g-saturation of an ideal I⊆R

hbghlyj

Let $R$ be a commutative ring and take $f, g \in R$. Consider the homomorphism $R \to R /(f) \otimes_R R_g$.
$$
\ker\big(R \to(R/(f))\otimes_R R_g\big)
=\ker\big(R \to(R/(f))_g\big)
=\{r\in R:\exists n\ge 0\text{ with }g^{n}r\in (f)\}.
$$
Equivalently,
$$
\ker=\bigcup_{n\ge 0}\big((f):g^{n}\big)
=(f)R_g\cap R,
$$
i.e. the $g$-saturation of the ideal $(f)$.

hbghlyj

If $\gcd(f,g)=1$, the $g$-saturation of $(f)$ is just $(f)$.
If $g^n\in(f)$, then in $R_g$ the element $g$ is a unit, so $g^n$ is a unit and hence $f$ becomes a unit too; thus $(R/(f))\otimes_R R_g=0$. So every $r\in R$ maps to $0$, i.e. the kernel is $R$.
Take $R=\mathbb Z$, the kernel is generated by the part of $f$ supported on primes not dividing $g$: if $f=\prod p^{a_p}$ and $g=\prod p^{b_p}$, then
$$
\ker\big(\mathbb Z \to (\mathbb Z/(f))\otimes \mathbb Z_g\big)
=\Big(\frac{f}{\prod_{p\mid g}p^{a_p}}\Big)
$$