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.
