Lemma 33.7.11. Let $k$ be a field. Let $X$ be a scheme over $k$. The following are equivalent

1. $X$ is geometrically connected,

2. for every finite separable field extension $k \subset k'$ the scheme $X_{k'}$ is connected.

Proof. It follows immediately from the definition that (1) implies (2). Assume that $X$ is not geometrically connected. Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. By Lemma 33.7.7 it follows that $X_{\overline{k}}$ is disconnected. Say $X_{\overline{k}} = \overline{U} \amalg \overline{V}$ with $\overline{U}$ and $\overline{V}$ open, closed, and nonempty.

Suppose that $W \subset X$ is any quasi-compact open. Then $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are open and closed in $W_{\overline{k}}$. In particular $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are quasi-compact, and by Lemma 33.7.9 both $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are defined over a finite subextension and invariant under an open subgroup of $\text{Gal}(\overline{k}/k)$. We will use this without further mention in the following.

Pick $W_0 \subset X$ quasi-compact open such that both $W_{0, \overline{k}} \cap \overline{U}$ and $W_{0, \overline{k}} \cap \overline{V}$ are nonempty. Choose a finite subextension $k \subset k' \subset \overline{k}$ and a decomposition $W_{0, k'} = U_0' \amalg V_0'$ into open and closed subsets such that $W_{0, \overline{k}} \cap \overline{U} = (U'_0)_{\overline{k}}$ and $W_{0, \overline{k}} \cap \overline{V} = (V'_0)_{\overline{k}}$. Let $H = \text{Gal}(\overline{k}/k') \subset \text{Gal}(\overline{k}/k)$. In particular $\sigma (W_{0, \overline{k}} \cap \overline{U}) = W_{0, \overline{k}} \cap \overline{U}$ and similarly for $\overline{V}$.

Having chosen $W_0$, $k'$ as above, for every quasi-compact open $W \subset X$ we set

$U_ W = \bigcap \nolimits _{\sigma \in H} \sigma (W_{\overline{k}} \cap \overline{U}), \quad V_ W = \bigcup \nolimits _{\sigma \in H} \sigma (W_{\overline{k}} \cap \overline{V}).$

Now, since $W_{\overline{k}} \cap \overline{U}$ and $W_{\overline{k}} \cap \overline{V}$ are fixed by an open subgroup of $\text{Gal}(\overline{k}/k)$ we see that the union and intersection above are finite. Hence $U_ W$ and $V_ W$ are both open and closed. Also, by construction $W_{\bar k} = U_ W \amalg V_ W$.

We claim that if $W \subset W' \subset X$ are quasi-compact open, then $W_{\overline{k}} \cap U_{W'} = U_ W$ and $W_{\overline{k}} \cap V_{W'} = V_ W$. Verification omitted. Hence we see that upon defining $U = \bigcup _{W \subset X} U_ W$ and $V = \bigcup _{W \subset X} V_ W$ we obtain $X_{\overline{k}} = U \amalg V$ is a disjoint union of open and closed subsets. It is clear that $V$ is nonempty as it is constructed by taking unions (locally). On the other hand, $U$ is nonempty since it contains $W_0 \cap \overline{U}$ by construction. Finally, $U, V \subset X_{\bar k}$ are closed and $H$-invariant by construction. Hence by Lemma 33.7.10 we have $U = (U')_{\bar k}$, and $V = (V')_{\bar k}$ for some closed $U', V' \subset X_{k'}$. Clearly $X_{k'} = U' \amalg V'$ and we see that $X_{k'}$ is disconnected as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).