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 $k'/k$ 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 $\overline{k}/k'/k$ 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. $\square$

Comment #7102 by WhatJiaranEatsTonight on

I think there is an ad hoc proof if $X$ is qcqs. It seems that $X$ is connected if and only if there does not exist nonzero global idempotent section. And by cohomology of base change, we know taking global sections commutes with change of base field. Hence $X$ is geometrically connected if and only if $\Gamma(X,\O_X)$ is a geometrically connected algebra. Hence lemma 0389 is a direct conclusion of 037S for the case that $X$ is qcqs.

Comment #7266 by on

Yes, I agree that works. But it is fun that the result also is true in the non-quasi-compact case, so I am going to leave as is for now.

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