Let $k$ be a field, let $I \triangleleft k[X_1,\dots,X_n]=S$ be an ideal and fix $f \in S$.
The saturated ideal of $I$ is $I^{sat}=I:f^\infty=\{g \in S \mid \exists m \in \mathbb{N} \ s.t. \ f^mg \in I \}=\displaystyle\bigcup_{i \geq 1} I:f^i$.
Prove that $I^{sat}=I:f^m \Leftrightarrow f^m=f^{m+1}$.
My attempt:
"$\Rightarrow$" Since we have the ascending chain $I:f \subseteq I:f^2 \subseteq \dots$ and $S$ is Noetherian, it follows that the $m$ that we are looking for is exactly the one that stops the chain, i.e. the one from which on all ideals in the chain are equal. From $I^{sat}=\displaystyle\bigcup_{i \geq 1} I:f^i$, we have that $I^{sat}=I:f^m$.
"$\Leftarrow$" We have to show that all of the ideals $I:f^q$ are in $I:f^m$, i.e. the chain stops after $m$ steps. We have to prove $\{g \in S \mid f^mg \in I \} = \{h \in S \mid f^{m+1}h \in I \}$. "$\subseteq$" is clear, from the chain.
What about the reverse inclusion? It seems like going around in circles, so it must be something easy that I don't see.
Thank you.