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According to wikipedia, the saturation with respect to $f$ of an ideal $I$ in $R$ is the ideal $I:f^\infty:=\{g\in R:\exists k\in\mathbb{N}, f^kg\in I\}$.

The important property of the saturation, which ensures that it removes from the algebraic set defined by the ideal I the irreducible components on which the polynomial f is zero is the following: The primary decomposition of $I:f^\infty$ consists in the components of the primary decomposition of $I$ that do not contain any power of $f$.

Does anyone have references/books about the proof of this important property?

Given any set $S$ of polynomials in $R$, we denote by $V(S)$ the zero set, or variety of $S$: $V(S) = \{(p_1,\cdots,p_n) \in k^n | f(p_1,\cdots,p_n) = 0\ \text{for all } f \in S\}$. If $\mathfrak{a}=\langle S\rangle$, then $V(\mathfrak{a})=V(S)$. If $S=\{f\}$, then $V(S)=V(f)$.

According to Proposition 3.13 of the note "Grobner Bases: a Tutorial---Mike Stillman":

We have $V(I:f^\infty)=\overline{V(I)\backslash V(f)}$.

Does anyone have references/books contain a proof of this proposition?

Finally,

can one prove the important property by the Proposition 3.13?

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    $\begingroup$ Another way of looking at this is by localization. If $f$ is nilpotent, clearly the saturation is all of $R$, so assume $f$ is not nilpotent. Then, we have the localization map $i:R\to R_f$ and the saturation is just $i^{-1} i(I)R_f$. Proposition 3.13 above is an easy consequence. $\endgroup$
    – Mohan
    Nov 16, 2019 at 15:48

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