Lemma 15.89.6. Let $R$ be a Noetherian local ring. Let $I \subset R$ be an ideal and let $E$ be a nonzero module over $R/I$. If $R/I$ has finite projective dimension and $E$ has finite projective dimension over $R/I$, then $E$ has finite projective dimension over $R$ and

$\text{pd}_ R(E) = \text{pd}_ R(R/I) + \text{pd}_{R/I}(E)$

Proof. We will use that, for a finite module, having finite projective dimension over $R$, resp. $R/I$ is the same as being a perfect module, see discussion following Definition 15.69.1. We see that $E$ has finite projective dimension over $R$ by Lemma 15.69.7. Thus we can apply Auslander-Buchsbaum (Algebra, Proposition 10.110.1) to see that

$\text{pd}_ R(E) + \text{depth}(E) = \text{depth}(R),\quad \text{pd}_{R/I}(E) + \text{depth}(E) = \text{depth}(R/I),$

and

$\text{pd}_ R(R/I) + \text{depth}(R/I) = \text{depth}(R)$

Note that in the first equation we take the depth of $E$ as an $R$-module and in the second as an $R/I$-module. However these depths are the same (this is trivial but also follows from Algebra, Lemma 10.71.10). This concludes the proof. $\square$

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