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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?

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    Yes, I read that thread just before posting mine. But I can't see any useful connection. $f$ seems arbitrary in my question, so it could be out of $I(V(I))$. This is my difficulty. If it were there, I think I solved it, as above. But again, what if $f \notin \sqrt{I}$? That's the point that puzzles me.2012-06-04

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$\tilde{I}$ is the kernel of the natural map $S[Y] \to (S/I)_f$ mapping $Y \mapsto f^{-1}$ (because of the natural isomorphism $R_f \cong R[Y]/(1-fY)$, which holds in general due to the universal properties). Hence, $\tilde{I} \cap S$ is the kernel of the natural map $S \to (S/I)_f$ which maps $s$ to the residue class of $s/1$. But from the description of the kernels of localizations as well as quotients, this now equals $I^{sat}$.

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    Right, it seems clear enough. Thank you.2012-06-04