Theorem 37.53.5 (Stein factorization; general case). Let $S$ be a scheme. Let $f : X \to S$ be a proper morphism. There exists a factorization

\[ \xymatrix{ X \ar[rr]_{f'} \ar[rd]_ f & & S' \ar[dl]^\pi \\ & S & } \]

with the following properties:

the morphism $f'$ is proper with geometrically connected fibres,

the morphism $\pi : S' \to S$ is integral,

we have $f'_*\mathcal{O}_ X = \mathcal{O}_{S'}$,

we have $S' = \underline{\mathop{\mathrm{Spec}}}_ S(f_*\mathcal{O}_ X)$, and

$S'$ is the normalization of $S$ in $X$, see Morphisms, Definition 29.53.3.

**Proof.**
We may apply Lemma 37.53.1 to get the morphism $f' : X \to S'$. Note that besides the conclusions of Lemma 37.53.1 we also have that $f'$ is separated (Schemes, Lemma 26.21.13) and finite type (Morphisms, Lemma 29.15.8). Hence $f'$ is proper. At this point we have proved all of the statements except for the statement that $f'$ has geometrically connected fibres.

We may assume that $S = \mathop{\mathrm{Spec}}(R)$ is affine. Set $R' = \Gamma (X, \mathcal{O}_ X)$. Then $S' = \mathop{\mathrm{Spec}}(R')$. Thus we may replace $S$ by $S'$ and assume that $S = \mathop{\mathrm{Spec}}(R)$ is affine $R = \Gamma (X, \mathcal{O}_ X)$. Next, let $s \in S$ be a point. Let $U \to S$ be an étale morphism of affine schemes and let $u \in U$ be a point mapping to $s$. Let $X_ U \to U$ be the base change of $X$. By Lemma 37.53.3 it suffices to show that the fibre of $X_ U \to U$ over $u$ is connected. By Cohomology of Schemes, Lemma 30.5.2 we see that $\Gamma (X_ U, \mathcal{O}_{X_ U}) = \Gamma (U, \mathcal{O}_ U)$. Hence we have to show: Given $S = \mathop{\mathrm{Spec}}(R)$ affine, $X \to S$ proper with $\Gamma (X, \mathcal{O}_ X) = R$ and $s \in S$ is a point, the fibre $X_ s$ is connected.

To do this it suffices to show that the only idempotents $e \in H^0(X_ s, \mathcal{O}_{X_ s})$ are $0$ and $1$ (we already know that $X_ s$ is nonempty by Lemma 37.53.1). By Derived Categories of Schemes, Lemma 36.32.7 after replacing $R$ by a principal localization we may assume $e$ is the image of an element of $R$. Since $R \to H^0(X_ s, \mathcal{O}_{X_ s})$ factors through $\kappa (s)$ we conclude.
$\square$

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