I am not sure whether the title is appropriate for this question or not. I am sorry if there is anyone who is confused with the title and the contents.

What I want to ask is the following: let $k$ be a field and $\bar{k}$ be a separable closure of $k$. Let $X$ be a 'good' scheme over $k$ (you can give any condition on $X$, for example, separated, geometrically connected, locally Noetherian, etc). Let $f:Y\to \bar{X}:=X\times_k\bar{k}$ be a (surjective) finite etale morphism of schemes defined over $\bar{k}$. In this situation, how can I get a `$k$-model' for $f$ and $Y$? What I mean by '$k$-module' is as follows: a scheme $Y^0$ and a (surjective) finite etale morphism $f^0:Y^0\to X$ both are defined over $k$ such that $f$ can be obtained by base-change from $f^0$.

Any reference related to this would be helpful for me. Thank you.

  • 1
    $\begingroup$ Why do you think this would be true? You can get a model over a finite extension of $k$, but certainly not over $k$. For example, take $X$ to be an elliptic curve with no rational points over $k$, and $f:Y\to \bar X$ a $2$-isogeny. $\endgroup$ – Piotr Achinger May 29 '14 at 18:53
  • $\begingroup$ @PiotrAchinger Could you let me now the explicit construction for a model over a finite extension of $k$? Maybe, the thing I am focusing now would be solved using that construction. $\endgroup$ – Kevin.lijh May 29 '14 at 20:20
  • $\begingroup$ This is the general "spreading out". The claim is that given a finite diagram of finite type schemes over $\bar k$, there exists a finite extension $k'$ of $k$ and a model of the diagram over $k'$. If your diagram consists of just a single affine (or projective) scheme $X$, write $X = Spec(\bar k[x_1, \ldots, x_n]/(f_1, \ldots, f_r))$ (resp. $Proj(...)$), and let $k'$ be the extension of $k$ generated in $\bar k$ by the coefficients of the $f_i$, then we have a model $X_0=Spec(k'[x_1, \ldots, x_n]/(f_1, \ldots, f_r))$ (resp. $Proj(\ldots)$) over $k'$. $\endgroup$ – Piotr Achinger May 29 '14 at 20:48
  • $\begingroup$ Since the $k$-model doesn't necessarily exist, it might help if you told us a bit more about what you really want. $\endgroup$ – S. Carnahan May 30 '14 at 14:15
  • $\begingroup$ @S.Carnahan To be precise, for a connected scheme $X$ over a field $k$, we may consider the universal etale cover $\tilde{\bar{X}}$ of $\bar{X}:=X\times_k\bar{k}$. Using the model of $\tilde{\bar{X}}\to \bar{X}$, I want to find a $k$-rational etale universal cover $\tilde{X}$ of $X$. $\endgroup$ – Kevin.lijh May 31 '14 at 22:51

That is often impossible. I am pretty certain all of the following has been explained here before, but here it is again.

Begin with a smooth, projective $k$-curve $P$ such that $\overline{P}$ is isomorphic to $\mathbb{P}^1_{\overline{k}}=\text{Proj}\ \overline{k}[s,t]$, yet $P$ is not isomorphic to $\mathbb{P}^1_k$, e.g., for $k=\mathbb{R}$, let $P$ be the smooth plane conic in $$\mathbb{P}^2_{\mathbb{R}} = \text{Proj} \ \mathbb{R}[x,y,z]$$ with defining equation $x^2 + y^2 + z^2$. Next, let $s$ be any nonzero global section in $H^0(P,\omega_{P/k}^\vee)$. In the example, by adjunction, $\omega_{P/k}^\vee$ equals $\mathcal{O}_{\mathbb{P}^2_{\mathbb{R}}}(1)|_P$, so let $s$ be the global section of $\mathcal{O}_{\mathbb{P}^2_{\mathbb{R}}}(1)$ corresponding to $z\in \mathbb{R}[x,y,z]_1$. Let $X$ be $D(s)$, the open subscheme of $P$ on which $s$ is a generator of $\omega_{P/k}^\vee$, i.e., in the example $X$ is isomorphic to $\text{Spec} \mathbb{R}[u,v]/\langle u^2 + v^2 +1 \rangle$, where $u=x/z$ and $v=y/z$.

There exists an isomorphism $$i: \mathbb{P}^1_{\overline{k}} \to \overline{P}$$ with $i^{-1}(\overline{D})$ equal to $D(st)$. Consider the finite flat morphism, $$F:\mathbb{P}^1_{\overline{k}} \to \mathbb{P}^1_{\overline{k}}, \ F^*(s)=s^2, \ F^*(t) = t^2.$$ Denote $F^{-1}(D(st))$ by $Y$. Assuming that $\text{char}(k)\neq 2$, the induced morphism, $$ F|_Y : Y \to D(st),$$ is finite and étale. Thus the composition $i\circ F|_Y$ is finite and étale. Denote this by $f$.

By way of contradiction, assume that there exists a $k$-morphis, $$f^0:Y^0\to X$$ whose basechange to $\overline{k}$ is isomorphic, as a scheme over $\overline{X}$ to $f=i\circ F|_Y$. By the theory of smooth curves over a field, there is a dense open immersion of $Y^0$ in a smooth, projective $k$-curve $Q^0$, and this open immersion is unique up to unique $k$-isomorphism. Moreover, by the valuative criterion of properness, the morphism $f^0$ extends to a $k$-morphism, $$ F^0:Q^0 \to P,$$ whose restriction to $Y_0$ equals $f^0$. Since $Y^0$ is dense in $Q^0$, and since $Q^0$ is separated, also the base change of $F^0$ to $\overline{k}$ is equivalent, as a scheme over $\overline{P}$, to $i\circ F$.

In particular, consider the coherent $\mathcal{O}_P$-module $\mathcal{L}$ that is the cokernel of $$ F_0^\#:\mathcal{O}_P \to F^0_*\mathcal{O}_{Q^0}.$$ The base change of $\mathcal{L}$ to $\mathcal{k}$ is isomorphic to $\mathcal{O}_{\mathbb{P}^1_{\overline{k}}}(-1)$. Thus, $\mathcal{L}^\vee$ is an invertible sheaf on $P$ whose base change is isomorphic to $\mathcal{O}_{\mathbb{P}^1_k}(1)$. But then the complete linear system of $\mathcal{L}^\vee$ defines a $k$-isomorphism of $P$ to $\mathbb{P}^1_k$, contrary to hypothesis. Therefore, there does not exist any $k$-morphism $f^0$ whose basechange to $\overline{k}$ equals $f$.

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