Question: Let $I$ be an ideal of $R$ (a ring with unity). 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 $M$.
Proof: Let $r \in R$ and $x, y \in M'$.
Note that $-y \in M'$ because $I^ly = \{0\}, l \in \mathbb{Z}_{>0}$ and $\{0\} = I^l0 = I^l(y - y) = I^ly + -I^ly = \{0\} + -I^ly = -I^ly $ hence $-y \in M'$
$I^kx = \{0\}$ and $I^ly = \{0\}$ for $k, l \in \mathbb{Z}_{>0}$
If $k \geq l$ $\Rightarrow$ $I^k(-y) = I^{k-l}I^l(-y) = I^{k-l}\{0\} = \{0\}$
Also $Ir \subseteq I$ so then $I^krx \subseteq I^kx = \{0\}$. This allows us to have $I^k(rx + (-y)) = I^krx + I^k(-y)) = \{0\} + \{0\} = \{0\}$
If $l > k$ we have a very similar argument.
Thus $M'$ is a submodule of $M$.
This is an exercise from Dummit and Foote with a note that says to use a previous exercise where we prove that the infinite union of an ascending chain submodules is a submodule. I did not use that theorem and am unsure if I have missed something crucial in this proof or if there is simply an alternative way to use it.