## 31.6 Morphisms and weakly associated points

Lemma 31.6.1. Let $f : X \to S$ be an affine morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then we have

$\text{WeakAss}_ S(f_*\mathcal{F}) \subset f(\text{WeakAss}_ X(\mathcal{F}))$

Proof. We may assume $X$ and $S$ affine, so $X \to S$ comes from a ring map $A \to B$. Then $\mathcal{F} = \widetilde M$ for some $B$-module $M$. By Lemma 31.5.2 the weakly associated points of $\mathcal{F}$ correspond exactly to the weakly associated primes of $M$. Similarly, the weakly associated points of $f_*\mathcal{F}$ correspond exactly to the weakly associated primes of $M$ as an $A$-module. Hence the lemma follows from Algebra, Lemma 10.66.11. $\square$

Lemma 31.6.2. Let $f : X \to S$ be an affine morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. If $X$ is locally Noetherian, then we have

$f(\text{Ass}_ X(\mathcal{F})) = \text{Ass}_ S(f_*\mathcal{F}) = \text{WeakAss}_ S(f_*\mathcal{F}) = f(\text{WeakAss}_ X(\mathcal{F}))$

Proof. We may assume $X$ and $S$ affine, so $X \to S$ comes from a ring map $A \to B$. As $X$ is locally Noetherian the ring $B$ is Noetherian, see Properties, Lemma 28.5.2. Write $\mathcal{F} = \widetilde M$ for some $B$-module $M$. By Lemma 31.2.2 the associated points of $\mathcal{F}$ correspond exactly to the associated primes of $M$, and any associated prime of $M$ as an $A$-module is an associated points of $f_*\mathcal{F}$. Hence the inclusion

$f(\text{Ass}_ X(\mathcal{F})) \subset \text{Ass}_ S(f_*\mathcal{F})$

follows from Algebra, Lemma 10.63.13. We have the inclusion

$\text{Ass}_ S(f_*\mathcal{F}) \subset \text{WeakAss}_ S(f_*\mathcal{F})$

by Lemma 31.5.3. We have the inclusion

$\text{WeakAss}_ S(f_*\mathcal{F}) \subset f(\text{WeakAss}_ X(\mathcal{F}))$

by Lemma 31.6.1. The outer sets are equal by Lemma 31.5.8 hence we have equality everywhere. $\square$

Lemma 31.6.3. Let $f : X \to S$ be a finite morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Then $\text{WeakAss}(f_*\mathcal{F}) = f(\text{WeakAss}(\mathcal{F}))$.

Proof. We may assume $X$ and $S$ affine, so $X \to S$ comes from a finite ring map $A \to B$. Write $\mathcal{F} = \widetilde M$ for some $B$-module $M$. By Lemma 31.5.2 the weakly associated points of $\mathcal{F}$ correspond exactly to the weakly associated primes of $M$. Similarly, the weakly associated points of $f_*\mathcal{F}$ correspond exactly to the weakly associated primes of $M$ as an $A$-module. Hence the lemma follows from Algebra, Lemma 10.66.13. $\square$

Lemma 31.6.4. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{G}$ be a quasi-coherent $\mathcal{O}_ S$-module. Let $x \in X$ with $s = f(x)$. If $f$ is flat at $x$, the point $x$ is a generic point of the fibre $X_ s$, and $s \in \text{WeakAss}_ S(\mathcal{G})$, then $x \in \text{WeakAss}(f^*\mathcal{G})$.

Proof. Let $A = \mathcal{O}_{S, s}$, $B = \mathcal{O}_{X, x}$, and $M = \mathcal{G}_ s$. Let $m \in M$ be an element whose annihilator $I = \{ a \in A \mid am = 0\}$ has radical $\mathfrak m_ A$. Then $m \otimes 1$ has annihilator $I B$ as $A \to B$ is faithfully flat. Thus it suffices to see that $\sqrt{I B} = \mathfrak m_ B$. This follows from the fact that the maximal ideal of $B/\mathfrak m_ AB$ is locally nilpotent (see Algebra, Lemma 10.25.1) and the assumption that $\sqrt{I} = \mathfrak m_ A$. Some details omitted. $\square$

Lemma 31.6.5. Let $K/k$ be a field extension. Let $X$ be a scheme over $k$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $y \in X_ K$ with image $x \in X$. If $y$ is a weakly associated point of the pullback $\mathcal{F}_ K$, then $x$ is a weakly associated point of $\mathcal{F}$.

Proof. This is the translation of Algebra, Lemma 10.66.19 into the language of schemes. $\square$

Here is a simple lemma where we find that pushforwards often have depth at least 2.

Lemma 31.6.6. Let $f : X \to S$ be a quasi-compact and quasi-separated morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Let $s \in S$.

1. If $s \not\in f(X)$, then $s$ is not weakly associated to $f_*\mathcal{F}$.

2. If $s \not\in f(X)$ and $\mathcal{O}_{S, s}$ is Noetherian, then $s$ is not associated to $f_*\mathcal{F}$.

3. If $s \not\in f(X)$, $(f_*\mathcal{F})_ s$ is a finite $\mathcal{O}_{S, s}$-module, and $\mathcal{O}_{S, s}$ is Noetherian, then $\text{depth}((f_*\mathcal{F})_ s) \geq 2$.

4. If $\mathcal{F}$ is flat over $S$ and $a \in \mathfrak m_ s$ is a nonzerodivisor, then $a$ is a nonzerodivisor on $(f_*\mathcal{F})_ s$.

5. If $\mathcal{F}$ is flat over $S$ and $a, b \in \mathfrak m_ s$ is a regular sequence, then $a$ is a nonzerodivisor on $(f_*\mathcal{F})_ s$ and $b$ is a nonzerodivisor on $(f_*\mathcal{F})_ s/a(f_*\mathcal{F})_ s$.

6. If $\mathcal{F}$ is flat over $S$ and $(f_*\mathcal{F})_ s$ is a finite $\mathcal{O}_{S, s}$-module, then $\text{depth}((f_*\mathcal{F})_ s) \geq \min (2, \text{depth}(\mathcal{O}_{S, s}))$.

Proof. Part (1) is Lemma 31.5.9. Part (2) follows from (1) and Lemma 31.5.8.

Proof of part (3). To show the depth is $\geq 2$ it suffices to show that $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{S, s}}(\kappa (s), (f_*\mathcal{F})_ s) = 0$ and $\mathop{\mathrm{Ext}}\nolimits ^1_{\mathcal{O}_{S, s}}(\kappa (s), (f_*\mathcal{F})_ s) = 0$, see Algebra, Lemma 10.72.5. Using the exact sequence $0 \to \mathfrak m_ s \to \mathcal{O}_{S, s} \to \kappa (s) \to 0$ it suffices to prove that the map

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{S, s}}(\mathcal{O}_{S, s}, (f_*\mathcal{F})_ s) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{S, s}}(\mathfrak m_ s, (f_*\mathcal{F})_ s)$

is an isomorphism. By flat base change (Cohomology of Schemes, Lemma 30.5.2) we may replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ and $X$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \times _ S X$. Denote $\mathfrak m \subset \mathcal{O}_ S$ the ideal sheaf of $s$. Then we see that

$\mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{S, s}}(\mathfrak m_ s, (f_*\mathcal{F})_ s) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ S}(\mathfrak m, f_*\mathcal{F}) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ X}(f^*\mathfrak m, \mathcal{F})$

the first equality because $S$ is local with closed point $s$ and the second equality by adjunction for $f^*, f_*$ on quasi-coherent modules. However, since $s \not\in f(X)$ we see that $f^*\mathfrak m = \mathcal{O}_ X$. Working backwards through the arguments we get the desired equality.

For the proof of (4), (5), and (6) we use flat base change (Cohomology of Schemes, Lemma 30.5.2) to reduce to the case where $S$ is the spectrum of $\mathcal{O}_{S, s}$. Then a nonzerodivisor $a \in \mathcal{O}_{S, s}$ deterimines a short exact sequence

$0 \to \mathcal{O}_ S \xrightarrow {a} \mathcal{O}_ S \to \mathcal{O}_ S/a \mathcal{O}_ S \to 0$

Since $\mathcal{F}$ is flat over $S$, we obtain an exact sequence

$0 \to \mathcal{F} \xrightarrow {a} \mathcal{F} \to \mathcal{F}/a\mathcal{F} \to 0$

Pushing forward we obtain an exact sequence

$0 \to f_*\mathcal{F} \xrightarrow {a} f_*\mathcal{F} \to f_*(\mathcal{F}/a\mathcal{F})$

This proves (4) and it shows that $f_*\mathcal{F}/ af_*\mathcal{F} \subset f_*(\mathcal{F}/a\mathcal{F})$. If $b$ is a nonzerodivisor on $\mathcal{O}_{S, s}/a\mathcal{O}_{S, s}$, then the exact same argument shows $b : \mathcal{F}/a\mathcal{F} \to \mathcal{F}/a\mathcal{F}$ is injective. Pushing forward we conclude

$b : f_*(\mathcal{F}/a\mathcal{F}) \to f_*(\mathcal{F}/a\mathcal{F})$

is injective and hence also $b : f_*\mathcal{F}/ af_*\mathcal{F} \to f_*\mathcal{F}/ af_*\mathcal{F}$ is injective. This proves (5). Part (6) follows from (4) and (5) and the definitions. $\square$

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