# can the projective dimension be read from any projective resolution?

Let $$P_{\bullet}, P'_{\bullet}$$ be two projective resolutions of an $$R$$-module $$M$$. Denote their differentials by $$d,d'$$ respectively. Define $$M_i = \operatorname{ker} d_{i-1}, M'_i = \operatorname{ker} d'_{i-1}$$ to be the $$i$$-th syzygies of $$M$$. Then by the generalized Schanuel Lemma, $$M_i, M_i'$$ are projectively equivalent, i.e. there exist projective modules $$E_i,E'_i$$ such that $$E' \oplus M_i \cong E_i \oplus M_i'$$.

Suppose that $$M_i$$ is projective. If $$R$$ is local, then every projective module is free. Hence $$M_i'$$ is a direct summand of a free module, thus projective. Hence, if $$M$$ has finite projective dimension equal to $$n$$, then given any projective resolution, we can read the projective dimension as the least index $$i$$ such that $$M_i$$ is projective.

Now suppose that $$R$$ is not local and $$M$$ has projective dimension equal to $$n$$. Let $$(P_{\bullet},d)$$ be a projective resolution of $$M$$. Then $$\operatorname{ker} d_{n-1}$$ is a direct summand of a projective module. But a direct summand of a projective module need not be projective.

However, if an $$R$$-module $$M$$ is of finite presentation, then it is projective if and only if $$M_p$$ is $$R_p$$-free for every prime ideal $$p$$ of $$R$$ (e.g. Theorem 7.12, Matsumura, CRT).

Hence, if $$M$$ is a finite $$R$$-module of projective dimension $$n$$, and $$P_{\bullet}$$ any projective resolution, where the modules $$P_i$$ are finitely-generated, then the $$n$$-th syzygy $$K_n$$ of this resolution will be a direct summand of a projective module. By localizing at $$p \in \operatorname{Spec} R$$ we get that $$(K_n)_p$$ is free and so Theorem 7.12 of Matsumura applies giving that $$K_n$$ is projective.

I seem to have proved that: the projective dimension of a finite module $$M$$ can be read from any projective resolution, where the appearing projective modules are finite, as the smallest $$i$$ such that the $$i$$-th syzygy is projective.

Question 1: Are my arguments above correct?

Question 2: can the projective dimension of any module over a commutative ring be read from any projective resolution?

Remark: The motivation for question 2 is the definition of projective dimension by Bruns and Herzog, CMR, page 16, which also suggests that the answer to question 2 is affirmative.

## 1 Answer

You can argue using the criteria in Matsumura, but there's a simpler approach: a direct summand of a projective is projective. If $P$ is projective, then there exists $Q$ such that $P \oplus Q$ is free. If $P = P_1 \oplus Q_1$, then $P_1 \oplus Q_1 \oplus Q$ is free, so $P_1$ is projective.

For question $2$, the answer is yes, and you may wish to prove the following fact: if $M$ is any $R$-module, then the projective dimension of $M$ is at most $n$ iff every $n^\text{th}$ syzygy of $M$ is projective.

• Great answer! Indeed, your first argument is very simply and elegant. And i see the truth of your second remark. Thanks! – Manos Jan 24 '14 at 21:26