Does anyone know an example of an integral scheme $X$ over a field $k$ such that $X_{\overline{k}}$ is connected but reducible? Does it make a difference if $k$ is perfect, or if we ask for $X_{\overline{k}}$ to be reduced as well?

6$\begingroup$ For the field $\mathbb{R}$, the affine $\mathbb{R}$scheme $\text{Spec} \ \mathbb{R}[x,y]/\langle x^2+y^2\rangle$ is integral and geometrically connected, but it is not geometrically irreducible. If $X_k$ is an integral, locally finite type $k$scheme that is geometrically connected and normal, then $X_{\overline{k}}$ is irreducible. For geometric irreduciblity it is irrelevant whether $k$ is perfect: the field extension $\overline{k}/k^{\text{sep}}$ is a universal homeomorphism. However, perfectness is relevant for reducedness. $\endgroup$– Jason StarrAug 22, 2017 at 10:44

$\begingroup$ Thanks, Jason! If you'll submit this as an answer, I'll accept it. $\endgroup$– dorebellAug 24, 2017 at 4:02
1 Answer
I am just posting my comment as an answer. For the field $\mathbb{R}$, the affine $\mathbb{R}$scheme $\text{Spec}\ \mathbb{R}[x,y]/\langle x^2+y^2\rangle$ is integral and geometrically connected, but it is not geometrically irreducible. If $X_k$ is an integral, locally finite type $k$scheme that is normal and geometrically connected, then $X_{\overline{k}}$ is irreducible. For geometric irreducibility, it is irrelevant whether $k$ is perfect: the field extension $\overline{k}/k^{\text{sep}}$ is a universal homeomorphism. However, perfectness is relevant for reducedness.

1$\begingroup$ I was surprised by the statement when $X$ is normal, and I couldn't find a proof online, so here's a sketch of the argument I worked out for future reference: We can work affinelocally on $X$, so assume $X = \mathrm{Spec}(A)$ for $A$ an integrally closed domain. It suffices to show that $A \otimes_k \overline{k}$ has a unique minimal prime ideal. Since field extensions are flat and $A$ injects into $K(X)$, it suffices to show that $K(X) \otimes_k \overline{k}$ has a unique minimal prime ideal. This is equivalent to $k$ being separably closed in $K(X)$. $\endgroup$– dorebellAug 24, 2017 at 22:03

1$\begingroup$ But an element of $K(X)$ which is algebraic over $k$ is certainly integral over $A$, so it is an element of $K(X)$. Let $k'$ be the separable closure of $k$ in $K(X)$; then we've shown that $X$ is naturally a $k'$scheme ($k'$ doesn't depend on the choice of affine open), and that it is geometrically irreducible over $k'$ (since $k'$ is separably closed in $K(X)$). $\endgroup$– dorebellAug 24, 2017 at 22:12

1$\begingroup$ We have \begin{align*} X \times_{\mathrm{Spec}\ k} \mathrm{Spec}\ \overline{k} &= (X \times_{\mathrm{Spec}\ k'} \mathrm{Spec}\ k') \times_{\mathrm{Spec}\ k} \mathrm{Spec}\ \overline{k}\\ &= X \times_{\mathrm{Spec}\ k'} \mathrm{Spec}\ (k' \otimes_k \overline{k})\\ &= X \times_{\mathrm{Spec}\ k'} \left( \sqcup_{k' \hookrightarrow \overline{k}} \mathrm{Spec}\ \overline{k}\right) \\ &= \sqcup_{k' \hookrightarrow \overline{k}} \left(X \times_{\mathrm{Spec}\ k'} \overline{k}\right) \end{align*} This is not connected, since $k'/k$ is a nontrivial separable extension. $\endgroup$– dorebellAug 24, 2017 at 22:14

1$\begingroup$ Another way to argue the normal case is by using that étale extensions of normal rings are normal rings (see Tag 033C). Thus, if $X$ is normal, then so is $X_{k^{\operatorname{sep}}}$. Since the latter is also connected by assumption, it is integral. Irreducibility is not changed when passing from $k^{\operatorname{sep}}$ to $\bar k$ (but reducedness can be). $\endgroup$ Aug 26, 2017 at 0:10