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Let $M$ be a module over a ring $R$.

Let $\operatorname{Ass}(M)$ be the set of annihilator ideals $\operatorname{Ann}(x)$, which are prime, so

$$\operatorname{Ass}(M) = \{\operatorname{Ann}(x) \mid \operatorname{Ann}(x)\text{ is prime}, x \in M\}.$$

Recall that $\operatorname{Ann}(x) = \{r \in R \mid rx=0\}$.

If $M_1$ and $M_2$ are two modules, I wish to prove that $$\operatorname{Ass}(M_1 \oplus M_2) = \operatorname{Ass}(M_1) \cup \operatorname{Ass}(M_2),$$ where $\oplus$ is direct sum and $\cup$ is ordinary union of sets.

I need to do this by considering an element of the left hand side and show it is in the right hand side, so nothing fancy. The direction from right to left is easy, since for any $m_1 \in M_1$ I have $\operatorname{Ann}(m_1) = \operatorname{Ann}(m_1,0)$, but the other direction causes me trouble.

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2 Answers 2

Let $\mathfrak{p} \in \textrm{Ass}(M_1 \oplus M_2)$. Suppose $\mathfrak{p} = \textrm{Ann}(m_1 + m_2)$ where $m_1 \in M_1$ and $m_2 \in M_2$. (I am considering $M_1$ and $M_2$ as submodules of $M_1 \oplus M_2$.) So for all $a$ in $\mathfrak{p}$, $a m_1 + a m_2 = 0$, so $a m_1 = 0$ and $a m_2 = 0$. Thus $\mathfrak{p} \subseteq \textrm{Ann}(m_1) \cap \textrm{Ann}(m_2)$.

Now, either $\mathfrak{p} = \textrm{Ann}(m_1)$ or not; if it is we are done, so suppose $\mathfrak{p} \ne \textrm{Ann}(m_1)$. Let $a \in \textrm{Ann}(m_1)$, $a \notin \mathfrak{p}$. Then, $$a m_1 + a m_2 = a m_2 \ne 0$$ since otherwise $a \in \textrm{Ann}(m_1 + m_2)$, which would contradict $a \notin \mathfrak{p}$. Thus $a \notin \textrm{Ann}(m_2)$. So indeed $\mathfrak{p} = \textrm{Ann}(m_1) \cap \textrm{Ann}(m_2)$. Let $a$ be as above, and let $b \in \textrm{Ann}(m_2)$; then $a b \in \textrm{Ann}(m_1) \cap \textrm{Ann}(m_2) = \mathfrak{p}$, so $b \in \mathfrak{p}$. Thus $\textrm{Ann}(m_2) = \mathfrak{p}$. Hence, either $\textrm{Ann}(m_1) = \mathfrak{p}$ or $\textrm{Ann}(m_2)$, so $$\textrm{Ass}(M_1 \oplus M_2) = \textrm{Ass}(M_1) \cup \textrm{Ass}(M_2)$$ as required.

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This is a nice fact to isolate: if $\mathfrak a_1, \ldots, \mathfrak a_n$ are ideals and $\mathfrak p = \bigcap \mathfrak a_n$ is prime, then $\mathfrak p = \mathfrak a_i$ for some $i$. If you know basic algebraic geometry (the notion of an irreducible variety), then you can form a geometric interpretation of this, which makes the fact easier to remember. –  Dylan Moreland Jan 21 '12 at 20:48
Nice answer! A tiny question, why does for all $a\in\mathfrak{p}$, $am_1+am_2=0$ imply $am_1=0$ and $am_2=0$? –  Buble Feb 20 '12 at 6:13
$a m_1 = 0$ because $a \in \textrm{Ann}(m_1)$. –  Zhen Lin Feb 20 '12 at 7:20
@Zhen Lin: Btw: This should work with arbitrary direct sums as well, right? By induction on the finite ones and in the infinite case using the fact that every element lives in finitely many components only... Or is there some catch I don't realize? –  PavelC May 22 '14 at 21:23

You could use the fact that for a submodule $N\subseteq M$ we have $Ass(M)\subseteq Ass(N)\cup Ass(M/N)$. This can be proved directly: Let $Ann(m)=p\in Ass(M)$ and $A/p\cap N\neq 0$, then there is $0\neq n\in A/p\cap N$. By $A/Ann(m)\cong mA$ you get $Ann(n)=p$ and $p\in Ass(N)$. If otherwise $A/p\cap N=0$ you can use that $p\in Ass (M)$ iff there is a injection $A/p\rightarrow M$ to show that $p\in Ass(M/N)$.

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Thanks. Though I was aiming at doing it even more directly. Say taking an element Ann(x_1,x_2) in Ass(M_1 (+) M_2), then r(x_1,x_2) = (0,0) for all r in Ann(x_1,x_2). Can I then somehow infer that this annihilator is in either Ass(M_1) or Ass(M_2)? –  Jon Nørgaard Jan 21 '12 at 19:30

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