**Proof.**
(This proof is the same as the proof of Lemma 37.24.4. We urge the reader to read that proof first.) Since the statement is about fibres it is clear that we may replace $Y$ by its reduction. Hence we may assume that $Y$ is integral, see Properties, Lemma 28.3.4. We may also assume that $Y = \mathop{\mathrm{Spec}}(A)$ is affine. Then $A$ 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 $X_ i$ and set $\mathcal{F}_ i = \mathcal{F}|_{X_ i}$. By assumption the generic fibre of $U_ i = X_ i \cap U$ contains $\text{Ass}_{X_{i, \eta }}(\mathcal{F}_{i, \eta })$. Thus it suffices to prove the result for the triples $(X_ i, \mathcal{F}_ i, U_ i)$, in other words we may assume that $X$ is affine.

Write $X = \mathop{\mathrm{Spec}}(B)$. Let $N$ be a finite $B$-module such that $\mathcal{F} = \widetilde{N}$. 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 $. Since $\text{Ass}_{X_\eta }(\mathcal{F}_\eta ) \subset U_\eta $ we see that $N_ K \to \bigoplus _ j (N_ K)_{g_ j}$ is injective. By Algebra, Lemma 10.24.4 this is equivalent to the injectivity of $N_ K \to \bigoplus \nolimits _{j = 1, \ldots , m} N_ K$, $n \mapsto (g_1n, \ldots , g_ mn)$. Let $I$ and $M$ be the kernel and cokernel of this map over $A$, i.e., such that we have an exact sequence

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

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

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

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

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

see Algebra, Lemma 10.39.12. Reversing the arguments above this means that $\bigcup D(g_ j \otimes 1)$ contains $\text{Ass}_{B \otimes _ A \kappa }(N \otimes _ A \kappa )$. As $\bigcup D(g_ j \otimes 1) = \bigcup D(g_ j)_\kappa \subset U_\kappa $ we obtain that $U_\kappa $ contains $\text{Ass}_{X \otimes \kappa }(\mathcal{F} \otimes \kappa )$ which is what we wanted to prove.
$\square$

## Comments (0)