Annihilator of annihilator of annihilator of a submodule

Question

Let $R$ be a commutative ring and M be an $R$-module. The question is whether for every submodule $N$ of $M$ the equality

$$\operatorname{Ann}(\operatorname{Ann}(\operatorname{Ann}(N)))=\operatorname{Ann}(N)$$

holds. Anyone can give me a proof or a counter example? Thank you.

Answer 1

If $\operatorname{Ann}$ means $\operatorname{Ann}_R$, then this is not true in general when $N$ is not an ideal.

For example, take $R=\mathbb{Z}$ and $M=N=\mathbb{Z}/2\mathbb{Z}$.

Then $\operatorname{Ann}(N)=2\mathbb{Z}$, $\operatorname{Ann}(\operatorname{Ann}(N))=0$, and $\operatorname{Ann}(\operatorname{Ann}(\operatorname{Ann}(N)))=\mathbb{Z}$.

Answer 2

As you've undoubtedly discovered in the other solutions, you need to specify your annihilators more clearly. The expression $\operatorname{Ann}(\operatorname{Ann}(N))$ could be legitimately interpreted as either $\operatorname{Ann}_R(\operatorname{Ann}_R(N))$ or $\operatorname{Ann}_M(\operatorname{Ann}_R(N))$. Jeremy Rickard's example clearly shows the first interpretation leads to disaster, but the second interpretation will work.

Annnihilators furnish an excellent example of ("antitone") Galois connections. Actually all such Galois connections have this property, so we'll be able to simplify notation and get a better result.

Let's denote the poset of submodules of $M$ by $\mathcal{S}$ and the poset of ideals of $R$ by $\mathcal{I}$.

We can define two maps

$F(-)=\operatorname{Ann}_R(-):\mathcal{S}\to \mathcal{I}$ and

$G(-)=\operatorname{Ann}_M(-):\mathcal{I}\to \mathcal{S}$.

1. It is easy to show that if $I\subseteq J$ are ideals of $R$, then $G(J)\subseteq G(I)$, and if $N\subseteq N'$ are submodules of $M$, $F(N')\subseteq F(N)$.

2. Just as easily, you can also show $I\subseteq FG(I)$ and $N\subseteq GF(N)$.

So, $F$ and $G$ create a Galois connection.

Claim: For a Galois connection using $F$ and $G$, we have both $FGF=F$ and $GFG=G$.

Proof: On one hand, we're assured that $F(N)\subseteq FG(F(N))$ by item 2 above. On the other hand, we know $N\subseteq GF(N)$ by item 2, and then applying $F$ to both sides and reversing order of containment (allowed by item 1) you have $F(N)\supseteq FGF(N)$. Thus $F(N)= FGF(N)$ for all $N$. The proof for $GFG=G$ is virtually identical.

Conclusion

We can also define a map $H(-)=\operatorname{Ann}_R:\mathcal I\to\mathcal I$, and $H$ with $H$ does indeed produce a Galois connection on the ideals of $R$. As such, $HHH=H$, but this is a totally different Galois connection from the $F-G$ one!

As we've seen in this analysis, we can show that $FGF=F$, but there is no expectation of $HHF(N)=F(N)$!