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

Comment #2234 by David Savitt on

Fussy remark: to see that a resolution exists as in the first sentence, I suggest adding a corollary to [00O5] to the effect that if $R$ is Noetherian and $M$ is finite and of projective dimension $d$, then $M$ has a projective resolution of length $d$ with all entries finite.

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