Ideals in $\left\{\frac{m}{2^n}\mid m\in \mathbb Z, n\in \mathbb N_0\right\}$

Question

Consider the subring $R=\left\{\frac{m}{2^n}\mid m\in \mathbb Z, n\in \mathbb N_0\right\}$ of $\mathbb Q$. Show that for any ideal $I$ of $R$ there is an ideal $J$ of $\mathbb Z$ with $I=\left\{\frac{m}{2^n}\mid m\in J, n\in \mathbb N_0\right\}$.

Any hints or suggestions at least to get started?

Answer 1

The construction of $J$ is simply $J = I \cap \Bbb Z$. It is routine to check that this is an ideal.

We claim that $I=\left\{ \frac{m}{2^n} ~\middle|~ m\in (I \cap \Bbb Z), n\in \mathbb N_0 \right\}$.

For $\subseteq$, note that if $\frac m{2^n} \in I$, then $m \in I$, so $m \in (I \cap \Bbb Z)$, so $\frac m{2^n} \in \left\{ \frac{m}{2^n} ~\middle|~ m\in (I \cap \Bbb Z), n\in \mathbb N_0 \right\}$. $\supseteq$ is clear.

---

More generally, referring to Stacks 10.9.16, if $S$ is a multiplicatively closed subset of a ring $R$, then we can form the ring $S^{-1}R$, and again every ideal in $S^{-1}R$ can be expressed in the form $S^{-1}I$ where $I$ is an ideal of $R$.

In this case, $R = \Bbb Z$ and $S = \{ 2^n \mid n \in \Bbb N \}$.