The Stacks project

Theorem 37.48.5 (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:

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

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

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

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

  5. $S'$ is the normalization of $S$ in $X$, see Morphisms, Definition 29.51.3.

Proof. We may apply Lemma 37.48.1 to get the morphism $f' : X \to S'$. Note that besides the conclusions of Lemma 37.48.1 we also have that $f'$ is separated (Schemes, Lemma 26.21.13) and finite type (Morphisms, Lemma 29.14.8). 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 ├ętale 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 37.48.3 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.

By Limits, Lemma 32.13.3 we can write $(X \to S) = \mathop{\mathrm{lim}}\nolimits (X_ i \to S_ i)$ with $X_ i \to S_ i$ proper and of finite presentation and $S_ i$ Noetherian. For $i$ large enough $S_ i$ is affine (Limits, Lemma 32.4.13). Say $S_ i = \mathop{\mathrm{Spec}}(R_ i)$. Let $R'_ i = \Gamma (X_ i, \mathcal{O}_{X_ i})$. Observe that we have ring maps $R_ i \to R_ i' \to R$. Namely, we have the first because $X_ i$ is a scheme over $R_ i$ and the second because we have $X \to X_ i$ and $R = \Gamma (X, \mathcal{O}_ X)$. Note that $R = \mathop{\mathrm{colim}}\nolimits R'_ i$ by Limits, Lemma 32.4.7. Then

\[ \xymatrix{ X \ar[d] \ar[r] & X_ i \ar[d] \\ S \ar[r] & S'_ i \ar[r] & S_ i } \]

is commutative with $S'_ i = \mathop{\mathrm{Spec}}(R'_ i)$. Let $s'_ i \in S'_ i$ be the image of $s$. We have $X_ s = \mathop{\mathrm{lim}}\nolimits X_{i, s'_ i}$ because $X = \mathop{\mathrm{lim}}\nolimits X_ i$, $S = \mathop{\mathrm{lim}}\nolimits S'_ i$, and $\kappa (s) = \mathop{\mathrm{colim}}\nolimits \kappa (s'_ i)$. Now let $X_ s = U \amalg V$ with $U$ and $V$ open and closed. Then $U, V$ are the inverse images of opens $U_ i, V_ i$ in $X_{i, s'_ i}$ (Limits, Lemma 32.4.11). By Theorem 37.48.4 the fibres of $X_ i \to S'_ i$ are connected, hence either $U$ or $V$ is empty. This finishes the proof. $\square$

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