## 10.110 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 10.110.1. 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 10.106.6 and Lemma 10.104.9. The last part follows from this and Lemma 10.109.12. $\square$

Lemma 10.110.2. 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 10.109.13 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 $\leq n$. 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 10.109.3 and the assumption all the modules $K_ n \otimes _ R R_{\mathfrak m}$ are projective. Hence by Lemma 10.78.2 the module $K_ n$ is finite projective. $\square$

Lemma 10.110.3. 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

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

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 10.20.1 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 10.109.7). By Lemma 10.102.2 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 10.71.4 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 10.110.4. 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 10.109.7). By Lemma 10.102.2 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 10.102.9 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 10.72.3 also $\dim (R) \geq n$. $\square$

Proposition 10.110.5. Let $(R, \mathfrak m, \kappa )$ be a Noetherian local ring. The following are equivalent

1. $\kappa$ has finite projective dimension as an $R$-module,

2. $R$ has finite global dimension,

3. $R$ is a regular local ring.

Moreover, in this case the global dimension of $R$ equals $\dim (R) = \dim _\kappa (\mathfrak m/\mathfrak m^2)$.

Proof. We have (3) $\Rightarrow$ (2) by Proposition 10.110.1. The implication (2) $\Rightarrow$ (1) is trivial. Assume (1). By Lemmas 10.110.3 and 10.110.4 we see that $\dim (R) \geq \dim _\kappa (\mathfrak m /\mathfrak m^2)$. Thus $R$ is regular, see Definition 10.60.10 and the discussion preceding it. Assume the equivalent conditions (1) – (3) hold. By Proposition 10.110.1 the global dimension of $R$ is at most $\dim (R)$ and by Lemma 10.110.3 it is at least $\dim _\kappa (\mathfrak m/\mathfrak m^2)$. Thus the stated equality holds. $\square$

Lemma 10.110.6. 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 10.110.5 and 10.110.1 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 10.109.13, and hence is a regular local ring. $\square$

By Lemma 10.110.6 it makes sense to make the following definition, because it does not conflict with the earlier definition of a regular local ring.

Definition 10.110.7. 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 10.110.8. Let $R$ be a Noetherian ring. The following are equivalent:

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

2. $R$ is a regular ring of dimension $n$,

3. 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

4. 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 follows from the discussion above. More precisely, it follows by combining Definition 10.110.7 with Lemma 10.110.2 and Proposition 10.110.5. $\square$

Lemma 10.110.9. 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 10.110.1, we see that $K_ d \otimes _ R S$ is a finite free $S$-module (see also Lemma 10.109.3). By Lemma 10.78.6 we see that $K_ d$ is a finite free $R$-module. Hence $\kappa$ has finite projective dimension and $R$ is regular by Proposition 10.110.5. $\square$

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