Lemma 33.7.17 (Gabber). Let $K/k$ be an extension of fields. Let $X$ be a scheme over $k$. Denote $p : X_ K \to X$ the projection morphism. Let $\overline{T} \subset X_ K$ be a connected component. Then $p(\overline{T})$ is a connected component of $X$.
Email from Ofer Gabber dated June 4, 2016
Proof.
When $K/k$ is finite this is Lemma 33.7.16. In general the proof is more difficult.
Let $T \subset X$ be the connected component of $X$ containing the image of $\overline{T}$. We may replace $X$ by $T$ (with the induced reduced subscheme structure). Thus we may assume $X$ is connected. Let $A = H^0(X, \mathcal{O}_ X)$. Let $L \subset A$ be the maximal weakly étale $k$-subalgebra, see More on Algebra, Lemma 15.105.2. Since $A$ does not have any nontrivial idempotents we see that $L$ is a field and a separable algebraic extension of $k$ by More on Algebra, Lemma 15.105.1. Observe that $L$ is also the maximal weakly étale $L$-subalgebra of $A$ (because any weakly étale $L$-algebra is weakly étale over $k$ by More on Algebra, Lemma 15.104.9). By Schemes, Lemma 26.6.4 we obtain a factorization $X \to \mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(k)$ of the structure morphism.
Let $L'/L$ be a finite separable extension. By Cohomology of Schemes, Lemma 30.5.3 we have
The maximal weakly étale $L'$-subalgebra of $A \otimes _ L L'$ is $L \otimes _ L L' = L'$ by More on Algebra, Lemma 15.105.4. In particular $A \otimes _ L L'$ does not have nontrivial idempotents (such an idempotent would generate a weakly étale subalgebra) and we conclude that $X \times _{\mathop{\mathrm{Spec}}(L)} \mathop{\mathrm{Spec}}(L')$ is connected. By Lemma 33.7.11 we conclude that $X$ is geometrically connected over $L$.
Let's give $\overline{T}$ the reduced induced scheme structure and consider the composition
The image is contained in a connected component of $\mathop{\mathrm{Spec}}(L \otimes _ k K)$. Since $K \to L \otimes _ k K$ is integral we see that the connected components of $\mathop{\mathrm{Spec}}(L \otimes _ k K)$ are points and all points are closed, see Algebra, Lemma 10.36.19. Thus we get a quotient field $L \otimes _ k K \to E$ such that $\overline{T}$ maps into $\mathop{\mathrm{Spec}}(E) \subset \mathop{\mathrm{Spec}}(L \otimes _ k K)$. Hence $i(\overline{T}) \subset \pi ^{-1}(\mathop{\mathrm{Spec}}(E))$. But
which is connected because $X$ is geometrically connected over $L$. Then we get the equality $\overline{T} = X \times _{\mathop{\mathrm{Spec}}(L)} \mathop{\mathrm{Spec}}(E)$ (set theoretically) and we conclude that $\overline{T} \to X$ is surjective as desired.
$\square$
\[ A \otimes _ L L' = H^0(X \times _{\mathop{\mathrm{Spec}}(L)} \mathop{\mathrm{Spec}}(L'), \mathcal{O}_{X \times _{\mathop{\mathrm{Spec}}(L)} \mathop{\mathrm{Spec}}(L')}) \]
\[ \overline{T} \xrightarrow {i} X_ K = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(K) \xrightarrow {\pi } \mathop{\mathrm{Spec}}(L \otimes _ k K) \]
\[ \pi ^{-1}(\mathop{\mathrm{Spec}}(E)) = (X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(K)) \times _{\mathop{\mathrm{Spec}}(L \otimes _ k K)} \mathop{\mathrm{Spec}}(E) = X \times _{\mathop{\mathrm{Spec}}(L)} \mathop{\mathrm{Spec}}(E) \]
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