Lemma 29.53.11. Let $f : X \to S$ be a quasi-compact, quasi-separated and universally closed morphisms of schemes. Then $f_*\mathcal{O}_ X$ is integral over $\mathcal{O}_ S$. In other words, the normalization of $S$ in $X$ is equal to the factorization

$X \longrightarrow \underline{\mathop{\mathrm{Spec}}}_ S(f_*\mathcal{O}_ X) \longrightarrow S$

of Constructions, Lemma 27.4.7.

Proof. The question is local on $S$, hence we may assume $S = \mathop{\mathrm{Spec}}(R)$ is affine. Let $h \in \Gamma (X, \mathcal{O}_ X)$. We have to show that $h$ satisfies a monic equation over $R$. Think of $h$ as a morphism as in the following commutative diagram

$\xymatrix{ X \ar[rr]_ h \ar[rd]_ f & & \mathbf{A}^1_ S \ar[ld] \\ & S & }$

Let $Z \subset \mathbf{A}^1_ S$ be the scheme theoretic image of $h$, see Definition 29.6.2. The morphism $h$ is quasi-compact as $f$ is quasi-compact and $\mathbf{A}^1_ S \to S$ is separated, see Schemes, Lemma 26.21.14. By Lemma 29.6.3 the morphism $X \to Z$ is dominant. By Lemma 29.41.7 the morphism $X \to Z$ is closed. Hence $h(X) = Z$ (set theoretically). Thus we can use Lemma 29.41.9 to conclude that $Z \to S$ is universally closed (and even proper). Since $Z \subset \mathbf{A}^1_ S$, we see that $Z \to S$ is affine and proper, hence integral by Lemma 29.44.7. Writing $\mathbf{A}^1_ S = \mathop{\mathrm{Spec}}(R[T])$ we conclude that the ideal $I \subset R[T]$ of $Z$ contains a monic polynomial $P(T) \in R[T]$. Hence $P(h) = 0$ and we win. $\square$

## Comments (2)

Comment #705 by Kestutis Cesnavicius on

Same here: quasi-compactness is superfluous due to http://stacks.math.columbia.edu/tag/04XU

Comment #714 by on

Yes, you are right. I am feeling too lazy to change it right now.

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