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Suppose $R$ is a regular ring and $F^{\bullet}: 0\to F^0 \to F^1 \to \dots \to F^d \to 0$ is a complex of finite rank free $R$-modules. Is is true that $\mathrm{projdim}H^i(F^{\bullet}) \leq d$ for all $i$?

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2 Answers 2

No if $R$ has dimension $>1$. Take a finitely presented module $M$ of projective dimension $>1$. Choose a finite presentation of $M$, i.e. a complex $0\rightarrow F^0\rightarrow F^1\rightarrow 0$ as in your question with $H^1(F^\bullet)=M$. Then you get a counterexample.

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Since this is false in an embarrassingly obvious way, let me point out my motivation for asking. Suppose $(R,\mathfrak{m})$ is a local Cohen-Macaulay ring and $F^{\bullet}$ is as in the question. Then one can show that every minimal prime in the $R$-support of $H^{\ast}(F^{\bullet})$ has height $\leq d$. This would follow from the stronger but false statement which I queried. In the situation of Fernando Muro's counterexample, this theorem says that if $R^{m}\overset{f}{\to } R^{n} \to M \to 0$ is a presentation, then $M$ and $\ker{f}$ can't both have dimension $ <d-1$.

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