37.53 Stein factorization
Stein factorization is the statement that a proper morphism f : X \to S with f_*\mathcal{O}_ X = \mathcal{O}_ S has connected fibres.
Lemma 37.53.1. Let S be a scheme. Let f : X \to S be a universally closed and quasi-separated 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 universally closed, quasi-compact, quasi-separated, and surjective,
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.
Formation of the factorization f = \pi \circ f' commutes with flat base change.
Proof.
By Morphisms, Lemma 29.41.8 the morphism f is quasi-compact. Hence the normalization S' of S in X is defined (Morphisms, Definition 29.53.3) and we have the factorization X \to S' \to S. By Morphisms, Lemma 29.53.11 we have (2), (4), and (5). The morphism f' is universally closed by Morphisms, Lemma 29.41.7. It is quasi-compact by Schemes, Lemma 26.21.14 and quasi-separated by Schemes, Lemma 26.21.13.
To show the remaining statements we may assume the base scheme S is affine, say S = \mathop{\mathrm{Spec}}(R). Then S' = \mathop{\mathrm{Spec}}(A) with A = \Gamma (X, \mathcal{O}_ X) an integral R-algebra. Thus it is clear that f'_*\mathcal{O}_ X is \mathcal{O}_{S'} (because f'_*\mathcal{O}_ X is quasi-coherent, by Schemes, Lemma 26.24.1, and hence equal to \widetilde{A}). This proves (3).
Let us show that f' is surjective. As f' is universally closed (see above) the image of f' is a closed subset V(I) \subset S' = \mathop{\mathrm{Spec}}(A). Pick h \in I. Then h|_ X = f^\sharp (h) is a global section of the structure sheaf of X which vanishes at every point. As X is quasi-compact this means that h|_ X is a nilpotent section, i.e., h^ n|X = 0 for some n > 0. But A = \Gamma (X, \mathcal{O}_ X), hence h^ n = 0. As every element of I is nilpotent, we conclude that V(I) = S' as desired.
By Cohomology of Schemes, Lemma 30.5.2 we see that formation of f_*\mathcal{O}_ X commutes with flat base change. Formation of the relative spectrum commutes with any base change by Constructions, Lemma 27.4.6. Thus formation of the factorization commutes with flat base change.
\square
Lemma 37.53.2. In Lemma 37.53.1 assume in addition that f is locally of finite type. Then for s \in S the fibre \pi ^{-1}(\{ s\} ) = \{ s_1, \ldots , s_ n\} is finite and the field extensions \kappa (s_ i)/\kappa (s) are finite.
Proof.
Recall that there are no specializations among the points of \pi ^{-1}(\{ s\} ), see Algebra, Lemma 10.36.20. As f' is surjective, we find that |X_ s| \to \pi ^{-1}(\{ s\} ) is surjective. Observe that X_ s is a quasi-separated scheme of finite type over a field (quasi-compactness was shown in the proof of the referenced lemma). Thus X_ s is Noetherian (Morphisms, Lemma 29.15.6). A topological argument (omitted) now shows that \pi ^{-1}(\{ s\} ) is finite. For each i we can pick a finite type point x_ i \in X_ s mapping to s_ i (Morphisms, Lemma 29.16.7). We conclude that \kappa (s_ i)/\kappa (s) is finite: x_ i can be represented by a morphism \mathop{\mathrm{Spec}}(k_ i) \to X_ s of finite type (by our definition of finite type points) and hence \mathop{\mathrm{Spec}}(k_ i) \to s = \mathop{\mathrm{Spec}}(\kappa (s)) is of finite type (as a composition of finite type morphisms), hence k_ i/\kappa (s) is finite (Morphisms, Lemma 29.16.1).
\square
Lemma 37.53.3. Let f : X \to S be a morphism of schemes. Let s \in S. Then X_ s is geometrically connected, if and only if for every étale neighbourhood (U, u) \to (S, s) the base change X_ U \to U has connected fibre X_ u.
Proof.
If X_ s is geometrically connected, then any base change of it is connected. On the other hand, suppose that X_ s is not geometrically connected. Then by Varieties, Lemma 33.7.11 we see that X_ s \times _{\mathop{\mathrm{Spec}}(\kappa (s))} \mathop{\mathrm{Spec}}(k) is disconnected for some finite separable field extension k/\kappa (s). By Lemma 37.35.2 there exists an affine étale neighbourhood (U, u) \to (S, s) such that \kappa (u)/\kappa (s) is identified with k/\kappa (s). In this case X_ u is disconnected.
\square
Theorem 37.53.4 (Stein factorization; Noetherian case). Let S be a locally Noetherian 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 finite,
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.
Let f = \pi \circ f' be the factorization of Lemma 37.53.1. 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. By Cohomology of Schemes, Proposition 30.19.1 we see that f_*\mathcal{O}_ X is a coherent \mathcal{O}_ S-module. Hence we see that \pi is finite, i.e., (2) holds.
This proves all but the most interesting assertion, namely that all the fibres of f' are geometrically connected. It is clear from the discussion above that we may replace S by S', and we may therefore assume that S is Noetherian, affine, f : X \to S is proper, and f_*\mathcal{O}_ X = \mathcal{O}_ S. Let s \in S be a point of S. We have to show that X_ s is geometrically connected. By Lemma 37.53.3 we see that it suffices to show X_ u is connected for every étale neighbourhood (U, u) \to (S, s). We may assume U is affine. Thus U is Noetherian (Morphisms, Lemma 29.15.6), the base change f_ U : X_ U \to U is proper (Morphisms, Lemma 29.41.5), and that also (f_ U)_*\mathcal{O}_{X_ U} = \mathcal{O}_ U (Cohomology of Schemes, Lemma 30.5.2). Hence after replacing (f : X \to S, s) by the base change (f_ U : X_ U \to U, u) it suffices to prove that the fibre X_ s is connected when f_*\mathcal{O}_ X = \mathcal{O}_ S. We can deduce this from Derived Categories of Schemes, Lemma 36.32.7 (by looking at idempotents in the structure sheaf of X_ s) but we will also give a direct argument below.
Namely, we apply the theorem on formal functions, more precisely Cohomology of Schemes, Lemma 30.20.7. It tells us that
\mathcal{O}^\wedge _{S, s} = (f_*\mathcal{O}_ X)_ s^\wedge = \mathop{\mathrm{lim}}\nolimits _ n H^0(X_ n, \mathcal{O}_{X_ n})
where X_ n is the nth infinitesimal neighbourhood of X_ s. Since the underlying topological space of X_ n is equal to that of X_ s we see that if X_ s = T_1 \amalg T_2 is a disjoint union of nonempty open and closed subschemes, then similarly X_ n = T_{1, n} \amalg T_{2, n} for all n. And this in turn means H^0(X_ n, \mathcal{O}_{X_ n}) contains a nontrivial idempotent e_{1, n}, namely the function which is identically 1 on T_{1, n} and identically 0 on T_{2, n}. It is clear that e_{1, n + 1} restricts to e_{1, n} on X_ n. Hence e_1 = \mathop{\mathrm{lim}}\nolimits e_{1, n} is a nontrivial idempotent of the limit. This contradicts the fact that \mathcal{O}^\wedge _{S, s} is a local ring. Thus the assumption was wrong, i.e., X_ s is connected, and we win.
\square
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
Here is an application.
Lemma 37.53.6. Let f : X \to S be a morphism of schemes. Assume
f is proper,
S is integral with generic point \xi ,
S is normal,
X is reduced,
every generic point of an irreducible component of X maps to \xi ,
we have H^0(X_\xi , \mathcal{O}) = \kappa (\xi ).
Then f_*\mathcal{O}_ X = \mathcal{O}_ S and f has geometrically connected fibres.
Proof.
Apply Theorem 37.53.5 to get a factorization X \to S' \to S. It is enough to show that S' = S. This will follow from Morphisms, Lemma 29.54.8. Namely, S' is reduced because X is reduced (Morphisms, Lemma 29.53.8). The morphism S' \to S is integral by the theorem cited above. Every generic point of S' lies over \xi by Morphisms, Lemma 29.53.9 and assumption (5). On the other hand, since S' is the relative spectrum of f_*\mathcal{O}_ X we see that the scheme theoretic fibre S'_\xi is the spectrum of H^0(X_\xi , \mathcal{O}) which is equal to \kappa (\xi ) by assumption. Hence S' is an integral scheme with function field equal to the function field of S. This finishes the proof.
\square
Here is another application.
Lemma 37.53.7. Let X \to S be a flat proper morphism of finite presentation. Let n_{X/S} be the function on S counting the numbers of geometric connected components of fibres of f introduced in Lemma 37.28.3. Then n_{X/S} is lower semi-continuous.
Proof.
Let s \in S. Set n = n_{X/S}(s). Note that n < \infty as the geometric fibre of X \to S at s is a proper scheme over a field, hence Noetherian, hence has a finite number of connected components. We have to find an open neighbourhood V of s such that n_{X/S}|_ V \geq n. Let X \to S' \to S be the Stein factorization as in Theorem 37.53.5. By Lemma 37.53.2 there are finitely many points s'_1, \ldots , s'_ m \in S' lying over s and the extensions \kappa (s'_ i)/\kappa (s) are finite. Then Lemma 37.42.1 tells us that after replacing S by an étale neighbourhood of s we may assume S' = V_1 \amalg \ldots \amalg V_ m as a scheme with s'_ i \in V_ i and \kappa (s'_ i)/\kappa (s) purely inseparable. Then the schemes X_{s_ i'} are geometrically connected over \kappa (s), hence m = n. The schemes X_ i = (f')^{-1}(V_ i), i = 1, \ldots , n are flat and of finite presentation over S. Hence the image of X_ i \to S is open (Morphisms, Lemma 29.25.10). Thus in a neighbourhood of s we see that n_{X/S} is at least n.
\square
Lemma 37.53.8. Let f : X \to S be a morphism of schemes. Assume
f is proper, flat, and of finite presentation, and
the geometric fibres of f are reduced.
Then the function n_{X/S} : S \to \mathbf{Z} counting the numbers of geometric connected components of fibres of f is locally constant.
Proof.
By Lemma 37.53.7 the function n_{X/S} is lower semincontinuous. For s \in S consider the \kappa (s)-algebra
A = H^0(X_ s, \mathcal{O}_{X_ s})
By Varieties, Lemma 33.9.3 and the fact that X_ s is geometrically reduced A is finite product of finite separable extensions of \kappa (s). Hence A \otimes _{\kappa (s)} \kappa (\overline{s}) is a product of \beta _0(s) = \dim _{\kappa (s)} H^0(E \otimes ^\mathbf {L} \kappa (s)) copies of \kappa (\overline{s}). Thus X_{\overline{s}} has \beta _0(s) = \dim _{\kappa (s)} A connected components. In other words, we have n_{X/S} = \beta _0 as functions on S. Thus n_{X/S} is upper semi-continuous by Derived Categories of Schemes, Lemma 36.32.1. This finishes the proof.
\square
A final application.
Lemma 37.53.9.reference Let (A, I) be a henselian pair. Let X \to \mathop{\mathrm{Spec}}(A) be separated and of finite type. Set X_0 = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I). Let Y \subset X_0 be an open and closed subscheme such that Y \to \mathop{\mathrm{Spec}}(A/I) is proper. Then there exists an open and closed subscheme W \subset X which is proper over A with W \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I) = Y.
Proof.
We will denote T \mapsto T_0 the base change by \mathop{\mathrm{Spec}}(A/I) \to \mathop{\mathrm{Spec}}(A). By Chow's lemma (in the form of Limits, Lemma 32.12.1) there exists a surjective proper morphism \varphi : X' \to X such that X' admits an immersion into \mathbf{P}^ n_ A. Set Y' = \varphi ^{-1}(Y). This is an open and closed subscheme of X'_0. Suppose the lemma holds for (X', Y'). Let W' \subset X' be the open and closed subscheme proper over A such that Y' = W'_0. By Morphisms, Lemma 29.41.7 W = \varphi (W') \subset X and Q = \varphi (X' \setminus W') \subset X are closed subsets and by Morphisms, Lemma 29.41.9 W is proper over A. The image of W \cap Q in \mathop{\mathrm{Spec}}(A) is closed. Since (A, I) is henselian, if W \cap Q is nonempty, then we find that W \cap Q has a point lying over \mathop{\mathrm{Spec}}(A/I). This is impossible as W'_0 = Y' = \varphi ^{-1}(Y). We conclude that W is an open and closed subscheme of X proper over A with W_0 = Y. Thus we reduce to the case described in the next paragraph.
Assume there exists an immersion j : X \to \mathbf{P}^ n_ A over A. Let \overline{X} be the scheme theoretic image of j. Since j is a quasi-compact morphism (Schemes, Lemma 26.21.14) we see that j : X \to \overline{X} is an open immersion (Morphisms, Lemma 29.7.7). Hence the base change j_0 : X_0 \to \overline{X}_0 is an open immersion as well. Thus j_0(Y) \subset \overline{X}_0 is open. It is also closed by Morphisms, Lemma 29.41.7. Suppose that the lemma holds for (\overline{X}, j_0(Y)). Let \overline{W} \subset \overline{X} be the corresponding open and closed subscheme proper over A such that j_0(Y) = \overline{W}_0. Then T = \overline{W} \setminus j(X) is closed in \overline{W}, hence has closed image in \mathop{\mathrm{Spec}}(A) by properness of \overline{W} over A. Since (A, I) is henselian, we find that if T is nonempty, then there is a point of T mapping into \mathop{\mathrm{Spec}}(A/I). This is impossible because j_0(Y) = \overline{W}_0 is contained in j(X). Hence \overline{W} is contained in j(X) and we can set W \subset X equal to the unique open and closed subscheme mapping isomorphically to \overline{W} via j. Thus we reduce to the case described in the next paragraph.
Assume X \subset \mathbf{P}^ n_ A is a closed subscheme. Then X \to \mathop{\mathrm{Spec}}(A) is a proper morphism. Let Z = X_0 \setminus Y. This is an open and closed subscheme of X_0 and X_0 = Y \amalg Z. Let X \to X' \to \mathop{\mathrm{Spec}}(A) be the Stein factorization as in Theorem 37.53.5. Let Y' \subset X'_0 and Z' \subset X'_0 be the images of Y and Z. Since the fibres of X \to Z are geometrically connected, we see that Y' \cap Z' = \emptyset . Hence X'_0 = Y' \amalg Z' as X \to X' is surjective. Since X' \to \mathop{\mathrm{Spec}}(A) is integral, we see that X' is the spectrum of an A-algebra integral over A. Recall that open and closed subsets of spectra correspond 1-to-1 with idempotents in the corresponding ring, see Algebra, Lemma 10.21.3. Hence by More on Algebra, Lemma 15.11.6 we see that we may write X' = W' \amalg V' with W' and V' open and closed and with Y' = W'_0 and Z' = V'_0. Let W be the inverse image in X to finish the proof.
\square
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