The Stacks project

Lemma 33.7.12. Let $k$ be a field. Let $\overline{k}/k$ be a (possibly infinite) Galois extension. Let $f : T \to X$ be a morphism of schemes over $k$. Assume $T_{\overline{k}}$ connected and $X_{\overline{k}}$ disconnected. Then $X$ is disconnected.

Proof. Write $X_{\overline{k}} = \overline{U} \amalg \overline{V}$ with $\overline{U}$ and $\overline{V}$ open and closed. Denote $\overline{f} : T_{\overline{k}} \to X_{\overline{k}}$ the base change of $f$. Since $T_{\overline{k}}$ is connected we see that $T_{\overline{k}}$ is contained in either $\overline{f}^{-1}(\overline{U})$ or $\overline{f}^{-1}(\overline{V})$. Say $T_{\overline{k}} \subset \overline{f}^{-1}(\overline{U})$.

Fix a quasi-compact open $W \subset X$. There exists a finite Galois subextension $\overline{k}/k'/k$ such that $\overline{U} \cap W_{\overline{k}}$ and $\overline{V} \cap W_{\overline{k}}$ come from quasi-compact opens $U', V' \subset W_{k'}$. Then also $W_{k'} = U' \amalg V'$. Consider

\[ U'' = \bigcap \nolimits _{\sigma \in \text{Gal}(k'/k)} \sigma (U'), \quad V'' = \bigcup \nolimits _{\sigma \in \text{Gal}(k'/k)} \sigma (V'). \]

These are Galois invariant, open and closed, and $W_{k'} = U'' \amalg V''$. By Lemma 33.7.10 we get open and closed subsets $U_ W, V_ W \subset W$ such that $U'' = (U_ W)_{k'}$, $V'' = (V_ W)_{k'}$ and $W = U_ W \amalg V_ W$.

We claim that if $W \subset W' \subset X$ are quasi-compact open, then $W \cap U_{W'} = U_ W$ and $W \cap V_{W'} = V_ W$. 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 $X = U \amalg V$. 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 $f(T)$ by construction. $\square$

Comments (3)

Comment #2055 by yogesh on

Just a comment to help future readers: It took me a minute to figure out the difference between and as it seemed after the first paragraph things with U and V are symmetrically defined. The is defined by an intersection, is defined by a union.

Comment #7104 by WhatJiaranEatsTonight on

I am sorry that I made some mistakes in the notation of my last comment. Could you please delete it?

I think the verification can be done as follows: Let and . Then is closed and contains the image of since the latter is contained in all . By the construction, we know that is exactly the unique composition such that and .

And by the construction, contains the image of . Thus .

Comment #7267 by on

OK, I am a bit confused as to where this argument should be inserted. I think the argument as we have it now does work, so I am going to leave it as is for now. OTOH at the end of the argument we do indeed get that the opens and that we construct base change to what you call and , so maybe your argument is quicker? Anybody else have an opinion or a suggested rewrite of the proof?

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