I was trying to show that on a ringed space $(X, \mathcal{O})$, that $T^n(\mathscr{F})$ is free of finite rank if $\mathscr{F}$ is a free $\mathcal{O}$-module of finite rank. We recall that $T^r(\mathscr{F})$ is the sheafification of the presheaf $t^r(\mathscr{F})$ defined by $U \mapsto \mathscr{F}(U)^{\otimes r }$ for open sets $U$.

For simplicity assume that $\mathscr{F} = \mathcal{O}^{\oplus n}$. Then, if $U \subseteq X$ is open and $e_1, \dots, e_n$ is the canonical basis of $\mathcal{O}(U)^{\oplus n}$, then the usual argument for modules shows that $t^n( \mathscr{F})(U)$ has a basis given by $e_{j_1} \otimes e_{j_2} \otimes \cdots \otimes e_{j_r}$ for all $r$-tuples formed from $\{1, \dots, n\}$. This yields a canonical isomorphism $t^n(\mathscr{F})(U) \to \mathcal{O}(U)^{\oplus n^r}$. Since basis elements restrict to basis elements, this map is compatible with restrictions to form an isomorphism $t^n(\mathscr{F}) \to \mathcal{O}^{\oplus n^r}$.

This would then imply that $t^n(\mathscr{F})$ is already a sheaf, which seems off since the tensor product(as presheaves) of sheaves is not in general a sheaf.

Is it always a sheaf in the free situation?



1 Answer 1


What you want to show is that $F$ and $G$ are two (quasi-?)coherent sheaves of $\mathcal O_X$-modules, then for each open affine $U \subset X$ we have $$(F \otimes_{\mathcal O_X} G)(U) = F(U) \otimes_{\mathcal O(U)} G(U).$$

In general, you will certainly have to sheafify. For example, $$\dim \Gamma(\mathbb P^n, \mathcal O(1)) = n+1$$ and $\mathcal O(1)^{\otimes 2} = \mathcal O(2)$, but $$\dim \Gamma(\mathbb P^n, \mathcal O(2)) = \binom{n+2}{n} = \frac{(n+2)(n+1)}{2}\neq (n+1)^2.$$

  • $\begingroup$ Thanks, and that is a good example to keep in mind. Full disclosure, this comes from Hartshorne's exercise II.5.16 which deals with the general case of ringed spaces rather than schemes. That's why I'm not using this fact about affines. $\endgroup$
    – Daniel
    Sep 9, 2021 at 15:03

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