Let $k$ be a field and $I\trianglelefteq k[X_1,\dots,X_n]=S$ an ideal, generated by $\{f_1,\dots,f_s\}$. Fix $f \in S$ and let $Y$ be a new indeterminate. Let $\tilde{I}=(f_1,\dots,f_s,1-fY)\trianglelefteq S[Y]$.
Let $I^{sat}=I:f^\infty=\{g \in S \mid \exists m \in \mathbb{N} \ s.t. \ f^mg \in I$ be the saturation of the ideal $I$.
Prove that $I^{sat}=\tilde{I} \cap S$.
My attempt: Since the first moment I saw this, it reminded me of the Rabinowitz trick used to prove the Nullstellensatz, since it involves introducing a new variable and constructing a clever ideal like $\tilde{I}$ here.
So, if $f \in \sqrt{I}$, then, from the Weak Nullstellensatz, $\tilde{I}$ is not proper, i.e. $\tilde{I}=S[Y] \Rightarrow \tilde{I} \cap S = S$. So we have to prove that $I^{sat}=S$. One inclusion is obvious. For the other, let $g \in S$. Since $f \in \sqrt{I} \Rightarrow f^m \in I$, for some positive integer $m$. But then $f^mg \in I$, so $g \in I^{sat}$.
But what if $f \notin \sqrt{I}$? Could it be possible? And if so, what next?