Let $R$ be a ring, commutative with $1$. Let $S$ be a multiplicatively closed subset of $R$, with $0\notin S,1\in R$. Let $R_S$ be the localization of $R$ at $S$. For every ideal $\mathfrak{a}\subseteq R$ of $R$, define
$$\mathfrak{a}^e:=f(\mathfrak{a})R_S$$
where $f:R\longrightarrow R_S$ is the canonical morphism sending $r$ to $\frac{r}{1}$ and $f(\mathfrak{a})R_S$ means the ideal generated by $f(\mathfrak{a})$ in $R_S$.
I want to show (or disprove) that every ideal $\mathfrak{b}$ of $R_S$ is extended, meaning that it is of the form
$$\mathfrak{b}=\mathfrak{a}^e$$
for some ideal $\mathfrak{a}$ in $R$.
Best Answer
Given $\mathfrak{b}\lhd R_S$, you want to show that $\mathfrak{b}=\mathfrak{b}^{ce}$, where $\mathfrak{b}^c$ is the contraction of $\mathfrak{b}$ to $R$. Clearly $\mathfrak{b}^{ce}\subset\mathfrak{b}$. For the reverse inclusion $$ \frac{r}{s}\in\mathfrak{b}\Rightarrow\frac{r}{1}=\frac{s}{1}\frac{r}{s}\in\mathfrak{b}\Rightarrow r\in\mathfrak{b}^c\Rightarrow\frac{r}{s}=\frac{1}{s}\frac{r}{1}\in(\mathfrak{b}^c)^e $$