Flat base change and the quasicoherator

Question

Flat base change seems to always be stated with a quasicompact, quasiseparated hypothesis--this is needed so that the pushforward of a quasicoherent sheaf is quasicoherent. However, we can remove this hypothesis if we use the quasicoherator.

Specifically, let $f : X\to S$ be a morphism of schemes, and let $f_* : \operatorname{Mod}(\mathcal{O}_X) \to \operatorname{Mod}(\mathcal{O}_S)$ be the usual pushforward.

There is a functor $f_{\#} : \operatorname{Qcoh}(X) \to \operatorname{Qcoh}(S)$ which sends $\mathcal{F}$ to the quasicoherator applied to $f_*(\mathcal{F}).$ (See https://stacks.math.columbia.edu/tag/08D6 for additional information about this coherator.)

Suppose that $j : U \to S$ is an open immersion. Is it true that in the Cartesian diagram $$\require{AMScd}\begin{CD} f^{-1}(U) @>>j'> X \\ @VVf'V @VVfV \\ U @>>j> S, \end{CD}$$ we have the analogue of flat base change -- that the natural map $j^*f_{\#} \to f'_{\#}(j')^*$ is an isomorphism?

I'm also interested if flat base change holds for any flat base change with the quasicoherator, but I'm more skeptical of that -- to be honest, I even am skeptical that this statement holds! Does anyone know of a relevant reference, a proof, or a counterexample?

Answer 1

First, we note that the coherator on an affine scheme is given by the associated quasi-coherent sheaf to the global section (cf. Thomason-Trobaugh, Appendix B. 14).

Let $A$ be a ring. Write $S :\overset{\mathrm{def}}{=} \mathrm{Spec}(A)$. For any element $f\in A$, write $S_f:\overset{\mathrm{def}}{=} \mathrm{Spec}(A_f)$. Write $\tilde{(-)}$ for the associated quasi-coherent sheaf to the $A$-module $(-)$.

Then, the functor $\Gamma(S, -)\tilde{ \ }: \mathsf{Mod}(\mathcal{O}_S) \to \mathsf{QCoh}(S)$ is right adjoint to the inclusion $\mathsf{QCoh}(S)\hookrightarrow \mathsf{Mod}(\mathcal{O}_S)$. Hence, it holds that $Q_S = \Gamma(S, -)\tilde{ \ }$.

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Here, we give a simple counterexample: Let $(R, \mathfrak{m})$ be a DVR. Write

• $S:\overset{\mathrm{def}}{=} \mathrm{Spec}(R)$, $K$ for the field of fractions of $R$, $U :\overset{\mathrm{def}}{=} \mathrm{Spec}(K)$, $j:U\hookrightarrow S$ for the natural open immersion,
• $X :\overset{\mathrm{def}}{=} \coprod_{i\in \mathbb{N}}\mathrm{Spec}(R/\mathfrak{m}^i)$, $F:\overset{\mathrm{def}}{=} \mathcal{O}_X$, and
• $f:X\to S$ for the natural morphism.

Then, $f_{\#}F$ is equal to the associated quasi-coherent sheaf to $\prod_{i\in \mathbb{N}}R/\mathfrak{m}^i$. Hence, $j^*f_{\#}F\neq 0$. On the other hand, since $X\times_{\mathrm{Spec}(K)} S = \emptyset$, the natural morphism $0\neq j^*f_{\#}F \to f_{\#}'{j'}^*F = 0$ is not an isomorphism.

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On a related note, we can prove the following claim:

Claim. Let $f:X\to S$ be a morphism of schemes and $F$ a quasi-coherent sheaf on $X$. Assume that for any open immersion $j:U \hookrightarrow S$, the natural morphism $j^*f_{\#}F\to f_{\#}'{j'}^*F$ is an isomorphism.

Then, $f_*F$ is a quasi-coherent sheaf on $S$.

Proof. Write $Q_{(-)}(-)$ for the coherator. Let $j:U\hookrightarrow S$ be an open immersion. Since $f'_{\#}(-) = Q_U(f'_*(-))$, it follows from the assumption that \begin{align} f_{\#}F(U) &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, j^*f_{\#}F) \xrightarrow{\sim} \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U, f'_{\#}{j'}^*F) \\ &= \mathrm{Hom}_{\mathsf{QCoh}(U)}(\mathcal{O}_U,Q_U(f_*'{j'}^*F)) = \mathrm{Hom}_{\mathsf{Mod}(\mathcal{O}_U)}(\mathcal{O}_U,f_*'{j'}^*F) \\ &= \Gamma(U, f_*'{j'}^*F) = F(f^{-1}(U)) = f_*F(U). \end{align} This implies that the natural morphism $f_{\#}F \to f_*F$ is an isomorphism of sheaves. Thus, in particular, $f_*F$ is quasi-coherent. ◻︎

By the above claim, we conclude that if $f_*F$ is not quasi-coherent, then there exists an open immersion $j:U\hookrightarrow S$ such that the natural morphism $j^*f_{\#}F\to f_{\#}'{j'}^*F$ is not an isomorphism.