Proof.
Let X_{red} denote the reduction of X. Then Z \cap X_{red} is a locally principal closed subscheme of X_{red}, see Divisors, Lemma 31.13.11. Hence we may assume that X is reduced. In other words X is integral, see Properties, Lemma 28.3.4. In this case the morphism X \to S factors through S_{red}, see Schemes, Lemma 26.12.7. Thus we may replace S by S_{red} and assume that S is integral too.
The assertion that f is dominant signifies that the generic point of X is mapped to \eta , see Morphisms, Lemma 29.8.6. Moreover, the scheme X_\eta is an integral scheme which is locally of finite type over the field \kappa (\eta ). Hence d = \dim (X_\eta ) \geq 0 is equal to \dim _\xi (X_\eta ) for every point \xi of X_\eta , see Algebra, Lemmas 10.114.4 and 10.114.5. In view of Morphisms, Lemma 29.28.4 and condition (5) we conclude that \dim _ x(X_ s) = d for every x \in X_ s.
In the Noetherian case the assertion can be proved as follows. If the lemma does not holds there exists x \in Z_ s which is a generic point of an irreducible component of Z_ s but not a generic point of any irreducible component of X_ s. Then we see that \dim _ x(Z_ s) \leq d - 1, because \dim _ x(X_ s) = d and in a neighbourhood of x in X_ s the closed subscheme Z_ s does not contain any of the irreducible components of X_ s. Hence after replacing X by an open neighbourhood of x we may assume that \dim _ z(Z_{f(z)}) \leq d - 1 for all z \in Z, see Morphisms, Lemma 29.28.4. Let \xi ' \in Z be a generic point of an irreducible component of Z and set s' = f(\xi ). As Z \not= X is locally principal we see that \dim (\mathcal{O}_{X, \xi }) = 1, see Algebra, Lemma 10.60.11 (this is where we use X is Noetherian). Let \xi \in X be the generic point of X and let \xi _1 be a generic point of any irreducible component of X_{s'} which contains \xi '. Then we see that we have the specializations
\xi \leadsto \xi _1 \leadsto \xi '.
As \dim (\mathcal{O}_{X, \xi }) = 1 one of the two specializations has to be an equality. By assumption s' \not= \eta , hence the first specialization is not an equality. Hence \xi ' = \xi _1 is a generic point of an irreducible component of X_{s'}. Applying Morphisms, Lemma 29.28.4 one more time this implies \dim _{\xi '}(Z_{s'}) = \dim _{\xi '}(X_{s'}) \geq \dim (X_\eta ) = d which gives the desired contradiction.
In the general case we reduce to the Noetherian case as follows. If the lemma is false then there exists a point x \in X lying over s such that x is a generic point of an irreducible component of Z_ s, but not a generic point of any of the irreducible components of X_ s. Let U \subset S be an affine neighbourhood of s and let V \subset X be an affine neighbourhood of x with f(V) \subset U. Write U = \mathop{\mathrm{Spec}}(A) and V = \mathop{\mathrm{Spec}}(B) so that f|_ V is given by a ring map A \to B. Let \mathfrak q \subset B, resp. \mathfrak p \subset A be the prime corresponding to x, resp. s. After possibly shrinking V we may assume Z \cap V is cut out by some element g \in B. Denote K the fraction field of A. What we know at this point is the following:
A \subset B is a finitely generated extension of domains,
the element g \otimes 1 is invertible in B \otimes _ A K,
d = \dim (B \otimes _ A K) = \dim (B \otimes _ A \kappa (\mathfrak p)),
g \otimes 1 is not a unit of B \otimes _ A \kappa (\mathfrak p), and
g \otimes 1 is not in any of the minimal primes of B \otimes _ A \kappa (\mathfrak p).
We are seeking a contradiction.
Pick elements x_1, \ldots , x_ n \in B which generate B over A. For a finitely generated \mathbf{Z}-algebra A_0 \subset A let B_0 \subset B be the A_0-subalgebra generated by x_1, \ldots , x_ n, denote K_0 the fraction field of A_0, and set \mathfrak p_0 = A_0 \cap \mathfrak p. We claim that when A_0 is large enough then (1) – (5) also hold for the system (A_0 \subset B_0, g, \mathfrak p_0).
We prove each of the conditions in turn. Part (1) holds by construction. For part (2) write (g \otimes 1) h = 1 for some h \otimes 1/a \in B \otimes _ A K. Write g = \sum a_ I x^ I, h = \sum a'_ I x^ I (multi-index notation) for some coefficients a_ I, a'_ I \in A. As soon as A_0 contains a and the a_ I, a'_ I then (2) holds because B_0 \otimes _{A_0} K_0 \subset B \otimes _ A K (as localizations of the injective map B_0 \to B). To achieve (3) consider the exact sequence
0 \to I \to A[X_1, \ldots , X_ n] \to B \to 0
which defines I where the second map sends X_ i to x_ i. Since \otimes is right exact we see that I \otimes _ A K, respectively I \otimes _ A \kappa (\mathfrak p) is the kernel of the surjection K[X_1, \ldots , X_ n] \to B \otimes _ A K, respectively \kappa (\mathfrak p)[X_1, \ldots , X_ n] \to B \otimes _ A \kappa (\mathfrak p). As a polynomial ring over a field is Noetherian there exist finitely many elements h_ j \in I, j = 1, \ldots , m which generate I \otimes _ A K and I \otimes _ A \kappa (\mathfrak p). Write h_ j = \sum a_{j, I}X^ I. As soon as A_0 contains all a_{j, I} we get to the situation where
B_0 \otimes _{A_0} K_0 \otimes _{K_0} K = B \otimes _ A K \quad \text{and}\quad B_0 \otimes _{A_0} \kappa (\mathfrak p_0) \otimes _{\kappa (\mathfrak p_0)} \kappa (\mathfrak p) = B \otimes _ A \kappa (\mathfrak p).
By either Morphisms, Lemma 29.28.3 or Algebra, Lemma 10.116.5 we see that the dimension equalities of (3) are satisfied. Part (4) is immediate. As B_0 \otimes _{A_0} \kappa (\mathfrak p_0) \subset B \otimes _ A \kappa (\mathfrak p) each minimal prime of B_0 \otimes _{A_0} \kappa (\mathfrak p_0) lies under a minimal prime of B \otimes _ A \kappa (\mathfrak p) by Algebra, Lemma 10.30.6. This implies that (5) holds. In this way we reduce the problem to the Noetherian case which we have dealt with above.
\square
Comments (2)
Comment #6699 by Sebastian Bozlee on
Comment #6702 by Johan on