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.


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.

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

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