Fpqc-morphisms are epimorphisms
问题内容
How to prove that for any faithfully flat quasi-compact (or, more generally, fpqc) morphisms are epimorphisms?
My attempt: Let $f:X\to Y$ be faithfully flat quasi-compact (or, more generally, fpqc) and suppose that $g_1,g_2: Y\to Z$ are two morphisms such that $g_1\circ f=g_2\circ f$. Since $f$ is surjective, we have $g_1=g_2$ set-theoretically and we denote such map by $g$. It then suffices to prove that $g_1^\flat=g_2^\flat:\mathscr O_Z\to g_*\mathscr O_Y$. By assumption, we have $(g_1\circ f)^\flat=(g_2\circ f)^\flat:\mathscr O_Z\to g_*\mathscr O_Y\to g_*f_*\mathscr O_X$. Hence it suffices to show that $g_*\mathscr O_Y\to g_*f_*\mathscr O_X$ is injective. Since $g_*$ is left exact, it suffices to show that $f^\flat:\mathscr O_Y\to f_*\mathscr O_X$ is injective. Then I got stuck here.
I found some proofs using the theory of fpqc-sheaves but I am not satisfied with such answer since it is an easy exercise in Görtz-Wedhorn's book Algebraic Geometry I-Schemes. I would like an elementary proof, i.e., I want to use the results in this book only. I don't know if one can prove the injectivity $f^\flat:\mathscr O_Y\to f_*\mathscr O_X$ directly. Can someone help me?
回答 (1)
First of all, it should be mentioned that fpqc morphisms are more than just epimorphisms: they are effective epimorphisms. (And since fpqc morphisms are stable under base change, it even follows that they are universally effective epimorphisms.) This means that fpqc morphisms satisfy the descent property. A proof can be found in the Stacks Project, for example.
But your question can be answered much more easily:
We need to show that for every open subset $V \subseteq Y$ the map $f^\sharp : \mathcal{O}_Y(V) \to \mathcal{O}_X(f^{-1}(V))$ is injective. Clearly, we may assume that $V$ is affine. Next, write $V$ as the $f$-image of a quasi-compact open subset of $X$, and choose a finite open affine covering $\bigcup_i U_i$ of it. Since $\bigcup_i U_i \subseteq f^{-1}(V)$, the ring $\mathcal{O}_X(f^{-1}(V))$ maps into $\prod_i \mathcal{O}_X(U_i)$, so it suffices to prove that $f^\sharp : \mathcal{O}_Y(V) \to \prod_i \mathcal{O}_X(U_i)$ is injective. Write $V = \mathrm{Spec}(A)$ and $U_i = \mathrm{Spec}(B_i)$. Then $B_i$ is a flat $A$-algebra. Moreover, since $f : \coprod_i U_i \to V$ is surjective, $\prod_i B_i$ is a faithfully flat $A$-algebra (SP/00HQ), and we need to prove that $A \to \prod_i B_i$ is injective.
So it all comes down to the affine case: If $A \to B$ is a faithfully flat ring map, then $A \to B$ is injective. For the sake of completeness, here is the direct proof (see SP/05CK for a generalization). If $I$ is the kernel of $A \to B$, then since $B$ is flat over $A$, the map $I \otimes_A B \to A \otimes_A B$ is injective. But it is also zero since $i \otimes b$ is mapped to $i \otimes b = 1 \otimes ib = 0$. Hence, $I \otimes_A B = 0$. Since $B$ is faithfully flat over $A$, we conclude $I = 0$.