Now I'm reading Cornell and Silverman's "Arithmetic Geometry" and I have trouble with a statement in this book. On page 39 of this book, the author says "Now $G\to Q$ is a surjection, consequently it is dominated by a covering in the fpqc topology. We may suppose this is of the form $X\to G\to Q$ and $X\to Q$ is fpqc". My question is this: To what extent this is true?

I think this is too strong to be true, since if it is true, then we can do some kind of 'surjective descent'. Assume this statement is true, and let $P$ be a class of morphisms of schemes which is stable under base change and fpqc-local. Let $Y\to Z$ be a surjective morphism of schemes. Then there exist morphisms $X_i\to Y$ such that $\left\{X_i\to Z\right\}$ is a fpqc covering. Let $Z'\to Z$ be a morphism such that the base change $Y\times_{Z}Z'\to Y$ is in $P$. Then since $P$ is stable under base change, $X_i\times_Z Z'\to X_i$ are also in $P$. Since $P$ is fpqc-local and $\left\{X_i\to Z\right\}$ is a fpqc covering, $Z'\to Z$ is also in $P$. Hence $P$ is in fact 'surjective-local', i.e. any class of morphism which is closed under base change and fpqc-local is in fact 'surjective-local'. In my narrow point of view, this is too strong to be true to any extent.

In fact, by using the above idea, we can construct a counterexample. $\text{Spec }k\to \text{Spec }k[x]/(x^2)$ is a surjective morphism which is not flat. Consider base change of this morphism by $\text{Spec }k\to \text{Spec }k[x]/(x^2)$. Since any morphism $X\to \text{Spec }k$ is faithfully flat (if $X$ is nonempty), the base change is faithfully flat. Then by the above argument, since we can let $P$ be `faithfully flat morphisms', $\text{Spec }k\to \text{Spec }k[x]/(x^2)$ is faithfully flat. This is contradiction.

Hence I want to ask what is the intention of the above statement. Despite I have a counterexample, I am asking this question since if the statement is true to any extent, then I think it will be very important (because of the 'surjective descent'). Thanks in advance.

  • $\begingroup$ Presumably there is no way to dominate $\mathbb A^1 \setminus \{0\} \cup \{0\} \to \mathbb A^1$ by an fpqc covering. This is maybe the nicest possible counterexample - everything smooth, finite type. Is there something in the context of the book that would rule out this kind of counterexample? $\endgroup$ – Will Sawin Apr 10 at 13:32
  • $\begingroup$ The author used this statement to prove the following theorem: If $G$ is a finite flat group scheme over $S$ ad $H$ is a (finite flat) subgroup scheme of $G$, then $G/H$ exists and it is flat over $S$. The statement is used to prove $G/H$ is flat. We assume every scheme is Noetherian, but there are no further assumptions up to this point. Since I found another proof which does not use the statement, so I can just skip this proof. However, I am curious whether this statement is a totally wrong statement or right in some good circumstances. $\endgroup$ – Daebeom Choi Apr 10 at 15:19
  • $\begingroup$ However, considering your counterexample, I think it will be very hard to find a reasonable setup where this statement is true.. $\endgroup$ – Daebeom Choi Apr 10 at 15:19
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    $\begingroup$ The author refers to Raynaud's paper in the Driebergen conference on local fields. Raynaud's argument there makes clever use of the equivalence relation defined by the action; this essential part is missing in Shatz's paper. $\endgroup$ – Laurent Moret-Bailly Apr 10 at 16:47
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    $\begingroup$ Bhargav Bhatt told me that the statement the author may have meant that this is true if $G \to Q$ induces a surjection on sheaves in the fpqc topology - i.e. if any map from a scheme $Y$ to $Q$ arises fpqc-locally on $Y$ from a map to $G$. This implication is clear on taking $Y = Q$. The idea is that at this point in the argument the sheaf of $Q$ is already known as the quotient sheaf and so this condition is automatic. $\endgroup$ – Will Sawin Apr 11 at 1:21

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