Proving a subring has only one maximal ideal.

Question

Let the ring $R=\{\frac{a}{b}|a,b\in\mathbb{Z},$ and $gcd(2,b)=1\}\subseteq\mathbb{Q}$. I have proven that $R$ is a subring and that the inverible elements of $R$ are $\frac{a}{b}$ for $a$ and $b$ odd. Now, how can I prove that $2R$ is the only maximal ideal of $R$?

Answer 1

Note that the set of non-invertible elements (that is $2R$), in this case, actually forms an ideal. Any maximal ideal of $R$ is included in this set (because a maximal ideal does not contain any invertible element!). So by maximality, it must equal $2R$.

(This is an example of a local ring. By definition, a local ring is a ring which contains only one maximal ideal. This is equivalent to saying that the set of non-invertible elements form an ideal - which is, in this case, the unique maximal ideal.)