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Let $K$ be a field. Let $X$ be a scheme over $K$. We denote by $K_s$ and by $\bar{K}$ the separable closure and the algebraic closure of $K$ respectively. By base change we have the schemes $X_{K_s}$ and $X_{\bar{K}}$. I have the following questions.

  1. Is it possible that $X_{K_s}$ and $X_{\bar{K}}$ are isomorphic as schemes over $K$?
  2. Is it possible to have a bijection between $X_{K_s}(K_s)$ and $X_{\bar{K}}(\bar{K})$?

In the case I need, for example, $X_\bar{K}\simeq\mathbb{G}_{m,\bar{K}}^n$, where $n$ is a positive integer and $\mathbb{G}_{m,\bar{K}}=$Spec$(\frac{\bar{K}[X,Y]}{(XY-1)})$. Is some of the previous points satisfied in this case?

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    $\begingroup$ If $K_s \not \cong \bar K$, then $X_{K_s}$ and $X_{\bar K}$ are schemes defined over different fields, so there is no hope for them to be isomorphic. $\endgroup$
    – Daniel Loughran
    Commented Jun 11, 2014 at 9:41
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    $\begingroup$ Moreover $X_{\bar{K}}(K_s)$ does not make sense (unless $K_s=\bar{K}$ ...). $\endgroup$
    – abx
    Commented Jun 11, 2014 at 10:25
  • $\begingroup$ Thank you for the observation. I edited the question. In point (1) I should have been clearer, I think that, anyway, there is no possibility for this to be true. In point (2) I corrected my mistake. Anyway, I am fearing that also this can never be true. $\endgroup$
    – Pgatti
    Commented Jun 11, 2014 at 11:19
  • $\begingroup$ Perhaps you are after the fact that base change along the map $X_{\overline{K}} \rightarrow X_{K_s}$ induces an equivalence of etale sites and moreover that pushforward and pullback along that map induce quasi-inverse equivalences of etale topoi. This is all valid for any surjective integral radiciel morphism of schemes (of which any base change of ${\rm{Spec}}(\overline{K}) \rightarrow {\rm{Spec}}(K_s)$ is an example). $\endgroup$
    – user27920
    Commented Jun 11, 2014 at 12:17
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    $\begingroup$ The answer to question 2 is obviously NO: If $X=\mathbf A^1$ is the affine line, then $X_K(K)=K$, for every field (even, ring) $K$. So $X_{K_s}(K_s)=K_s$ and $X_{\overline K}(\overline K)=\overline K$... $\endgroup$
    – ACL
    Commented Jun 11, 2014 at 14:03

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Let $X$ be a scheme over $K$ with $X_{\bar K}$ isomorphic to a power of the multiplicative group and you want to know if it is possible for $X_{\bar K}$ and $X_{K_s}$ to be isomorphic as $K$-schemes. The answer is no, unless $K$ is perfect. Suppose otherwise and, after replacing $K$ by a finite extension we can assume that $X$ is already isomorphic to a power of the multiplicative group and the hypothesis is that $K_s[x_1^{\pm 1},\ldots,x_m^{\pm 1}]$ is isomorphic to ${\bar K}[x_1^{\pm 1},\ldots,x_m^{\pm 1}]$ as $K$-algebras. If that's so, then the group of units of the two rings are isomorphic as groups and the elements of $K_s^*,{\bar K}^*$ respectively are the units which are divisible by all $n, (n,p)=1$, where $p$ is the characteristic. But, if $K$ is not perfect, there is an element of $K^*$ which is not arbitrarily divisible by powers of $p$ in $K_s^*$.

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