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I apologize in advance if this question is basic.

If $P_{\bullet}$ is a perfect complex over say a ring $R$ such that

  1. $H_{i}(P_{\bullet})=0 $ if $i\neq n$
  2. $H_{i}(P_{\bullet})=E$ if $i=n$

is $E$ a finitely generated $R$-module ?

What can we say about the homology of a generic perfect complex in general?

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    $\begingroup$ For any perfect complex $P_\bullet$, the lowest nonvanishing homology $H_n(P_\bullet)$ is a finitely presented $R$-module [Lurie, Spectral Algebraic Geometry, Corollary 7.2.4.5]. $\endgroup$
    – user147129
    Commented Mar 28, 2020 at 16:47

2 Answers 2

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Yes. Let $... 0\to P_r \to ... \to P_0 \to 0 ...$ be a complex of projective modules of finite type and denote by $Z_*$ the cycles. If $n=0$ it is clear. If not, $0\to Z_1\to P_1\to P_0\to 0$ is exact and so $Z_1$ is projective and of finite type. Then if $n=1$, $H_1(P)$ is of finite type. If $n\neq 1$, $0\to Z_2\to P_2\to Z_1\to 0$ is exact. And so on.

So the "last" nonzero homology module is of finite type.

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It is even finitely presented

See Lemma 14.1.27 of the book Derived Categories (also available at the arXiv at https://arxiv.org/abs/1610.09640).

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  • $\begingroup$ Dear Amnon: the downvotes might have been confusing, but I suspect they are connected with an unwritten rule at MO to avoid the appearance of author self-promotion, particularly with regard to books for sale. $\endgroup$
    – Todd Trimble
    Commented Mar 28, 2020 at 23:35
  • $\begingroup$ Thanks for the heads up! I'm glad the arxiv version saved the day. $\endgroup$
    – Amnon Yekutieli
    Commented Mar 29, 2020 at 19:30

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