doing some self-study in module theory, and I've run into a difficult (for me) exercise:
"(Let $R$ be a ring with unity, and let M be a left $R$-module.)
Let $I$ be an ideal of $R$. Let $M'$ be the subset of elements $a$ of $M$ that are annihilated by some power, $I^k$, of the ideal $I$, where the power may depend on $a$. Prove that $M'$ is a submodule of $M$."
So here is about where I'm at now: I know that the annihilator of an ideal (i.e. the set of elements of the module that annihilate it) is a submodule, I know that $I^{k+1} \subseteq I^k$, and I know that the union of a (possibly infinite) ascending chain of submodules is itself a submodule (the book suggests I use the last result here).
I've also found an explicit formulation of the annihilator of $I^k$, for all such $k$ that that actually exists here, and I'm fairly certain it's correct.
(Essentially, we define $f:M' \rightarrow \mathbb{Z}_+,\ f(a)$ is minimal $k$ such that $ra = 0\: \forall\: r \in I^k$. We then order the elements of $f(M')$, calling the smallest $k_1$, the next smallest $k_2$, etc. I'm fairly certain the annihilator of $I^{k_i}$ is given by $A_i=f^{-1}(k_i) \cup \{0_M\}$ for all $i$.)
However, $M'$ is not a union of an ascending chain of $A_i$, unless there is a maximal index for the $A_i$'s (call it j), in which case $M'=\bigcup\limits_{i=0}^{j-i} A_{j-1}$, and is thus a submodule. It seems like, because neither the ring $R$ nor the ideal $I$ nor $M'$ are known to be finite, the only way I can guarantee this maximal index exists is by applying Zorn's lemma. And I believe I can, because we can order the sets $A_i$ by reverse inclusion into a single chain, and they are bounded above by $\{0\}$.
And if that's necessary that's fine, but is there a way to do this without choice? (Not that I have anything against choice but it seems like too big a hammer for too small a problem in this case.)
$M’$ is indeed the union of an infinite increasing sequence of sub modules of $M$. Precisely, $M’ = \bigcup\limits_{n \in \mathbb{N}} A_n$, where $A_n = \{x \in M \mid I^n x = 0\}$ is the set of elements of $M$ annihilated by $I^n$. It’s easy to verify that $A$ is an increasing sequence. I’m not sure why you would need anything elaborate here, and I think you’re overthinking things.