Lemma 29.53.13. Let $f : Y \to X$ be a quasi-compact and quasi-separated morphism of schemes. Let $X'$ be the normalization of $X$ in $Y$. Assume

$Y$ is a normal scheme,

quasi-compact opens of $Y$ have finitely many irreducible components.

Then $X'$ is a disjoint union of integral normal schemes. Moreover, the morphism $Y \to X'$ is dominant and induces a bijection of irreducible components.

**Proof.**
Let $U \subset X$ be an affine open. Consider the inverse image $U'$ of $U$ in $X'$. Set $V = f^{-1}(U)$. By Lemma 29.53.6 we $V \to U' \to U$ is the normalization of $U$ in $V$. Say $U = \mathop{\mathrm{Spec}}(A)$. Then $V$ is quasi-compact, and hence has a finite number of irreducible components by assumption. Hence $V = \coprod _{i = 1, \ldots n} V_ i$ is a finite disjoint union of normal integral schemes by Properties, Lemma 28.7.5. By Lemma 29.53.10 we see that $U' = \coprod _{i = 1, \ldots , n} U_ i'$, where $U'_ i$ is the normalization of $U$ in $V_ i$. By Properties, Lemma 28.7.9 we see that $B_ i = \Gamma (V_ i, \mathcal{O}_{V_ i})$ is a normal domain. Note that $U_ i' = \mathop{\mathrm{Spec}}(A_ i')$, where $A_ i' \subset B_ i$ is the integral closure of $A$ in $B_ i$, see Lemma 29.53.1. By Algebra, Lemma 10.37.2 we see that $A_ i' \subset B_ i$ is a normal domain. Hence $U' = \coprod U_ i'$ is a finite union of normal integral schemes and hence is normal.

As $X'$ has an open covering by the schemes $U'$ we conclude from Properties, Lemma 28.7.2 that $X'$ is normal. On the other hand, each $U'$ is a finite disjoint union of irreducible schemes, hence every quasi-compact open of $X'$ has finitely many irreducible components (by a topological argument which we omit). Thus $X'$ is a disjoint union of normal integral schemes by Properties, Lemma 28.7.5. It is clear from the description of $X'$ above that $Y \to X'$ is dominant and induces a bijection on irreducible components $V \to U'$ for every affine open $U \subset X$. The bijection of irreducible components for the morphism $Y \to X'$ follows from this by a topological argument (omitted).
$\square$

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