maximal ideals in a ring (subring of rational numbers) given by $2$ primes

Question

Consider the ring $R=\{\frac{a}{b} : p_1\nmid b,p_2\nmid b\}\subset \mathbb{Q}$ for distinct primes $p_1,p_2$.

Question is to prove that $I=\{\frac{a}{b}\in R : p_1\mid a \}$ is a maximal ideal of $R$.

Let $\frac{a}{b}\in R\setminus I$ i.e., $p_1\nmid a$. Then, we have to prove that $\frac{a}{b}$ with $I$ generates the whole ring.

Suppose $p_2\nmid a$ then $\frac{b}{a}$ is an element of $R$ and so $\frac{a}{b}$ is a unit and we have nothing to prove.

So, we can suppose that $p_2\mid a$. We then want $\frac{m}{n}\in R$ such that $$1-\frac{ma}{nb}\in I$$

i.e., Given $a,b$ such that $p_1,p_2\nmid b$ and $p_1\nmid a, p_2\mid a$, we want $m,n\in \mathbb{Z}$ such that $p_1\mid ma\pm nb$

I got stuck at this point. Could not think beyond this.

Answer 1

We construct a homomorphism from $R$ onto a field, with kernel $I$, then the result follows from the isomorphism theorem. Let $\varphi: R \rightarrow \mathbb{F}_{p_1}$ be given by $\frac{a}{b} \mapsto ab^{-1}$, where we write $\frac{a}{b}$ such that $\gcd(a,b) = 1$(note that $\varphi$ is well defined). This surely is a homomorphism, $1 \mapsto 1$, it respects the multiplicative structure and the additive structure. $ab^{-1} = 0$ in $\mathbb{F}_{p_1}$ if and only if $a = 0 $ in $\mathbb{F}_{p_1}$ because $\mathbb{F}_{p_1}$ is a field. So the kernel is exactly $I$. Hope this helps. It seems odd to me because I did not use $p_2$, but I hope that doesn't matter.

Answer 2

The ring $R$ is the ring of fractions $S^{-1}\mathbf Z$, where $S$ is the multiplicative set $$S=(\mathbf Z\setminus p_1\mathbf Z)\cap (\mathbf Z\setminus p_2\mathbf Z).$$ Now the non-zero prime ideals of $S^{-1}\mathbf Z$ correspond bijectively to the non-zero prime ideals of $\mathbf Z$ which do not meet $S$, via the correspondence $p\mathbf Z\longmapsto S^{-1} p \mathbf Z=p S^{-1}\mathbf Z =pR$, i.e. to the prime ideals $p\mathbf Z \subset p_1\mathbf Z\cup p_2\mathbf Z$. This implies $p=p_1$ or $p=p_2$.

Added: a proof without the notion of ring of fractions:

If you don't have this tool, it's not hard to adapt the previous arguments: consider the canonical injection $i\colon\mathbf Z\longrightarrow R$, which we identify to the inclusion $\;\mathbf Z\subset R$. Observe that, as $R\neq\mathbf Q$, a maximal ideal $\mathfrak p$ of $R$ is non-zero. Now $\mathfrak p\cap\mathbf Z$ is a (non-zero) maximal ideal of $\mathbf Z$, generated by a prime number $p$. This prime is one of $\;\{p_1,p_2\}$, since all other primes $p$ are units in $R$.

Suppose $\mathfrak p\cap\mathbf Z=p_1\mathbf Z$. Then $p_1R\subset \mathfrak p$. Conversely, for any element $a/b\in\mathfrak p$, $\;b\cdot(a/b)\in\mathfrak p\cap\mathbf Z$, hence $\;a=p_1a'\;$ for some integer $a'$, so $\;a/b=p_1\cdot(a'/b)\in p_1 R$.

This proves the non-zero prime ideals $\mathfrak p$ of $R$ are $\;(\mathfrak p\cap\mathbf Z)R=p_1R$ or $\;p_2R\;$.