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

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 10.19.1 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

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. $\square$

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## Comments (2)

Comment #2771 by Darij Grinberg on

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