**Proof.**
Let $f : Y \to X$ be as in (29.54.0.1). The scheme $X^\nu $ is a disjoint union of normal integral schemes because $Y$ is normal and every affine open of $Y$ has finitely many irreducible components, see Lemma 29.53.13. This proves (1). Alternatively one can deduce (1) from Lemmas 29.54.2 and 29.54.3.

The morphism $\nu $ is integral by Lemma 29.53.4. By Lemma 29.53.13 the morphism $Y \to X^\nu $ induces a bijection on irreducible components, and by construction of $Y$ this implies that $X^\nu \to X$ induces a bijection on irreducible components. By construction $f : Y \to X$ is dominant, hence also $\nu $ is dominant. Since an integral morphism is closed (Lemma 29.44.7) this implies that $\nu $ is surjective. This proves (2).

Suppose that $\alpha : X' \to X$ is as in (3). It is clear that $X'$ satisfies the assumptions under which the normalization is defined. Let $f' : Y' \to X'$ be the morphism (29.54.0.1) constructed starting with $X'$. As $\alpha $ is locally birational it is clear that $Y' = Y$ and $f = \alpha \circ f'$. Hence the factorization $X^\nu \to X' \to X$ exists and $X^\nu \to X'$ is the normalization of $X'$ by Lemma 29.53.4. This proves (3).

Let $g : Z \to X$ be a morphism whose domain is a normal scheme and such that every irreducible component dominates an irreducible component of $X$. By Lemma 29.54.2 we have $X^\nu = X_{red}^\nu $ and by Schemes, Lemma 26.12.7 $Z \to X$ factors through $X_{red}$. Hence we may replace $X$ by $X_{red}$ and assume $X$ is reduced. Moreover, as the factorization is unique it suffices to construct it locally on $Z$. Let $W \subset Z$ and $U \subset X$ be affine opens such that $g(W) \subset U$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $W = \mathop{\mathrm{Spec}}(B)$, with $g|_ W$ given by $\varphi : A \to B$. We will use the results of Lemma 29.54.3 freely. Let $\mathfrak p_1, \ldots , \mathfrak p_ t$ be the minimal primes of $A$. As $Z$ is normal, we see that $B$ is a normal ring, in particular reduced. Moreover, by assumption any minimal prime $\mathfrak q \subset B$ we have that $\varphi ^{-1}(\mathfrak q)$ is a minimal prime of $A$. Hence if $x \in A$ is a nonzerodivisor, i.e., $x \not\in \bigcup \mathfrak p_ i$, then $\varphi (x)$ is a nonzerodivisor in $B$. Thus we obtain a canonical ring map $Q(A) \to Q(B)$. As $B$ is normal it is equal to its integral closure in $Q(B)$ (see Algebra, Lemma 10.37.12). Hence we see that the integral closure $A' \subset Q(A)$ of $A$ maps into $B$ via the canonical map $Q(A) \to Q(B)$. Since $\nu ^{-1}(U) = \mathop{\mathrm{Spec}}(A')$ this gives the canonical factorization $W \to \nu ^{-1}(U) \to U$ of $\nu |_ W$. We omit the verification that it is unique.
$\square$

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