A new proof of Serre’s homological characterization of regular local rings

Abstract

We give a new proof of Serre’s result that a Noetherian local ring is regular if and only if it has finite global dimension. Our proof avoids the explicit construction of a Koszul complex.

1 Background

The proof that localization at a prime preserves regularity relies on Serre’s homological characterization of regularity (see [2, Théorème 5]). This was the first application of homological methods to prove a non-homological result and inspired the use of homological techniques in algebra. In this paper, we present a new proof of Serre’s theorem using basic notions from the theory of derived categories.

By a ring, we mean a commutative ring with a unit. Let $\dim A$ denote the Krull dimension of a ring A. Recall that a Noetherian local ring $(A,{\mathfrak m},k)$ with maximal ideal ${\mathfrak m}$ and residue field k is said to be regular if $\dim _k {\mathfrak m}/{\mathfrak m}^2 = \dim A.$ Let ${{\mathrm{gl.dim}}}A$ denote the global dimension of A.

An alternate proof of the following theorem is the main result of this paper.

Theorem 1.1

(Serre  [2, Théorème 3]) Let A be a Noetherian local ring. The following are equivalent:

• (1) A is regular;

• (2) ${{\mathrm{gl.dim}}}{A} = \dim A$;

• (3) ${{\mathrm{gl.dim}}}{A} < \infty$.

Serre’s original proof of Theorem 1.1 shows the difficult step, namely that (3) implies (1), as follows. One first proves that ${{\mathrm{gl.dim}}}{A} < \infty$ implies that ${{\mathrm{gl.dim}}}{A} \le \dim A$ using the interaction between regular sequences and projective dimension. One then uses the Koszul complex to prove that $\dim _k {\mathfrak m}/{\mathfrak m}^2 \le {{\mathrm{gl.dim}}}{A}$. The chain of inequalities ${{\mathrm{gl.dim}}}A \le \dim A \le \dim _k {\mathfrak m}/{\mathfrak m}^2 \le {{\mathrm{gl.dim}}}A$ then implies the result. The argument also shows that (3) implies (2).

We prove Theorem 1.1 without having to construct a Koszul complex, adapting an approach suggested by Dennis Gaitsgory. We first prove the existence of an element $f \in {\mathfrak m}\backslash {\mathfrak m}^2$ that is not a zero-divisor whenever ${\mathfrak m}$ is not associated. By proving $k \otimes ^{\mathbf {L}}_A A/f$ splits in D(A / f), we prove that ${{\mathrm{gl.dim}}}{A} = {{\mathrm{gl.dim}}}{A/f} + 1$. This allows us to deal with the case of A regular and the case of $0< {{\mathrm{gl.dim}}}{A} < \infty$ uniformly to conclude the proof of Theorem 1.1.

A proof using similar ideas is independently presented in [4, Theorem 1.4.4].

1.1 Notation and conventions

All complexes are graded homologically. For a ring R, let D(R) denote the derived category of R-modules. Let ${-}\otimes ^{\mathbf {L}}_R{-}$ denote the derived tensor product.

2 Proof of Theorem 1.1

The proof relies on the following technical result whose proof is deferred to Sect.  3.

Proposition 2.1

Let $(A,{\mathfrak m},k)$ be a local ring. If $f \in {\mathfrak m}\backslash {\mathfrak m}^2$ is not a zero-divisor, then ${{\mathrm{gl.dim}}}{A} = {{\mathrm{gl.dim}}}{A/f} + 1$.

In proving Serre’s theorem, we will use the following easy fact.

Lemma 2.2

If A is a Noetherian local ring with ${{\mathrm{gl.dim}}}{A} = 0$, then A is a field.

Proof

Because ${{\mathrm{gl.dim}}}{A} = 0,$ every finitely generated A-module is projective. Because A is local, every finitely generated projective A-module is free. Therefore, every finitely generated A-module is free, which implies that A is a field. $\square$

In order to apply Proposition 2.1 in the proof of the inductive step, we will need the following two lemmas. The former is a form of prime avoidance, where one of the ideals is taken to be ${\mathfrak m}^2$, and the rest are taken to be the associated prime ideals, whose proof can be found in [3, Tag 00DS]). The latter is immediate from [2, Proposition 4], but we provide a proof for completeness.

Proposition 2.3

(Prime Avoidance [3, Tag 00DS]) Let $(A,{\mathfrak m})$ be a Noetherian local ring. If ${\mathfrak m}$ is not associated, then there exists $f \in {\mathfrak m}\backslash {\mathfrak m}^2$ that is not a zero-divisor.

Lemma 2.4

If $(A,{\mathfrak m})$ is a Noetherian local ring with $0< {{\mathrm{gl.dim}}}{A} < \infty$, then ${\mathfrak m}$ is not an associated prime of A.

Proof

Let $n = {{\mathrm{gl.dim}}}{A}$. Assume for sake of deriving a contradiction that ${\mathfrak m}$ is associated. There exists an element $a \in A$ with ${\text {Ann}}(a) = {\mathfrak m}$, so that there exists a nonzero homomorphism $\theta \in {{\mathrm{Hom}}}(k,A)$ sending 1 to a. Consider the long exact sequence of ${{\mathrm{Tor}}}$ associated to the short exact sequence

where $M = {{\mathrm{coker}}}\theta$. We obtain an exact sequence

The first term vanishes because ${{\mathrm{gl.dim}}}{A} = n$ and the third term vanishes because A is projective and $n > 0$. Therefore, the middle term vanishes as well. This contradicts the fact that ${{\mathrm{Tor}}}_n^A (k,k) \ne 0$ for a Noetherian local ring A of global dimension n with residue field k (see, for example, [1, Theorem 8.55(i)]). $\square$

We are now ready to prove Serre’s theorem.

Proof of Theorem 1.1

It is clear that (2) implies (3). We will prove that (1) implies (2) and that (3) implies (1).

2.1 Step 1: $(1) \implies (2)$

We proceed by induction on $\dim A$. Suppose first that A is a regular local ring with $\dim A = 0$. Then, A is a field, so ${{\mathrm{gl.dim}}}{A} = 0$.

We now prove the inductive step. Let n be a positive integer and assume that ${{\mathrm{gl.dim}}}{A} = n-1$ for all regular local rings A with $\dim A = n-1$. Let $(A,{\mathfrak m})$ be a regular local ring with $\dim A = n$. Because $\dim A \ge 1$, there exists an element $f \in {\mathfrak m}\backslash {\mathfrak m}^2$. Because A is regular local, it an integral domain, so f is not a zero-divisor. Krull’s principal ideal theorem ensures that A / f is a regular local ring of Krull dimension $n-1$. The induction hypothesis guarantees that ${{\mathrm{gl.dim}}}{A} = n-1$. Proposition 2.1 yields that ${{\mathrm{gl.dim}}}{A} = n$, which completes the proof of the inductive step.

2.2 Step 2: $(3) \implies (1)$

We proceed by induction on ${{\mathrm{gl.dim}}}{A}$. Suppose first that ${{\mathrm{gl.dim}}}{A} = 0$. Lemma 2.2 implies that A is a field, hence regular.

We now prove the inductive step. Let n be a positive integer and assume that all Noetherian local rings A with ${{\mathrm{gl.dim}}}{A} = n-1$ are regular. Let $(A,{\mathfrak m})$ be a Noetherian local ring with ${{\mathrm{gl.dim}}}{A} = n$. Proposition 2.3 and Lemma 2.4 ensure that there exists an element $f \in {\mathfrak m}\backslash {\mathfrak m}^2$ that is not a zero-divisor. Proposition 2.1 implies that ${{\mathrm{gl.dim}}}{A/f} = n-1$. The inductive hypothesis guarantees that A / f is regular. Because $f \in {\mathfrak m}$ is not a zero-divisor, the local ring A is regular as well, which completes the proof of the inductive step. $\square$

3 Proof of Proposition 2.1

The key technical tool in the proof of Proposition 2.1 is the following lemma. Recall that if R is a ring, we use D(R) to denote the derived category of R-modules.

Lemma 3.1

Let $(A,{\mathfrak m})$ be a local ring. If $f \in {\mathfrak m}\backslash {\mathfrak m}^2$ is not a zero-divisor, then $k \otimes ^{\mathbf {L}}_A A/f$ and $k \oplus k[1]$ are isomorphic in D(A / f).

Proof of Proposition 2.1 assuming Lemma 3.1

Recall the isomorphism

$$\begin{aligned} k \otimes ^{\mathbf {L}}_A k \cong \left( k \otimes ^{\mathbf {L}}_A A/f\right) \otimes ^{\mathbf {L}}_{A/f} k \end{aligned}$$

in D(A / f) (see, for example, [3, Tag 06Y5], ). Lemma 3.1 yields that

$$\begin{aligned} k \otimes ^{\mathbf {L}}_A k \cong \left( k \otimes ^{\mathbf {L}}_A A/f\right) \otimes ^{\mathbf {L}}_{A/f} k \cong \left( k \oplus k[1]\right) \otimes ^{\mathbf {L}}_{A/f} k = \left( k \otimes ^{\mathbf {L}}_{A/f} k\right) \oplus \left( k \otimes ^{\mathbf {L}}_{A/f} k\right) [1]. \end{aligned}$$

Taking homology yields that

$$\begin{aligned} {{\mathrm{Tor}}}_i^A(k,k) \cong {{\mathrm{Tor}}}_{i-1}^{A/f}(k,k) \oplus {{\mathrm{Tor}}}_i^{A/f}(k,k) \end{aligned}$$

for all i. Recall the equality

$$\begin{aligned} {{\mathrm{gl.dim}}}{B} = \sup \left\{ i \mid {{\mathrm{Tor}}}_i^B(k,k) \not = 0\right\} \end{aligned}$$

in a Noetherian local ring B with residue field k. It follows that

$$\begin{aligned} {{\mathrm{gl.dim}}}{A}&= \sup \left\{ i \mid {{\mathrm{Tor}}}_i^A(k,k) \not = 0\right\} = 1 + \sup \left\{ i \mid {{\mathrm{Tor}}}_i^{A/f}(k,k) \not = 0\right\} \\&= 1 + {{\mathrm{gl.dim}}}{A/f}, \end{aligned}$$

as desired. $\square$

The strategy of the proof of Lemma 3.1 is to find a two-term complex $C_\bullet$ of A / f-modules representing $k \otimes ^{\mathbf {L}}_A A/f$ in D(A / f) and then produce a quasi-isomorphism from $C_\bullet$ to $k \oplus k[1]$. The description of $C_\bullet$, which we give in Lemma 3.2, follows from the fact that ${{\mathrm{Tor}}}_i^A(k,A/f) = 0$ for $i \ge 2$. The existence of the second quasi-isomorphism relies crucially on the assumption that $f \notin {\mathfrak m}^2$.

Lemma 3.2

Let $(A,{\mathfrak m},k)$ be a Noetherian local ring and suppose that $f \in {\mathfrak m}$ is not a zero-divisor. The complex

represents $k \otimes ^{\mathbf {L}}_A A/f$ in D(A / f), where $i: {\mathfrak m}\rightarrow A$ is the inclusion. Furthermore, we have $H_1(C_\bullet ) \cong H_0(C_\bullet ) = k.$

Proof

The short exact sequence of modules

yields an exact triple

in D(A / f). We will show the first two terms of the above sequence lie in the heart of D(A / f). Note that $A\otimes ^{\mathbf {L}}_A A/f \cong A \otimes _A A/f \cong A/f$ lies in the heart. To prove that ${\mathfrak m}\otimes ^{\mathbf {L}}_A A/f$ lies in the heart, it suffices to prove that ${{\mathrm{Tor}}}_i^A({\mathfrak m},A/f) = 0$ for all $i \ge 0$. We have a resolution of A / f given by

Therefore, ${{\mathrm{Tor}}}_i^A({\mathfrak m},A/f)$ is the ith homology of the complex

Because f is not a zero-divisor on ${\mathfrak m}$, we obtain that ${{\mathrm{Tor}}}_1^A({\mathfrak m},A/f) = 0$, and it is manifestly true that ${{\mathrm{Tor}}}_i^A({\mathfrak m},A/f) = 0$ for $i > 1$. It follows that ${\mathfrak m}\otimes ^{\mathbf {L}}_A A/f \cong {\mathfrak m}\otimes _A A/f$ lies in the heart.

The discussion of the previous paragraph yields that

$$\begin{aligned} {{\mathrm{Cone}}}\left( i \otimes _A A/f\right) \cong k \otimes ^{\mathbf {L}}_A A/f \end{aligned}$$

in D(A / f). The complex $C_\bullet$ represents $k \otimes ^{\mathbf {L}}_A A/f$ in D(A / f) because ${\mathfrak m}\otimes _A A/f$ and $A \otimes _A A/f$ (the domain and codomain of $i \otimes _A A/f$) are complexes concentrated in degree 0.

Finally, to see that $H_1(C_\bullet ) \cong H_0(C_\bullet ) = k,$ simply tensor the resolution $F_\bullet$ of A / f by k and recall that $H_i(C_\bullet ) = {{\mathrm{Tor}}}_i^A(k,A/f)$. $\square$

Proof of Lemma 3.1

We first prove the existence of a splitting $\theta$ of the canonical inclusion

Note that $f \otimes 1 \in \ker \left( i \otimes _A A/f\right)$ and $f \otimes 1 \not = 0$. Lemma 3.2 ensures that $H_1(C_\bullet ) \cong k$, so that $f \otimes 1$ generates $H_1(C_\bullet )$. Let $\zeta$ be the composite

$$\begin{aligned} \zeta : {\mathfrak m}\otimes _A A/f \cong {\mathfrak m}/f{\mathfrak m}\rightarrow {\mathfrak m}/{\mathfrak m}^2. \end{aligned}$$

Note that $\zeta (f \otimes 1) \not = 0$ because $f \notin {\mathfrak m}^2$. Because f generates the simple A / f-module $H_1(C_{\bullet }),$ the composite $\zeta \circ \psi$ is an injection from $H_1(C_\bullet )$ to ${\mathfrak m}/{\mathfrak m}^2$. Because both $H_1(C_\bullet )$ and ${\mathfrak m}/{\mathfrak m}^2$ are vector spaces over k, there exists a map $\xi : {\mathfrak m}/{\mathfrak m}^2 \rightarrow H_1(C_\bullet )$ splitting the inclusion $\zeta \circ \psi$. We can then let $\theta = \xi \circ \zeta$ to split the inclusion $\psi$.

The map $\theta$ fits into a commutative square

with surjective horizontal maps, which yields a quasi-isomorphism from $C_\bullet$ to $H_0(C_\bullet ) \oplus H_1(C_\bullet )[1]$. It follows from Lemma 3.2 that $k \otimes ^{\mathbf {L}}_A A/f \cong k \oplus k[1]$ in D(A / f), as desired. $\square$