Regular Local Rings and Homological Dimension — Examples

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Cohen–Macaulayness over a finite regular local base

Example

Let $A ↪ B$ be an injective finite local map of nonzero Noetherian local rings, with $A$ regular. Then $B$ is Cohen–Macaulay if and only if it is free as an $A$ -module.

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

auslander buchsbaum formula: For a nonzero finite module $M$ of finite projective dimension over a nonzero Noetherian local ring $R$ , ${pd}_{R} M + {depth}_{R} M = depth R$ . Consequently such an $M$ with $depth M = depth R$ is free.

[F2]

auslander buchsbaum serre regularity criterion: For a nonzero Noetherian local ring $(R, m, k)$ the following are equivalent: $R$ is regular; ${pd}_{R} k < \infty$ ; $gldim R < \infty$ ; and every finite $R$ -module has finite projective dimension. When these hold, $gldim R = {pd}_{R} k = \dim R$ . A nonzero finite module over regular local $R$ is maximal Cohen–Macaulay (depth $\dim R$ ) if and only if it is free.

[F3]

Injective integral extensions preserve Krull dimension: Assume the Axiom of Choice. Let $A \subseteq B$ be an injective integral extension of nonzero commutative rings. Then $\dim A = \dim B$ .

[F4]

Every system of parameters is regular in a Cohen--Macaulay module: Every system of parameters of a nonzero finite Cohen--Macaulay module over a Noetherian local ring is a regular sequence on that module.

[F5]

One regular system of parameters implies Cohen--Macaulayness: Let $0 \neq M$ be finite over a Noetherian local ring. If one system of parameters for $M$ is $M$ -regular, then $M$ is Cohen--Macaulay. Here a system of parameters for $M$ means a tuple $x_{1}, \dots, x_{d}$ in the maximal ideal, where $d = \dim {Supp}_{R} (M)$ , such that $M / (x_{1}, \dots, x_{d}) M$ has finite length.

[F6]

regular local rings are domains and cohen macaulay: A regular local ring $R$ of dimension $d$ is a domain and Cohen–Macaulay. For every regular system $(x_{1}, \dots, x_{d})$ , the tuple is $R$ -regular and $R / (x_{1}, \dots, x_{c})$ is regular local of dimension $d - c$ for all $0 \leq c \leq d$ .

[F7]

Depth is bounded by support dimension: For every nonzero finite module $M$ over a Noetherian local ring $R$ , $0 \leq {depth}_{R} (M) \leq \dim {Supp}_{R} (M) .$ The nonzero hypothesis is essential for this formulation: under the adopted convention ${depth}_{R} (0) = + \infty$ , whereas the empty support has no nonnegative Krull dimension.

[F8]

Assuming the Axiom of Choice, Nakayama's lemma: Assume the Axiom of Choice. Let $R$ be a commutative ring, let $I ⊴ R$ satisfy $I \subseteq J (R)$ , and let $M$ be a finitely generated left $R$ -module. If $I M = M$ , then $M = 0$ .

Verification

1.1

Finite injectivity makes the extension integral and gives $d = \dim A = \dim B$ . Choose a regular system $x_{1}, \dots, x_{d}$ of $A$ . The quotient $C = B / m_{A} B$ is a finite-dimensional algebra over $k_{A}$ and a nonzero local ring. Its descending powers of the maximal ideal stabilize as vector subspaces; at stabilization Nakayama makes that power zero. Thus $\sqrt{m_{A} B} = m_{B}$ , and the images of the $x_{i}$ are parameters of $B$ .

F3F6F8

2.1

If $B$ is Cohen–Macaulay, those parameters are $B$ -regular. Regarded as an $A$ -module, $B$ consequently has depth at least $d$ . The support-dimension bound gives depth at most $d$ (its annihilator is zero by injectivity). Homological regularity makes its projective dimension over $A$ finite; Auslander–Buchsbaum gives projective dimension zero and freeness.

F4F7F2F1step 1.1

3.1

Conversely, if $B$ is free over $A$ , the regular parameter sequence of $A$ remains injective successively on the finite direct sums describing $B$ and its successive quotients. The terminal quotient is nonzero. Since this is a system of parameters of $B$ , the regular-parameter criterion makes $B$ Cohen–Macaulay. If $d = 0$ , the sequence is empty and $A$ is a field; all steps remain valid.

F6F5step 1.1∎

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fields as regular local rings

Example

Every field $k$ is regular local with dimension and embedding dimension zero. Conversely every zero-dimensional regular local ring is a field. Its regular system is empty and its residue field has a free resolution concentrated in degree zero.

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

embedding dimension and regular local ring: For a nonzero commutative Noetherian local ring $(R, m, k)$ , define $edim R = \dim_{k} (m / m^{2})$ . The ring is regular local when $edim R = \dim R$ . The cotangent space is intrinsic, and is finite-dimensional because $m$ is finitely generated.

[F2]

Assuming the Axiom of Choice, Nakayama's lemma: Assume the Axiom of Choice. Let $R$ be a commutative ring, let $I ⊴ R$ satisfy $I \subseteq J (R)$ , and let $M$ be a finitely generated left $R$ -module. If $I M = M$ , then $M = 0$ .

Verification

1.1

A field has the single prime and maximal ideal $(0)$ , so its dimension and cotangent dimension are both zero. It is regular by definition. The empty tuple generates its maximal ideal and $k \overset{1}{\to} k$ is its length-zero augmented resolution.

F1algebra

2.1

For a zero-dimensional regular local ring, $m / m^{2} = 0$ . Nakayama applied to the finite ideal $m$ gives $m = 0$ . Every element outside the maximal ideal is a unit, so the ring is a field.

F2step 1.1∎

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dvrs as regular local rings

Example

A DVR $R$ with uniformizer $t$ and residue field $k$ is regular local of dimension one, with regular system $(t)$ and ${gr}_{(t)} R ≅ k [T]$ .

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

one dimensional regular local rings are dvrs: A nonzero Noetherian local ring of dimension one is regular if and only if it is a discrete valuation ring. Fields are excluded from the term DVR.

[F2]

associated graded ring of a regular local ring: If $(R, m, k)$ is regular local of dimension $d$ , any cotangent basis induces a graded isomorphism $k [X_{1}, \dots, X_{d}] ≅ {gr}_{m} R$ . Conversely, if the associated graded ring of a nonzero Noetherian local ring is isomorphic as a graded $k$ -algebra to $k [X_{1}, \dots, X_{d}]$ with standard grading, then $R$ is regular of dimension $d$ .

Verification

1.1

The DVR equivalence gives dimension-one regularity. Its maximal ideal is $(t)$ and $t \notin (t^{2})$ , since otherwise cancellation would make the nonunit $t$ a unit. Thus its cotangent basis is the class of $t$ .

F1algebra

2.1

For every $n \geq 0$ , multiplication by $t^{n}$ identifies $k$ with $(t^{n}) / (t^{n + 1})$ : injectivity follows by cancellation and surjectivity by principality. Products of these classes are powers of the degree-one class, so the graded map $k [T] \to {gr}_{(t)} R$ is an isomorphism, also as given by the regular graded theorem.

F2step 1.1algebra∎

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localised polynomial ring regular

Example

For a field $k$ and integers $0 \leq r \leq n$ , the ring $R = k [x_{1}, \dots, x_{n}]_{(x_{1}, \dots, x_{r})}$ is regular local of dimension $r$ , with residue field $k (x_{r + 1}, \dots, x_{n})$ and regular system $(x_{1}, \dots, x_{r})$ .

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

localisation and polynomial extension of regular rings: Localizations and finite polynomial extensions of a commutative regular Noetherian ring are regular. Regularity can equivalently be tested at maximal ideals. For every nonzero such ring, $gldim R = \dim R$ , allowing infinity. More generally, for a finite module over any commutative Noetherian ring, projective dimension is the supremum of its prime-local projective dimensions. Dedekind domains and their finite polynomial extensions are regular.

[F2]

dimension at most embedding dimension: Every nonzero commutative Noetherian local ring $R$ satisfies $\dim R \leq edim R < \infty$ .

Verification

1.1

A field is regular; polynomial extension and localization make $R$ regular. The quotient by the indicated prime inverts every nonzero polynomial in the remaining variables, hence is $k (x_{r + 1}, \dots, x_{n})$ . The maximal ideal is generated by the first $r$ variables.

F1algebra

2.1

The strict coordinate chain $(0) ⊊ (x_{1}) ⊊ \dots ⊊ (x_{1}, \dots, x_{r})$ survives in the localization and has length $r$ . The $r$ generators bound cotangent dimension by $r$ , and the embedding bound gives $r \leq \dim R \leq edim R \leq r$ . Thus those generators are minimal and form regular parameters. If $r = 0$ , the ring is the fraction field; if $r = n$ , the residue field is $k$ , including $n = r = 0$ .

F2step 1.1algebra∎

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formal power series ring regular

Example

For every field $k$ and integer $n \geq 0$ , $k [⁣ [x_{1}, \dots, x_{n}] ⁣]$ is a Noetherian regular local ring of dimension $n$ , with maximal ideal generated by the variables and residue field $k$ .

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

localisation and polynomial extension of regular rings: Localizations and finite polynomial extensions of a commutative regular Noetherian ring are regular. Regularity can equivalently be tested at maximal ideals. For every nonzero such ring, $gldim R = \dim R$ , allowing infinity. More generally, for a finite module over any commutative Noetherian ring, projective dimension is the supremum of its prime-local projective dimensions. Dedekind domains and their finite polynomial extensions are regular.

[F2]

completion preserves regular local rings: A nonzero Noetherian local ring $R$ is regular if and only if its maximal-adic completion $\hat{R}$ is regular.

[F3]

completion preserves embedding dimension: For a nonzero Noetherian local ring $(R, m, k)$ , its maximal-adic completion $\hat{R}$ has maximal ideal $\hat{m} = m \hat{R}$ , residue field $k$ , and a canonical isomorphism $m / m^{2} ≅ \hat{m} / {\hat{m}}^{2}$ . In particular their embedding dimensions agree.

[F4]

Completion of a Noetherian local ring is local with the same residue field: Assume the Axiom of Choice. Let $(R, m)$ be a Noetherian local ring, and let $\hat{R}$ be its $m$ -adic completion. 1. $\hat{R}$ is a Noetherian local ring with maximal ideal $m \hat{R}$ . 2. The residue field is unchanged: $\hat{R} / m \hat{R} ≅ R / m .$ 3. The completion map $R \to \hat{R}$ is faithfully flat.

[F5]

Completion preserves dimension and Hilbert-Samuel data: Assume the Axiom of Choice. Let $(R, m)$ be a Noetherian local ring, let $M \neq 0$ be a finitely generated $R$ -module, and let $\hat{R}$ , $\hat{M}$ denote the $m$ -adic completions. 1. For every $n \geq 0$ , $\hat{M} / m^{n + 1} \hat{M} ≅ M / m^{n + 1} M .$ In particular the Hilbert-Samuel functions of $M$ and $\hat{M}$ agree. 2. The Hilbert-Samuel multiplicity of $M$ equals that of $\hat{M}$ . 3. The support dimensions of $M$ and $\hat{M}$ are equal.

[F6]

dimension at most embedding dimension: Every nonzero commutative Noetherian local ring $R$ satisfies $\dim R \leq edim R < \infty$ .

Verification

1.1

Define a series by coefficients $a_{α} \in k$ for $α \in N^{n}$ , with coefficientwise addition and convolution multiplication. For each fixed multi-index only finitely many pairs sum to it, so multiplication is defined and associative by finite reindexing. Compatible truncations in total degrees below $q$ , for all $q \geq 1$ , identify this ring with the inverse limit of $k [x_{1}, \dots, x_{n}] / (x_{1}, \dots, x_{n})^{q}$ . In these quotients every polynomial with nonzero constant term has a finite geometric-series inverse, so the same inverse limit is the completion of the coordinate local polynomial ring.

givenalgebra

2.1

That local polynomial ring is regular: it is a localization of a polynomial ring over a field, and the coordinate prime chain and $n$ maximal-ideal generators give dimension $n$ . Completion is Noetherian local, preserves dimension and embedding dimension, and preserves regularity. Its maximal ideal is generated by the variable images and its residue field is $k$ . For $n = 0$ , the index set $N^{0}$ has one element and the ring is just $k$ .

F1F2F3F4F5F6step 1.1∎

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dual numbers not regular

Example

For every field $k$ , the dual-number ring $R = k [ε] / (ε^{2})$ is local with $\dim R = 0$ and $edim R = 1$ , so is not regular.

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

embedding dimension and regular local ring: For a nonzero commutative Noetherian local ring $(R, m, k)$ , define $edim R = \dim_{k} (m / m^{2})$ . The ring is regular local when $edim R = \dim R$ . The cotangent space is intrinsic, and is finite-dimensional because $m$ is finitely generated.

Verification

1.1

Every element has a unique form $a + b ε$ . It is a unit precisely when $a \neq 0$ , with inverse $a^{- 1} - b a^{- 2} ε$ . Hence the unique maximal ideal is $(ε)$ . Every prime contains the nilpotent $ε$ , so this is the only prime and the dimension is zero.

givenalgebra

2.1

The square of the maximal ideal is zero and the class of $ε$ is a nonzero $k$ -basis of it. Therefore the cotangent dimension is one, strictly larger than Krull dimension. The ring is finite-dimensional over $k$ and hence Noetherian, so the regularity definition applies and fails.

F1step 1.1algebra∎

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cusp local ring not regular

Example

For every field $k$ , the cusp local ring $R = (k [x, y] / (y^{2} - x^{3}))_{(x, y)}$ has dimension one and embedding dimension two, hence is not regular.

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

quotient and lifting regularity across a regular element: Let $(R, m)$ be nonzero Noetherian local. If $x \in m$ is a nonzerodivisor and $R / (x)$ is regular, then $R$ is regular and $x \notin m^{2}$ . For every nonzerodivisor $x \in m$ , $\dim (R / (x)) = \dim R - 1$ . If $R$ is regular and $0 \neq x \in m$ , then $R / (x)$ is regular if and only if $x \notin m^{2}$ .

[F2]

associated graded polynomial surjection: Let $(R, m, k)$ be nonzero Noetherian local and let $x_{1}, \dots, x_{e}$ lift a basis of $m / m^{2}$ . There is a surjective graded $k$ -algebra map $ϕ : k [X_{1}, \dots, X_{e}] \to {gr}_{m} R$ , determined by $X_{i} \mapsto x_{i} + m^{2}$ , with every variable of degree one.

[F3]

dimension at most embedding dimension: Every nonzero commutative Noetherian local ring $R$ satisfies $\dim R \leq edim R < \infty$ .

Verification

1.1

The quotient $A = k [x, y] / (y^{2} - x^{3})$ has unique representatives $a (x) + y b (x)$ by division by the monic polynomial in $y$ . Under $x \mapsto t^{2}$ , $y \mapsto t^{3}$ , the two summands have even and odd powers of $t$ , respectively; their vanishing forces both to be zero. Hence $A$ embeds in $k [t]$ and is a domain in every characteristic. The origin ideal remains a proper nonzero maximal ideal after localization.

givenalgebra

2.1

The ambient local ring $S = k [x, y]_{(x, y)}$ has dimension two and cotangent basis $x, y$ : the coordinate chain gives dimension at least two, and its two maximal-ideal generators give the reverse bound. The nonzero $f = y^{2} - x^{3}$ is a nonzerodivisor in this polynomial domain; the dimension-drop argument of the regular-element quotient theorem gives $\dim S / (f) = 1$ . Since $f \in (x, y)^{2}$ , quotienting adds no linear cotangent relation, so the embedding dimension stays two. This proves the claim over any field, including characteristics two and three.

F1F2F3step 1.1algebra∎

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betti numbers from a koszul resolution

Example

For $R = k [x, y]_{(x, y)}$ , the augmented complex $0 \to R \overset{c \mapsto (- y c, x c)}{\to} R^{2} \overset{(a, b) \mapsto x a + y b}{\to} R \to k \to 0$ is a minimal free resolution. Thus $β (k) = (1, 2, 1)$ , with all higher Betti numbers zero.

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

regular local residue field koszul resolution: For a regular local ring $(R, m, k)$ of dimension $d$ , the Koszul complex on any regular system of parameters is a minimal free resolution of $k$ of length $d$ .

[F2]

betti number is rank in minimal resolution: For every minimal degreewise finite free resolution $F_{∙} \to M$ of a finite module over a nonzero Noetherian local ring, $β_{i}^{R} (M) = {rank}_{R} F_{i}$ for all $i \geq 0$ .

[F3]

localisation and polynomial extension of regular rings: Localizations and finite polynomial extensions of a commutative regular Noetherian ring are regular. Regularity can equivalently be tested at maximal ideals. For every nonzero such ring, $gldim R = \dim R$ , allowing infinity. More generally, for a finite module over any commutative Noetherian ring, projective dimension is the supremum of its prime-local projective dimensions. Dedekind domains and their finite polynomial extensions are regular.

Verification

1.1

The coordinate local ring is regular of dimension two: the coordinate chain and two generators give the dimension. Its variables are regular parameters. The displayed maps are exactly its two-variable Koszul maps; their composition is $- x y c + y x c = 0$ and the Koszul theorem gives exactness.

F3F1algebra

2.1

Every entry is in $(x, y)$ , so the resolution is minimal. Its ranks in degrees zero, one, two are $1, 2, 1$ and it is zero above two. The rank formula gives the asserted Betti numbers, independently of the characteristic.

F2step 1.1∎

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residue field infinite projective dimension singular

Example

Let $k$ be a field. For $R = k [ε] / (ε^{2})$ , its residue field $k$ has an infinite minimal free resolution with one copy of $R$ in every degree and every positive differential multiplication by $ε$ . Consequently $β_{i}^{R} (k) = 1$ for all $i \geq 0$ and ${pd}_{R} k = \infty$ .

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

projective dimension from last nonzero betti number: For a nonzero finite module $M$ over a nonzero Noetherian local ring, ${pd}_{R} M = \sup {i \geq 0 : β_{i}^{R} (M) \neq 0}$ , allowing infinity. For each integer $q \geq 0$ , ${pd}_{R} M \leq q$ if and only if ${Tor}_{q + 1}^{R} (k, M) = 0$ .

[F2]

betti number is rank in minimal resolution: For every minimal degreewise finite free resolution $F_{∙} \to M$ of a finite module over a nonzero Noetherian local ring, $β_{i}^{R} (M) = {rank}_{R} F_{i}$ for all $i \geq 0$ .

Verification

1.1

The ring is local with maximal ideal $(ε)$ . For multiplication by $ε$ , the image and kernel both equal $(ε)$ : $ε (a + b ε) = a ε$ . The augmentation $R \to k$ has that same kernel. Thus the infinite augmented complex is exact in every degree and all positive matrix entries are in the maximal ideal.

givenalgebra

2.1

The rank formula gives $β_{i} (k) = 1$ in every degree. These nonzero Betti numbers are unbounded in degree, so the projective-dimension criterion gives infinity. A finite initial truncation would have a nonzero left kernel and is not a finite resolution.

F2F1step 1.1∎

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embedding dimension versus dimension node

Example

For every field $k$ , the split node $R = (k [x, y] / (x y))_{(x, y)}$ is reduced and has dimension one and embedding dimension two. It is neither a domain nor regular.

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F2]

embedding dimension and regular local ring: For a nonzero commutative Noetherian local ring $(R, m, k)$ , define $edim R = \dim_{k} (m / m^{2})$ . The ring is regular local when $edim R = \dim R$ . The cotangent space is intrinsic, and is finite-dimensional because $m$ is finitely generated.

Verification

1.1

In $k [x, y]$ , $(x y) = (x) \cap (y)$ , as divisibility of monomials shows. It is therefore radical. Every prime of the quotient contains $x$ or $y$ ; in the origin localization either branch is the local line $k [t]_{(t)}$ , with prime chain of length one. Thus $R$ is reduced of dimension one. Both $x$ and $y$ survive and their product is zero, so it is not a domain.

givenalgebra

2.1

The relation $x y$ is quadratic, so $m / m^{2}$ has independent basis $\overset{ˉ}{x}, \overset{ˉ}{y}$ . Its dimension two strictly exceeds the dimension one just computed, and the definition makes $R$ nonregular. Localization does not change these cotangent classes since denominators have nonzero constant term. This works also in characteristic two.

F2step 1.1algebra∎

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associated graded polynomial map singular kernel

Example

For the cusp local ring $R = (k [x, y] / (y^{2} - x^{3}))_{(x, y)}$ with maximal ideal $m$ , the associated graded ring is $k [X, Y] / (Y^{2})$ . Thus the polynomial map defined by the cotangent classes has kernel exactly $(Y^{2})$ .

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

associated graded polynomial surjection: Let $(R, m, k)$ be nonzero Noetherian local and let $x_{1}, \dots, x_{e}$ lift a basis of $m / m^{2}$ . There is a surjective graded $k$ -algebra map $ϕ : k [X_{1}, \dots, X_{e}] \to {gr}_{m} R$ , determined by $X_{i} \mapsto x_{i} + m^{2}$ , with every variable of degree one.

Verification

1.1

In the ambient coordinate local ring $S = k [x, y]_{(x, y)}$ , the associated graded ring is $k [X, Y]$ : a rational function with denominator of nonzero constant term has initial form equal to its numerator initial form divided by that constant. This identifies each graded piece and respects products. Therefore orders add on products of nonzero elements of $S$ . In particular for $f = y^{2} - x^{3}$ , $in (f) = Y^{2}$ , and $in (h f) = in (h) Y^{2}$ for every nonzero $h \in S$ .

givenalgebra

2.1

The degree- $n$ kernel of $gr S \to gr (S / (f))$ consists of classes of $a \in n^{n}$ with $a \in (f) + n^{n + 1}$ . Write $a = h f + b$ with $b \in n^{n + 1}$ . If its degree- $n$ class is nonzero, it is exactly the initial form of $h f$ , hence a multiple of $Y^{2}$ . Conversely every homogeneous multiple of $Y^{2}$ is the initial form of a polynomial multiple of $f$ . Thus the graded kernel is exactly $(Y^{2})$ and the surjective polynomial map of the cotangent-basis lemma has the stated quotient.

F1step 1.1algebra∎

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minimal resolution unit cancellation

Example

Over $R = k [x]_{(x)}$ , the free resolution $0 \to R^{2} \overset{diag (x, 1)}{\to} R^{2} \to R / (x) \to 0$ , with augmentation $(a, b) \mapsto a m o d x$ , contracts to the minimal resolution $0 \to R \overset{x}{\to} R \to R / (x) \to 0$ .

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

minimal free resolution differentials land in maximal ideal: For an augmented degreewise finite free resolution over a nonzero Noetherian local ring $(R, m)$ , minimality means that every positive differential matrix has entries in $m$ . Equivalently no positive differential admits a unit pivot, or a nonzero two-term identity direct summand. A unit pivot can be cancelled without changing the resolved module.

Verification

1.1

The displayed diagonal map is injective since $R$ is a domain. Its image consists exactly of pairs whose first coordinate lies in $(x)$ , which is the augmentation kernel. Hence the complex is exact.

givenalgebra

2.1

The second coordinates form the two-term identity summand, whose identity homotopy contracts it. Removing it leaves multiplication by $x$ on the first coordinates. Since $x$ belongs to the maximal ideal, this remaining resolution is minimal by the unit-cancellation criterion.

F1step 1.1∎

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betti numbers residue field regular ring

Example

For $R = k [x, y, z]_{(x, y, z)}$ , the residue field has Betti numbers $(1, 3, 3, 1)$ and projective dimension three.

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

regular local residue field koszul resolution: For a regular local ring $(R, m, k)$ of dimension $d$ , the Koszul complex on any regular system of parameters is a minimal free resolution of $k$ of length $d$ .

[F2]

betti number is rank in minimal resolution: For every minimal degreewise finite free resolution $F_{∙} \to M$ of a finite module over a nonzero Noetherian local ring, $β_{i}^{R} (M) = {rank}_{R} F_{i}$ for all $i \geq 0$ .

[F3]

regular local residue field projective dimension dimension: For a regular local ring $(R, m, k)$ of dimension $d$ , ${pd}_{R} k = d$ and $β_{i}^{R} (k) = (\binom{d}{i})$ for $0 \leq i \leq d$ , with $β_{i}^{R} (k) = 0$ for $i > d$ .

[F4]

dimension at most embedding dimension: Every nonzero commutative Noetherian local ring $R$ satisfies $\dim R \leq edim R < \infty$ .

Verification

1.1

The coordinate chain of prime ideals gives dimension at least three, while the three generators of the maximal ideal give embedding dimension at most three and hence dimension at most three. Thus $R$ is regular of dimension three with parameters $x, y, z$ . Their Koszul complex is a minimal resolution.

F1F4algebra

2.1

The exterior bases have ranks $1, 3, 3, 1$ in degrees zero through three and zero above. The rank and projective-dimension formulas give these Betti numbers and projective dimension three, since the top rank is one.

F2F3step 1.1∎

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auslander buchsbaum first syzygy

Example

For $R = k [x, y]_{(x, y)}$ with maximal ideal $m$ , ${pd}_{R} k = 2$ , ${depth}_{R} k = 0$ , and its first syzygy satisfies ${pd}_{R} m = 1$ and ${depth}_{R} m = 1$ .

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

auslander buchsbaum syzygy projective dimension: Let $0 \to K \to F_{0} \to M \to 0$ be the initial minimal presentation of a nonzero finite module over a nonzero Noetherian local ring. If $0 < n = pd M < \infty$ , then $K \neq 0$ and $pd K = n - 1$ .

[F2]

auslander buchsbaum formula: For a nonzero finite module $M$ of finite projective dimension over a nonzero Noetherian local ring $R$ , ${pd}_{R} M + {depth}_{R} M = depth R$ . Consequently such an $M$ with $depth M = depth R$ is free.

[F3]

regular local residue field projective dimension dimension: For a regular local ring $(R, m, k)$ of dimension $d$ , ${pd}_{R} k = d$ and $β_{i}^{R} (k) = (\binom{d}{i})$ for $0 \leq i \leq d$ , with $β_{i}^{R} (k) = 0$ for $i > d$ .

[F4]

localisation and polynomial extension of regular rings: Localizations and finite polynomial extensions of a commutative regular Noetherian ring are regular. Regularity can equivalently be tested at maximal ideals. For every nonzero such ring, $gldim R = \dim R$ , allowing infinity. More generally, for a finite module over any commutative Noetherian ring, projective dimension is the supremum of its prime-local projective dimensions. Dedekind domains and their finite polynomial extensions are regular.

Verification

1.1

The ring is regular local of dimension two, by polynomial regularity and the coordinate chain and generator count. The residue-field computation gives projective dimension two. Its depth is zero because every maximal-ideal element kills the nonzero module $k$ . The presentation $0 \to m \to R \to k \to 0$ is minimal.

F4F3algebra

2.1

The syzygy theorem gives $pd m = 1$ . The ring depth is two, since $x, y$ is a regular sequence and depth is bounded by dimension. Auslander–Buchsbaum gives $depth m = 2 - 1 = 1$ . Concretely its minimal resolution is $0 \to R \overset{c \mapsto (- y c, x c)}{\to} R^{2} \to m \to 0$ : reducing a relation modulo $x$ shows $b = x c$ , and then cancellation gives $a = - y c$ .

F1F2step 1.1algebra∎

ExampleConstruction: AI-generatedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

completion regularity invariance

Example

The ring $R = k [x, y]_{(x, y)}$ and its completion $k [⁣ [x, y] ⁣]$ both have dimension and embedding dimension two. For every $q \geq 1$ , their quotients by the $q$ th powers of the maximal ideals agree and have basis the monomials of total degree less than $q$ .

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

completion preserves embedding dimension: For a nonzero Noetherian local ring $(R, m, k)$ , its maximal-adic completion $\hat{R}$ has maximal ideal $\hat{m} = m \hat{R}$ , residue field $k$ , and a canonical isomorphism $m / m^{2} ≅ \hat{m} / {\hat{m}}^{2}$ . In particular their embedding dimensions agree.

[F2]

completion preserves regular local rings: A nonzero Noetherian local ring $R$ is regular if and only if its maximal-adic completion $\hat{R}$ is regular.

[F3]

Completion of a Noetherian local ring is local with the same residue field: Assume the Axiom of Choice. Let $(R, m)$ be a Noetherian local ring, and let $\hat{R}$ be its $m$ -adic completion. 1. $\hat{R}$ is a Noetherian local ring with maximal ideal $m \hat{R}$ . 2. The residue field is unchanged: $\hat{R} / m \hat{R} ≅ R / m .$ 3. The completion map $R \to \hat{R}$ is faithfully flat.

[F4]

Completion preserves dimension and Hilbert-Samuel data: Assume the Axiom of Choice. Let $(R, m)$ be a Noetherian local ring, let $M \neq 0$ be a finitely generated $R$ -module, and let $\hat{R}$ , $\hat{M}$ denote the $m$ -adic completions. 1. For every $n \geq 0$ , $\hat{M} / m^{n + 1} \hat{M} ≅ M / m^{n + 1} M .$ In particular the Hilbert-Samuel functions of $M$ and $\hat{M}$ agree. 2. The Hilbert-Samuel multiplicity of $M$ equals that of $\hat{M}$ . 3. The support dimensions of $M$ and $\hat{M}$ are equal.

[F5]

dimension at most embedding dimension: Every nonzero commutative Noetherian local ring $R$ satisfies $\dim R \leq edim R < \infty$ .

Verification

1.1

Degree truncation identifies series modulo $(x, y)^{q}$ with polynomials modulo that ideal. In the truncated polynomial ring, every denominator allowed in $R$ is a unit, by a finite geometric-series expansion of its nonconstant part. Thus both quotients have the stated monomial basis, and their inverse limit is $k [⁣ [x, y] ⁣]$ , identifying it as the maximal-adic completion.

givenalgebra

2.1

The coordinate chain and two maximal-ideal generators give $\dim R = edim R = 2$ . The cotangent-completion theorem preserves embedding dimension and the completion dimension theorem preserves dimension independently. The completion is Noetherian local and regular. At $q = 1$ the quotient is $k$ ; at $q = 2$ the basis $1, x, y$ exhibits the two cotangent classes.

F1F4F3F2F5step 1.1∎

ExampleConstruction: AI-generatedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

hypersurface regularity at a rational point

Example

Let $k$ be any field, $a \in k^{n}$ , and $0 \neq f \in k [x_{1}, \dots, x_{n}]$ with $f (a) = 0$ . The local hypersurface ring at $a$ is regular if and only if at least one formal partial derivative $\partial f / \partial x_{i}$ is nonzero at $a$ .

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

localisation and polynomial extension of regular rings: Localizations and finite polynomial extensions of a commutative regular Noetherian ring are regular. Regularity can equivalently be tested at maximal ideals. For every nonzero such ring, $gldim R = \dim R$ , allowing infinity. More generally, for a finite module over any commutative Noetherian ring, projective dimension is the supremum of its prime-local projective dimensions. Dedekind domains and their finite polynomial extensions are regular.

[F2]

quotient and lifting regularity across a regular element: Let $(R, m)$ be nonzero Noetherian local. If $x \in m$ is a nonzerodivisor and $R / (x)$ is regular, then $R$ is regular and $x \notin m^{2}$ . For every nonzerodivisor $x \in m$ , $\dim (R / (x)) = \dim R - 1$ . If $R$ is regular and $0 \neq x \in m$ , then $R / (x)$ is regular if and only if $x \notin m^{2}$ .

Verification

1.1

Translate coordinates $u_{i} = x_{i} - a_{i}$ . The ambient local ring $S = k [u_{1}, \dots, u_{n}]_{(u_{1}, \dots, u_{n})}$ is regular, with cotangent basis the $u_{i}$ . It is a domain, so the nonzero polynomial $f$ is a nonzerodivisor. The quotient regularity criterion says $S / (f)$ is regular exactly when $f \notin m_{S}^{2}$ . The hypotheses cannot hold for n=0, because then f is a nonzero constant.

F1F2

2.1

Monomial expansion after translation gives $f \equiv \sum_{i} (\partial f / \partial x_{i}) (a) u_{i} (m o d m_{S}^{2})$ ; the constant term vanishes. Independence of the cotangent basis makes this class nonzero precisely when at least one coefficient is nonzero. This proves both implications over every characteristic. It is a rational-point hypersurface statement and makes no assertion about smoothness over arbitrary residue-field extensions.

step 1.1algebra∎

ExampleConstruction: AI-generatedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local ambient cover minimal dimension

Example

If a nonzero Noetherian local ring $A$ is a quotient of at least one regular local ring, then the least dimension of a regular local ring surjecting onto $A$ is $edim A$ .

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

regular local regular quotient ideal is parameter generated: Let $(R, m, k)$ be regular local of dimension $d$ and $I \subseteq m$ . The following are equivalent: $R / I$ is regular; $I$ is generated by an initial part of a regular system of parameters; and $\dim_{k} ((I + m^{2}) / m^{2}) = d - \dim (R / I)$ .

[F2]

regular local quotient by parameter is regular: Let $(R, m, k)$ be regular local of dimension $d$ , and let $x \in m ∖ m^{2}$ . Then $R / (x)$ is regular local, of dimension and embedding dimension $d - 1$ .

[F3]

embedding dimension is minimal maximal ideal generator number: For a nonzero Noetherian local ring $(R, m, k)$ , $edim R$ is the least number of generators of $m$ .

Verification

1.1

For a surjection $R \to A = R / I$ of local rings the maximal ideal of $A$ is $m_{R} / I$ and the residue fields agree. Thus its cotangent space is the quotient $m_{R} / (I + m_{R}^{2})$ . If $R$ is regular of dimension $d$ , this gives $edim A \leq d$ .

F3algebra

2.1

For a supplied regular cover put $c = \dim_{k} ((I + m_{R}^{2}) / m_{R}^{2})$ . Lift a basis to $x_{1}, \dots, x_{c} \in I$ and extend their cotangent classes to a basis. The construction in the parameter-generated quotient lemma and repeated parameter reduction show that $R / (x_{1}, \dots, x_{c})$ is regular of dimension $d - c = edim A$ . It still surjects onto $A$ , so the lower bound is attained. If $c = 0$ retain the original cover; if $c = d$ the new cover is the residue field.

F1F2step 1.1∎

ExampleConstruction: AI-generatedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular flat local map with singular closed fibre

Example

For every field $k$ , the local map $k [s]_{(s)} \to k [t]_{(t)}$ , $s \mapsto t^{2}$ , is finite free of rank two between regular DVRs. Its closed fibre is $k [t] / (t^{2})$ and is not regular.

Facts & Assumptions

Given: The objects and hypotheses in the example. We work with the Axiom of Choice; cited dependent-choice and resolution-existence hypotheses are retained.

[F1]

localisation and polynomial extension of regular rings: Localizations and finite polynomial extensions of a commutative regular Noetherian ring are regular. Regularity can equivalently be tested at maximal ideals. For every nonzero such ring, $gldim R = \dim R$ , allowing infinity. More generally, for a finite module over any commutative Noetherian ring, projective dimension is the supremum of its prime-local projective dimensions. Dedekind domains and their finite polynomial extensions are regular.

[F2]

embedding dimension and regular local ring: For a nonzero commutative Noetherian local ring $(R, m, k)$ , define $edim R = \dim_{k} (m / m^{2})$ . The ring is regular local when $edim R = \dim R$ . The cotangent space is intrinsic, and is finite-dimensional because $m$ is finitely generated.

Verification

1.1

Before localization every polynomial in $t$ has a unique expression $a (t^{2}) + t b (t^{2})$ , so $k [t]$ is free on $1, t$ over $k [s]$ . After localizing the base at $(s)$ , call the resulting rank-two free algebra $C$ . If $h (t)$ has nonzero constant term, write $h = a (s) + t b (s)$ . Then $(a + t b) (a - t b) = a (s)^{2} - s b (s)^{2}$ is a unit of the base, since its constant term is $a (0)^{2} \neq 0$ . Thus every such $h$ is a unit of $C$ , proving $C = k [t]_{(t)}$ , also in characteristic two.

givenalgebra

2.1

The source and target are regular one-dimensional coordinate local rings (their elements are units times powers of their variable, giving DVRs). The maximal ideal contracts correctly, and freeness makes the map flat. Modulo the source maximal ideal the fibre is $k [t] / (t^{2})$ , whose only prime is $(t)$ , with zero square and one-dimensional cotangent space. Its Krull dimension is zero, so it is singular by the regularity definition.

F1F2step 1.1algebra∎

Regular Local Rings and Homological Dimension — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Cohen–Macaulayness over a finite regular local base

Example

Facts & Assumptions

Verification

fields as regular local rings

Example

Facts & Assumptions

Verification

dvrs as regular local rings

Example

Facts & Assumptions

Verification

localised polynomial ring regular

Example

Facts & Assumptions

Verification

formal power series ring regular

Example

Facts & Assumptions

Verification

dual numbers not regular

Example

Facts & Assumptions

Verification

cusp local ring not regular

Example

Facts & Assumptions

Verification

betti numbers from a koszul resolution

Example

Facts & Assumptions

Verification

residue field infinite projective dimension singular

Example

Facts & Assumptions

Verification

embedding dimension versus dimension node

Example

Facts & Assumptions

Verification

associated graded polynomial map singular kernel

Example

Facts & Assumptions

Verification

minimal resolution unit cancellation

Example

Facts & Assumptions

Verification

betti numbers residue field regular ring

Example

Facts & Assumptions

Verification

auslander buchsbaum first syzygy

Example

Facts & Assumptions

Verification

completion regularity invariance

Example

Facts & Assumptions

Verification

hypersurface regularity at a rational point

Example

Facts & Assumptions

Verification

regular local ambient cover minimal dimension

Example

Facts & Assumptions

Verification

regular flat local map with singular closed fibre

Example

Facts & Assumptions

Verification

Sources