Lemma 37.52.6. Let $f : X \to S$ be a morphism of schemes. Assume

$f$ is proper,

$S$ is integral with generic point $\xi $,

$S$ is normal,

$X$ is reduced,

every generic point of an irreducible component of $X$ maps to $\xi $,

we have $H^0(X_\xi , \mathcal{O}) = \kappa (\xi )$.

Then $f_*\mathcal{O}_ X = \mathcal{O}_ S$ and $f$ has geometrically connected fibres.

**Proof.**
Apply Theorem 37.52.5 to get a factorization $X \to S' \to S$. It is enough to show that $S' = S$. This will follow from Morphisms, Lemma 29.54.8. Namely, $S'$ is reduced because $X$ is reduced (Morphisms, Lemma 29.53.8). The morphism $S' \to S$ is integral by the theorem cited above. Every generic point of $S'$ lies over $\xi $ by Morphisms, Lemma 29.53.9 and assumption (5). On the other hand, since $S'$ is the relative spectrum of $f_*\mathcal{O}_ X$ we see that the scheme theoretic fibre $S'_\xi $ is the spectrum of $H^0(X_\xi , \mathcal{O})$ which is equal to $\kappa (\xi )$ by assumption. Hence $S'$ is an integral scheme with function field equal to the function field of $S$. This finishes the proof.
$\square$

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