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. $\square$

Comment #2771 by on

Sorry for a trivial nitpick, but on the first line of the proof, a comma is missing before "x_n", and some explanation for the "clearly" might be useful (why do the x_i generate the maximal ideal? it follows from Tag 00DV part (8), but this perhaps should be said).

Comment #5578 by Rankeya on

I am confused by the construction of the map $\alpha$. The Koszul complex $K_\bullet(R,x_\bullet)$ is not a resolution of $\kappa$. So why do we get a map from $K_\bullet(R,x_\bullet) \rightarrow F_\bullet$? The citation for the existence of $\alpha$ is Lemma 00LS, which implies that there exists a map $F_\bullet \rightarrow K_\bullet(R,\x_\bullet)$ and not the other way around.

Comment #5579 by Rankeya on

There is a tex error in my previous comment. I meant to say that Lemma 00LS implies the existence of a map $F_\bullet \rightarrow K_\bullet(R,x_\bullet)$ and not the other way around.

Comment #5580 by on

@Rankeya You are right, the reference doesn't work. Damn! But the arrow is certainly in the right direction, see for example Lemma 13.19.6.

Comment #5581 by Rankeya on

Please disregard my earlier comments. I guess I was confused by the reference to Lemma 00LS because the convention for a resolution of a module is a complex acyclic except in degree 0 (Def 00LQ), but you don't need $K_\bullet(R,x_\bullet)$ to be a 'resolution' to get the map $\alpha$; only $F_\bullet$ has to be a resolution.

So may be one should restate 00LS with weaker assumptions.

Comment #5582 by Rankeya on

I sent the last comment without refreshing the page and didn't realize you replied to my earlier comments with the appropriate citation fix. Thanks and sorry!

Comment #7594 by Omri Zemer on

Shouldn't the definition of the differential in the Koszul complex include $(-1)^{a+1}$ instead if the written $(-1)^{i+1}$?

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