Lemma 33.9.4. Let $X$ be a proper scheme over a field $k$. Set $A = H^0(X, \mathcal{O}_ X)$. The fibres of the canonical morphism $X \to \mathop{\mathrm{Spec}}(A)$ are geometrically connected.

**Proof.**
Set $S = \mathop{\mathrm{Spec}}(A)$. The canonical morphism $X \to S$ is the morphism corresponding to $\Gamma (S, \mathcal{O}_ S) = A = \Gamma (X, \mathcal{O}_ X)$ via Schemes, Lemma 26.6.4. The $k$-algebra $A$ is a finite product $A = \prod A_ i$ of local Artinian $k$-algebras finite over $k$, see Lemma 33.9.3. Denote $s_ i \in S$ the point corresponding to the maximal ideal of $A_ i$. Choose an algebraic closure $\overline{k}$ of $k$ and set $\overline{A} = A \otimes _ k \overline{k}$. Choose an embedding $\kappa (s_ i) \to \overline{k}$ over $k$; this determines a $\overline{k}$-algebra map

Consider the base change

of $X$ to $\overline{S} = \mathop{\mathrm{Spec}}(\overline{A})$. By Cohomology of Schemes, Lemma 30.5.2 we have $\Gamma (\overline{X}, \mathcal{O}_{\overline{X}}) = \overline{A}$. If $\overline{s}_ i \in \mathop{\mathrm{Spec}}(\overline{A})$ denotes the $\overline{k}$-rational point corresponding to $\sigma _ i$, then we see that $\overline{s}_ i$ maps to $s_ i \in S$ and $\overline{X}_{\overline{s}_ i}$ is the base change of $X_{s_ i}$ by $\mathop{\mathrm{Spec}}(\sigma _ i)$. Thus we see that it suffices to prove the lemma in case $k$ is algebraically closed.

Assume $k$ is algebraically closed. In this case $\kappa (s_ i)$ is algebraically closed and we have to show that $X_{s_ i}$ is connected. The product decomposition $A = \prod A_ i$ corresponds to a disjoint union decomposition $\mathop{\mathrm{Spec}}(A) = \coprod \mathop{\mathrm{Spec}}(A_ i)$, see Algebra, Lemma 10.21.2. Denote $X_ i$ the inverse image of $\mathop{\mathrm{Spec}}(A_ i)$. It follows from Lemma 33.9.3 part (2) that $A_ i = \Gamma (X_ i, \mathcal{O}_{X_ i})$. Observe that $X_{s_ i} \to X_ i$ is a closed immersion inducing an isomorphism on underlying topological spaces (because $\mathop{\mathrm{Spec}}(A_ i)$ is a singleton). Hence if $X_{s_ i}$ isn't connected, then neither is $X_ i$. So either $X_ i$ is empty and $A_ i = 0$ or $X_ i$ can be written as $U \amalg V$ with $U$ and $V$ open and nonempty which would imply that $A_ i$ has a nontrivial idempotent. Since $A_ i$ is local this is a contradiction and the proof is complete. $\square$

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