## 37.24 Relative assassins

Lemma 37.24.1. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $\xi \in \text{Ass}_{X/S}(\mathcal{F})$ and set $Z = \overline{\{ \xi \} } \subset X$. If $f$ is locally of finite type and $\mathcal{F}$ is a finite type $\mathcal{O}_ X$-module, then there exists a nonempty open $V \subset Z$ such that for every $s \in f(V)$ the generic points of $V_ s$ are elements of $\text{Ass}_{X/S}(\mathcal{F})$.

Proof. We may replace $S$ by an affine open neighbourhood of $f(\xi )$ and $X$ by an affine open neighbourhood of $\xi$. Hence we may assume $S = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(B)$ and that $f$ is given by the finite type ring map $A \to B$, see Morphisms, Lemma 29.15.2. Moreover, we may write $\mathcal{F} = \widetilde{M}$ for some finite $B$-module $M$, see Properties, Lemma 28.16.1. Let $\mathfrak q \subset B$ be the prime corresponding to $\xi$ and let $\mathfrak p \subset A$ be the corresponding prime of $A$. By assumption $\mathfrak q \in \text{Ass}_ B(M \otimes _ A \kappa (\mathfrak p))$, see Algebra, Remark 10.65.6 and Divisors, Lemma 31.2.2. With this notation $Z = V(\mathfrak q) \subset \mathop{\mathrm{Spec}}(B)$. In particular $f(Z) \subset V(\mathfrak p)$. Hence clearly it suffices to prove the lemma after replacing $A$, $B$, and $M$ by $A/\mathfrak pA$, $B/\mathfrak pB$, and $M/\mathfrak pM$. In other words we may assume that $A$ is a domain with fraction field $K$ and $\mathfrak q \subset B$ is an associated prime of $M \otimes _ A K$.

At this point we can use generic flatness. Namely, by Algebra, Lemma 10.118.3 there exists a nonzero $g \in A$ such that $M_ g$ is flat as an $A_ g$-module. After replacing $A$ by $A_ g$ we may assume that $M$ is flat as an $A$-module.

In this case, by Algebra, Lemma 10.65.4 we see that $\mathfrak q$ is also an associated prime of $M$. Hence we obtain an injective $B$-module map $B/\mathfrak q \to M$. Let $Q$ be the cokernel so that we obtain a short exact sequence

$0 \to B/\mathfrak q \to M \to Q \to 0$

of finite $B$-modules. After applying generic flatness Algebra, Lemma 10.118.3 once more, this time to the $B$-module $Q$, we may assume that $Q$ is a flat $A$-module. In particular we may assume the short exact sequence above is universally injective, see Algebra, Lemma 10.39.12. In this situation $(B/\mathfrak q) \otimes _ A \kappa (\mathfrak p') \subset M \otimes _ A \kappa (\mathfrak p')$ for any prime $\mathfrak p'$ of $A$. The lemma follows as a minimal prime $\mathfrak q'$ of the support of $(B/\mathfrak q) \otimes _ A \kappa (\mathfrak p')$ is an associated prime of $(B/\mathfrak q) \otimes _ A \kappa (\mathfrak p')$ by Divisors, Lemma 31.2.9. $\square$

Lemma 37.24.2. Let $f : X \to Y$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $U \subset X$ be an open subscheme. Assume

1. $f$ is of finite type,

2. $\mathcal{F}$ is of finite type,

3. $Y$ is irreducible with generic point $\eta$, and

4. $\text{Ass}_{X_\eta }(\mathcal{F}_\eta )$ is not contained in $U_\eta$.

Then there exists a nonempty open subscheme $V \subset Y$ such that for all $y \in V$ the set $\text{Ass}_{X_ y}(\mathcal{F}_ y)$ is not contained in $U_ y$.

Proof. Let $Z \subset X$ be the scheme theoretic support of $\mathcal{F}$, see Morphisms, Definition 29.5.5. Then $Z_\eta$ is the scheme theoretic support of $\mathcal{F}_\eta$ (Morphisms, Lemma 29.25.14). Hence the generic points of irreducible components of $Z_\eta$ are contained in $\text{Ass}_{X_\eta }(\mathcal{F}_\eta )$ by Divisors, Lemma 31.2.9. Hence we see that $Z_\eta \cap U_\eta = \emptyset$. Thus $T = Z \setminus U$ is a closed subset of $Z$ with $T_\eta = \emptyset$. If we endow $T$ with the induced reduced scheme structure then $T \to Y$ is a morphism of finite type. By Lemma 37.23.1 there is a nonempty open $V \subset Y$ with $T_ V = \emptyset$. Then $V$ works. $\square$

Lemma 37.24.3. Let $f : X \to Y$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $U \subset X$ be an open subscheme. Assume

1. $f$ is of finite type,

2. $\mathcal{F}$ is of finite type,

3. $Y$ is irreducible with generic point $\eta$, and

4. $\text{Ass}_{X_\eta }(\mathcal{F}_\eta ) \subset U_\eta$.

Then there exists a nonempty open subscheme $V \subset Y$ such that for all $y \in V$ we have $\text{Ass}_{X_ y}(\mathcal{F}_ y) \subset U_ y$.

Proof. (This proof is the same as the proof of Lemma 37.23.4. 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. $\square$

Lemma 37.24.4. Let $f : X \to S$ be a morphism which is locally of finite type. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module of finite type. Let $U \subset X$ be an open subscheme. Let $g : S' \to S$ be a morphism of schemes, let $f' : X' = X_{S'} \to S'$ be the base change of $f$, let $g' : X' \to X$ be the projection, set $\mathcal{F}' = (g')^*\mathcal{F}$, and set $U' = (g')^{-1}(U)$. Finally, let $s' \in S'$ with image $s = g(s')$. In this case

$\text{Ass}_{X_ s}(\mathcal{F}_ s) \subset U_ s \Leftrightarrow \text{Ass}_{X'_{s'}}(\mathcal{F}'_{s'}) \subset U'_{s'}.$

Proof. This follows immediately from Divisors, Lemma 31.7.3. See also Divisors, Remark 31.7.4. $\square$

Lemma 37.24.5. Let $f : X \to Y$ be a morphism of finite presentation. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module of finite presentation. Let $U \subset X$ be an open subscheme such that $U \to Y$ is quasi-compact. Then the set

$E = \{ y \in Y \mid \text{Ass}_{X_ y}(\mathcal{F}_ y) \subset U_ y\}$

is locally constructible in $Y$.

Proof. Let $y \in Y$. We have to show that there exists an open neighbourhood $V$ of $y$ in $Y$ such that $E \cap V$ is constructible in $V$. Thus we may assume that $Y$ is affine. Write $Y = \mathop{\mathrm{Spec}}(A)$ and $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as a directed limit of finite type $\mathbf{Z}$-algebras. By Limits, Lemma 32.10.1 we can find an $i$ and a morphism $f_ i : X_ i \to \mathop{\mathrm{Spec}}(A_ i)$ of finite presentation whose base change to $Y$ recovers $f$. After possibly increasing $i$ we may assume there exists a quasi-coherent $\mathcal{O}_{X_ i}$-module $\mathcal{F}_ i$ of finite presentation whose pullback to $X$ is isomorphic to $\mathcal{F}$, see Limits, Lemma 32.10.2. After possibly increasing $i$ one more time we may assume there exists an open subscheme $U_ i \subset X_ i$ whose inverse image in $X$ is $U$, see Limits, Lemma 32.4.11. By Lemma 37.24.4 it suffices to prove the lemma for $f_ i$. Thus we reduce to the case where $Y$ is the spectrum of a Noetherian ring.

We will use the criterion of Topology, Lemma 5.16.3 to prove that $E$ is constructible in case $Y$ is a Noetherian scheme. To see this let $Z \subset Y$ be an irreducible closed subscheme. We have to show that $E \cap Z$ either contains a nonempty open subset or is not dense in $Z$. This follows from Lemmas 37.24.2 and 37.24.3 applied to the base change $(X, \mathcal{F}, U) \times _ Y Z$ over $Z$. $\square$

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