The Stacks project

Proof. Apply Theorem 76.36.5 to get a factorization $X \to Y' \to Y$. It is enough to show that $Y' = Y$. It suffices to show that $Y' \times _ Y V \to V$ is an isomorphism, where $V \to Y$ is an étale morphism and $V$ an affine integral scheme, see Spaces over Fields, Lemma 72.4.5. The formation of $Y'$ commutes with étale base change, see Morphisms of Spaces, Lemma 67.48.4. The generic points of $X \times _ Y V$ lie over the generic points of $X$ (Decent Spaces, Lemma 68.20.1) hence map to the generic point of $V$ by assumption (5). Moreover, condition (6) is preserved under the base change by $V \to Y$, for example by flat base change (Cohomology of Spaces, Lemma 69.11.2). Thus it suffices to prove the lemma in case $Y$ is a normal integral affine scheme.

Assume $Y$ is a normal integral affine scheme. We will show $Y' \to Y$ is an isomorphism by an application of Morphisms, Lemma 29.54.8. Namely, $Y'$ is reduced because $X$ is reduced (Morphisms of Spaces, Lemma 67.48.6). The morphism $Y' \to Y$ is integral by the theorem cited above. Since $Y$ is decent and $X \to Y$ is separated, we see that $X$ is decent too; to see this use Decent Spaces, Lemmas 68.17.2 and 68.17.5. By assumption (5), Morphisms of Spaces, Lemma 67.48.7, and Decent Spaces, Lemma 68.20.1 we see that every generic point of an irreducible component of $|Y'|$ maps to $\xi $. On the other hand, since $Y'$ is the relative spectrum of $f_*\mathcal{O}_ X$ we see that the scheme theoretic fibre $Y'_\xi $ is the spectrum of $H^0(X_\xi , \mathcal{O})$ which is equal to $\kappa (\xi )$ by assumption. Hence $Y'$ is an integral scheme with function field equal to the function field of $Y$. This finishes the proof. $\square$