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
\sigma _ i : \overline{A} = A \otimes _ k \overline{k} \to \kappa (s_ i) \otimes _ k \overline{k} \to \overline{k}
Consider the base change
\xymatrix{ \overline{X} \ar[r] \ar[d] & X \ar[d] \\ \overline{S} \ar[r] & S }
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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