Jacobson radical of $R=\{\frac{a}{b}:a,b \in \Bbb Z,b \neq0\text{ and }p\nmid b\}$.

Question

I want to find the Jacobson radical, $J(R)$, of $R=\{\frac{a}{b}:a,b \in \Bbb Z,b \neq0\text{ and }p\nmid b\}$.

Here my idea:

One could use the definition $J(R)$ $=$ {$x \in R;\,\,\forall y \in R: 1-xy \in R^{\times}$}.

Let $\frac{a}{b}$ be an element of $R$. Then, $\frac{a}{b}$ is an element of $J(R)$ if $1-\frac{a}{b} \frac{c}{d} \in R^{\times}$ for $\frac{c}{d} \in R$.

It holds that $1-\frac{a}{b} \frac{c}{d} = \frac{bd-ac}{bd}$, whereby $\frac{bd-ac}{bd}$ is an element of $R^{\times}$ if $bd-ac \neq 0$.

At this point, I don' t know how to continue.

I would be really happy if someone could help me.

Answer 1

It may be easier to use the equivalent definition that $J(R)$ is the intersection of all maximal ideals.

I assume that $p$ is prime. Note that your $R = \mathbb{Z}_{(p)}$ is the localization of $\mathbb{Z}$ at the prime ideal $(p)$.
$R$ is a local ring (why this is the case see e.g. this thread) with the unique maximal ideal $(p)\mathbb{Z}_{(p)}$.

$J(R)$ equals the intersection of all maximal ideals. We only have one, so $J(R) = (p)\mathbb{Z}_{(p)}$

Answer 2

Assuming you know nothing about the structure of ideals in the ring $R$, a necessary condition for $a/b\in J(R)$ is that $a/1\in J(R)$, so $$ 1-\frac{a}{1}\frac{c}{1}=\frac{1-ac}{1} $$ is invertible, for every integer $c$. If $p\nmid a$, then there are integers $x$ and $y$ so that $1=ax+py$ and obviously $$ 1-\frac{a}{1}\frac{x}{1}=\frac{py}{1} $$ is not invertible. Thus $p\mid a$ is a necessary condition for $a/b\in J(R)$.

Can you know show it is sufficient?

If $a=pa'$, then, for every $c/d\in R$,

$\displaystyle 1-\frac{a}{b}\frac{c}{d}=\frac{bd-pa'c}{bd}$

and $p\nmid(bd-pa'c)$