1
$\begingroup$

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?

$\endgroup$
2
  • 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. $\endgroup$ Aug 22, 2017 at 10:44
  • $\begingroup$ Thanks, Jason! If you'll submit this as an answer, I'll accept it. $\endgroup$
    – dorebell
    Aug 24, 2017 at 4:02

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
4
  • 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 affine-locally 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 $K(X)$. $\endgroup$
    – dorebell
    Aug 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 $K(X)$). $\endgroup$
    – dorebell
    Aug 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 non-trivial separable extension. $\endgroup$
    – dorebell
    Aug 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.