Symmetric Powers Commute with Pullback of (Locally Free) Sheaves

algebraic-geometry

This question was asked here and is exercise II 5.16 e) in Hartshorne. It is a small remark after Theorem 16.3.7 in Vakil's FOAG, which the author says we should be able to prove as a Corollary immediately. Its proof is omitted on Stacks Project. This is self-study.

We seek to show symmetric/exterior powers commute with pulling back locally free sheaves. More precisely, if $\mathcal F$ is a locally free sheaf on, say, scheme $Y$, and $\pi: X \to Y$ is a morphism of schemes, then

$$\pi^* \operatorname{Sym}^n \mathcal F \simeq \operatorname{Sym}^n \pi^* \mathcal F$$

where $\pi^*$ is the pullback functor and $\operatorname{Sym}^n$ means taking the tensor product $T^n$ and modding out by the two-sided ideal $I$ "where tensor multiplication is commutative."

Things I am allowed to assume: $T^n$ commutes with $\pi^*$, and (by induction), $n = 2$. I also can assume that $\pi^*$ is right-exact, so if we take the exact sequence

$$0 \to I \to T^2 \mathcal F \to \operatorname{Sym}^2 \mathcal F \to 0$$

we know that

$$\pi^*I \to \pi^*T^2 \mathcal F \simeq T^2 \pi^* \mathcal F \to \pi^*\operatorname{Sym}^2 \mathcal F \to 0$$

is exact. This leads me to believe we merely need to show that $\pi^*I$ is the kernel of $T^2 \pi^* \mathcal F \to \pi^*\operatorname{Sym}^2 \mathcal F$. I suspect I need to use the fact that $\mathcal F$ is locally free somehow, but am unsure how to do this.

I have found two completely unconvincing solutions in Hartshorne solution documents, one of which merely states that it is obvious that $\pi^*I$ is the required kernel, and the other of which just says the overall claim is true by universal properties. I would prefer to stay away from the latter-style argument, because in FOAG, we have thus far never established the universal properties of symmetric powers.

Best Answer

Since everything involved is local, we may replace $X$ and $Y$ by affine schemes $\operatorname{Spec} B$ and $\operatorname{Spec} A$ and $\mathcal{F}\cong \mathcal{O}_Y^n\cong \widetilde{A}^n$. Then $I$ is the sheaf corresponding to the $A$-module $J$ generated by $\{e_i\otimes e_j - e_j\otimes e_i\}$ as $e_i,e_j$ run over the standard basis vectors of $A^n$. The pullback of the sheaf $I$ to $X$ corresponds to taking the sheaf associated to $J\otimes_A B$, which is the $B$-module generated by $\{e_i\otimes e_j-e_j\otimes e_i\}$ as $e_i,e_j$ run over the standard basis vectors of $B^n$. But this is exactly the kernel of $T^2\mathcal{O}_X^n\to \operatorname{Sym}^2\mathcal{O}_X^n$, so you have that $\pi^*\operatorname{Sym}\mathcal{O}_Y^n \cong \operatorname{Sym}^2 \mathcal{O}_X^n \cong \operatorname{Sym}^2\pi^*\mathcal{O}_Y$.

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