TITLE: Nonaffines must base change to nonaffines
QUESTION [1 upvotes]: Let $F \subseteq E$ be fields, and let $X$ be an affine scheme of finite type over $E$.  There is a general problem of finding affine schemes $X_0$ over $F$ such that $X \cong_E X_0 \otimes_F E$.  For example, if $E/F$ is Galois, then the $F$-isomorphism classes of such $X_0$ are parameterized by the cohomology set $H^1(\textrm{Gal}(E/F), \textrm{Aut}_E(X))$.
My question is, is it possible for there to exist a nonaffine scheme $X_0$ over $F$ such that $X \cong X_0 \times_F E$?  In general, I don't expect nonaffine schemes to become affine after base change, but I don't know.  
Attempt: Use Serre's affine criterion.  If $\mathcal F$ is any quasicoherent sheaf of $\mathcal O_{X_0}$-modules, then $H^1(X_0, \mathcal F) \otimes_F E = H^1(X, \pi_X^{\ast}(\mathcal F))  = 0$, where $\pi_X: X \rightarrow X_0$ is the projection.  Hence $H^1(X_0,\mathcal F) = 0$.  The problem is I don't know that $X_0$ is Noetherian, or separated.

REPLY [0 votes]: We can use this version 01XF of Serre's criterion. We have that $X_{0}$ is quasi-compact by e.g. 02KQ. The argument in our case is that the morphism $\pi_{X} : X \to X_{0}$ is a quasi-compact, surjective morphism and $X$ is quasi-compact so $X_{0}$ is quasi-compact (just point-set topology). One shows that $X_{0}$ is quasi-separated in the same way. The isomorphism $H^{1}(X_{0},\mathcal{F}) \otimes_{F} E \simeq H^{1}(X,\pi_{X}^{\ast}(\mathcal{F}))$ follows from flat base change 02KH.
Here is a general result: 02L5 If $S' \to S$ is a faithfully flat morphism of affine schemes and $X \to S$ is an $S$-scheme such that $X' := X \times_{S} S'$ is affine, then $X$ is affine.
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