Maximal Ideals in the Ring of Rationals whose denominators are not divisible by a prime $p$.

Let $R_p$ be that subset of $\mathbb{Q}$ consisting of rationals $a/b$ where $b$ is not divisible by $p$ a prime. It is true that $R_p$ is a commutative ring. What are all of its maximal ideals? I know that $I_p := \{a/b \in R_p : p | a \}$ is an ideal of $R_p$. Furthermore, it is true that $R_p / I_p$ is isomorphic to the field $\mathbb{Z}/(p\mathbb{Z})$, a field. Therefore, $I_p$ is certainly a maximal ideal. But is it true that $I_p$ is the unique maximal ideal? I suspect so, but cannot prove or disprove this.

-
No quotient of $R_p$ can contain $\mathbb{Q}$, so the quotient by any maximal ideal necessarily has positive characteristic. Since $R_p$ contains $\frac{1}{q}$ for every $q \neq p$, the quotient by any maximal ideal necessarily has characteristic $p$. Can you finish from here? –  Qiaochu Yuan Feb 21 '12 at 0:47

Let $\frac{a}{b}\in R_p$, with $\gcd(b,p)=1$. Prove that if $\gcd(a,p)=1$, then $\frac{a}{b}$ is a unit in $R_p$, and so cannot be contained in any proper ideal of $R_p$.
Conclude that if $I$ is a proper ideal of $R_p$, then every element of $I$ can be written with a denominator that is prime to $p$, and the numerator divisible by $p$. Deduce that if $I$ is a proper ideal, then $I\subseteq I_p$.
(For extra points, use the idea above to describe all ideals of $R_p$).