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.


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 $\textrm{Ann}(m_2)=\mathfrak{p}$, so $$\textrm{Ass}(M_1 \oplus M_2) = \textrm{Ass}(M_1) \cup \textrm{Ass}(M_2)$$ as required.

  • 2
    $\begingroup$ 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. $\endgroup$ – Dylan Moreland Jan 21 '12 at 20:48
  • $\begingroup$ Nice answer! A tiny question, why does for all $a\in\mathfrak{p}$, $am_1+am_2=0$ imply $am_1=0$ and $am_2=0$? $\endgroup$ – Buble Feb 20 '12 at 6:13
  • $\begingroup$ $a m_1 = 0$ because $a \in \textrm{Ann}(m_1)$. $\endgroup$ – Zhen Lin Feb 20 '12 at 7:20
  • 1
    $\begingroup$ @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? $\endgroup$ – Pavel Čoupek May 22 '14 at 21:23
  • $\begingroup$ @PavelČoupek You're right, this works for arbitrary direct sums as well. $\endgroup$ – Brahadeesh Jun 16 at 11:31

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

  • $\begingroup$ 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)? $\endgroup$ – Jon Nørgaard Jan 21 '12 at 19:30

This is just a restructuring of @ZhenLin's answer in light of the comments below it by @DylanMoreland and @PavelČoupek.

Let $M$ be an $A$-module that is a finite direct sum of its submodules $M_i$, that is, let $M = \bigoplus M_i$. If $\mathfrak{p}_i$ is an associated prime of $M_i$, then there exists $m_i \in M_i$ such that $\mathfrak{p}_i = \mathrm{ann}_A(m_i)$. But since $m_i \in M$ as well, this shows that $\mathfrak{p}_i$ is an associated prime of $M$. Hence, $\mathrm{Ass}(M) \supset \bigcup \mathrm{Ass}(M_i)$.

Conversely, suppose that $\mathfrak{p}$ is an associated prime of $M$. We will show that $\mathfrak{p} = \bigcap \mathrm{ann}_A(m_i)$.

Let $m = \sum m_i$, $m_i \in M_i$, be an element such that $\mathfrak{p} = \mathrm{ann}_A(m)$. In particular, $\mathfrak{p}$ annihilates $m_i$ for each $i$, so $\mathfrak{p} \subset \mathrm{ann}_A(m_i)$ for each $i$. Hence, $\mathfrak{p} \subset \bigcap \mathrm{ann}_A(m_i)$.

For the reverse inclusion, we first observe that $\prod \mathrm{ann}_A(m_i) \subset \mathfrak{p}$. This is because if $a_i \in \mathrm{ann}_A(m_i)$ for each $i$ and $a = \prod a_i$, then $am = 0$. Hence, every element of $\prod \mathrm{ann}_A(m_i)$ annihilates $m$, and thus belongs to $\mathfrak{p}$. Since $\mathfrak{p}$ is prime, $\mathrm{ann}_A(m_i) \subset \mathfrak{p}$ for some $i$. Thus, $\bigcap \mathrm{ann}_A(m_i) \subset \mathfrak{p}$.

Hence, $\mathfrak{p} = \bigcap \mathrm{ann}_A(m_i)$. Now, we prove the observation mentioned by @DylanMoreland, namely that this implies $\mathfrak{p} = \mathrm{ann}_A(m_i)$ for some $i$.

To this end, wlog assume that $\mathfrak{p} \neq \mathrm{ann}_A(m_i)$ for all $i \neq 1$. We will show that $\mathfrak{p} = \mathrm{ann}_A(m_1)$. Since we have already shown that $\mathfrak{p} \subset \mathrm{ann}_A(m_1)$, we only need to show the reverse inclusion. So, let $a_1 \in \mathrm{ann}_A(m_1)$. Choose $a_i \in \mathrm{ann}_A(m_i) \setminus \mathfrak{p}$ for each $i \neq 1$. As before, let $a = \prod a_i$. Then, $a \in \prod \mathrm{ann}_A(m_i) \subset \bigcap \mathrm{ann}_A(m_i) = \mathfrak{p}$. Hence, $a_1 \in \mathfrak{p}$, so that $\mathrm{ann}_A(m_1) \subset \mathfrak{p}$.

Thus, $\mathfrak{p} = \mathrm{ann}_A(m_1)$, and so $\mathrm{Ass}(M) \subset \bigcup \mathrm{Ass}(M_i)$.

The same proof works for arbitrary direct sums as well. $\mathrm{Ass}(M) \supset \bigcup \mathrm{Ass}(M_i)$ by the same argument as before. To prove the reverse inclusion, let $\mathfrak{p}$ be an associated prime of $M$. Let $m = \sum m_i$, $m_i \in M_i$, be an element such that $\mathfrak{p} = \mathrm{ann}_A(m)$. In particular, $m_i = 0$ for all but finitely many indices $i$. So, $m$ is an element of a finite direct sum. Now, we run through the same proof as before for this finite direct sum to prove that $\mathfrak{p}$ is an associated prime of one of the $M_i$.

Hence, the result $$\operatorname{Ass}\left(\bigoplus M_i\right) = \bigcup \operatorname{Ass}(M_i)$$ is true for arbitrary direct sums.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.