The Stacks project

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. $\square$

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. $\square$

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

with maps given by

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

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

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

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. $\square$

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 $\dim (R) \geq n$. $\square$

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. $\square$

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. $\square$

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. $\square$

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. $\square$