The Stacks project

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10.109 Regular rings and global dimension

We can use the material on rings of finite global dimension to give another characterization of regular local rings.

Proposition 10.109.1. Let $R$ be a regular local ring of dimension $d$. Every finite $R$-module $M$ of depth $e$ has a finite free resolution

\[ 0 \to F_{d-e} \to \ldots \to F_0 \to M \to 0. \]

In particular a regular local ring has global dimension $\leq d$.

Proof. The first part holds in view of Lemma 10.105.6 and Lemma 10.103.9. The last part follows from this and Lemma 10.108.12. $\square$

Lemma 10.109.2. Let $R$ be a Noetherian ring. Then $R$ has finite global dimension if and only if there exists an integer $n$ such that for all maximal ideals $\mathfrak m$ of $R$ the ring $R_{\mathfrak m}$ has global dimension $\leq n$.

Proof. We saw, Lemma 10.108.13 that if $R$ has finite global dimension $n$, then all the localizations $R_{\mathfrak m}$ have finite global dimension at most $n$. Conversely, suppose that all the $R_{\mathfrak m}$ have global dimension $n$. Let $M$ be a finite $R$-module. Let $0 \to K_ n \to F_{n-1} \to \ldots \to F_0 \to M \to 0$ be a resolution with $F_ i$ finite free. Then $K_ n$ is a finite $R$-module. According to Lemma 10.108.3 and the assumption all the modules $K_ n \otimes _ R R_{\mathfrak m}$ are projective. Hence by Lemma 10.77.2 the module $K_ n$ is finite projective. $\square$

Lemma 10.109.3. Suppose that $R$ is a Noetherian local ring with maximal ideal $\mathfrak m$ and residue field $\kappa $. In this case the projective dimension of $\kappa $ is $\geq \dim _\kappa \mathfrak m / \mathfrak m^2$.

Proof. Let $x_1 , \ldots , x_ n$ be elements of $\mathfrak m$ whose images in $\mathfrak m / \mathfrak m^2$ form a basis. Consider the Koszul complex on $x_1, \ldots , x_ n$. This is the complex

\[ 0 \to \wedge ^ n R^ n \to \wedge ^{n-1} R^ n \to \wedge ^{n-2} R^ n \to \ldots \to \wedge ^ i R^ n \to \ldots \to R^ n \to R \]

with maps given by

\[ e_{j_1} \wedge \ldots \wedge e_{j_ i} \longmapsto \sum _{a = 1}^ i (-1)^{i + 1} x_{j_ a} e_{j_1} \wedge \ldots \wedge \hat e_{j_ a} \wedge \ldots \wedge e_{j_ i} \]

It is easy to see that this is a complex $K_{\bullet }(R, x_{\bullet })$. Note that the cokernel of the last map of $K_{\bullet }(R, x_{\bullet })$ is $\kappa $ by Lemma 10.19.1 part (8).

If $\kappa $ has finite projective dimension $d$, then we can find a resolution $F_{\bullet } \to \kappa $ by finite free $R$-modules of length $d$ (Lemma 10.108.7). By Lemma 10.101.2 we may assume all the maps in the complex $F_{\bullet }$ have the property that $\mathop{\mathrm{Im}}(F_ i \to F_{i-1}) \subset \mathfrak m F_{i-1}$, because removing a trivial summand from the resolution can at worst shorten the resolution. By Lemma 10.70.4 we can find a map of complexes $\alpha : K_{\bullet }(R, x_{\bullet }) \to F_{\bullet }$ inducing the identity on $\kappa $. We will prove by induction that the maps $\alpha _ i : \wedge ^ i R^ n = K_ i(R, x_{\bullet }) \to F_ i$ have the property that $\alpha _ i \otimes \kappa : \wedge ^ i \kappa ^ n \to F_ i \otimes \kappa $ are injective. This shows that $F_ n \not= 0$ and hence $d \geq n$ as desired.

The result is clear for $i = 0$ because the composition $R \xrightarrow {\alpha _0} F_0 \to \kappa $ is nonzero. Note that $F_0$ must have rank $1$ since otherwise the map $F_1 \to F_0$ whose cokernel is a single copy of $\kappa $ cannot have image contained in $\mathfrak m F_0$.

Next we check the case $i = 1$ as we feel that it is instructive; the reader can skip this as the induction step will deduce the $i = 1$ case from the case $i = 0$. We saw above that $F_0 = R$ and $F_1 \to F_0 = R$ has image $\mathfrak m$. We have a commutative diagram

\[ \begin{matrix} R^ n & = & K_1(R, x_{\bullet }) & \to & K_0(R, x_{\bullet }) & = & R \\ & & \downarrow & & \downarrow & & \downarrow \\ & & F_1 & \to & F_0 & = & R \end{matrix} \]

where the rightmost vertical arrow is given by multiplication by a unit. Hence we see that the image of the composition $R^ n \to F_1 \to F_0 = R$ is also equal to $\mathfrak m$. Thus the map $R^ n \otimes \kappa \to F_1 \otimes \kappa $ has to be injective since $\dim _\kappa (\mathfrak m / \mathfrak m^2) = n$.

Let $i \geq 1$ and assume injectivity of $\alpha _ j \otimes \kappa $ has been proved for all $j \leq i - 1$. Consider the commutative diagram

\[ \begin{matrix} \wedge ^ i R^ n & = & K_ i(R, x_{\bullet }) & \to & K_{i-1}(R, x_{\bullet }) & = & \wedge ^{i-1} R^ n \\ & & \downarrow & & \downarrow & & \\ & & F_ i & \to & F_{i-1} & & \end{matrix} \]

We know that $\wedge ^{i-1} \kappa ^ n \to F_{i-1} \otimes \kappa $ is injective. This proves that $\wedge ^{i-1} \kappa ^ n \otimes _{\kappa } \mathfrak m/\mathfrak m^2 \to F_{i-1} \otimes \mathfrak m/\mathfrak m^2$ is injective. Also, by our choice of the complex, $F_ i$ maps into $\mathfrak mF_{i-1}$, and similarly for the Koszul complex. Hence we get a commutative diagram

\[ \begin{matrix} \wedge ^ i \kappa ^ n & \to & \wedge ^{i-1} \kappa ^ n \otimes \mathfrak m/\mathfrak m^2 \\ \downarrow & & \downarrow \\ F_ i \otimes \kappa & \to & F_{i-1} \otimes \mathfrak m/\mathfrak m^2 \end{matrix} \]

At this point it suffices to verify the map $\wedge ^ i \kappa ^ n \to \wedge ^{i-1} \kappa ^ n \otimes \mathfrak m/\mathfrak m^2$ is injective, which can be done by hand. $\square$

Lemma 10.109.4. Let $R$ be a Noetherian local ring. Suppose that the residue field $\kappa $ has finite projective dimension $n$ over $R$. In this case $\dim (R) \geq n$.

Proof. Let $F_{\bullet }$ be a finite resolution of $\kappa $ by finite free $R$-modules (Lemma 10.108.7). By Lemma 10.101.2 we may assume all the maps in the complex $F_{\bullet }$ have to property that $\mathop{\mathrm{Im}}(F_ i \to F_{i-1}) \subset \mathfrak m F_{i-1}$, because removing a trivial summand from the resolution can at worst shorten the resolution. Say $F_ n \not= 0$ and $F_ i = 0$ for $i > n$, so that the projective dimension of $\kappa $ is $n$. By Proposition 10.101.9 we see that $\text{depth}_{I(\varphi _ n)}(R) \geq n$ since $I(\varphi _ n)$ cannot equal $R$ by our choice of the complex. Thus by Lemma 10.71.3 also $\dim (R) \geq n$. $\square$

Proposition 10.109.5. A Noetherian local ring whose residue field has finite projective dimension is a regular local ring. In particular a Noetherian local ring of finite global dimension is a regular local ring.

Proof. By Lemmas 10.109.3 and 10.109.4 we see that $\dim (R) \geq \dim _\kappa (\mathfrak m /\mathfrak m^2)$. Thus the result follows immediately from Definition 10.59.9. $\square$

Lemma 10.109.6. A Noetherian local ring $R$ is a regular local ring if and only if it has finite global dimension. In this case $R_{\mathfrak p}$ is a regular local ring for all primes $\mathfrak p$.

Proof. By Propositions 10.109.5 and 10.109.1 we see that a Noetherian local ring is a regular local ring if and only if it has finite global dimension. Furthermore, any localization $R_{\mathfrak p}$ has finite global dimension, see Lemma 10.108.13, and hence is a regular local ring. $\square$

By Lemma 10.109.6 it makes sense to make the following definition, because it does not conflict with the earlier definition of a regular local ring.

Definition 10.109.7. A Noetherian ring $R$ is said to be regular if all the localizations $R_{\mathfrak p}$ at primes are regular local rings.

It is enough to require the local rings at maximal ideals to be regular. Note that this is not the same as asking $R$ to have finite global dimension, even assuming $R$ is Noetherian. This is because there is an example of a regular Noetherian ring which does not have finite global dimension, namely because it does not have finite dimension.

Lemma 10.109.8. Let $R$ be a Noetherian ring. The following are equivalent:

  1. $R$ has finite global dimension $n$,

  2. there exists an integer $n$ such that all the localizations $R_{\mathfrak m}$ at maximal ideals are regular of dimension $\leq n$ with equality for at least one $\mathfrak m$, and

  3. there exists an integer $n$ such that all the localizations $R_{\mathfrak p}$ at prime ideals are regular of dimension $\leq n$ with equality for at least one $\mathfrak p$.

Proof. This is a reformulation of Lemma 10.109.2 in view of the discussion surrounding Definition 10.109.7. See especially Propositions 10.109.1 and 10.109.5. $\square$

Lemma 10.109.9. Let $R \to S$ be a local homomorphism of local Noetherian rings. Assume that $R \to S$ is flat and that $S$ is regular. Then $R$ is regular.

Proof. Let $\mathfrak m \subset R$ be the maximal ideal and let $\kappa = R/\mathfrak m$ be the residue field. Let $d = \dim S$. Choose any resolution $F_\bullet \to \kappa $ with each $F_ i$ a finite free $R$-module. Set $K_ d = \mathop{\mathrm{Ker}}(F_{d - 1} \to F_{d - 2})$. By flatness of $R \to S$ the complex $0 \to K_ d \otimes _ R S \to F_{d - 1} \otimes _ R S \to \ldots \to F_0 \otimes _ R S \to \kappa \otimes _ R S \to 0$ is still exact. Because the global dimension of $S$ is $d$, see Proposition 10.109.1, we see that $K_ d \otimes _ R S$ is a finite free $S$-module (see also Lemma 10.108.3). By Lemma 10.77.5 we see that $K_ d$ is a finite free $R$-module. Hence $\kappa $ has finite projective dimension and $R$ is regular by Proposition 10.109.5. $\square$


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