Regular Local Rings and Homological Dimension

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

embedding dimension and regular local ring

Definition

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.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

embedding dimension is minimal maximal ideal generator number

Statement

For a nonzero Noetherian local ring $(R, m, k)$ , $edim R$ is the least number of generators of $m$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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, minimal generators over a local ring are exactly residue-field bases: Assume the Axiom of Choice. Let $(R, m)$ be a local ring with residue field $k = R / m$ , and let $M$ be a finitely generated left $R$ -module. A finite generating set $x_{1}, \dots, x_{r}$ of $M$ is minimal if and only if the images of $x_{1}, \dots, x_{r}$ in $M / m M$ form a $k$ -basis. In particular every minimal generating set of $M$ has the same cardinality.

[F3]

Assuming the Axiom of Choice, generators modulo an ideal in the Jacobson radical lift to generators: 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 elements $x_{1}, \dots, x_{r} \in M$ generate $M / I M$ , then $x_{1}, \dots, x_{r}$ generate $M$ .

Proof

1.1

Write $e = \dim_{k} m / m^{2}$ . Lift a basis to $x_{1}, \dots, x_{e} \in m$ . Since $m$ is finite and $m = J (R)$ , Nakayama gives $m = (x_{1}, \dots, x_{e})$ . If $e = 0$ , the same assertion gives $m = 0$ .

F1F3

2.1

Any generating tuple of $m$ spans its quotient by $m^{2}$ , so its length is at least $e$ . The lifted basis is a minimal generating tuple by the local generator criterion. Thus the least length is $e$ .

F2step 1.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

dimension at most embedding dimension

Statement

Every nonzero commutative Noetherian local ring $R$ satisfies $\dim R \leq edim R < \infty$ .

Facts & Assumptions

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

[F1]

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 $m$ .

[F2]

Krull's height theorem: Let $R$ be a Noetherian commutative ring, let $I = (x_{1}, \dots, x_{n})$ be an ideal generated by $n \geq 1$ elements, and let $p$ be a prime ideal minimal over $I$ . Then $ht (p) \leq n$ .

Proof

1.1

Let $e = edim R$ . The maximal ideal has a generating tuple of length $e$ . If $e = 0$ , it is zero; then every nonzero element is a unit and $R$ is a field of dimension zero.

F1

2.1

If $e \geq 1$ , the maximal ideal is minimal over itself, so its height is at most $e$ by the height theorem. Every prime chain in a local ring can be extended to end at its maximal ideal; hence $\dim R = ht m \leq e$ .

F2step 1.1algebra∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular system of parameters

Definition

In a regular local ring $(R, m, k)$ of dimension $d$ , an ordered minimal generating tuple $(x_{1}, \dots, x_{d})$ of $m$ is a regular system of parameters. The tuple is empty when $d = 0$ . This definition concerns generators of the maximal ideal; the regular-sequence property is a theorem, not part of the definition.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular system of parameters equivalent basis

Statement

Let $(R, m, k)$ be a nonzero Noetherian local ring of dimension $d$ , and let $x = (x_{1}, \dots, x_{d}) \in m^{d}$ . Then $x$ is a regular system of parameters if and only if its classes form a $k$ -basis of $m / m^{2}$ . In particular every lift of a cotangent basis in a regular local ring generates $m$ and is a system of parameters.

Facts & Assumptions

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

[F1]

regular system of parameters: In a regular local ring $(R, m, k)$ of dimension $d$ , an ordered minimal generating tuple $(x_{1}, \dots, x_{d})$ of $m$ is a regular system of parameters. The tuple is empty when $d = 0$ . This definition concerns generators of the maximal ideal; the regular-sequence property is a theorem, not part of the definition.

[F2]

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 $m$ .

[F3]

Assuming the Axiom of Choice, generators modulo an ideal in the Jacobson radical lift to generators: 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 elements $x_{1}, \dots, x_{r} \in M$ generate $M / I M$ , then $x_{1}, \dots, x_{r}$ generate $M$ .

Proof

1.1

If $x$ is a regular system, it minimally generates $m$ in a regular ring, whose cotangent dimension is $d$ . Its $d$ spanning classes therefore form a basis.

F1F2

2.1

Conversely, a basis of length $d$ makes the embedding dimension $d$ , so $R$ is regular. Nakayama lifts the spanning classes to generators of $m$ , and no generator can be removed since its class is independent. Their ideal has radical $m$ and length $d$ , which is exactly the parameter condition. For $d = 0$ , Nakayama gives $m = 0$ and the empty tuple has the same property.

F3F1step 1.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

associated graded polynomial surjection

Statement

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.

Facts & Assumptions

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

[F1]

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 $m$ .

[F2]

The associated graded ring and associated graded module of an ideal-adic filtration: Let $R$ be a commutative ring, let $I \subset R$ be an ideal, and let $M$ be an $R$ -module. The associated graded ring of the $I$ -adic filtration is ${gr}_{I} (R) : = ⨁_{n \geq 0} I^{n} / I^{n + 1} .$ Multiplication is induced by multiplication in $R$ : $(a + I^{m + 1}) (b + I^{n + 1}) = a b + I^{m + n + 1} .$ The associated graded module is ${gr}_{I} (M) : = ⨁_{n \geq 0} I^{n} M / I^{n + 1} M,$ viewed as a graded ${gr}_{I} (R)$ -module by $(a + I^{m + 1}) (x + I^{n + 1} M) = a x + I^{m + n + 1} M .$

[F3]

Assuming the Axiom of Choice, Nakayama's lemma: If $I$ is contained in the Jacobson radical of a commutative ring and $M$ is finite with $I M = M$ , then $M = 0$ .

Proof

1.1

The degree-zero part is $R / m = k$ . On $m^{n} / m^{n + 1}$ the action of $R$ factors through $k$ , since $m m^{n} \subseteq m^{n + 1}$ . The graded multiplication therefore defines the displayed polynomial map. Altering a representative by $m^{n + 1}$ changes a product of degrees $n, j$ by $m^{n + j + 1}$ , so multiplication and the map are well-defined.

F2givenalgebra

2.1

Put $N = (x_{1}, \dots, x_{e}) \subseteq m$ . The basis hypothesis says $m = N + m^{2}$ , so the finite module $m / N$ satisfies $m (m / N) = m / N$ . Nakayama gives $m = N$ . Expanding products now shows that degree- $n$ monomials in the $x_{i}$ generate $m^{n}$ over $R$ ; reducing coefficients modulo $m$ spans the degree- $n$ quotient over $k$ . Thus every graded component is in the image. When $e = 0$ , the same Nakayama argument gives $m = 0$ and the map is the identity on $k$ .

F1F3step 1.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local graded surjection has zero kernel

Statement

For the graded map $ϕ : k [X_{1}, \dots, X_{e}] ↠ {gr}_{m} R$ defined by a cotangent basis in a nonzero Noetherian local ring, if $e = \dim R$ , then $\ker ϕ = 0$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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.

[F2]

The degree of the Hilbert-Samuel polynomial equals the dimension of the support: Assume the Axiom of Choice. Let $(R, m)$ be a Noetherian local ring, let $M \neq 0$ be a finite $R$ -module, and let $I$ be an ideal of definition for $M$ . Then the Hilbert-Samuel polynomial $P_{I, M}$ has degree $\deg P_{I, M} = \dim Supp (M) .$

[F3]

A polynomial ring in finitely many indeterminates over an integral domain is an integral domain: If $R$ is an integral domain, then $R [x_{1}, \dots, x_{n}]$ is an integral domain for every $n \in N$ , including $n = 0$ .

Proof

1.1

If $e = 0$ , the map is the identity of $k$ . Suppose $e \geq 1$ . If the homogeneous kernel were nonzero, it would contain a nonzero homogeneous polynomial $f$ of degree $a \geq 1$ , because the degree-zero map is injective.

F1given

2.1

Put $P = k [X_{1}, \dots, X_{e}]$ . Since $P$ is a domain, multiplication by $f$ injects $P (- a)$ into $P$ . Counting monomials of total degree at most $n$ , for $n \geq a$ , gives $\sum_{j = 0}^{n} \dim_{k} (P / (f))_{j} = (\binom{n + e}{e}) - (\binom{n - a + e}{e})$ . This polynomial has degree $e - 1$ : the terms of degree $e$ cancel.

F3step 1.1algebra

3.1

The surjection $P / (f) \to {gr}_{m} R$ bounds $ℓ_{R} (R / m^{n + 1})$ by that count, because its filtration factors are precisely the graded pieces. But its eventual Hilbert–Samuel polynomial has degree $\dim R = e$ and positive leading coefficient (it is eventually positive). A polynomial of degree $e$ cannot be bounded by one of degree $e - 1$ for all large $n$ . Thus no such $f$ exists.

F2step 2.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

associated graded ring of a regular local ring

Statement

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 $d$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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.

[F2]

regular local graded surjection has zero kernel: For the graded map $ϕ : k [X_{1}, \dots, X_{e}] ↠ {gr}_{m} R$ defined by a cotangent basis in a nonzero Noetherian local ring, if $e = \dim R$ , then $\ker ϕ = 0$ .

[F3]

The degree of the Hilbert-Samuel polynomial equals the dimension of the support: Assume the Axiom of Choice. Let $(R, m)$ be a Noetherian local ring, let $M \neq 0$ be a finite $R$ -module, and let $I$ be an ideal of definition for $M$ . Then the Hilbert-Samuel polynomial $P_{I, M}$ has degree $\deg P_{I, M} = \dim Supp (M) .$

Proof

1.1

In a regular local ring the cotangent dimension equals $d$ . The polynomial map is surjective and has zero kernel, hence is the claimed isomorphism.

F1F2

2.1

Conversely, the degree-one component of a supplied graded isomorphism has dimension $d$ , so $edim R = d$ . Its cumulative graded dimensions are $(\binom{n + d}{d})$ , also when $d = 0$ , where the count is one. These are the lengths of $R / m^{n + 1}$ . The Hilbert–Samuel dimension theorem gives $\dim R = d$ , proving regularity.

F3givenalgebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local hilbert samuel multiplicity one

Statement

For a regular local ring $(R, m, k)$ of dimension $d$ and every integer $n \geq 0$ , $ℓ_{R} (R / m^{n + 1}) = (\binom{n + d}{d})$ . Consequently $e_{m} (R) = 1$ .

Facts & Assumptions

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

[F1]

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 $d$ .

[F2]

Hilbert-Samuel multiplicity as the factorial-scaled leading coefficient: Let $(R, m)$ be a Noetherian local ring, let $M$ be a finite $R$ -module, and let $I$ be an ideal of definition for $M$ . If $M = 0$ , define $e_{I} (M) : = 0.$ If $M \neq 0$ , let $P_{I, M}$ be the eventual Hilbert-Samuel polynomial from thm-existence-of-hilbert-samuel-polynomial, and let $d = \deg P_{I, M}$ . Because $I \subseteq m$ and $M \neq 0$ , Nakayama's lemma makes $M / I^{n + 1} M$ nonzero for every $n$ , so $P_{I, M}$ is not the zero polynomial and $d$ is defined. The Hilbert-Samuel multiplicity of $M$ with respect to $I$ is $e_{I} (M) : = d! \cdot (leading coefficient of P_{I, M}) .$ Equivalently, when $M \neq 0$ and $P_{I, M} (n) = \frac{e_{I} (M)}{d!} n^{d} + lower-degree terms,$ then $e_{I} (M)$ is the integer scaling the top term.

Proof

1.1

The filtration of $R / m^{n + 1}$ has factors $m^{j} / m^{j + 1}$ for $0 \leq j \leq n$ . The graded polynomial description identifies their total dimension with the number of monomials in $d$ variables of degree at most $n$ . Introducing a slack exponent identifies these with $(d + 1)$ -tuples of nonnegative integers summing to $n$ , counted by $(\binom{n + d}{d})$ .

F1algebra

2.1

The leading coefficient is $1 / d!$ , so multiplying it by $d!$ gives multiplicity one. If $d = 0$ , the only monomial is $1$ , and the constant polynomial has leading coefficient one and $0! = 1$ . The formula also gives length one at $n = 0$ .

F2step 1.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local domain induction

Statement

Every regular local ring is an integral domain.

Facts & Assumptions

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

[F1]

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 $d$ .

[F2]

The Krull intersection is the $(1 - a)$ -torsion submodule, and it vanishes in the Jacobson-radical case: The first clause below is choice-free; the second uses the published Jacobson-radical unit criterion and therefore inherits its Axiom-of-Choice boundary. Let $R$ be a Noetherian commutative ring, let $I \subset R$ be an ideal, and let $M$ be a finite $R$ -module. Put $K : = ⋂_{n \geq 0} I^{n} M .$ Then: 1. $K$ is exactly the set of elements $m \in M$ for which $(1 - a) m = 0$ for some $a \in I$ ; 2. if $I \subseteq J (R)$ , then $K = 0$ .

Proof

1.1

The maximal-adic filtration is separated by Krull intersection. For each nonzero $a \in R$ there is therefore a largest integer $r \geq 0$ with $a \in m^{r}$ ; its class $in (a)$ in degree $r$ is nonzero.

F2

2.1

For nonzero $a, b$ of orders $r, s$ , their initial classes have nonzero product in the graded polynomial ring, which is a domain: multiplying leading monomials proves this over the field $k$ . This product is the class of $a b$ in $m^{r + s} / m^{r + s + 1}$ , so $a b \neq 0$ . The same argument includes $r = 0$ , $s = 0$ , and dimension zero.

F1step 1.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local parameter is nonzerodivisor

Statement

In a positive-dimensional regular local ring, every member of a regular system of parameters is a nonzerodivisor.

Facts & Assumptions

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

[F1]

regular system of parameters equivalent basis: Let $(R, m, k)$ be a nonzero Noetherian local ring of dimension $d$ , and let $x = (x_{1}, \dots, x_{d}) \in m^{d}$ . Then $x$ is a regular system of parameters if and only if its classes form a $k$ -basis of $m / m^{2}$ . In particular every lift of a cotangent basis in a regular local ring generates $m$ and is a system of parameters.

[F2]

regular local domain induction: Every regular local ring is an integral domain.

Proof

1.1

The class of any member $x$ is a member of a cotangent basis and hence is nonzero. In particular $x \neq 0$ .

F1

2.1

The ring is a domain, so multiplication by this nonzero $x$ is injective. This proves the assertion for every member of the supplied tuple.

F2step 1.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local quotient by parameter is regular

Statement

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 $d - 1$ .

Facts & Assumptions

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

[F1]

regular system of parameters equivalent basis: Let $(R, m, k)$ be a nonzero Noetherian local ring of dimension $d$ , and let $x = (x_{1}, \dots, x_{d}) \in m^{d}$ . Then $x$ is a regular system of parameters if and only if its classes form a $k$ -basis of $m / m^{2}$ . In particular every lift of a cotangent basis in a regular local ring generates $m$ and is a system of parameters.

[F2]

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

[F3]

Local dimension is the minimal number of generators of an ideal with maximal radical: Let $(R, m)$ be a finite-dimensional Noetherian local ring of dimension $d < \infty$ . Then $d$ is the least integer $n$ for which there exists an $n$ -generated ideal $J \subseteq R$ with $\sqrt{J} = m$ .

Proof

1.1

Extend the nonzero class of $x$ to a basis of $m / m^{2}$ and lift it. The resulting $d$ elements generate $m$ ; their last $d - 1$ images generate the maximal ideal of $S = R / (x)$ . Hence $edim S \leq d - 1$ . The hypotheses force $d \geq 1$ and $S \neq 0$ .

F1given

2.1

Let $t = \dim S$ , which is finite by the embedding bound. Lift $t$ radical generators of the maximal ideal of $S$ . Together with $x$ they generate an ideal of $R$ with radical $m$ , so $d \leq t + 1$ . Combining $d - 1 \leq t \leq edim S \leq d - 1$ proves all the assertions, including the field quotient when $d = 1$ .

F3F2step 1.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

quotient and lifting regularity across a regular element

Statement

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 $x \notin m^{2}$ .

Facts & Assumptions

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

[F1]

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 $d - 1$ .

[F2]

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

[F3]

Local dimension is the minimal number of generators of an ideal with maximal radical: Let $(R, m)$ be a finite-dimensional Noetherian local ring of dimension $d < \infty$ . Then $d$ is the least integer $n$ for which there exists an $n$ -generated ideal $J \subseteq R$ with $\sqrt{J} = m$ .

[F4]

Zero divisors on a module over a Noetherian ring are the union of its associated primes: Let $R$ be a Noetherian commutative ring and let $M$ be a left $R$ -module. Then the set of zero divisors on $M$ is $⋃_{p \in {Ass}_{R} (M)} p .$ If $M$ is finitely generated, this is a finite union.

[F5]

regular local domain induction: Every regular local ring is an integral domain.

[F6]

Minimal support primes of a finite module are associated: Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$ -module. If $p$ is minimal in ${Supp}_{R} (M)$ , then $p \in {Ass}_{R} (M) .$

Proof

1.1

For a nonzerodivisor $x \in m$ , put $d = \dim R$ and $t = \dim R / (x)$ . Both dimensions are finite by the embedding bound. Lifting $t$ radical generators gives $d \leq t + 1$ . Every prime chain containing $x$ can be extended strictly downwards by a minimal prime of $R$ : minimal primes are associated, hence omit $x$ by the zero-divisor criterion. Thus $t + 1 \leq d$ , and $t = d - 1$ .

F2F3F4F6

2.1

If $R / (x)$ is regular, lift its $d - 1$ maximal-ideal generators and adjoin $x$ . This gives $edim R \leq d$ , and the embedding bound makes it equality. If $x$ were in $m^{2}$ , cotangent reduction would leave dimension unchanged, giving $edim (R / (x)) = d$ , contrary to $d - 1$ .

F2step 1.1algebra

3.1

In a regular local ring, a nonzero $x$ is a nonzerodivisor: the domain property is exactly F5. Thus the preceding implication applies. In the other direction, $x \notin m^{2}$ makes the quotient regular by the parameter-quotient lemma. There is no $0 \neq x \in m$ when the regular ring has dimension zero; in dimension one the regular quotient is a field.

F1step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local rings are domains and cohen macaulay

Statement

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 $0 \leq c \leq d$ .

Facts & Assumptions

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

[F1]

regular local domain induction: Every regular local ring is an integral domain.

[F2]

regular local parameter is nonzerodivisor: In a positive-dimensional regular local ring, every member of a regular system of parameters is a nonzerodivisor.

[F3]

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 $d - 1$ .

[F4]

Cohen--Macaulay local modules and rings: Let $(R, m)$ be a Noetherian local ring and let $M$ be a nonzero finite $R$ -module. The module $M$ is Cohen--Macaulay when $depth (M) = \dim {Supp}_{R} (M) .$ The zero module is excluded from this local definition. The local ring $R$ is Cohen--Macaulay when it is Cohen--Macaulay as an $R$ -module.

[F5]

Regular Sequence On A Module: Let $R$ be a commutative unital ring, let $M$ be an $R$ -module, and let $x = (x_{1}, \dots, x_{n})$ be a finite ordered sequence in $R$ . The sequence is $M$ -regular when $M / (x_{1}, \dots, x_{i - 1}) M \neq 0$ and multiplication by $x_{i}$ is injective on it for every $i$ , and $M / (x) M \neq 0$ .

[F6]

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.

[F7]

Depth with respect to an ideal: For a finite module $M$ with $I M \neq M$ , ${depth}_{I} (M)$ is the supremum of the lengths of $M$ -regular sequences in $I$ ; for a local ring depth means depth with respect to its maximal ideal.

Proof

1.1

The ring is a domain. Successively apply the parameter-quotient lemma: after $c$ quotients the remaining cotangent classes form a basis, and the quotient is regular of dimension $d - c$ . This starts with $c = 0$ and ends with $R / m = k \neq 0$ .

F1F3

2.1

At each nonterminal stage the next parameter is a nonzerodivisor. All the quotients are nonzero, so the tuple satisfies the definition of a regular sequence. Its length is $d$ , and the depth definition therefore gives $depth R \geq d$ ; the support-dimension bound gives $depth R \leq d$ . Thus the Cohen–Macaulay definition holds. For $d = 0$ , the empty tuple and the field $R$ give the same conclusion.

F2F4F5F6F7step 1.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

one dimensional regular local rings are dvrs

Statement

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.

Facts & Assumptions

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

[F1]

regular local domain induction: Every regular local ring is an integral domain.

[F2]

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 $m$ .

[F3]

Equivalent characterizations of a DVR: Let $R$ be a nonfield domain. The following are equivalent. 1. $R$ is a discrete valuation ring. 2. $R$ is a Noetherian valuation ring. 3. $R$ is a one-dimensional Noetherian local integrally closed domain. 4. $R$ is a local principal ideal domain with nonzero maximal ideal.

[F4]

The Krull intersection is the $(1 - a)$ -torsion submodule, and it vanishes in the Jacobson-radical case: The first clause below is choice-free; the second uses the published Jacobson-radical unit criterion and therefore inherits its Axiom-of-Choice boundary. Let $R$ be a Noetherian commutative ring, let $I \subset R$ be an ideal, and let $M$ be a finite $R$ -module. Put $K : = ⋂_{n \geq 0} I^{n} M .$ Then: 1. $K$ is exactly the set of elements $m \in M$ for which $(1 - a) m = 0$ for some $a \in I$ ; 2. if $I \subseteq J (R)$ , then $K = 0$ .

Proof

1.1

If $R$ is regular of dimension one, it is a domain and $m = (t)$ for a nonzero nonunit $t$ . Krull intersection gives for any $a \neq 0$ a largest $n$ with $a \in (t^{n})$ ; writing $a = t^{n} u$ , maximality makes $u$ a unit.

F1F2F4

2.1

In a nonzero ideal choose an element with least such exponent $n$ . Every other nonzero element has exponent at least $n$ , so the ideal is $(t^{n})$ . The zero ideal is principal as well. Thus $R$ is a local PID with nonzero maximal ideal, and the stated DVR equivalence applies.

F3step 1.1algebra

3.1

Conversely, a DVR is a nonfield local PID of dimension one. Its maximal ideal $(t)$ is nonzero, and $t \notin (t^{2})$ by cancellation in a domain. Therefore its embedding dimension is one and it is regular.

F3F2algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local regular quotient ideal is parameter generated

Statement

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 $\dim_{k} ((I + m^{2}) / m^{2}) = d - \dim (R / I)$ .

Facts & Assumptions

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

[F1]

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 $d - 1$ .

[F2]

regular local domain induction: Every regular local ring is an integral domain.

[F3]

regular system of parameters equivalent basis: Let $(R, m, k)$ be a nonzero Noetherian local ring of dimension $d$ , and let $x = (x_{1}, \dots, x_{d}) \in m^{d}$ . Then $x$ is a regular system of parameters if and only if its classes form a $k$ -basis of $m / m^{2}$ . In particular every lift of a cotangent basis in a regular local ring generates $m$ and is a system of parameters.

Proof

1.1

Put $c = \dim_{k} ((I + m^{2}) / m^{2})$ . The cotangent space of $R / I$ is $m / (I + m^{2})$ and has dimension $d - c$ . Consequently the numerical equality is precisely the definition of regularity of $R / I$ .

givenalgebra

2.1

If $R / I$ is regular, choose $x_{1}, \dots, x_{c} \in I$ lifting a basis of that subspace and extend their classes to a cotangent basis of $R$ . Put $S = R / (x_{1}, \dots, x_{c})$ . Repeated parameter reduction makes $S$ regular of dimension $d - c$ , and the extended tuple is a regular system.

F3F1step 1.1

3.1

The ring $S$ is a domain. If the kernel $J$ of $S \to R / I$ were nonzero, any prime chain in $S / J$ would lift to a chain of nonzero primes of $S$ , to which $(0)$ can be prepended. Hence $\dim (S / J) \leq \dim S - 1$ , contradicting equality of dimensions. Thus $I = (x_{1}, \dots, x_{c})$ . Conversely, repeated parameter reduction makes every such quotient regular. This includes $c = 0$ , when $I = 0$ , and $c = d$ , when $I = m$ and the quotient is $k$ .

F2F1step 2.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

finite local modules admit minimal free resolutions

Statement

Every finite module $M$ over a nonzero Noetherian local ring $(R, m, k)$ has an augmented resolution $\dots \to F_{1} \to F_{0} \to M \to 0$ by finite-rank free modules, with $d_{i} (F_{i}) \subseteq m F_{i - 1}$ for $i > 0$ . Such a resolution is called minimal; it need not be bounded. This extends the bounded terminology without changing it.

Facts & Assumptions

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

[F1]

Minimal Free Resolution Over A Local Ring: A finite free resolution $F_{∙} \to N$ over local $(R, m)$ is minimal when $d_{i} (F_{i}) \subseteq m F_{i - 1}$ for every $i > 0$ .

[F2]

Assuming the Axiom of Choice, minimal generators over a local ring are exactly residue-field bases: Assume the Axiom of Choice. Let $(R, m)$ be a local ring with residue field $k = R / m$ , and let $M$ be a finitely generated left $R$ -module. A finite generating set $x_{1}, \dots, x_{r}$ of $M$ is minimal if and only if the images of $x_{1}, \dots, x_{r}$ in $M / m M$ form a $k$ -basis. In particular every minimal generating set of $M$ has the same cardinality.

[F3]

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 $M = 0$ .

[F4]

Finite generation, ACC, and maximal-condition characterizations of Noetherian modules: For a left $R$ -module $M$ , the following are equivalent: every submodule is finitely generated; every ascending chain of submodules stabilizes; and every nonempty family of submodules has a maximal member. The implication from ACC to the maximal condition uses dependent choice; the other displayed implications are choice-free. See def-noetherian-module.

Proof

1.1

Choose a basis of $M / m M$ and lift it to $x_{1}, \dots, x_{r} \in M$ . If $N = \sum_{i} R x_{i}$ , then $M = N + m M$ , so the finite module $M / N$ satisfies $m (M / N) = M / N$ . Nakayama gives $M = N$ . Thus the corresponding map $F_{0} = R^{r} \to M$ is onto. A relation among the $x_{i}$ has all coefficients in $m$ , because their residue classes are independent. Hence $K_{0} = \ker (F_{0} \to M) \subseteq m F_{0}$ .

F2F3algebra

2.1

Every kernel is finite by Noetherianity. Repeating the same construction on $K_{0}$ and on each successive kernel produces an exact augmented complex whose differential images lie in the required maximal-ideal multiples. Dependent Choice suffices for the infinite recursive selections; the cited Nakayama results are used with their AC ledger. If a kernel is zero, choose zero modules thereafter; for $M = 0$ choose the zero complex. The bounded case agrees with the prior definition.

F4F3F1step 1.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

minimal free resolution differentials land in maximal ideal

Statement

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.

Facts & Assumptions

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

[F1]

finite local modules admit minimal free resolutions: Every finite module $M$ over a nonzero Noetherian local ring $(R, m, k)$ has an augmented resolution $\dots \to F_{1} \to F_{0} \to M \to 0$ by finite-rank free modules, with $d_{i} (F_{i}) \subseteq m F_{i - 1}$ for $i > 0$ . Such a resolution is called minimal; it need not be bounded. This extends the bounded terminology without changing it.

Proof

1.1

The condition that an image lie in $m F_{i - 1}$ is exactly that all matrix entries lie in $m$ , and is basis-independent. Since the complement of $m$ consists of units, failure supplies a unit entry. This is the minimality convention of the existence lemma.

F1given

2.1

Move that entry to the first position, scale it to $1$ , and clear its row and column by elementary basis changes. The matrix becomes $diag (1, D)$ . The identities $d_{i - 1} d_{i} = d_{i} d_{i + 1} = 0$ force adjacent maps to vanish on or into the isolated coordinates; for $i = 1$ the augmentation also vanishes there. Hence these coordinates form the direct summand $0 \to R \overset{1}{\to} R \to 0$ . Deleting it preserves exactness. Conversely an identity summand cannot have zero residue differential, while every matrix with entries in $m$ does.

step 1.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

minimal free resolution reduces to zero differential

Statement

If $F_{∙} \to M$ is a minimal degreewise finite free resolution over a nonzero Noetherian local ring $(R, m, k)$ , every differential of the unaugmented complex $k \otimes_{R} F_{∙}$ is zero.

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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.

[F2]

The balanced Tor bifunctor: For a right $R$ -module $N$ , a left $R$ -module $M$ , and $i \geq 0$ , define ${Tor}_{i}^{R} (N, M)$ to be either $H_{i} (N \otimes_{R} P_{∙})$ for a projective resolution of $M$ or $H_{i} (Q_{∙} \otimes_{R} M)$ for a projective resolution of $N$ , identified by the preceding natural balance isomorphism. On maps it uses the homology maps induced by comparison maps; coherence makes this a well-defined covariant bifunctor.

Proof

1.1

Every positive differential matrix has entries in $m$ . Tensoring with $k = R / m$ reduces those entries to zero.

F1

2.1

Thus the unaugmented residue complex has zero differential in every degree, including its map from degree zero to zero. Its homology in degree $i$ is $k \otimes_{R} F_{i}$ , and it computes ${Tor}_{i}^{R} (k, M)$ . The assertion includes the zero complex.

F2step 1.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

betti numbers of a finite local module

Definition

For a finite module $M$ over a nonzero Noetherian local ring $(R, m, k)$ and an integer $i \geq 0$ , its Betti number is $β_{i}^{R} (M) = \dim_{k} {Tor}_{i}^{R} (k, M)$ . The action factors through $k$ , and a degreewise finite free resolution makes this dimension finite. Tor is resolution-independent. This extends the Koszul rank notation: whenever a minimal Koszul resolution exists, the rank formula identifies these numbers with its Koszul Betti numbers. For $M = 0$ all Betti numbers are zero.

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

betti number is rank in minimal resolution

Statement

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 $i \geq 0$ .

Facts & Assumptions

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

[F1]

betti numbers of a finite local module: For a finite module $M$ over a nonzero Noetherian local ring $(R, m, k)$ and an integer $i \geq 0$ , its Betti number is $β_{i}^{R} (M) = \dim_{k} {Tor}_{i}^{R} (k, M)$ . The action factors through $k$ , and a degreewise finite free resolution makes this dimension finite. Tor is resolution-independent. This extends the Koszul rank notation: whenever a minimal Koszul resolution exists, the rank formula identifies these numbers with its Koszul Betti numbers. For $M = 0$ all Betti numbers are zero.

[F2]

minimal free resolution reduces to zero differential: If $F_{∙} \to M$ is a minimal degreewise finite free resolution over $(R, m, k)$ , every differential of the unaugmented complex $k \otimes_{R} F_{∙}$ is zero.

Proof

1.1

The residue complex has zero differentials and computes ${Tor}_{i}^{R} (k, M)$ , so this Tor group is $k \otimes_{R} F_{i}$ .

F2

2.1

Its vector-space dimension equals the finite free rank of $F_{i}$ . By definition this is $β_{i}^{R} (M)$ . This holds in degree zero, in all higher degrees, and for zero terms, including the zero module.

F1step 1.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

minimal free resolutions unique up to chain isomorphism

Statement

Any two minimal degreewise finite free resolutions of a finite module over a nonzero Noetherian local ring are augmentation-preservingly chain-isomorphic, in general noncanonically. In particular their ranks agree in every degree.

Facts & Assumptions

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

[F1]

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 $i \geq 0$ .

[F2]

Projective comparison maps exist: Assume the Axiom of Dependent Choice. Let $u : A \to B$ be a morphism, and let $P_{∙} \to A$ and $Q_{∙} \to B$ be projective resolutions. Then there exists an augmentation-preserving chain map $f_{∙} : P_{∙} \to Q_{∙}$ lifting $u$ .

[F3]

Projective comparison maps are unique up to chain homotopy: Assume the Axiom of Dependent Choice. Any two augmentation-preserving maps between projective resolutions lifting the same object morphism are chain-homotopic.

[F4]

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 $M = 0$ .

Proof

1.1

Choose comparison maps $f : F \to G$ and $g : G \to F$ lifting the identity on $M$ . Their composites are homotopic to the identities. The cited comparison assertions use DC and supplied resolution data.

F2F3

2.1

After reduction modulo $m$ , all differentials vanish, so the homotopy identities become ${\overset{ˉ}{g}}_{i} {\overset{ˉ}{f}}_{i} = 1$ and ${\overset{ˉ}{f}}_{i} {\overset{ˉ}{g}}_{i} = 1$ . The finite ranks agree, also by the Betti rank theorem. Nakayama makes each $f_{i}$ surjective; equivalently its square matrix has determinant nonzero modulo $m$ , hence unit. Its adjugate gives an inverse over $R$ . These inverses form a chain map since $f$ does. Rank zero causes no difficulty: the unique map between zero modules is invertible.

F1F4step 1.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedaudited 2026-09-07Open item page →

projective dimension from last nonzero betti number

Statement

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 ${Tor}_{q + 1}^{R} (k, M) = 0$ .

Facts & Assumptions

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

[F1]

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 $i \geq 0$ .

[F2]

Projective dimension of an object: Assume projective resolutions are supplied or exist in the relevant class. The projective dimension of $M$ is $pd (M) = \inf {d \geq 0 : M has a projective resolution of length d},$ with value $\infty$ if this set is empty. A length-zero projective resolution exists exactly when $M$ is projective.

[F3]

Projective dimension at most n iff the nth syzygy is projective: Let $A$ be an abelian category with enough projectives, fix a projective resolution $P_{∙} \to M$ , and let $n \geq 1$ . Then $pd (M) \leq n ⟺ Ω_{P}^{n} (M) is projective .$ In particular, the condition is independent of the chosen projective resolution.

[F4]

A finite flat module over a local ring is free: The standard theorem holds over arbitrary local rings; the proof written here is the Noetherian local case. Let $(R, m)$ be a Noetherian local ring and let $M$ be a finite flat $R$ -module. Then $M$ is free.

[F5]

Projective left and right modules are flat over an arbitrary ring: Every projective left or right module over an arbitrary ring is flat on its appropriate side.

[F6]

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 $M = 0$ .

[F7]

The balanced Tor bifunctor: For a right $R$ -module $N$ , a left $R$ -module $M$ , and $i \geq 0$ , define ${Tor}_{i}^{R} (N, M)$ to be either $H_{i} (N \otimes_{R} P_{∙})$ for a projective resolution of $M$ or $H_{i} (Q_{∙} \otimes_{R} M)$ for a projective resolution of $N$ , identified by the preceding natural balance isomorphism. On maps it uses the homology maps induced by comparison maps; coherence makes this a well-defined covariant bifunctor.

[F8]

minimal free resolution reduces to zero differential: Reducing a minimal degreewise finite free resolution modulo $m$ gives the zero differential, so ${Tor}_{i}^{R} (k, M) ≅ k \otimes_{R} F_{i}$ .

[F9]

finite local modules admit minimal free resolutions: Every finite module $M$ over a nonzero Noetherian local ring $(R, m, k)$ has an augmented resolution $\dots \to F_{1} \to F_{0} \to M \to 0$ by finite-rank free modules, with $d_{i} (F_{i}) \subseteq m F_{i - 1}$ for $i > 0$ . Such a resolution is called minimal; it need not be bounded. This extends the bounded terminology without changing it.

Proof

1.1

Choose a minimal degreewise finite free resolution $F_{∙} \to M$ by [F9]. By [F7], the zero differential in [F8] identifies ${Tor}_{q + 1}^{R} (k, M)$ with $F_{q + 1} / m F_{q + 1}$ . Its vanishing and Nakayama give $F_{q + 1} = 0$ . Exactness then gives $\ker (F_{q} \to F_{q - 1}) = 0$ (using the augmentation when $q = 0$ ), so the truncated complex is a length- $q$ free resolution. Also $F_{q + 2} = \ker d_{q + 2} = im d_{q + 3} \subseteq m F_{q + 2}$ , so Nakayama prevents a restart, and the same argument applies successively in every subsequent degree.

F2F6F7F8F9algebra

1.2

Conversely, if $pd M \leq q$ , a projective resolution of length at most $q$ computes Tor and gives zero in every degree above $q$ . The syzygy criterion also gives a finite free terminating resolution: for $q \geq 1$ its finite projective syzygy is flat and hence free; for $q = 0$ apply the same freeness result directly to $M$ .

F3F5F4F2F7

2.1

Since $M \neq 0$ , Nakayama gives $β_{0} (M) > 0$ . The two implications show that the last nonzero degree equals projective dimension when finite; if there is no finite bound, nonzero Betti degrees are unbounded and both sides are infinite.

F1F6step 1.1step 1.2∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

auslander buchsbaum syzygy projective dimension

Statement

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 $pd K = n - 1$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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 ${Tor}_{q + 1}^{R} (k, M) = 0$ .

Proof

1.1

The minimal resolution of $M$ has last nonzero term $F_{n}$ by the Betti criterion. Truncating it gives a minimal resolution $\dots \to F_{2} \to F_{1} \to K \to 0$ . If $K = 0$ , the initial presentation would make $M$ free, contrary to $n > 0$ .

F1

2.1

The truncated resolution has last nonzero term $F_{n}$ in degree $n - 1$ , so the same criterion gives $pd K = n - 1$ . For $n = 1$ this says that $K$ is a nonzero finite free module.

F1step 1.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

auslander buchsbaum base case free module

Statement

If a nonzero finite module $M$ over a nonzero Noetherian local ring $R$ has projective dimension zero, then it is finite free of positive rank and ${depth}_{R} M = depth R$ .

Facts & Assumptions

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

[F1]

Projective dimension of an object: Assume projective resolutions are supplied or exist in the relevant class. The projective dimension of $M$ is $pd (M) = \inf {d \geq 0 : M has a projective resolution of length d},$ with value $\infty$ if this set is empty. A length-zero projective resolution exists exactly when $M$ is projective.

[F2]

A finite flat module over a local ring is free: The standard theorem holds over arbitrary local rings; the proof written here is the Noetherian local case. Let $(R, m)$ be a Noetherian local ring and let $M$ be a finite flat $R$ -module. Then $M$ is free.

[F3]

Projective left and right modules are flat over an arbitrary ring: Every projective left or right module over an arbitrary ring is flat on its appropriate side.

[F4]

Depth as the first nonzero Ext degree: Let $R$ be Noetherian, let $M$ be finite, and let $I$ lie in the Jacobson radical. Then ${depth}_{I} (M) = \inf {i \geq 0 : {Ext}_{R}^{i} (R / I, M) \neq 0},$ where the infimum of the empty set is $\infty$ .

Proof

1.1

Projective dimension zero means projective. A projective module is flat, and the finite-flat theorem for Noetherian local rings makes $M$ finite free, say $R^{r}$ . Nonzeroness forces $r \geq 1$ .

F1F3F2

2.1

Ext into a finite direct sum is the finite direct sum of the corresponding Ext groups, as follows by applying Hom to a resolution. Thus ${Ext}_{R}^{i} (k, R^{r}) = {Ext}_{R}^{i} (k, R)^{r}$ has the same first nonzero degree as ${Ext}_{R}^{i} (k, R)$ . The Ext-depth criterion gives equality of depths, including depth zero.

F4step 1.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

minimal free matrix induces zero on residue ext

Statement

For a nonzero commutative Noetherian local ring $(R, m, k)$ , let $α : R^{s} \to R^{t}$ be a map between finite free modules all of whose matrix entries lie in $m$ . Then ${Ext}_{R}^{i} (k, α) = 0$ for every $i \geq 0$ .

Facts & Assumptions

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

[F1]

The balanced Ext bifunctor: Assume the Axiom of Dependent Choice. Let $A$ be an abelian category with enough projectives and enough injectives, and fix supplied projective and injective resolution data on all objects of $A$ . For each $n \geq 0$ , define ${Ext}_{A}^{n} (M, N)$ to mean either ${Ext}_{P}^{n} (M, N)$ or ${Ext}_{I}^{n} (M, N)$ , identified by the natural comparison isomorphism already proved. This notation is justified by the comparison theorem, its independence of comparison data, its two-variable naturality, and its change-of-resolution cocycle law; it is not a definition by equality of the two complexes.

[F2]

Module categories have enough injectives: Assume the Axiom of Choice. For every unital ring $R$ and every left $R$ -module $M$ , there is an injective left $R$ -module $I$ and a monomorphism $M \to I$ . Thus left $R$ -modules have enough injectives. For commutative $R$ , one explicit functorial target is $J (M) = {(R^{(M^{\lor})})}^{\lor} ≅ \prod_{ϕ \in M^{\lor}} R^{\lor},$ where $X^{\lor} = {Hom}_{Z} (X, Q / Z)$ ; the embedding is $M \to M^{\lor \lor} \to J (M)$ . Here $X^{\lor}$ is a left $R$ -module by $(r ϕ) (x) = ϕ (r x)$ .

Proof

1.1

Choose an injective resolution $I^{∙}$ of $R$ . Finite direct sums $(I^{∙})^{s}$ and $(I^{∙})^{t}$ are injective resolutions of the free modules. The same coefficient matrix defines a chain map between them extending $α$ . Enough injectives is used with AC, and the balanced Ext convention with its supplied data and DC.

F2F1

2.1

For $a \in m$ and $h : k \to I^{j}$ , $a h (z) = h (a z) = 0$ . Consequently that matrix induces the zero map on every term of ${Hom}_{R} (k, I^{∙})$ . It therefore induces zero on cohomology in every degree. This includes $i = 0$ and $s = 0$ or $t = 0$ .

F1step 1.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

auslander buchsbaum projective dimension one

Statement

If $M$ is a nonzero finite module of projective dimension one over a nonzero Noetherian local ring $R$ , then $depth R \geq 1$ and $depth M = depth R - 1$ .

Facts & Assumptions

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

[F1]

minimal free matrix induces zero on residue ext: For a nonzero commutative Noetherian local ring $(R, m, k)$ , let $α : R^{s} \to R^{t}$ be a map between finite free modules all of whose matrix entries lie in $m$ . Then ${Ext}_{R}^{i} (k, α) = 0$ for every $i \geq 0$ .

[F2]

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 ${Tor}_{q + 1}^{R} (k, M) = 0$ .

[F3]

Depth as the first nonzero Ext degree: Let $R$ be Noetherian, let $M$ be finite, and let $I$ lie in the Jacobson radical. Then ${depth}_{I} (M) = \inf {i \geq 0 : {Ext}_{R}^{i} (R / I, M) \neq 0},$ where the infimum of the empty set is $\infty$ .

[F4]

The long exact Ext sequence in the second variable: Assume the Axiom of Dependent Choice. Let $A$ be abelian with enough projectives and enough injectives, and fix supplied projective and injective resolution data on all its objects. For $0 \to N^{'} \to N \to N^{''} \to 0$ and every $M$ , there is a natural exact sequence $0 \to Hom (M, N^{'}) \to Hom (M, N) \to Hom (M, N^{''}) \overset{δ^{0}}{\to} {Ext}^{1} (M, N^{'}) \to {Ext}^{1} (M, N) \to \dots,$ where $δ^{q} : {Ext}^{q} (M, N^{''}) \to {Ext}^{q + 1} (M, N^{'})$ ; it is natural in the short exact sequence and contravariantly natural in $M$ .

[F5]

finite local modules admit minimal free resolutions: Every finite module over a nonzero Noetherian local ring has a degreewise finite minimal free resolution.

Proof

1.1

Choose the minimal resolution supplied by [F5]. Since ${pd}_{R} M = 1$ , [F2] says that its last nonzero term is $F_{1} ≅ R^{s}$ with $s > 0$ and $F_{i} = 0$ for $i \geq 2$ . Thus it gives a minimal exact sequence $0 \to R^{s} \overset{α}{\to} R^{t} \to M \to 0$ . Write $r = depth R$ . The map on every ${Ext}_{R}^{i} (k, -)$ induced by $α$ is zero. If $r = 0$ , the injection $Hom (k, R^{s}) \to Hom (k, R^{t})$ would be zero with nonzero source, impossible. Hence $r \geq 1$ .

F1F2F3F4F5

2.1

For $i < r - 1$ , the adjacent Ext terms for the free modules vanish, so ${Ext}_{R}^{i} (k, M) = 0$ . At $i = r - 1$ , exactness and the zero map in degree $r$ identify ${Ext}_{R}^{r - 1} (k, M)$ with ${Ext}_{R}^{r} (k, R^{s}) \neq 0$ . Thus the first nonzero Ext degree is $r - 1$ , proving the depth formula, also for $r = 1$ .

F4F3F1step 1.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

auslander buchsbaum first syzygy depth

Statement

In a minimal presentation $0 \to K \to F \to M \to 0$ of a nonzero finite module over a nonzero Noetherian local ring, let $n = pd M \geq 2$ be finite. If $depth K = depth R - (n - 1)$ , then $depth M = depth K - 1$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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]

The three Depth Lemma inequalities: Let $(R, m)$ be a Noetherian local ring and $0 \to A \to B \to C \to 0$ a short exact sequence of finite $R$ -modules. With $a = {depth}_{R} (A)$ , $b = {depth}_{R} (B)$ , and $c = {depth}_{R} (C)$ , $b \geq \min {a, c}, a \geq \min {b, c + 1}, c \geq \min {a - 1, b} .$ The last inequality is vacuous when $a = 0$ .

Proof

1.1

Write $a = depth K$ , $b = depth F = depth R$ , and $c = depth M$ . The equality for $F$ follows directly since an element is injective on a nonzero finite direct sum of $R$ exactly when it is injective on $R$ , also after successive quotients. The hypothesis gives $a < b$ . The syzygy is nonzero and finite of projective dimension $n - 1$ .

F1given

2.1

The depth inequality $a \geq \min (b, c + 1)$ forces $c + 1 \leq a$ , since $b > a$ . In particular $a \geq 1$ . The other inequality $c \geq \min (a - 1, b) = a - 1$ now applies and yields $c = a - 1$ . This uses only the stated depth hypothesis on $K$ , not the formula being proved by induction later.

F2step 1.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

auslander buchsbaum formula

Statement

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.

Facts & Assumptions

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

[F1]

auslander buchsbaum base case free module: If a nonzero finite module $M$ over a nonzero Noetherian local ring $R$ has projective dimension zero, then it is finite free of positive rank and ${depth}_{R} M = depth R$ .

[F2]

auslander buchsbaum projective dimension one: If $M$ is a nonzero finite module of projective dimension one over a nonzero Noetherian local ring $R$ , then $depth R \geq 1$ and $depth M = depth R - 1$ .

[F3]

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 $pd K = n - 1$ .

[F4]

auslander buchsbaum first syzygy depth: In a minimal presentation $0 \to K \to F \to M \to 0$ of a nonzero finite module over a nonzero Noetherian local ring, let $n = pd M \geq 2$ be finite. If $depth K = depth R - (n - 1)$ , then $depth M = depth K - 1$ .

Proof

1.1

Induct on $n = pd M$ . For $n = 0$ the module is nonzero finite free and has the ring depth. For $n = 1$ the separate minimal-matrix argument proves the formula.

F1F2

2.1

For $n \geq 2$ , take the first syzygy $K$ in a minimal presentation. It is nonzero finite with projective dimension $n - 1$ , so the inductive assertion gives $depth K = depth R - (n - 1)$ . The conditional syzygy-depth lemma then gives $depth M = depth R - n$ . This completes the induction.

F3F4step 1.1

3.1

If $depth M = depth R$ , the formula forces projective dimension zero, and the base-case theorem gives freeness. The nonzero and finite-projective-dimension hypotheses are retained throughout.

F1step 2.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

global dimension is detected on cyclic modules

Statement

For a unital ring $R$ , its left global dimension equals $\sup_{I} {pd}_{R} (R / I)$ over all left ideals $I$ , and equals the supremum of the injective dimensions of all left modules. The equalities allow infinity; in the commutative Noetherian case the cyclic modules are finite.

Facts & Assumptions

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

[F1]

Left and right global dimension of a ring: For a ring $R$ , define $l.gl.dim R = \sup {{pd}_{R} M : M is a left R -module},$ and $r.gl.dim R = \sup {{pd}_{R^{o p}} M : M is a right R -module} .$ They are separately defined extended natural numbers; their equality is not part of the notation.

[F2]

Baer's criterion for injective modules: Assume the Axiom of Choice. A left $R$ -module $I$ is injective if and only if every homomorphism $f : J \to I$ from a left ideal $J \leq R$ extends to a homomorphism $R \to I$ . The forward implication is choice-free. The converse uses AC through Zorn's lemma.

[F3]

Injective dimension at most n iff higher Ext vanishes: Assume enough injectives. For an object $N$ and $n \geq 0$ , $id (N) \leq n$ if and only if ${Ext}^{k} (M, N) = 0$ for every object $M$ and every $k > n$ .

[F4]

Projective dimension at most n iff higher Ext vanishes: Assume the Axiom of Dependent Choice. In an abelian category with enough projectives and enough injectives, fix supplied projective and injective resolution data on all objects. Let $M$ be an object and $n \geq 0$ . The following are equivalent: 1. $pd (M) \leq n$ ; 2. ${Ext}^{k} (M, N) = 0$ for every object $N$ and every $k > n$ ; 3. ${Ext}^{n + 1} (M, N) = 0$ for every object $N$ .

[F5]

Ext dimension shifting in the second variable: Assume the Axiom of Dependent Choice. Let $A$ be abelian with enough projectives and enough injectives, and fix supplied projective and injective resolution data on all its objects. If $0 \to N \to I \to Σ N \to 0$ is an injective copresentation, then for $q \geq 1$ there are natural isomorphisms ${Ext}^{q + 1} (M, N) ≅ {Ext}^{q} (M, Σ N)$ ; its low-degree part is $0 \to Hom (M, N) \to Hom (M, I) \to Hom (M, Σ N) \to {Ext}^{1} (M, N) \to 0$ .

[F6]

Module categories have enough injectives: Assume the Axiom of Choice. For every unital ring $R$ and every left $R$ -module $M$ , there is an injective left $R$ -module $I$ and a monomorphism $M \to I$ . Thus left $R$ -modules have enough injectives. For commutative $R$ , one explicit functorial target is $J (M) = {(R^{(M^{\lor})})}^{\lor} ≅ \prod_{ϕ \in M^{\lor}} R^{\lor},$ where $X^{\lor} = {Hom}_{Z} (X, Q / Z)$ ; the embedding is $M \to M^{\lor \lor} \to J (M)$ . Here $X^{\lor}$ is a left $R$ -module by $(r ϕ) (x) = ϕ (r x)$ .

Proof

1.1

Fix $n \geq 0$ and suppose every $R / I$ has projective dimension at most $n$ . For any left module $N$ , choose an injective resolution and let $C$ be its $n$ th cosyzygy, with $C = N$ when $n = 0$ . Dimension shifting gives ${Ext}^{1} (R / I, C) = {Ext}^{n + 1} (R / I, N) = 0$ . The last vanishing follows from the projective-dimension Ext criterion.

F6F5F4

2.1

To apply Baer, any map $I \to C$ extends to $R$ : its pushout with $I ↪ R$ yields an extension of $R / I$ by $C$ , whose Ext class is zero and hence splits. Therefore $C$ is injective by Baer. The truncated injective resolution gives $id N \leq n$ , so all ${Ext}^{j} (M, N)$ vanish for $j > n$ and arbitrary $M, N$ .

F2F3step 1.1

3.1

The projective-dimension criterion now gives $pd M \leq n$ for every module $M$ . Conversely such a global bound applies to all cyclic modules and forces every injective dimension at most $n$ by the same Ext criterion. Thus all three bounds are equivalent for each finite $n$ , proving equality of their extended suprema. This includes the zero ring, whose only module has dimension zero under the adopted resolution convention.

F4F3F1step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

local global dimension equals residue field projective dimension

Statement

For a nonzero Noetherian local ring $(R, m, k)$ , $gldim R = {pd}_{R} k$ , allowing infinity. If this common value is $n < \infty$ , every $R$ -module has projective dimension at most $n$ .

Facts & Assumptions

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

[F1]

global dimension is detected on cyclic modules: For a unital ring $R$ , its left global dimension equals $\sup_{I} {pd}_{R} (R / I)$ over all left ideals $I$ , and equals the supremum of the injective dimensions of all left modules. The equalities allow infinity; in the commutative Noetherian case the cyclic modules are finite.

[F2]

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 ${Tor}_{q + 1}^{R} (k, M) = 0$ .

[F3]

Tor is symmetric over a commutative ring: If $R$ is commutative and $M, N$ are $R$ -modules, then ${Tor}_{i}^{R} (M, N) ≅ {Tor}_{i}^{R} (N, M)$ naturally.

Proof

1.1

The lower bound is immediate because $k$ is an $R$ -module. If its projective dimension is infinite this already proves the equality. Otherwise let $n = pd k$ . Compute Tor using a length- $n$ resolution of $k$ and use symmetry to get ${Tor}_{n + 1}^{R} (k, M) = 0$ for every module $M$ .

F3given

2.1

For each nonzero finite $M$ , the minimal-resolution criterion gives $pd M \leq n$ ; the zero module is projective as well. In particular every cyclic module has that bound. Cyclic detection extends it to all modules and hence bounds global dimension by $n$ . This includes $n = 0$ .

F2F1step 1.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

positive depth ring has regular minimal generator

Statement

If a nonzero Noetherian local ring $(R, m, k)$ has positive depth, then some $x \in m ∖ m^{2}$ is a nonzerodivisor. The residue field need not be infinite.

Facts & Assumptions

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

[F1]

The local depth-zero associated-prime criterion: Let $(R, m)$ be a Noetherian local ring and let $M \neq 0$ be a finite $R$ -module. Then $depth (M) = 0 ⟺ m \in {Ass}_{R} (M) .$

[F2]

Finite modules over Noetherian rings have finitely many associated primes: Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$ -module. Then ${Ass}_{R} (M)$ is a finite set.

[F3]

Zero divisors on a module over a Noetherian ring are the union of its associated primes: Let $R$ be a Noetherian commutative ring and let $M$ be a left $R$ -module. Then the set of zero divisors on $M$ is $⋃_{p \in {Ass}_{R} (M)} p .$ If $M$ is finitely generated, this is a finite union.

[F4]

An ideal contained in a finite union of prime ideals lies in one of them: Let $R$ be a commutative ring, let $I ⊴ R$ be an ideal, and let $p_{1}, \dots, p_{n}$ be prime ideals with $n \geq 1$ . If $I \subseteq p_{1} \cup \dots \cup p_{n},$ then $I \subseteq p_{i}$ for some $i$ .

[F5]

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 $M = 0$ .

Proof

1.1

The associated primes are finite, none is $m$ , and their union is the set of zero divisors. Discard primes contained in others to obtain an antichain $p_{1}, \dots, p_{s}$ . Prime avoidance chooses $a \in m$ outside their union (if the list is empty this restriction is vacuous). If $a \notin m^{2}$ , take $x = a$ .

F1F2F3F4

2.1

If $a \in m^{2}$ , Nakayama and positive depth give $m \neq m^{2}$ ; choose $b \in m ∖ m^{2}$ . If $b$ avoids every retained prime take $x = b$ . Otherwise divide them into the nonempty class $T$ containing $b$ and the class $U$ not containing it. For each $p \in U$ , antichain incomparability and prime avoidance give $c_{p} \in p$ outside all primes of $T$ . Put $c = \prod_{p \in U} c_{p}$ , with empty product $1$ .

F5F4step 1.1

3.1

Then $x = b + a c$ is outside $m^{2}$ , since $a c \in m^{2}$ . At a prime of $T$ , $b$ lies in the prime and $a c$ does not. At a prime of $U$ , $a c$ lies in the prime and $b$ does not. Thus $x$ avoids every associated prime and is a nonzerodivisor. No infinite-field argument was used.

F3step 2.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular element reduction preserves minimal resolution

Statement

Let $(R, m, k)$ be nonzero Noetherian local, let $M$ be a nonzero finite module, and let $x \in m$ be a nonzerodivisor on both $R$ and $M$ . Reducing a minimal free resolution of $M$ modulo $x$ gives a minimal free resolution of $M / x M$ over $S = R / (x)$ . Moreover ${pd}_{S} (M / x M) = {pd}_{R} M$ , including infinity. For $M = 0$ the zero-complex assertion also holds, with both projective dimensions zero.

Facts & Assumptions

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

[F1]

finite local modules admit minimal free resolutions: Every finite module $M$ over a nonzero Noetherian local ring $(R, m, k)$ has an augmented resolution $\dots \to F_{1} \to F_{0} \to M \to 0$ by finite-rank free modules, with $d_{i} (F_{i}) \subseteq m F_{i - 1}$ for $i > 0$ . Such a resolution is called minimal; it need not be bounded. This extends the bounded terminology without changing it.

[F2]

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 ${Tor}_{q + 1}^{R} (k, M) = 0$ .

[F3]

The balanced Tor bifunctor: For a right $R$ -module $N$ , a left $R$ -module $M$ , and $i \geq 0$ , define ${Tor}_{i}^{R} (N, M)$ to be either $H_{i} (N \otimes_{R} P_{∙})$ for a projective resolution of $M$ or $H_{i} (Q_{∙} \otimes_{R} M)$ for a projective resolution of $N$ , identified by the preceding natural balance isomorphism. On maps it uses the homology maps induced by comparison maps; coherence makes this a well-defined covariant bifunctor.

[F4]

Tor is symmetric over a commutative ring: If $R$ is commutative and $M, N$ are $R$ -modules, then ${Tor}_{i}^{R} (M, N) ≅ {Tor}_{i}^{R} (N, M)$ naturally.

[F5]

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 $M = 0$ .

Proof

1.1

The complex $0 \to R \overset{x}{\to} R \to S \to 0$ resolves $S$ . Tensoring it with $M$ has no positive homology since multiplication by $x$ is injective on $M$ . Balance and symmetry of Tor show that reducing any free resolution of $M$ modulo $x$ has no positive homology and degree-zero homology $M / x M$ .

F3F4

2.1

A minimal degreewise finite resolution exists, and its matrices reduce to entries in $m / (x)$ . Its finite ranks do not change on reduction to the nonzero local ring $S$ . Nakayama gives $M / x M \neq 0$ , so the last-nonzero-Betti criterion identifies both projective dimensions with the same last nonzero rank, or infinity if ranks persist arbitrarily far. For $M = 0$ choose the zero complex on both sides.

F1F2F5step 1.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

residue field splits off reduced maximal ideal

Statement

Let $(R, m, k)$ be nonzero Noetherian local and $x \in m ∖ m^{2}$ a nonzerodivisor. Over $S = R / (x)$ the sequence $0 \to (x) / (x m) \to m / x m \to m / (x) \to 0$ splits, and $(x) / (x m) ≅ k$ . Consequently finite ${pd}_{R} k$ implies finite ${pd}_{S} k$ .

Facts & Assumptions

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

[F1]

regular element reduction preserves minimal resolution: Let $(R, m, k)$ be nonzero Noetherian local, let $M$ be a nonzero finite module, and let $x \in m$ be a nonzerodivisor on both $R$ and $M$ . Reducing a minimal free resolution of $M$ modulo $x$ gives a minimal free resolution of $M / x M$ over $S = R / (x)$ . Moreover ${pd}_{S} (M / x M) = {pd}_{R} M$ , including infinity. For $M = 0$ the zero-complex assertion also holds, with both projective dimensions zero.

[F2]

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 $pd K = n - 1$ .

[F3]

Projective dimension at most n iff higher Ext vanishes: Assume the Axiom of Dependent Choice. In an abelian category with enough projectives and enough injectives, fix supplied projective and injective resolution data on all objects. Let $M$ be an object and $n \geq 0$ . The following are equivalent: 1. $pd (M) \leq n$ ; 2. ${Ext}^{k} (M, N) = 0$ for every object $N$ and every $k > n$ ; 3. ${Ext}^{n + 1} (M, N) = 0$ for every object $N$ .

Proof

1.1

The sequence is the quotient sequence for $x m \subseteq (x) \subseteq m$ ; $x$ kills every term. Multiplication by $x$ identifies $R / m$ with $(x) / (x m)$ because cancellation is valid. Choose a $k$ -linear functional on $m / m^{2}$ taking the class of $x$ to $1$ . Composing with $m / x m \to m / m^{2}$ gives an $S$ -linear retraction onto $k$ . Thus the sequence splits.

givenalgebra

2.1

If ${pd}_{R} k$ is finite, it is positive: projectivity of $k$ would split $R \to k$ , giving a nontrivial idempotent unless $m = 0$ , impossible here. Its first minimal syzygy $m$ therefore has finite projective dimension. The element $x$ acts injectively on this ideal, so reduction gives finite ${pd}_{S} (m / x m)$ . Ext is additive on a finite direct sum, and the Ext criterion shows that its summand $k$ has finite projective dimension.

F2F1F3step 1.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

finite residue field projective dimension forces depth equals dimension

Statement

If the residue field of a nonzero Noetherian local ring $R$ has finite projective dimension, then $R$ is regular and $depth R = \dim R = edim R$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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]

positive depth ring has regular minimal generator: If a nonzero Noetherian local ring $(R, m, k)$ has positive depth, then some $x \in m ∖ m^{2}$ is a nonzerodivisor. The residue field need not be infinite.

[F3]

residue field splits off reduced maximal ideal: Let $(R, m, k)$ be nonzero Noetherian local and $x \in m ∖ m^{2}$ a nonzerodivisor. Over $S = R / (x)$ the sequence $0 \to (x) / (x m) \to m / x m \to m / (x) \to 0$ splits, and $(x) / (x m) ≅ k$ . Consequently finite ${pd}_{R} k$ implies finite ${pd}_{S} k$ .

[F4]

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 $x \notin m^{2}$ .

[F5]

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 $0 \leq c \leq d$ .

[F6]

Depth drops by one after quotienting by a regular element: Let $R$ be Noetherian, let $M$ be finite, let $I$ lie in the Jacobson radical, and let $x \in I$ be $M$ -regular. Then ${depth}_{I} (M / x M) = {depth}_{I} (M) - 1.$

Proof

1.1

Induct on the finite integer $r = depth R$ . The residue field has depth zero, since every member of $m$ kills it. If $r = 0$ , Auslander–Buchsbaum gives $pd k = 0$ and its freeness consequence makes $k$ nonzero free. A nonzero free module has zero annihilator, so $m = 0$ and $R = k$ is a field.

F1

2.1

For $r > 0$ choose a nonzerodivisor $x \in m ∖ m^{2}$ . The splitting lemma makes ${pd}_{R / (x)} k$ finite, and the regular-element depth formula gives depth $r - 1$ for the quotient. Depth of this annihilated module over $R$ equals its depth over $R / (x)$ : lift sequences from the quotient or project sequences from $R$ ; multiplication and all successive quotients are identical.

F2F3F6step 1.1

3.1

The inductive assertion makes $R / (x)$ regular. Lifting across the nonzerodivisor makes $R$ regular; its regular parameters make it Cohen–Macaulay, so depth equals dimension, and regularity equates that dimension with embedding dimension. This completes the induction.

F4F5step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local residue field koszul resolution

Statement

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 $d$ .

Facts & Assumptions

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

[F1]

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 $0 \leq c \leq d$ .

[F2]

Koszul Complex Resolves A Regular Quotient: If $M$ is finite free and $x$ is $M$ -regular, then $K (x; M)$ is a finite free resolution of $M / (x) M$ .

[F3]

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.

Proof

1.1

The parameters form an $R$ -regular sequence and generate $m$ . Koszul acyclicity for a finite free coefficient module gives a free resolution of $R / m = k$ .

F1F2

2.1

Every differential entry is a parameter up to sign and hence lies in $m$ , so the resolution is minimal. Its degree- $i$ module is $⋀^{i} R^{d}$ , zero for $i > d$ and rank one in degree $d$ . When $d = 0$ , it is just $R = k$ in degree zero, the Koszul complex on the empty tuple.

F3step 1.1algebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local residue field projective dimension dimension

Statement

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 $i > d$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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]

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 ${Tor}_{q + 1}^{R} (k, M) = 0$ .

[F3]

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 $i \geq 0$ .

[F4]

Complete Intersection Betti Numbers Binomial: For a length- $n$ regular sequence in the maximal ideal of a local ring, the minimal Koszul resolution has $β_{i}^{K} = (\binom{n}{i})$ for $0 \leq i \leq n$ and $0$ otherwise.

Proof

1.1

The minimal Koszul resolution has degree- $i$ rank $(\binom{d}{i})$ , and is zero above $d$ . The Koszul rank formula and the general minimal-resolution rank formula identify these with the stated Betti numbers.

F1F4F3

2.1

The top rank $(\binom{d}{d}) = 1$ is nonzero, so the projective-dimension criterion gives exactly $d$ , not merely an upper bound. For $d = 0$ the sole rank is $β_{0} (k) = 1$ .

F2step 1.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

auslander buchsbaum serre regularity criterion

Statement

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.

Facts & Assumptions

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

[F1]

finite residue field projective dimension forces depth equals dimension: If the residue field of a nonzero Noetherian local ring $R$ has finite projective dimension, then $R$ is regular and $depth R = \dim R = edim R$ .

[F2]

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 $i > d$ .

[F3]

local global dimension equals residue field projective dimension: For a nonzero Noetherian local ring $(R, m, k)$ , $gldim R = {pd}_{R} k$ , allowing infinity. If this common value is $n < \infty$ , every $R$ -module has projective dimension at most $n$ .

[F4]

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.

[F5]

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 $0 \leq c \leq d$ .

[F6]

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 $M = 0$ .

Proof

1.1

Regularity gives $pd k = \dim R$ through the Koszul computation. Residue-field projective dimension equals global dimension, so this also bounds every module. Conversely finite projective dimension for every finite module applies to $k$ , and finite projective dimension for $k$ forces regularity. These implications prove the four-way equivalence and the numerical equalities, also for dimension zero.

F2F3F1

2.1

Over a regular local ring every finite module has finite projective dimension and $depth R = \dim R$ . Auslander–Buchsbaum therefore makes depth $\dim R$ equivalent to projective dimension zero for a nonzero finite module, hence equivalent to freeness. Conversely a nonzero finite free module has the ring depth.

F4F5step 1.1

3.1

The useful freeness-lifting argument can also be seen directly. If $x \in m$ is injective on a finite $M$ and $M / x M$ is free over $R / (x)$ , lift a basis to a surjection $F \to M$ by Nakayama, with finite kernel $K$ . A relation has coefficients divisible by $x$ , so it is $x v$ ; injectivity on $M$ implies $v \in K$ . Thus $K = x K$ , and Nakayama gives $K = 0$ . This includes a zero quotient basis, when $M = 0$ .

F6algebra∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

localisations of regular local rings are regular

Statement

Every prime localization $R_{p}$ of a regular local ring $R$ is regular, and $edim R_{p} = ht p$ .

Facts & Assumptions

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

[F1]

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.

[F2]

Localisation of modules is exact: If $0 ⟶ M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''} ⟶ 0$ is a short exact sequence of $R$ -modules, then $0 ⟶ S^{- 1} M^{'} \overset{S^{- 1} f}{\to} S^{- 1} M \overset{S^{- 1} g}{\to} S^{- 1} M^{''} ⟶ 0$ is a short exact sequence of $S^{- 1} R$ -modules.

[F3]

Height equals local dimension: Let $R$ be a commutative ring and let $p \in Spec (R)$ . Then $ht (p) = \sup {n \geq 0 : p_{0} ⊊ \dots ⊊ p_{n} = p is a strict chain of prime ideals in R}$ . The supremum is allowed to be infinite.

Proof

1.1

The finite module $R / p$ has a finite projective resolution by the homological regularity criterion. Localizing preserves exactness; projective modules remain projective because their splittings as summands of free modules localize. The resulting resolution resolves $(R / p)_{p} = k (p)$ .

F1F2

2.1

The residue field of $R_{p}$ thus has finite projective dimension, and the same criterion makes this local ring regular. Prime chains in the localization correspond exactly to prime chains below $p$ , so its dimension is $ht p$ ; regularity gives its embedding dimension. At height zero the localization is a field.

F1F3step 1.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular noetherian ring

Definition

A commutative Noetherian ring $R$ is regular if for every prime ideal $p$ , the local ring $R_{p}$ is regular local. This includes the zero ring vacuously. The maximal-localization test is proved in the localization and polynomial-extension theorem.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

flat local ascent of regularity

Statement

For a flat local map $(R, m) \to (S, n)$ of nonzero Noetherian local rings: if $R$ and $S / m S$ are regular, then $S$ is regular. Conversely, regularity of $S$ implies regularity of $R$ .

Facts & Assumptions

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

[F1]

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 $0 \leq c \leq d$ .

[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 $x \notin m^{2}$ .

[F3]

Flat and faithfully flat modules and ring homomorphisms: Let $R$ be a commutative ring and let $M$ be an $R$ -module. The module $M$ is flat if the functor $- \otimes_{R} M$ preserves exact sequences: whenever $A \to B \to C$ is exact, so is $A \otimes_{R} M ⟶ B \otimes_{R} M ⟶ C \otimes_{R} M .$ Since tensoring is always right exact (thm-right-exactness-of-tensor-products, def-exact-and-short-exact-sequences-of-modules), the definition asks for the remaining left-hand exactness. Its equivalent formulation as preservation of injections is proved separately rather than built into the definition. The module $M$ is faithfully flat if a sequence of $R$ -modules is exact exactly when its tensor with $M$ is exact. For a unital ring homomorphism $f : R \to S$ (def-ring-homomorphism) between commutative rings, $S$ is an $R$ -module by $r \cdot s = f (r) s$ . The map $f$ is flat, respectively faithfully flat, when this $R$ -module is flat, respectively faithfully flat.

[F4]

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.

[F5]

finite local modules admit minimal free resolutions: Every finite module $M$ over a nonzero Noetherian local ring $(R, m, k)$ has an augmented resolution $\dots \to F_{1} \to F_{0} \to M \to 0$ by finite-rank free modules, with $d_{i} (F_{i}) \subseteq m F_{i - 1}$ for $i > 0$ . Such a resolution is called minimal; it need not be bounded. This extends the bounded terminology without changing it.

[F6]

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 ${Tor}_{q + 1}^{R} (k, M) = 0$ .

Proof

1.1

Suppose the base and closed fibre are regular. A regular system $x_{1}, \dots, x_{d}$ of $R$ is a regular sequence generating $m$ . Tensor the successive injective multiplication maps on $R / (x_{1}, \dots, x_{i - 1})$ with the flat module $S$ . This gives injective multiplication by each image $x_{i}$ on the corresponding quotient of $S$ . These quotients are nonzero because their defining ideals lie in $n$ .

F1F3

2.1

The terminal quotient is the regular closed fibre. Repeatedly lift regularity across those nonzerodivisors to get regularity of $S$ . If $d = 0$ , the fibre is $S$ and the implication is immediate.

F2step 1.1

3.1

For descent, choose a degreewise finite minimal resolution of $k_{R}$ and tensor it with $S$ . Flatness preserves its exactness, locality puts all differential entries in $n$ , and $S / m S$ is a nonzero finite $S$ -module. If $S$ is regular, its finite global dimension forces this minimal resolution to terminate by the Betti criterion. A term $S^{r}$ is zero only if $r = 0$ , so the original resolution over $R$ terminates as well. Finite ${pd}_{R} k_{R}$ gives regularity of $R$ . This argument also covers global dimension zero.

F5F3F4F6∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

polynomial local regularity fibre step

Statement

For a prime $q \subset R [t]$ with contraction $p \subset R$ , the closed fibre of $R_{p} \to R [t]_{q}$ is $k (p) [t]$ localized at a prime. That prime is either zero, giving a field, or generated by an irreducible polynomial, giving a DVR. In both cases the fibre is regular.

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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]

Every Euclidean domain is a principal ideal domain: Every Euclidean domain is a principal ideal domain.

[F3]

For every field $F$ , $F [x]$ is a Euclidean domain with degree as Euclidean function: For every field $F$ , the ring $F [x]$ is a Euclidean domain with Euclidean function $δ (f) = \deg f$ on nonzero polynomials.

Proof

1.1

Localize first at $R ∖ p$ , quotient by $p R_{p}$ , and then localize at the image of $q$ . Fractions and the quotient relation identify the fibre with $k (p) [t]_{\overset{ˉ}{q}}$ . Over this field the polynomial ring is Euclidean and hence a PID.

F3F2algebra

2.1

In a PID every nonzero prime is generated by an irreducible $f$ and is maximal. Localization at it is a nonfield local PID; every nonzero element is a unit times $f^{n}$ , so the exponent gives a discrete valuation and the localization is a DVR. The zero-prime localization is the rational function field. DVRs are regular by the one-dimensional theorem, and a field has zero maximal ideal and dimension zero, hence is regular.

F1step 1.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

localisation and polynomial extension of regular rings

Statement

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.

Facts & Assumptions

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

[F1]

localisations of regular local rings are regular: Every prime localization $R_{p}$ of a regular local ring $R$ is regular, and $edim R_{p} = ht p$ .

[F2]

regular noetherian ring: A commutative Noetherian ring $R$ is regular if for every prime ideal $p$ , the local ring $R_{p}$ is regular local. This includes the zero ring vacuously. The maximal-localization test is proved in the localization and polynomial-extension theorem.

[F3]

flat local ascent of regularity: For a flat local map $(R, m) \to (S, n)$ of nonzero Noetherian local rings: if $R$ and $S / m S$ are regular, then $S$ is regular. Conversely, regularity of $S$ implies regularity of $R$ ; it need not imply regularity of the closed fibre.

[F4]

polynomial local regularity fibre step: For a prime $q \subset R [t]$ with contraction $p \subset R$ , the closed fibre of $R_{p} \to R [t]_{q}$ is $k (p) [t]$ localized at a prime. That prime is either zero, giving a field, or generated by an irreducible polynomial, giving a DVR. In both cases the fibre is regular.

[F5]

If $R$ is Noetherian then $R [x_{1}, \dots, x_{n}]$ is Noetherian for every $n \in N$ : Let $R$ be a Noetherian commutative ring. Then the iterated polynomial ring $R [x_{1}, \dots, x_{n}]$ of def-multivariate-polynomial-ring-by-iteration is Noetherian for every $n \in N$ . The index starts at $0$ , where the published definition sets $R [x_{1}, \dots, x_{0}] = R$ and the assertion is the hypothesis itself.

[F6]

Localisation of modules is exact: If $0 ⟶ M^{'} \overset{f}{\to} M \overset{g}{\to} M^{''} ⟶ 0$ is a short exact sequence of $R$ -modules, then $0 ⟶ S^{- 1} M^{'} \overset{S^{- 1} f}{\to} S^{- 1} M \overset{S^{- 1} g}{\to} S^{- 1} M^{''} ⟶ 0$ is a short exact sequence of $S^{- 1} R$ -modules.

[F7]

A finite flat module over a Noetherian ring is finite projective: Let $R$ be a Noetherian commutative ring and let $M$ be a finite flat $R$ -module. Then $M$ is finite projective.

[F8]

A module is flat if and only if all prime localizations are flat, equivalently all maximal localizations are flat: Let $R$ be a commutative ring and let $M$ be an $R$ -module. The following are equivalent: 1. $M$ is flat over $R$ . 2. $M_{p}$ is flat over $R_{p}$ for every prime ideal $p \subset R$ . 3. $M_{m}$ is flat over $R_{m}$ for every maximal ideal $m \subset R$ .

[F9]

Projective dimension at most n iff the nth syzygy is projective: Let $A$ be an abelian category with enough projectives, fix a projective resolution $P_{∙} \to M$ , and let $n \geq 1$ . Then $pd (M) \leq n ⟺ Ω_{P}^{n} (M) is projective .$ In particular, the condition is independent of the chosen projective resolution.

[F10]

global dimension is detected on cyclic modules: For a unital ring $R$ , its left global dimension equals $\sup_{I} {pd}_{R} (R / I)$ over all left ideals $I$ , and equals the supremum of the injective dimensions of all left modules. The equalities allow infinity; in the commutative Noetherian case the cyclic modules are finite.

[F11]

Localizing a Dedekind domain at a nonzero prime gives a DVR: Let $R$ be a Dedekind domain and let $p \subset R$ be a nonzero prime ideal. Then $R_{p}$ is a discrete valuation ring.

[F12]

Projective left and right modules are flat over an arbitrary ring: Every projective left or right module over an arbitrary ring is flat on its appropriate side.

[F13]

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.

Proof

1.1

If maximal localizations are regular, choose a maximal ideal above any prime and use transitivity of localization and regular-local localization to get regularity at that prime. The reverse implication follows by selecting the maximal primes. Localizing a regular ring again has only such prime-local rings, so is regular; the zero ring and a localization that becomes zero satisfy this vacuously.

F1F2

1.2

For a finite module $M$ over any Noetherian $R$ , localization of a projective resolution gives ${pd}_{R_{p}} M_{p} \leq {pd}_{R} M$ . Conversely suppose every local dimension is at most a fixed $n < \infty$ . If $n \geq 1$ , form a partial finite free resolution of length $n$ by successively taking finite generators of finite kernels. Its $n$ th syzygy is projective at every prime by the syzygy criterion. It is therefore flat locally, hence globally, and finite flat implies projective. The syzygy criterion gives ${pd}_{R} M \leq n$ . If $n = 0$ , apply the local-flat and finite-projective argument to $M$ itself. Thus the supremum formula holds, including infinity and $M = 0$ .

F6F9F12F8F7

2.1

The module $R [t]$ is free over $R$ on the monomials, hence flat. Tensoring followed by localization is exact, so at a prime $q$ over $p$ the map $R_{p} \to R [t]_{q}$ is flat and local. The base is regular and its closed fibre is regular by the fibre computation; flat-local ascent gives regularity of the target. Polynomial Noetherianity and finite iteration prove the assertion for any finite number of variables, including zero.

F12F6F4F3F5step 1.1

2.2

For regular nonzero $R$ , local homological regularity gives ${pd}_{R_{p}} k (p) = ht p$ . Applying the preceding lower bound to $R / p$ gives global dimension at least every height, hence at least $\dim R$ . If $d = \dim R$ is finite, all localized finite modules have projective dimension at most $d$ ; the preceding upper bound and cyclic detection give global dimension at most $d$ . If $d = \infty$ , the lower bounds already give equality.

F13F10step 1.2

3.1

A Dedekind domain has DVR localizations at its nonzero primes and its fraction field at the zero prime. These are regular, so the domain and its finite polynomial extensions are regular by the preceding results. The equality involving Krull dimension was stated only for nonzero rings, avoiding an undefined dimension for the empty spectrum.

F11step 1.1step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

completion preserves embedding dimension

Statement

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.

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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]

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.

[F3]

Completion commutes with finite quotients and induced submodules: Assume the Axiom of Choice. Let $R$ be a Noetherian commutative ring, let $I \subseteq R$ be an ideal, and let $N \subseteq M$ be finitely generated $R$ -modules. 1. The natural map $\hat{M} / \hat{N} ⟶ \hat{M / N}$ is an isomorphism. 2. Under the natural map $\hat{N} \to \hat{M}$ , the image of $\hat{N}$ is the $\hat{R}$ -submodule $N \hat{R} \subseteq \hat{M}$ . In particular, for every ideal $J \subseteq R$ , $\hat{J M} ≅ J \hat{M} .$ 3. For every $n \geq 0$ , $\hat{M} / I^{n} \hat{M} ≅ M / I^{n} M .$

Proof

1.1

The completion theorem makes $\hat{R}$ Noetherian local with maximal ideal $m \hat{R}$ and residue field $k$ . Finite-quotient compatibility identifies $R / m^{2}$ with $\hat{R} / m^{2} \hat{R}$ compatibly with their maps to $k$ .

F2F3

2.1

The kernels of those maps to $k$ are the two cotangent spaces, since $(m \hat{R})^{2} = m^{2} \hat{R}$ . The induced isomorphism is $k$ -linear and canonical, so their dimensions agree by the embedding-dimension definition. For $m = 0$ both spaces are zero.

F1step 1.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

completion preserves regular local rings

Statement

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

Facts & Assumptions

Given: The objects and hypotheses in the statement. 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 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.

Proof

1.1

Completion preserves the embedding dimension. Applied to the nonzero finite module $R$ , the completion dimension theorem also gives $\dim \hat{R} = \dim R$ , because the support of a ring over itself is its entire spectrum.

F1F2

2.1

Thus $edim R = \dim R$ holds exactly when $edim \hat{R} = \dim \hat{R}$ . These are the two regularity conditions. The argument also applies when the common dimension or embedding dimension is zero.

step 1.1algebra∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

normal noetherian ring

Definition

A commutative Noetherian ring $R$ is normal if every prime localization $R_{p}$ is an integrally closed domain. This is a local condition and does not require $R$ itself to be a domain. The zero ring satisfies it vacuously. For a domain, integrally closed means that every element of its fraction field integral over it belongs to it.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

serre r k and s k conditions

Definition

For a commutative Noetherian ring $R$ and an integer $j \geq 0$ , condition $(R_{j})$ means that $R_{p}$ is regular whenever $ht p \leq j$ . Condition $(S_{j})$ means that $depth R_{p} \geq \min {j, \dim R_{p}}$ for every prime $p$ . A finite module $M$ satisfies $(S_{j})$ if ${depth}_{R_{p}} M_{p} \geq \min {j, \dim {Supp}_{R_{p}} M_{p}}$ for every prime in its support. Outside the support the condition is vacuous, consistent with depth of the zero module being $+ \infty$ and the empty support having no nonnegative dimension. Thus the zero module satisfies all $(S_{j})$ conditions, and the zero ring satisfies both families vacuously.

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local ring satisfies r one

Statement

Every regular local ring satisfies $(R_{1})$ . Its height-zero localizations are fields, and its height-one localizations are DVRs.

Facts & Assumptions

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

[F1]

serre r k and s k conditions: For a commutative Noetherian ring $R$ and an integer $j \geq 0$ , condition $(R_{j})$ means that $R_{p}$ is regular whenever $ht p \leq j$ . Condition $(S_{j})$ means that $depth R_{p} \geq \min {j, \dim R_{p}}$ for every prime $p$ . A finite module $M$ satisfies $(S_{j})$ if ${depth}_{R_{p}} M_{p} \geq \min {j, \dim {Supp}_{R_{p}} M_{p}}$ for every prime in its support. Outside the support the condition is vacuous, consistent with depth of the zero module being $+ \infty$ and the empty support having no nonnegative dimension. Thus the zero module satisfies all $(S_{j})$ conditions, and the zero ring satisfies both families vacuously.

[F2]

localisations of regular local rings are regular: Every prime localization $R_{p}$ of a regular local ring $R$ is regular, and $edim R_{p} = ht p$ .

[F3]

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.

[F4]

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 $m$ .

Proof

1.1

Every prime localization is regular, and its dimension is the height of the prime. Thus at all heights at most one it is regular, which is precisely $(R_{1})$ .

F2F1

2.1

At height one the DVR equivalence applies. At height zero regularity makes the cotangent space zero; the generator-number formula makes the maximal ideal zero, hence the local ring is a field.

F3F2F4step 1.1∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local ring satisfies s two

Statement

Every regular local ring satisfies $(S_{j})$ for every integer $j \geq 0$ , in particular $(S_{2})$ .

Facts & Assumptions

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

[F1]

serre r k and s k conditions: For a commutative Noetherian ring $R$ and an integer $j \geq 0$ , condition $(R_{j})$ means that $R_{p}$ is regular whenever $ht p \leq j$ . Condition $(S_{j})$ means that $depth R_{p} \geq \min {j, \dim R_{p}}$ for every prime $p$ . A finite module $M$ satisfies $(S_{j})$ if ${depth}_{R_{p}} M_{p} \geq \min {j, \dim {Supp}_{R_{p}} M_{p}}$ for every prime in its support. Outside the support the condition is vacuous, consistent with depth of the zero module being $+ \infty$ and the empty support having no nonnegative dimension. Thus the zero module satisfies all $(S_{j})$ conditions, and the zero ring satisfies both families vacuously.

[F2]

localisations of regular local rings are regular: Every prime localization $R_{p}$ of a regular local ring $R$ is regular, and $edim R_{p} = ht p$ .

[F3]

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 $0 \leq c \leq d$ .

Proof

1.1

At every prime the local ring is regular, hence Cohen–Macaulay. Its depth therefore equals its dimension.

F2F3

2.1

For every $j \geq 0$ , that dimension is at least its minimum with $j$ , which is the $(S_{j})$ inequality. The inequality includes $j = 0$ and local dimension zero.

F1step 1.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

normal domain implies r one

Statement

Every commutative Noetherian integrally closed domain satisfies $(R_{1})$ .

Facts & Assumptions

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

[F1]

serre r k and s k conditions: For a commutative Noetherian ring $R$ and an integer $j \geq 0$ , condition $(R_{j})$ means that $R_{p}$ is regular whenever $ht p \leq j$ . Condition $(S_{j})$ means that $depth R_{p} \geq \min {j, \dim R_{p}}$ for every prime $p$ . A finite module $M$ satisfies $(S_{j})$ if ${depth}_{R_{p}} M_{p} \geq \min {j, \dim {Supp}_{R_{p}} M_{p}}$ for every prime in its support. Outside the support the condition is vacuous, consistent with depth of the zero module being $+ \infty$ and the empty support having no nonnegative dimension. Thus the zero module satisfies all $(S_{j})$ conditions, and the zero ring satisfies both families vacuously.

[F2]

Height-one localizations of normal Noetherian domains are DVRs: Let $R$ be a Noetherian integrally closed domain, and let $p$ be a prime ideal of height $1$ . Then the localisation $R_{p}$ is a discrete valuation ring.

[F3]

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.

Proof

1.1

A height-one localization is a DVR by the normal-domain height-one theorem, and therefore regular by the DVR equivalence.

F2F3

2.1

The only height-zero prime of a domain is $(0)$ ; its localization is the fraction field, which is regular. These two cases give the definition of $(R_{1})$ , including a field, which has no height-one primes.

F1step 1.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

normal domain implies s two

Statement

Every commutative Noetherian integrally closed domain satisfies $(S_{2})$ .

Facts & Assumptions

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

[F1]

serre r k and s k conditions: For a commutative Noetherian ring $R$ and an integer $j \geq 0$ , condition $(R_{j})$ means that $R_{p}$ is regular whenever $ht p \leq j$ . Condition $(S_{j})$ means that $depth R_{p} \geq \min {j, \dim R_{p}}$ for every prime $p$ . A finite module $M$ satisfies $(S_{j})$ if ${depth}_{R_{p}} M_{p} \geq \min {j, \dim {Supp}_{R_{p}} M_{p}}$ for every prime in its support. Outside the support the condition is vacuous, consistent with depth of the zero module being $+ \infty$ and the empty support having no nonnegative dimension. Thus the zero module satisfies all $(S_{j})$ conditions, and the zero ring satisfies both families vacuously.

[F2]

A domain is integrally closed if and only if its prime localisations are, equivalently if and only if its maximal localisations are: Assume the Axiom of Choice. Let $A$ be a domain. Then the following are equivalent: 1. $A$ is integrally closed. 2. For every prime ideal $p$ of $A$ , the localisation $A_{p}$ is integrally closed. 3. For every maximal ideal $m$ of $A$ , the localisation $A_{m}$ is integrally closed.

[F3]

The local depth-zero associated-prime criterion: Let $(R, m)$ be a Noetherian local ring and let $M \neq 0$ be a finite $R$ -module. Then $depth (M) = 0 ⟺ m \in {Ass}_{R} (M) .$

[F4]

Depth drops by one after quotienting by a regular element: Let $R$ be Noetherian, let $M$ be finite, let $I$ lie in the Jacobson radical, and let $x \in I$ be $M$ -regular. Then ${depth}_{I} (M / x M) = {depth}_{I} (M) - 1.$

[F5]

Krull's height theorem: Let $R$ be a Noetherian commutative ring, let $I = (x_{1}, \dots, x_{n})$ be an ideal generated by $n \geq 1$ elements, and let $p$ be a prime ideal minimal over $I$ . Then $ht (p) \leq n$ .

Proof

1.1

Localize at any prime. The resulting ring $A$ is again an integrally closed domain. In dimension zero it is a field and the required bound is zero; in positive dimension a nonzero element of the maximal ideal is a nonzerodivisor, so its depth is at least one. It remains to consider $\dim A \geq 2$ .

F2F1

2.1

If such an $A$ had depth one, choose $0 \neq a \in m_{A}$ . The regular-element depth formula and the depth-zero criterion supply $b \notin (a)$ with ${Ann}_{A / (a)} (\overset{ˉ}{b}) = m_{A}$ . Put $u = b / a \in Frac A ∖ A$ . Then $u m_{A} \subseteq A$ .

F4F3step 1.1

3.1

If $u m_{A} \subseteq m_{A}$ , take finite generators $z_{1}, \dots, z_{s}$ of the nonzero ideal $m_{A}$ and write $u z_{i} = \sum_{j} c_{i j} z_{j}$ with $c_{i j} \in A$ . The adjugate identity for $u I - C$ gives $\det (u I - C) z_{i} = 0$ for all $i$ . Some $z_{i} \neq 0$ in a domain, hence $\det (u I - C) = 0$ . This is a monic equation for $u$ , contradicting integral closedness.

step 2.1algebra

4.1

Otherwise there exists $t \in m_{A}$ with $u t$ a unit. For every $s \in m_{A}$ , $s = t (u s) / (u t)$ belongs to $(t)$ , so $m_{A} = (t)$ . The height theorem with one generator gives $\dim A \leq 1$ , again impossible. Therefore depth is at least two at every prime of height at least two; with the low-dimensional cases this is $(S_{2})$ .

F5F1step 1.1step 3.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

r one s two intersection of height one localisations

Statement

If $R$ is a commutative Noetherian domain satisfying $(S_{2})$ , then inside its fraction field $K$ one has $R = ⋂_{ht p = 1} R_{p}$ . For a field the empty intersection is interpreted as $K = R$ .

Facts & Assumptions

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

[F1]

serre r k and s k conditions: For a commutative Noetherian ring $R$ and an integer $j \geq 0$ , condition $(R_{j})$ means that $R_{p}$ is regular whenever $ht p \leq j$ . Condition $(S_{j})$ means that $depth R_{p} \geq \min {j, \dim R_{p}}$ for every prime $p$ . A finite module $M$ satisfies $(S_{j})$ if ${depth}_{R_{p}} M_{p} \geq \min {j, \dim {Supp}_{R_{p}} M_{p}}$ for every prime in its support. Outside the support the condition is vacuous, consistent with depth of the zero module being $+ \infty$ and the empty support having no nonnegative dimension. Thus the zero module satisfies all $(S_{j})$ conditions, and the zero ring satisfies both families vacuously.

[F2]

Every submodule of a finite module over a Noetherian ring has a minimal primary decomposition: Assume Dependent Choice. Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$ -module. Every submodule $N \subseteq M$ has a finite primary decomposition. After deleting redundant components and combining equal radicals, one obtains a minimal primary decomposition. When $N = M$ , the decomposition is the empty intersection, interpreted as $M$ . In particular, every ideal of a Noetherian ring has a minimal primary decomposition.

[F3]

The radicals in a minimal primary decomposition are exactly the associated primes of the quotient: Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let $N = Q_{1} \cap \dots \cap Q_{r}$ be a minimal primary decomposition in which each $Q_{i}$ is $p_{i}$ -primary. Assume each $p_{i}$ is a prime ideal. Then ${Ass}_{R} (M / N) = {p_{1}, \dots, p_{r}} .$

[F4]

Depth drops by one after quotienting by a regular element: Let $R$ be Noetherian, let $M$ be finite, let $I$ lie in the Jacobson radical, and let $x \in I$ be $M$ -regular. Then ${depth}_{I} (M / x M) = {depth}_{I} (M) - 1.$

[F5]

The local depth-zero associated-prime criterion: Let $(R, m)$ be a Noetherian local ring and let $M \neq 0$ be a finite $R$ -module. Then $depth (M) = 0 ⟺ m \in {Ass}_{R} (M) .$

[F6]

Associated primes commute with localization for finite modules: Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let $S \subseteq R$ be multiplicative. Then ${Ass}_{S^{- 1} R} (S^{- 1} M) = {S^{- 1} p : p \in {Ass}_{R} (M), p \cap S = \emptyset} .$

[F7]

A nonzero module over a Noetherian ring has an associated prime: Let $R$ be a Noetherian commutative ring and let $M$ be a nonzero left $R$ -module. Then ${Ass}_{R} (M)$ is nonempty.

Proof

1.1

Let $0 \neq a \in R$ be a nonunit. If $p \in Ass (R / (a))$ , localization and the depth-zero criterion make $R_{p} / a R_{p}$ depth zero. Since $a$ is regular, the depth formula gives $depth R_{p} = 1$ . Condition $(S_{2})$ forces $ht p \leq 1$ , and $a \in p$ , $a \neq 0$ force equality.

F6F5F4F1

2.1

Choose a minimal primary decomposition $(a) = ⋂_{i} Q_{i}$ , with radicals $p_{i}$ . Those radicals are associated to $R / (a)$ , hence have height one. If $b / a$ belongs to every height-one localization, then for each $i$ there is $s_{i} \notin p_{i}$ with $s_{i} b \in (a) \subseteq Q_{i}$ . Primaryness gives $b \in Q_{i}$ , hence $b \in (a)$ and $b / a \in R$ .

F2F3step 1.1

3.1

If $a$ is a unit, membership is immediate without a primary decomposition. The inclusion from $R$ into every localization is automatic. If the height-one family is empty, a nonzero nonunit would yield an associated prime of its nonzero quotient and hence a height-one prime by the preceding argument; thus $R$ is a field and the stipulated empty intersection is correct.

F7step 1.1step 2.1algebra∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

r one s two integral element membership

Statement

A commutative Noetherian $(S_{2})$ domain whose height-one localizations are DVRs is integrally closed.

Facts & Assumptions

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

[F1]

r one s two intersection of height one localisations: If $R$ is a commutative Noetherian domain satisfying $(S_{2})$ , then inside its fraction field $K$ one has $R = ⋂_{ht p = 1} R_{p}$ . For a field the empty intersection is interpreted as $K = R$ .

[F2]

Valuation rings are integrally closed: Every valuation ring is an integrally closed domain.

Proof

1.1

Let $u \in Frac R$ satisfy a monic equation over $R$ . At each height-one prime the same equation is monic over $R_{p}$ . A DVR is a valuation ring, hence integrally closed, so $u \in R_{p}$ .

F2given

2.1

The height-one intersection theorem now gives $u \in R$ . If there are no height-one primes, its empty-intersection convention says $R$ is already the fraction field; the conclusion remains valid. Since $u$ was arbitrary, $R$ is integrally closed.

F1step 1.1∎

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

serre normality criterion two directions

Statement

A commutative Noetherian domain is normal if and only if it satisfies $(R_{1})$ and $(S_{2})$ . Equivalently its integral closedness is characterized by these two conditions.

Facts & Assumptions

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

[F1]

normal domain implies r one: Every commutative Noetherian integrally closed domain satisfies $(R_{1})$ .

[F2]

normal domain implies s two: Every commutative Noetherian integrally closed domain satisfies $(S_{2})$ .

[F3]

r one s two integral element membership: A commutative Noetherian $(S_{2})$ domain whose height-one localizations are DVRs is integrally closed.

[F4]

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.

[F5]

A domain is integrally closed if and only if its prime localisations are, equivalently if and only if its maximal localisations are: Assume the Axiom of Choice. Let $A$ be a domain. Then the following are equivalent: 1. $A$ is integrally closed. 2. For every prime ideal $p$ of $A$ , the localisation $A_{p}$ is integrally closed. 3. For every maximal ideal $m$ of $A$ , the localisation $A_{m}$ is integrally closed.

Proof

1.1

For a domain, normality is equivalent to integral closedness by local normality. An integrally closed Noetherian domain satisfies $(R_{1})$ and $(S_{2})$ by the two normal-domain lemmas.

F5F1F2

2.1

Conversely, $(R_{1})$ makes every height-one localization one-dimensional regular local and hence a DVR. With $(S_{2})$ , the integral-element membership lemma makes $R$ integrally closed, and local normality makes it normal. Fields satisfy both conditions and are included.

F4F3F5∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

serre r zero s one characterises reducedness

Statement

For a finite module $M$ over a commutative Noetherian ring, $(S_{1})$ is equivalent to every associated prime being minimal in $Supp M$ . For the ring itself, this means no embedded associated primes. A commutative Noetherian ring is reduced if and only if it satisfies $(R_{0})$ and $(S_{1})$ .

Facts & Assumptions

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

[F1]

serre r k and s k conditions: For a commutative Noetherian ring $R$ and an integer $j \geq 0$ , condition $(R_{j})$ means that $R_{p}$ is regular whenever $ht p \leq j$ . Condition $(S_{j})$ means that $depth R_{p} \geq \min {j, \dim R_{p}}$ for every prime $p$ . A finite module $M$ satisfies $(S_{j})$ if ${depth}_{R_{p}} M_{p} \geq \min {j, \dim {Supp}_{R_{p}} M_{p}}$ for every prime in its support. Outside the support the condition is vacuous, consistent with depth of the zero module being $+ \infty$ and the empty support having no nonnegative dimension. Thus the zero module satisfies all $(S_{j})$ conditions, and the zero ring satisfies both families vacuously.

[F2]

The local depth-zero associated-prime criterion: Let $(R, m)$ be a Noetherian local ring and let $M \neq 0$ be a finite $R$ -module. Then $depth (M) = 0 ⟺ m \in {Ass}_{R} (M) .$

[F3]

Zero divisors on a module over a Noetherian ring are the union of its associated primes: Let $R$ be a Noetherian commutative ring and let $M$ be a left $R$ -module. Then the set of zero divisors on $M$ is $⋃_{p \in {Ass}_{R} (M)} p .$ If $M$ is finitely generated, this is a finite union.

[F4]

A nonzero module over a Noetherian ring has an associated prime: Let $R$ be a Noetherian commutative ring and let $M$ be a nonzero left $R$ -module. Then ${Ass}_{R} (M)$ is nonempty.

[F5]

A Noetherian ring has finitely many minimal prime ideals: Let $R$ be a Noetherian commutative ring. Then $R$ has only finitely many minimal prime ideals. This theorem inherits only the dependent-choice cost already recorded in the cited Noetherian-induction corollary.

[F6]

A radical ideal in a Noetherian ring is the intersection of its minimal primes: Assume Dependent Choice. Let $R$ be a Noetherian commutative ring and let $I ⊴ R$ be a radical ideal. Then there exist finitely many prime ideals $p_{1}, \dots, p_{m}$ minimal over $I$ such that $I = p_{1} \cap \dots \cap p_{m} .$ When $I = R$ , this is the empty intersection.

[F7]

Associated primes commute with localization for finite modules: Let $R$ be a Noetherian commutative ring, let $M$ be a finitely generated left $R$ -module, and let $S \subseteq R$ be multiplicative. Then ${Ass}_{S^{- 1} R} (S^{- 1} M) = {S^{- 1} p : p \in {Ass}_{R} (M), p \cap S = \emptyset} .$

[F8]

An ideal contained in a finite union of prime ideals lies in one of them: Let $R$ be a commutative ring, let $I ⊴ R$ be an ideal, and let $p_{1}, \dots, p_{n}$ be prime ideals with $n \geq 1$ . If $I \subseteq p_{1} \cup \dots \cup p_{n},$ then $I \subseteq p_{i}$ for some $i$ .

[F9]

Minimal support primes of a finite module are associated: Let $R$ be a Noetherian commutative ring and let $M$ be a finitely generated left $R$ -module. If $p$ is minimal in ${Supp}_{R} (M)$ , then $p \in {Ass}_{R} (M) .$

Proof

1.1

At a prime in the support, depth zero is equivalent to that prime being associated, by localization of associated primes and the local depth-zero criterion. Such an associated prime violates $(S_{1})$ exactly when the localized support has positive dimension, namely when there is a strictly smaller support prime. Thus $(S_{1})$ is equivalent to all associated primes being minimal in support. Minimal support primes are associated as well. For $M = 0$ both conditions are vacuous.

F7F2F1F9

2.1

If $R$ is reduced, its minimal primes $p_{1}, \dots, p_{s}$ are finite and have intersection zero. An element outside their union is a nonzerodivisor, since its product with $b$ being zero forces $b$ into every $p_{i}$ . Conversely, for $a \in p_{i}$ , choose $b \in ⋂_{j \neq i} p_{j} ∖ p_{i}$ by taking a product of elements of $p_{j} ∖ p_{i}$ . Then $b \neq 0$ and $a b = 0$ . Thus zero divisors are exactly this finite union. An associated prime is contained in that union and hence in one minimal prime by prime avoidance; it must equal it. This proves $(S_{1})$ .

F5F6F3F8step 1.1

3.1

At a minimal prime, localization of a reduced ring is reduced and has only one prime ideal. Its nilradical, the intersection of its primes, is therefore that maximal ideal and is zero. It is a field, so $(R_{0})$ holds. Conversely suppose $(R_{0})$ and $(S_{1})$ hold. If the nilradical $N$ were nonzero, choose an associated prime of $N$ ; its annihilator witness in $N \subset R$ makes it associated to $R$ , hence minimal by $(S_{1})$ . The witness survives there, but $(R_{0})$ makes that localization a field and annihilates all nilpotents, a contradiction. Hence $N = 0$ . The zero ring satisfies the assertions vacuously.

F4F1step 1.1step 2.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

total ring of fractions

Definition

For a nonzero commutative ring $R$ , let $S$ be the set of its nonzerodivisors, meaning elements whose multiplication maps on $R$ are injective. Its total ring of fractions is $Q (R) = S^{- 1} R$ . The set $S$ is multiplicative since composites of injective multiplication maps are injective. The natural map $R \to Q (R)$ is injective: $a / 1 = 0$ implies $s a = 0$ for some $s \in S$ , hence $a = 0$ . Set $Q (0) = 0$ . For a domain this recovers the fraction field; for a ring with zero divisors it need not be a field.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

reduced noetherian total fractions and normal components

Statement

For a reduced commutative Noetherian ring $R$ with minimal primes $p_{1}, \dots, p_{s}$ , there is a canonical isomorphism $Q (R) ≅ \prod_{i = 1}^{s} Frac (R / p_{i})$ . The following are equivalent: $R$ is normal; $R$ is integrally closed in $Q (R)$ ; and $R$ is a finite product of normal domains. For $R = 0$ this is the empty product.

Facts & Assumptions

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

[F1]

total ring of fractions: For a nonzero commutative ring $R$ , let $S$ be the set of its nonzerodivisors, meaning elements whose multiplication maps on $R$ are injective. Its total ring of fractions is $Q (R) = S^{- 1} R$ . The set $S$ is multiplicative since composites of injective multiplication maps are injective. The natural map $R \to Q (R)$ is injective: $a / 1 = 0$ implies $s a = 0$ for some $s \in S$ , hence $a = 0$ . Set $Q (0) = 0$ . For a domain this recovers the fraction field; for a ring with zero divisors it need not be a field.

[F2]

normal noetherian ring: A commutative Noetherian ring $R$ is normal if every prime localization $R_{p}$ is an integrally closed domain. This is a local condition and does not require $R$ itself to be a domain. The zero ring satisfies it vacuously. For a domain, integrally closed means that every element of its fraction field integral over it belongs to it.

[F3]

A Noetherian ring has finitely many minimal prime ideals: Let $R$ be a Noetherian commutative ring. Then $R$ has only finitely many minimal prime ideals. This theorem inherits only the dependent-choice cost already recorded in the cited Noetherian-induction corollary.

[F4]

A radical ideal in a Noetherian ring is the intersection of its minimal primes: Assume Dependent Choice. Let $R$ be a Noetherian commutative ring and let $I ⊴ R$ be a radical ideal. Then there exist finitely many prime ideals $p_{1}, \dots, p_{m}$ minimal over $I$ such that $I = p_{1} \cap \dots \cap p_{m} .$ When $I = R$ , this is the empty intersection.

[F5]

A domain is integrally closed if and only if its prime localisations are, equivalently if and only if its maximal localisations are: Assume the Axiom of Choice. Let $A$ be a domain. Then the following are equivalent: 1. $A$ is integrally closed. 2. For every prime ideal $p$ of $A$ , the localisation $A_{p}$ is integrally closed. 3. For every maximal ideal $m$ of $A$ , the localisation $A_{m}$ is integrally closed.

[F6]

Chinese remainder theorem for pairwise comaximal ideals: Let $R$ be a commutative ring and let $I_{1}, \dots, I_{r}$ be pairwise comaximal ideals, where $r \geq 1$ . Then the canonical map $R ⟶ \prod_{i = 1}^{r} R / I_{i}, x ⟼ (x + I_{1}, \dots, x + I_{r})$ is surjective, its kernel is $⋂_{i = 1}^{r} I_{i}$ , and $⋂_{i = 1}^{r} I_{i} = \prod_{i = 1}^{r} I_{i} .$ Equivalently, $R / \prod_{i = 1}^{r} I_{i} ≅ \prod_{i = 1}^{r} R / I_{i} .$

[F7]

An ideal contained in a finite union of prime ideals lies in one of them: Let $R$ be a commutative ring, let $I ⊴ R$ be an ideal, and let $p_{1}, \dots, p_{n}$ be prime ideals with $n \geq 1$ . If $I \subseteq p_{1} \cup \dots \cup p_{n},$ then $I \subseteq p_{i}$ for some $i$ .

Proof

1.1

For $R \neq 0$ , the finite minimal-prime intersection is zero. If $a$ avoids every minimal prime, $a b = 0$ forces $b = 0$ . If $a \in p_{i}$ , a product of elements in $p_{j} ∖ p_{i}$ for $j \neq i$ supplies nonzero $b$ with $a b = 0$ . Thus the nonzerodivisors are the complement of the union of the minimal primes. Prime avoidance implies that the primes surviving in $Q (R)$ are exactly these minimal primes.

F3F4F7F1

2.1

The surviving primes of the reduced ring $Q (R)$ are finitely many distinct maximal ideals with intersection zero. CRT decomposes $Q (R)$ as their residue fields. Localization at the corresponding minimal prime of $R$ is reduced with only the zero prime, hence is a field, and is the fraction field of $R / p_{i}$ . This identifies each factor and the canonical map.

F6step 1.1

3.1

If $R$ is integrally closed in $Q (R)$ , it contains every coordinate idempotent $e_{i}$ , since each solves $T^{2} - T = 0$ . Thus $R = \prod e_{i} R$ , with $e_{i} R ≅ R / p_{i}$ . For an element integral over one factor, put it in that coordinate and zero in the other coordinates. A monic equation in the factor, multiplied by $T$ if necessary and with coefficients lifted to that coordinate, gives a monic equation over the product ring; integral closedness puts it in $R$ . Each factor is integrally closed, hence a normal domain by local normality.

F5step 2.1algebra

4.1

If $R$ is normal, no prime can contain two distinct minimal primes: localization would give two distinct minimal primes in a domain. Hence the minimal primes are pairwise comaximal. CRT gives $R = \prod R / p_{i}$ ; the localizations of a component are the corresponding localizations of $R$ , so the components are normal domains. Conversely a finite product of normal domains has normal prime localizations, and a monic equation in its total fractions is coordinatewise integral, so the product is integrally closed there. For $R = 0$ all assertions hold directly without applying CRT to an empty family.

F2F6F5step 2.1∎

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

depth two excludes finite punctured extension

Statement

Let $(R, m)$ be reduced Noetherian local with $depth R \geq 2$ . If $R \subseteq B \subseteq Q (R)$ is a finite intermediate ring and ${Supp}_{R} (B / R) \subseteq {m}$ , then $B = R$ .

Facts & Assumptions

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

[F1]

total ring of fractions: For a nonzero commutative ring $R$ , let $S$ be the set of its nonzerodivisors, meaning elements whose multiplication maps on $R$ are injective. Its total ring of fractions is $Q (R) = S^{- 1} R$ . The set $S$ is multiplicative since composites of injective multiplication maps are injective. The natural map $R \to Q (R)$ is injective: $a / 1 = 0$ implies $s a = 0$ for some $s \in S$ , hence $a = 0$ . Set $Q (0) = 0$ . For a domain this recovers the fraction field; for a ring with zero divisors it need not be a field.

[F2]

The three Depth Lemma inequalities: Let $(R, m)$ be a Noetherian local ring and $0 \to A \to B \to C \to 0$ a short exact sequence of finite $R$ -modules. With $a = {depth}_{R} (A)$ , $b = {depth}_{R} (B)$ , and $c = {depth}_{R} (C)$ , $b \geq \min {a, c}, a \geq \min {b, c + 1}, c \geq \min {a - 1, b} .$ The last inequality is vacuous when $a = 0$ .

[F3]

The local depth-zero associated-prime criterion: Let $(R, m)$ be a Noetherian local ring and let $M \neq 0$ be a finite $R$ -module. Then $depth (M) = 0 ⟺ m \in {Ass}_{R} (M) .$

[F4]

For a finite module, support is the set of primes containing the annihilator: If $M$ is a finitely generated left $R$ -module, then ${Supp}_{R} (M) = {p : {Ann}_{R} (M) \subseteq p} .$

[F5]

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 $M = 0$ .

Proof

1.1

Choose the first element $x \in m$ of an $R$ -regular sequence of length two. It is a unit in $Q (R)$ , so it acts injectively on $B$ . Since $B$ is nonzero finite and $x \in m$ , Nakayama gives $B / x B \neq 0$ , hence ${depth}_{R} B \geq 1$ . Applied to $0 \to R \to B \to C \to 0$ , the depth lemma gives $depth C \geq \min (depth R - 1, depth B) \geq 1$ if $C \neq 0$ .

F1F2F5

2.1

If $C \neq 0$ and its support is contained in the closed point, the support-annihilator theorem gives $\sqrt{Ann C} = m$ . Finitely many generators of $m$ each have a power in the annihilator; expanding monomials gives $m^{N} C = 0$ for some $N$ . A last nonzero power $m^{j} C$ contains a nonzero element killed by $m$ , making $m$ associated and $depth C = 0$ . This contradicts the preceding bound. Thus $C = 0$ and $B = R$ , including the case of empty support.

F4F3step 1.1algebra∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

serre normality criterion

Statement

For every commutative Noetherian ring $R$ , including rings with zero divisors and the zero ring, $R$ is normal if and only if it satisfies $(R_{1})$ and $(S_{2})$ .

Facts & Assumptions

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

[F1]

normal noetherian ring: A commutative Noetherian ring $R$ is normal if every prime localization $R_{p}$ is an integrally closed domain. This is a local condition and does not require $R$ itself to be a domain. The zero ring satisfies it vacuously. For a domain, integrally closed means that every element of its fraction field integral over it belongs to it.

[F2]

serre normality criterion two directions: A commutative Noetherian domain is normal if and only if it satisfies $(R_{1})$ and $(S_{2})$ . Equivalently its integral closedness is characterized by these two conditions.

[F3]

serre r zero s one characterises reducedness: For a finite module $M$ over a commutative Noetherian ring, $(S_{1})$ is equivalent to every associated prime being minimal in $Supp M$ . For the ring itself, this means no embedded associated primes. A commutative Noetherian ring is reduced if and only if it satisfies $(R_{0})$ and $(S_{1})$ .

[F4]

reduced noetherian total fractions and normal components: For a reduced commutative Noetherian ring $R$ with minimal primes $p_{1}, \dots, p_{s}$ , there is a canonical isomorphism $Q (R) ≅ \prod_{i = 1}^{s} Frac (R / p_{i})$ . The following are equivalent: $R$ is normal; $R$ is integrally closed in $Q (R)$ ; and $R$ is a finite product of normal domains. For $R = 0$ this is the empty product.

[F5]

depth two excludes finite punctured extension: Let $(R, m)$ be reduced Noetherian local with $depth R \geq 2$ . If $R \subseteq B \subseteq Q (R)$ is a finite intermediate ring and ${Supp}_{R} (B / R) \subseteq {m}$ , then $B = R$ .

[F6]

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.

[F7]

Valuation rings are integrally closed: Every valuation ring is an integrally closed domain.

[F8]

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 $m$ .

[F9]

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

Proof

1.1

If $R$ is normal, each prime localization is a normal domain. The domain criterion gives the required depth bound there and regularity when its dimension is at most one. Thus $R$ satisfies both conditions. The zero ring satisfies all three conditions vacuously.

F1F2

2.1

Conversely $(R_{1})$ and $(S_{2})$ imply $(R_{0})$ and $(S_{1})$ , so $R$ is reduced. Both conditions pass to prime localizations, since prime chains below a prime and successive localizations are unchanged. It is enough to prove that every reduced Noetherian local ring $A$ satisfying them is a normal domain. Induct on its finite dimension $d$ . For $d \leq 1$ , $(R_{1})$ makes $A$ regular; in dimension zero its maximal ideal is zero by the generator formula, so it is a field; in dimension one it is a DVR, hence an integrally closed domain by the valuation theorem.

F3F6F7F8F9step 1.1

3.1

Let $d \geq 2$ and $u \in Q (A)$ be integral over $A$ . A monic equation shows $B = A [u]$ is finite, generated by finitely many powers of $u$ . For a nonmaximal prime $p$ of $A$ , the dimension of $A_{p}$ is less than $d$ (append the maximal ideal to any chain below $p$ ), so it is a normal domain by induction. A nonzerodivisor of $A$ remains a nonzerodivisor after localization: clear denominators in the equation it kills. Thus $Q (A)_{p}$ embeds into $Q (A_{p})$ . The image of $u$ is integral and belongs to $A_{p}$ , giving $B_{p} = A_{p}$ .

step 2.1algebra

4.1

Therefore $B / A$ is supported only at the maximal ideal. Since $(S_{2})$ gives depth at least two, finite-extension rigidity implies $B = A$ . Every integral element of $Q (A)$ lies in $A$ . The total-fraction component theorem now makes $A$ a finite product of normal domains. A nonzero local ring has no idempotents except zero and one: one of $e, 1 - e$ is a unit, forcing the other to vanish. Thus the product has a single factor and $A$ is a normal domain. This completes the local induction and hence the global converse.

F5F4step 3.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

regular local rings are normal

Statement

Every regular local ring is an integrally closed domain. Every commutative regular Noetherian ring is normal and is a finite product of regular domains, with the zero ring corresponding to the empty product.

Facts & Assumptions

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

[F1]

regular local domain induction: Every regular local ring is an integral domain.

[F2]

regular local ring satisfies r one: Every regular local ring satisfies $(R_{1})$ . Its height-zero localizations are fields, and its height-one localizations are DVRs.

[F3]

regular local ring satisfies s two: Every regular local ring satisfies $(S_{j})$ for every integer $j \geq 0$ , in particular $(S_{2})$ .

[F4]

serre normality criterion: For every commutative Noetherian ring $R$ , including rings with zero divisors and the zero ring, $R$ is normal if and only if it satisfies $(R_{1})$ and $(S_{2})$ .

[F5]

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.

[F6]

reduced noetherian total fractions and normal components: For a reduced commutative Noetherian ring $R$ with minimal primes $p_{1}, \dots, p_{s}$ , there is a canonical isomorphism $Q (R) ≅ \prod_{i = 1}^{s} Frac (R / p_{i})$ . The following are equivalent: $R$ is normal; $R$ is integrally closed in $Q (R)$ ; and $R$ is a finite product of normal domains. For $R = 0$ this is the empty product.

Proof

1.1

A regular local ring is a domain and satisfies $(R_{1})$ and $(S_{2})$ . Serre normality therefore makes it normal; at its maximal ideal the localization is the ring itself, so it is integrally closed.

F1F2F3F4

2.1

For a regular Noetherian ring, every prime localization is regular local, hence an integrally closed domain by the preceding argument. It is therefore normal. The normal-component theorem expresses it as a finite product of normal domains; each factor is regular since its prime localizations are those of the product. The zero ring is the empty product. No factoriality assertion is made.

F5F4F6step 1.1∎

Regular Local Rings and Homological Dimension

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

embedding dimension and regular local ring

Definition

embedding dimension is minimal maximal ideal generator number

Statement

Facts & Assumptions

Proof

dimension at most embedding dimension

Statement

Facts & Assumptions

Proof

regular system of parameters

Definition

regular system of parameters equivalent basis

Statement

Facts & Assumptions

Proof

associated graded polynomial surjection

Statement

Facts & Assumptions

Proof

regular local graded surjection has zero kernel

Statement

Facts & Assumptions

Proof

associated graded ring of a regular local ring

Statement

Facts & Assumptions

Proof

regular local hilbert samuel multiplicity one

Statement

Facts & Assumptions

Proof

regular local domain induction

Statement

Facts & Assumptions

Proof

regular local parameter is nonzerodivisor

Statement

Facts & Assumptions

Proof

regular local quotient by parameter is regular

Statement

Facts & Assumptions

Proof

quotient and lifting regularity across a regular element

Statement

Facts & Assumptions

Proof

regular local rings are domains and cohen macaulay

Statement

Facts & Assumptions

Proof

one dimensional regular local rings are dvrs

Statement

Facts & Assumptions

Proof

regular local regular quotient ideal is parameter generated

Statement

Facts & Assumptions

Proof

finite local modules admit minimal free resolutions

Statement

Facts & Assumptions

Proof

minimal free resolution differentials land in maximal ideal

Statement

Facts & Assumptions

Proof

minimal free resolution reduces to zero differential

Statement

Facts & Assumptions

Proof

betti numbers of a finite local module

Definition

betti number is rank in minimal resolution