All ideals in $\mathbb{Z}[\frac{1}{2}]$ are finitely-generated

Question

The question comes from a problem, which asks to find all prime and maximal ideals in $\mathbb{Z}[\frac{1}{2}] := \{\frac{a}{2^k} : k\in\mathbb{Z} , a\in 2\mathbb{Z}+1 \} $. Well I solved it by assuiming that $\mathbb{Z}[\frac{1}{2}]$ is a $PID$, but I didn't succeed to prove that all the ideals are finitely generated (for those which are finitely generated I already demonstrated that they are generated by one element).

Answer 1

Take some ideal $I \leq \mathbb{Z}[\frac{1}{2}]$. Now consider the following ideal of the integers $J_I \leq \mathbb{Z}$: $$ J_I = <\{a \in \mathbb{Z} : \exists k \in \mathbb{Z}, \; \; \frac{a}{2^k} \in I \}>.$$ (Here the reason we're taking the ideal generated by this set is because technically according to the definition you gave, we can divide by $2^k$ for some negative $k$.)

Note that $J_I$ is an ideal of $\mathbb{Z}$, hence it is generated by an element $a^* \in \mathbb{Z}$. We show that $a^*$ generates $I$ (just multiply by $\frac{1}{2^k}$ for the appropriate $k$). Obviously $a^* \in I$, so $(a^*) \subseteq I$. On the other hand, take any element $i = \frac{b}{2^k} \in I$. It's Numerator can be written as $ b = a^*z$ for some $z \in \mathbb{Z}$. Hence $a^*\frac{z}{2^k} = i$ and thus $I \subseteq (a^*)$.

We conclude that every ideal is generated by one element.