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.

  • $\begingroup$ Why don't you try to use the characterisation of the invertible elements of fraction rings? You are basically asking for the Jacobson radical of the localisation of $\Bbb Z$ at $(p)$. (assuming $p$ is prime). $\endgroup$
    – user228113
    Jul 17, 2016 at 15:38

2 Answers 2


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 $J(R) = (p)\mathbb{Z}_{(p)}$

  • $\begingroup$ The OP should have know the answer from his own previous question. $\endgroup$
    – user26857
    Jul 17, 2016 at 18:18

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

  • $\begingroup$ Thank you very much. You helped me a lot. $\endgroup$
    – Peter123
    Jul 18, 2016 at 21:12

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