## Higher annihilators of an ideal

Let $R$ be a ring with 1, let $M$ be a left $R$-module, and let $I$ be an ideal of $R$. Prove that $\bigcup_{k \in \mathbb{N}^+} \mathsf{Ann}_M(I^k)$ is a submodule of $M$.

We claim that for each $k \geq 1$, $\mathsf{Ann}_M(I^k) \subseteq \mathsf{Ann}_M(I^{k+1})$. To see this, let $m \in \mathsf{Ann}_M(I^k)$ and let $a \in I^{k+1}$. Now by definition, $a$ is a finite sum of $k+1$-fold products of elements in $I$. Note that every $k$-fold product of elements in $I$ is annihilated my $M$, so that in fact $a \cdot m = 0$. Thus $m \in \mathsf{Ann}_M(I^{k+1})$.

So $\bigcup_{k \in \mathbb{N}^+} \mathsf{Ann}_M(I^k)$ is the union of a chain of submodules. By this previous exercise, it is a submodule of $M$.

• Yun  On November 13, 2011 at 11:50 pm

Perhaps I’m misreading this, but should you be taking a union instead of an intersection of the annihilators?

• nbloomf  On November 14, 2011 at 8:07 am

You’re right. This isn’t the first time I’ve mixed up ‘cap’ and ‘cup’…

Thanks!

• versus  On November 16, 2011 at 2:55 pm

What do you mean by fold product?

• nbloomf  On November 17, 2011 at 10:42 am

A $k$-fold product is just a product with $k$ factors. For example, if $I$ is an ideal, then the ideal product $I^2$ consists of all possible finite sums of 2-fold products of elements from $I$.

• Gobi Ree  On February 16, 2012 at 7:23 pm

Typo in the problem: \cap should be \cup

• nbloomf  On March 17, 2012 at 2:54 pm

Thanks!

Looking back, it seems I’ve fixed that typo before.