes \(c\) a unit. Thus \((a)=\mathfrak p\) is prime. \(\square\)

See [Stacks, Tags 034R and 0AFT].

Lemma 4.2 (Nagata's criterion, one prime). Let \(A\) be a Noetherian domain and \(x\) a prime element. If \(A_x\) is a UFD, then \(A\) is a UFD.

Proof. Let \(a\) be irreducible. If it is associated to \(x\), it is already prime. Otherwise \(x\nmid a\). First, \(a\) is not a unit in \(A_x\): an equality \(ab=x^n\) would force \(x\mid b\), and repeated cancellation would give \(ab'=1\), contrary to \(a\) being a nonunit.

If \(a=(b/x^r)(c/x^s)\), then \(x^{r+s}a=bc\) in the domain. Primality of \(x\) lets us remove one factor of \(x\) from \(b\) or \(c\) and cancel it, repeatedly. We reach \(a=b'c'\), with the original localized factors differing from \(b'\) and \(c'\) by units that are powers of \(x\). One of \(b',c'\) is a unit, so \(a\) remains irreducible in \(A_x\), hence prime there.

If \(a\mid uv\) in \(A\), primality in \(A_x\) implies, say, \(x^n u=ab\). Since \(x\) is prime and does not divide \(a\), it divides \(b\); repeat and cancel all \(n\) factors to obtain \(u=ab'\). Thus \(a\) is prime in \(A\). Lemma 4.1 finishes the proof. \(\square\)

The same argument can be iterated for finitely many prime elements; Exercise 7.5 spells out that extension. The essential hypothesis is primality before inversion. [Stacks, Tag 0AFU.]

5. From free resolutions to principal ideals

We need a precise reason that the rank-one projectives appearing after inversion are free.

Lemma 5.1 (rank-one determinant). Let \(P\) be a finite projective module over any ring \(A\), free of rank one at every prime. If

\[ P\oplus A^a\simeq A^{a+1}, \tag{13} \]

then \(P\simeq A\).

Proof. The map

\[ P\longrightarrow\bigwedge^{a+1}(P\oplus A^a), \quad p\longmapsto p\wedge e_1\wedge\cdots\wedge e_a \tag{14} \]

is an isomorphism. After localization at every prime, choose a basis of the rank-one summand; (14) maps a basis to a basis of the top exterior power. Its kernel and cokernel therefore vanish by local detection. On the other hand (13) identifies that exterior power with \(\bigwedge^{a+1}A^{a+1}\simeq A\). The argument also works when \(a=0\). \(\square\)

Proposition 5.2. Let \(R\) be regular local and \(f\in R\). Every finite projective \(R_f\)-module is stably free. In particular every such module of constant rank one is free.

Proof. Put \(A=R_f\), and take a finite presentation \(A^u\xrightarrow{B}A^v\to P\to0\). Finite projectives are finitely presented, as proved in the flatness lesson. Multiply \(B\) by a common power of \(f\) to obtain a matrix over \(R\). Its cokernel \(M\) is finite over \(R\), and \(M_f\simeq P\), because that power is a unit in \(A\). Theorem 2.2 supplies a finite resolution of \(M\) by finite free \(R\)-modules. Localizing gives

\[ 0\to G_n\to\cdots\to G_1\to G_0\to P\to0, \tag{15} \]

with all \(G_i\) finite free over \(A\).

Every short exact sequence in (15) splits. First split \(G_0\to P\); its kernel \(K_0\) is projective. Then split \(G_1\to K_0\), and continue. The decompositions \(G_0\simeq P\oplus K_0\), \(G_i\simeq K_{i-1}\oplus K_i\), and \(G_n\simeq K_{n-1}\) give an actual isomorphism

\[ P\oplus\bigoplus_{i\text{ odd}}G_i \simeq\bigoplus_{i\text{ even}}G_i. \tag{16} \]

Both sums of \(G_i\)'s are free, proving stable freeness. For constant rank one their ranks differ by one, as seen after localization; Lemma 5.1 applies. If \(A\) is the zero ring, all its modules are zero and the assertion holds directly. \(\square\)

The rank-one conclusion is also expressed as \(\operatorname{Pic}(R_f)=0\), where the Picard group consists of isomorphism classes of rank-one finite projective modules under tensor product. The proof above supplies the concrete split-resolution argument for [Stacks, Tag 0AFZ].

Theorem 5.3 (Auslander–Buchsbaum factoriality). Every regular local ring is a UFD.

Proof. By Theorem 1.1, \(R\) is a domain. Induct on its dimension \(D\). Dimension zero gives a field. For \(D>0\), choose \(x\in\mathfrak m\setminus\mathfrak m^2\). Corollary 1.2 makes \(R/xR\) regular, hence a domain. Thus \(x\) is a prime element.

Let \(A=R_x\), and take a height-one prime \(\mathfrak p\subset A\). It comes from a prime \(\mathfrak q\subset R\) avoiding \(x\); \(\mathfrak q\) also has height one. Indeed every prime below \(\mathfrak q\) avoids \(x\), so localization preserves all chains ending there.

At a prime \(\mathfrak r\) of \(R\) avoiding \(x\), the stalk of the ideal \(\mathfrak p\) is \(\mathfrak qR_{\mathfrak r}\). If \(\mathfrak q\nsubseteq\mathfrak r\), this is the whole ring. Otherwise it is a height-one prime ideal in \(R_{\mathfrak r}\), by the same chain correspondence. Since \(x\in\mathfrak m\), such \(\mathfrak r\) is nonmaximal. Its height is strictly less than \(D\): a chain of length \(\operatorname{ht}\mathfrak r\) ending at \(\mathfrak r\), followed by \(\mathfrak m\), has one more step. Theorem 3.2 makes \(R_{\mathfrak r}\) regular, and the induction hypothesis makes it a UFD. Lemma 4.1 makes \(\mathfrak qR_{\mathfrak r}\) principal. It is nonzero in a domain, so it is free of rank one.

Consequently \(\mathfrak p\), as an \(A\)-module, is free of rank one at every prime. It is finitely presented since \(A\) is Noetherian. Flatness is local, and a finitely presented flat module is projective, by Theorems 3.3 and 5.3 of the flatness lesson. Proposition 5.2 and Lemma 5.1 now give \(\mathfrak p\simeq A\); the image of a basis element generates the ideal. Every height-one prime of \(A\) is therefore principal, so \(A\) is a UFD. Apply Nagata's criterion to \(x\) to conclude that \(R\) is a UFD. \(\square\)

This is the full factoriality theorem of [Stacks, Tag 0AG0]. The finite free resolution in Proposition 5.2 is the step that upgrades locally principal ideals to principal ideals; local regularity alone would not supply that upgrade.

6. Coordinate rings and a singular cone

Formal coordinates. The ring \(k[[t_1,\ldots,t_n]]\) is Noetherian by Solution 7.3 of the finiteness lesson. Its maximal ideal consists of series with zero constant coefficient. Monomials of total degree \(j\) give a basis of its \(j\)-th graded piece: discarding terms of higher degree realizes the identification. Thus its graded ring is \(k[T_1,\ldots,T_n]\), its cumulative length is \({N+n-1\choose n}\) modulo the \(N\)-th power, and the dimension theorem gives dimension \(n\). Theorem 1.1 makes it regular. No completeness theorem is needed for this example.

Mixed characteristic. Let \(p\) be any prime integer. The domain

\[ B=\mathbb Z_{(p)}[t]_{(p,t)} \tag{17} \]

has nonzerodivisor \(p\), with quotient \(\mathbb F_p[t]_{(t)}\), regular of dimension one. Proposition 1.3 shows that \(B\) is regular of dimension two, with regular system \(p,t\).

For a ramified example put

\[ C=\mathbb Z_{(p)}[u]/(u^2-p). \tag{18} \]

The polynomial is irreducible over \(\mathbb Q\): a rational square has even \(p\)-adic exponent, whereas \(p\) has exponent one. The monic quotient is free of rank two over \(\mathbb Z_{(p)}\) and embeds in \(\mathbb Q[u]/(u^2-p)\), so is a domain. It is finite integral of dimension one, by integral dimension invariance. Every maximal ideal contracts to \((p)\), and its quotient modulo \(p\) is \(\mathbb F_p[u]/(u^2)\), which has just one maximal ideal. Hence \(C\) is local, with maximal ideal \((p,u)=(u)\). It is regular of dimension one. This argument applies also to \(p=2\).

The quadric cone. Over any field set

\[ Q=\bigl(k[x,y,z]/(xy-z^2)\bigr)_{(x,y,z)}. \tag{19} \]

The affine quotient embeds in \(k[s,t]\) by \(x\mapsto s^2\), \(y\mapsto t^2\), \(z\mapsto st\). To check injectivity, the monic equation in \(z\) gives the basis \(x^a y^b z^\epsilon\), \(a,b\geq0\), \(\epsilon=0,1\). Their images are distinct monomials, with exponent pairs \((2a+\epsilon,2b+\epsilon)\). Thus \(Q\) is a domain. It is the quotient of the regular local ring \(k[x,y,z]_{(x,y,z)}\), of dimension three, by a nonzero equation, so it has dimension two. The equation has no linear term, leaving the independent cotangent classes of \(x,y,z\); its embedding dimension is three. It is singular.

Moreover \(x,y,z\in\mathfrak m_Q\setminus\mathfrak m_Q^2\) are irreducible: a product of two nonunits belongs to \(\mathfrak m_Q^2\). Their distinct linear classes also make them pairwise nonassociated. The equality

\[ xy=z\cdot z \tag{20} \]

therefore contradicts unique factorization. There is no characteristic restriction in this calculation. This hypersurface is nevertheless Cohen–Macaulay by Corollary 6.2 of the depth lesson. Its normality will be established in Discrete valuation rings, normal rings and Serre's criterion.

7. Exercises

Exercise 7.1 (easy: one linear coordinate). For regular local \(R\) and \(x\in\mathfrak m\setminus\mathfrak m^2\), prove that \(R/xR\) is regular of dimension \(D-1\). Conversely, if \(D>0\) and \(R/(f)\) is regular of dimension \(D-1\), show that \(f\in\mathfrak m\setminus\mathfrak m^2\).

Exercise 7.2 (easy: dimension one). Show that a regular local ring of dimension one is a discrete valuation ring. Construct the valuation and describe every nonzero ideal, without appealing to factoriality.

Exercise 7.3 (medium: the cone). Show that (19) is neither regular nor a UFD, in every characteristic. Explain why the fact that \(Q_x\) is a UFD does not make Nagata's criterion apply to \(x\).

Exercise 7.4 (medium: rank-one cancellation). Over an arbitrary ring, prove that a stably free finite projective module of constant rank one is free. Exhibit the exterior-power isomorphism, including the case where no stabilization is needed.

Exercise 7.5 (hard: several prime elements). Prove Nagata's criterion for a multiplicative set generated by finitely many prime elements of a Noetherian domain. Give the full argument for one prime and justify iteration when some prime generators become units or associates after earlier inversions.

Exercise 7.6 (hard: a rank-one ideal with infinite homological cost). In the cone ring \(Q\), put \(P=(x,z)\). Show that \(P\) is height one, is not principal, and has rank one but two minimal generators. Prove \(\operatorname{pd}_Q(Q/P)=\infty\), using Auslander–Buchsbaum and the finite-projective local theorem.

8. Solutions

Solution 7.1. Extend \(\overline x\) to a cotangent basis. Its lifts minimally generate \(\mathfrak m\), so (3) gives the assertion. Conversely \(f\) is a nonunit since its quotient is a nonzero local ring, so \(f\in\mathfrak m\). The cotangent space of the quotient is \(\mathfrak m/(\mathfrak m^2+(f))\). Regularity and its given dimension require this space to have dimension \(D-1\), whereas \(\mathfrak m/\mathfrak m^2\) has dimension \(D\). Thus the linear class of \(f\) is nonzero. In particular, quotienting by a nonzero element of \(\mathfrak m^2\) produces dimension \(D-1\) and embedding dimension \(D\), hence a singular quotient.

Solution 7.2. Theorem 1.1 makes \(R\) a domain, and regularity gives \(\mathfrak m=(\pi)\), with \(\pi\ne0\). Krull intersection gives \(\bigcap_n(\pi^n)=0\). Therefore every nonzero \(a\in R\) belongs to a unique largest power \((\pi^n)\). Write \(a=\pi^n u\). Maximality makes \(u\notin(\pi)=\mathfrak m\), so \(u\) is a unit. Set \(v(a)=n\), and for nonzero fractions set \(v(a/b)=v(a)-v(b)\). Cancellation and the domain property give well-definedness, \(v(ab)=v(a)+v(b)\), and \(v(a+b)\geq\min(v(a),v(b))\) when the sum is nonzero. Every fraction of nonnegative valuation lies in \(R\), and \(v(\pi)=1\), so this is a discrete valuation with value group \(\mathbb Z\). For a nonzero ideal choose an element of least valuation \(n\); it is a unit times \(\pi^n\), and all ideal elements are divisible by \(\pi^n\). The ideal equals \((\pi^n)\). This proves the DVR assertion directly.

Solution 7.3. The monomial embedding, dimension calculation and cotangent calculation in Section 6 show that \(Q\) is a domain of dimension two with embedding dimension three. Its three coordinate classes are pairwise nonassociated irreducibles, and (20) violates uniqueness. Inverting \(x\) solves \(y=z^2/x\), so the affine ring after inversion is \(k[x,x^{-1},z]\); \(Q_x\) is a further localization of that UFD. A localization of a UFD is a UFD because factorizations remain factorizations after deleting prime factors that become units; the remaining prime factors remain prime. But \(x\) is not prime in \(Q\): modulo \(x\), the nonzero class of \(z\) has square zero. Nonvanishing follows also from its surviving cotangent class. The missing prime-element hypothesis is exactly what prevents Nagata's criterion from applying.

Solution 7.4. Write \(P\oplus A^a\simeq A^b\). Localization at any prime gives \(b=a+1\), unless \(A\) is the zero ring, in which case the conclusion is immediate. Use the map (14). Localization makes it the isomorphism from a free rank-one module to the top exterior power of a free module of rank \(a+1\). Local detection proves it is an isomorphism globally. Taking the top exterior power of the displayed stable isomorphism identifies its target with \(A\), hence \(P\simeq A\). For \(a=0\), the map is the identity \(P\to\bigwedge^1P\); the same proof includes that case. Rank one is crucial to this exterior-power cancellation.

Solution 7.5. For a single prime \(x\), take an irreducible \(a\) not associated to it. If \(a\) became a unit, an equation \(ab=x^n\) could be cancelled repeatedly using primality of \(x\) and \(x\nmid a\), making \(a\) a unit already. A factorization in the localization yields \(x^{r+s}a=bc\); distribute each prime factor \(x\) to one numerator and cancel. The resulting factorization of \(a\) has a unit factor, so \(a\) stays irreducible. When the localization is a UFD, \(a\) becomes prime; for \(a\mid uv\), a resulting equation \(x^n u=ab\) or \(x^n v=ab\) can again be cancelled, because every \(x\) must divide \(b\). Thus \(a\) is prime before inversion. Associates of \(x\) were prime from the start, and Noetherianity supplies factorizations, proving the one-prime criterion completely.

Now invert \(x_1,\ldots,x_r\) successively. In each intermediate localization the image of a later prime element is either a prime element or a unit: its quotient is the corresponding localization of the original domain \(A/(x_j)\), which is a domain or the zero ring. If it is a unit, that inversion changes nothing. Otherwise the one-prime criterion applies. Starting with the final UFD and descending through the intermediate rings proves each one a UFD. Associate generators cause no problem, because after one is inverted its associates are units. The opposite implication, from a UFD to its localization, follows from the factorization argument in Solution 7.3.

Solution 7.6. The affine prime \((x,z)\) has quotient \(k[y]\); the finite-type domain height formula gives height \(2-1=1\). Localization at the origin preserves its height, so \(P\) has height one and \(Q/P\simeq k[y]_{(y)}\). The classes of \(x,z\) in \(P/\mathfrak m_QP\) are independent: a linear relation over \(k\) would lie in \(\mathfrak m_QP\subseteq\mathfrak m_Q^2\), contradicting their independent cotangent classes. They generate \(P\), so its minimal generator number is two. A principal ideal has at most one, proving nonprincipality. Since \(P\ne0\) is an ideal in a domain, tensoring with the fraction field makes it a one-dimensional vector space: its rank is one.

The ring \(Q\) has depth two by its hypersurface Cohen–Macaulay calculation. The module \(Q/P\) has depth one: its image of \(y\) is a nonzerodivisor, the quotient by \(y\) is \(k\), and its support has dimension one, bounding depth. Suppose its projective dimension were finite. Auslander–Buchsbaum would give

\[ \operatorname{pd}_Q(Q/P)=2-1=1. \tag{21} \]

In the free presentation \(0\to P\to Q\to Q/P\to0\), the syzygy criterion of Theorem 1.3 in the projective-dimension lesson then makes \(P\) projective. It is finite, hence free over the local ring, by Proposition 2.1 of that lesson. Rank one forces it to be free of rank one, contradicting its two minimal generators. The supposed finite projective dimension is impossible.

References and proof scope

The six regular-local results, the supporting factoriality lemmas and all six exercises are proved here using the cited earlier results. Normality of the cone and general Serre conditions belong to the next lesson. Smoothness over a field, which requires a specified base field and behaves differently over imperfect fields, is treated later; it is not used in the regularity proofs above.

The official Stacks project is the maintained reference. The tag links below use AI Integrated Stacks Project, an unofficial edition with AI-proposed corrections and AI-written additions.