Lemma 37.24.4. Let $f : X \to Y$ be a finite type morphism of schemes. Assume $Y$ irreducible with generic point $\eta$. Let $U \subset X$ be an open subscheme such that $U_\eta$ is scheme theoretically dense in $X_\eta$. Then there exists a nonempty open $V \subset Y$ such that $U_ y$ is scheme theoretically dense in $X_ y$ for all $y \in V$.

Proof. Let $Y' \subset Y$ be the reduction of $Y$. Let $X' = Y' \times _ Y X$ and $U' = Y' \times _ Y U$. As $Y' \to Y$ induces a bijection on points, and as $U' \to U$ and $X' \to X$ induce isomorphisms of scheme theoretic fibres, we may replace $Y$ by $Y'$ and $X$ by $X'$. Thus we may assume that $Y$ is integral, see Properties, Lemma 28.3.4. We may also replace $Y$ by a nonempty affine open. In other words we may assume that $Y = \mathop{\mathrm{Spec}}(A)$ where $A$ is a domain with fraction field $K$.

As $f$ is of finite type we see that $X$ is quasi-compact. Write $X = X_1 \cup \ldots \cup X_ n$ for some affine opens $X_ i$. By Morphisms, Definition 29.7.1 we see that $U_ i = X_ i \cap U$ is an open subscheme of $X_ i$ such that $U_{i, \eta }$ is scheme theoretically dense in $X_{i, \eta }$. Thus it suffices to prove the result for the pairs $(X_ i, U_ i)$, in other words we may assume that $X$ is affine.

Write $X = \mathop{\mathrm{Spec}}(B)$. Note that $B_ K$ is Noetherian as it is a finite type $K$-algebra. Hence $U_\eta$ is quasi-compact. Thus we can find finitely many $g_1, \ldots , g_ m \in B$ such that $D(g_ j) \subset U$ and such that $U_\eta = D(g_1)_\eta \cup \ldots \cup D(g_ m)_\eta$. The fact that $U_\eta$ is scheme theoretically dense in $X_\eta$ means that $B_ K \to \bigoplus _ j (B_ K)_{g_ j}$ is injective, see Morphisms, Example 29.7.4. By Algebra, Lemma 10.24.4 this is equivalent to the injectivity of $B_ K \to \bigoplus \nolimits _{j = 1, \ldots , m} B_ K$, $b \mapsto (g_1b, \ldots , g_ mb)$. Let $M$ be the cokernel of this map over $A$, i.e., such that we have an exact sequence

$0 \to I \to B \xrightarrow {(g_1, \ldots , g_ m)} \bigoplus \nolimits _{j = 1, \ldots , m} B \to M \to 0$

After replacing $A$ by $A_ h$ for some nonzero $h$ we may assume that $B$ is a flat, finitely presented $A$-algebra, and that $M$ is flat over $A$, see Algebra, Lemma 10.118.3. The flatness of $B$ over $A$ implies that $B$ is torsion free as an $A$-module, see More on Algebra, Lemma 15.22.9. Hence $B \subset B_ K$. By assumption $I_ K = 0$ which implies that $I = 0$ (as $I \subset B \subset B_ K$ is a subset of $I_ K$). Hence now we have a short exact sequence

$0 \to B \xrightarrow {(g_1, \ldots , g_ m)} \bigoplus \nolimits _{j = 1, \ldots , m} B \to M \to 0$

with $M$ flat over $A$. Hence for every homomorphism $A \to \kappa$ where $\kappa$ is a field, we obtain a short exact sequence

$0 \to B \otimes _ A \kappa \xrightarrow {(g_1 \otimes 1, \ldots , g_ m \otimes 1)} \bigoplus \nolimits _{j = 1, \ldots , m} B \otimes _ A \kappa \to M \otimes _ A \kappa \to 0$

see Algebra, Lemma 10.39.12. Reversing the arguments above this means that $\bigcup D(g_ j \otimes 1)$ is scheme theoretically dense in $\mathop{\mathrm{Spec}}(B \otimes _ A \kappa )$. As $\bigcup D(g_ j \otimes 1) = \bigcup D(g_ j)_\kappa \subset U_\kappa$ we obtain that $U_\kappa$ is scheme theoretically dense in $X_\kappa$ which is what we wanted to prove. $\square$

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