The Stacks project

Theorem 36.48.5 (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 28.51.3.

Proof. We may apply Lemma 36.48.1 to get the morphism $f' : X \to S'$. Note that besides the conclusions of Lemma 36.48.1 we also have that $f'$ is separated (Schemes, Lemma 25.21.13) and finite type (Morphisms, Lemma 28.14.8). 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 36.48.3 it suffices to show that the fibre of $X_ U \to U$ over $u$ is connected. By Cohomology of Schemes, Lemma 29.5.2 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 31.13.3 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 31.4.13). 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 31.4.7. 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 31.4.11). By Theorem 36.48.4 the fibres of $X_ i \to S'_ i$ are connected, hence either $U$ or $V$ is empty. This finishes the proof. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03H2. Beware of the difference between the letter 'O' and the digit '0'.