**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 29.8.5. 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