Taken from: [1, Exercise 9.42].
Exercise 9.42
Let
\[0 \longrightarrow L \overset{f}{\longrightarrow} M \overset{g}{\longrightarrow} N \longrightarrow 0\]be a short exact sequence of modules and homomorphisms over the commutative Noetherian ring $R$. Prove that
\[\operatorname{Ass}(L) \subseteq \operatorname{Ass}(M) \subseteq \operatorname{Ass}(L) \cup \operatorname{Ass}(N).\]Proof
The first inclusion is almost immediate. Let $\mathfrak{p} \in \operatorname{Ass}(L)$. Then $\mathfrak{p} = \operatorname{Ann}(\ell)$ for some nonzero $\ell \in L$. Let $r \in \operatorname{Ann}(\ell)$, then $0 = f(0) = f(r\ell) = r f(\ell)$, hence $r \in \operatorname{Ann}(f(\ell))$. Conversely, if we take $r \in \operatorname{Ann}(f(\ell))$, we have that $0 = r f(\ell) = f(r \ell)$. Since $f$ is injective, it follows that $r \ell = 0$, that is, $r \in \operatorname{Ann}(\ell)$. By the injectivity of $f$ we also have that $f(\ell) \neq 0$, so
\[\mathfrak{p} = \operatorname{Ann}(\ell) = \operatorname{Ann}(f(\ell)) \in \operatorname{Ass}(M),\]and thus, $\operatorname{Ass}(L) \subseteq \operatorname{Ass}(M)$.
Now let us prove the inclusion on the right. Let $\mathfrak{p} \in \operatorname{Ass}(M)$, that is, $\mathfrak{p} = \operatorname{Ann}(m)$ for some nonzero $m \in M$. We consider two cases:
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If $Rm \cap \operatorname{Im}(f) = 0$, the restriction of $g$ to the cyclic submodule $Rm\subseteq M$ is still injective. Thus $Rm\cong g(Rm)\subseteq N$ and $g(m)\neq0$. Then
\[\begin{align*} \operatorname{Ann}(g(m)) &=\{r\in R : rg(m) = 0\}\\ &=\{r\in R : g(rm)=0\} \\ &=\{r\in R: rm\in\ker(g)\} \\ &=\{r\in R: rm\in\operatorname{Im}(f)\}\\ &=\{r\in R: rm=0\}\\ &=\mathfrak{p}. \end{align*}\]Hence $\mathfrak{p}=\operatorname{Ann}\bigl(g(m)\bigr)\in\operatorname{Ass}(N)$.
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If $Rm \cap\operatorname{Im}(f)\neq 0$, there exists a nonzero element in the intersection. Concretely, pick $0\neq m_0\in Rm \cap\operatorname{Im}(f)$. Since $m_0\in Rm$, we can write $m_0 = am$ for some $a\in R$. Since $m_0\in\operatorname{Im}(f)$, there is some $\ell\in L$ with $f(\ell)=m_0$.
By an analogous argument to the first inclusion, the injectivity of $f$ guarantees that
\[\operatorname{Ann}(m_0)=\operatorname{Ann}\bigl(f(\ell)\bigr)=\operatorname{Ann}(\ell).\]It is clear that $\mathfrak{p} = \operatorname{Ann}(m) \subseteq \operatorname{Ann}(am) = \operatorname{Ann}(m_0)$; the reverse inclusion also holds: take $s \in \operatorname{Ann}(m_0)$, then $0 = s m_0 = (s a) m$, that is, $s a \in \mathfrak{p}$. Moreover, $a \notin \mathfrak{p}$, for otherwise $m_0 = a m = 0$. Therefore, $s \in \mathfrak{p}$ by primality of $\mathfrak{p}$. That is, $\mathfrak{p} = \operatorname{Ann}(m_0) = \operatorname{Ann}(\ell) \in \operatorname{Ass}(L)$.
From both cases, it follows that $\mathfrak{p} \in \operatorname{Ass}(L) \cup \operatorname{Ass}(N)$. $\blacksquare$
References
- Sharp, Rodney Y. (2001). Steps in Commutative Algebra (2nd ed.). Cambridge University Press. http://dx.doi.org/10.1017/CBO9780511623684