I'm a little confused by the definition of perfect complex on a scheme $X$, namely a complex $E^{\cdot}$ of $\mathcal{O}_X$-modules such that for every $x\in X$ there exists $U\ni x$ open such that $E^{\cdot}|_U$ is quasi-isomorphic to a bounded complex of locally free $\mathcal{O}_X$-modules of finite type.
My question could be silly, but is that true that one could just define a perfect complex as a complex locally quasi-isomorphic to a bounded complex of free $\mathcal{O}_X$-modules of finite type?
Given a perfect $E^{\cdot}$, I would argue that one can consider $U\ni x$ such that $E^{\cdot}|_U$ is quasi-isomorphic to $F^{\cdot}$, which is a bounded complex of locally free sheaves over $U$, hence find for every $k$ an open $U_k\ni x$ such that $F^k|_{U_k}$ is free, and taking finite intersection over those $k$ such that $F^k\neq 0$, I find another open $V=\bigcap U_k$ neighborhood of $x$ such that $F^{\cdot}|_V$ is in fact a complex of free $\mathcal{O}_X$-modules
Is this reasoning correct or am I wrong somewere? It sounds wierd to me because the definition is usually given talking abound locally free modules. If I'm wrong, could you suggest a counterexample? Thanks in advance
Best Answer
Yes, this is correct. The use of "locally free" in the definition may stem from the fact that for modules over a ring, you can define a perfect complex simply as one that is quasi-isomorphic to a bounded complex of locally free modules of finite type. That is, in the affine case, there's no need to restrict to an open set $U$ if you say "locally free" instead of "free".