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. $\square$

Comment #2055 by yogesh on

Just a comment to help future readers: It took me a minute to figure out the difference between $U, U', U''$ and $V, V', V''$ as it seemed after the first paragraph things with U and V are symmetrically defined. The $U''$ is defined by an intersection, $V''$ 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 $\overline U_1=\bigcap_{\sigma\in Gal(\overline{k}/k)}\sigma(\overline U)$ and $\overline V_1=\bigcup_{\sigma\in Gal(\overline{k}/k)}\sigma(\overline V)$. Then $\overline U_1$ is closed and contains the image of $T_{\overline k}$ since the latter is contained in all $\sigma(\overline U)$. By the construction, we know that $W = U_W\sqcup V_W$ is exactly the unique composition such that $(U_W)_{\overline k}\subset \overline U_1$ and $(V_W)_{\overline k}\subset \overline V_1$.

And by the construction, $U_{\overline k}=\overline U_1$ contains the image of $T$. Thus $V\neq \varnothing$.

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 $U$ and $V$ that we construct base change to what you call $\overline{U}_1$ and $\overline{V}_1$, so maybe your argument is quicker? Anybody else have an opinion or a suggested rewrite of the proof?

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