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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 $R$.

Federica Maggioni
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2 Answers2

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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 $$ \frac{r}{s}\in\mathfrak{b}\Rightarrow\frac{r}{1}=\frac{s}{1}\frac{r}{s}\in\mathfrak{b}\Rightarrow r\in\mathfrak{b}^c\Rightarrow\frac{r}{s}=\frac{1}{s}\frac{r}{1}\in(\mathfrak{b}^c)^e $$

Warren Moore
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Hint $\ $ Every ideal $\rm\:I\subset S^{-1}R\:$ can be generated by elements of $\rm\,R,\,$ since $\rm\: r/s\in I\iff r\in I.$

Math Gems
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  • ok, so take $I$ in $S^{-1}R$ an ideal, then $I=(T)$ for some $T$ subset of $S^{-1}R$, but $T$ comes from $T'$ subset of $R$, hence the original $I$ is $I=(T')^{e}$... – Federica Maggioni Apr 28 '13 at 17:36
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    @Federica Yes. In fact, for this purpose, we don't need to be efficient in our choice of generators of $\rm,I,,$ so we can simply take the whole set $\rm,I,$ as a set of generators. – Math Gems Apr 28 '13 at 17:43