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Let $(X,\mathcal{A})$ be a ringed space. A complex $\mathcal{S}^{\bullet}$ of $\mathcal{A}$-modules is $\textit{perfect}$ if for any point $x\in X$, there exists an open neighborhood $U$ of $x$ and a bounded complex of finite rank free sheaves of $\mathcal{A}$-modules $\mathcal{E}^{\bullet}_{U}$ on $U$ such that the restriction $\mathcal{S}^{\bullet}|_U$ is quasi-isomorphic to $\mathcal{E}^{\bullet}_{U}$.

Moreover $\mathcal{S}^{\bullet}$ is $\textit{strictly perfect}$ if there exists a complex of finite rank locally free sheaves of $\mathcal{A}$-modules $\mathcal{E}^{\bullet}$ on $X$ such that $\mathcal{S}^{\bullet}$ is quasi-isomorphic to $\mathcal{E}^{\bullet}$.

Now I think the following statement is true:

Let $X$ be an affine scheme and $\mathcal{A}=\mathcal{O}_X$ be the sheaf of regular functions on $X$. Then any perfect complex on $(X,\mathcal{A})$ is actually a strictly perfect complex.

Is there any simple proof of the above statement?

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  • $\begingroup$ Your statement is false in general, a finitely generated non-free projective module concentrated in a fixed degree is a counterexample. $\endgroup$
    – Fernando Muro
    Commented Mar 9, 2015 at 22:40
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    $\begingroup$ I might be mistaken, but I'd have thought that the usual definition of "strictly perfect" (as found in, say, SGA6) would replace your "complex of finite rank free sheaves" with a complex of finite rank locally free sheaves. This would rule out Fernando's counterexample. $\endgroup$
    – Steven Landsburg
    Commented Mar 9, 2015 at 23:26
  • $\begingroup$ Thank you for pointing out! That is a typo and I have made correction in the definition of strictly perfect complex. $\endgroup$
    – Zhaoting Wei
    Commented Mar 10, 2015 at 2:43
  • $\begingroup$ It looks like if strictly perfect implies perfect on quasi-compact schemes on the nose, and affine schemes are quasi-compact. Please, correct me itf I'm missing something. $\endgroup$
    – Fernando Muro
    Commented Mar 10, 2015 at 6:48
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    $\begingroup$ Here's something that is true: On any scheme with an ample family of line bundles (e.g. quasi-projective, or affine, or separated regular Noetherian) every perfect complex is quasi-isomorphic to a strictly perfect complex. I'm not sure if that's what you meant by "is" in your claim, but the reference is SGA6 II.2.2.8 or Thomason-Trobaugh 2.3.1. $\endgroup$
    – Dylan Wilson
    Commented Mar 10, 2015 at 16:01

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