Let $p$ be prime and $R=\{\frac{a}{b}:a,b \in \Bbb Z,b \neq0\text{ and }p\nmid b\}$.

As an exercise, I have to prove that $R$ is a subring of $\Bbb Q$.

My idea:

With $a = 1$ and $b = 1$, $\frac{a}{b} = 1$ is an element of $R$.

Let $\frac{a}{b}$ and $\frac{c}{d}$ be elements of $R$. Then, $\frac{a}{b} + \frac{c}{d} = \frac{ad+bc}{bd}$ is an element of $R$.

Furthermore, $\frac{a}{b}\frac{c}{d} = \frac{ac}{bd}$ is an element of $R$ as $ac$, $bd$ $\in$ $\Bbb Z$, $bd \neq 0$ and $p \nmid bd$.

It follows that $R$ is a subring of $\Bbb Q$.

Is this proof correct?

As a next step in the exercise, I have to prove that all ideals of $R$ are of the form $Rp^n$ with $n \in \Bbb N_0$.

Unfortunately, I don't have an idea to prove that.

Could someone give me a hint please?

  • $\begingroup$ Hint: every other prime $\,q\ne p\,$ is a unit in $\,R\,$ since $\,1/q\in R.\ $ $\endgroup$ Jul 16, 2016 at 16:53

2 Answers 2


The first part of your proof is OK. For the second part you may want to do it via the following steps:

  1. Consider the map $\phi: \mathbb{Z}\to R$ that sends $n\mapsto n/1$. Prove that if $I$ is an ideal of $R$, then $\phi^{-1}(I)$ is an ideal of $\mathbb{Z}$. Denote it as $I^c:=\phi^{-1}(I)$.
  2. Prove that If $I$ is an ideal of $R$, then $$ I=\left\{\frac{a}{b}\in R\mid a\in I^c\right\}:=R I^c $$ i.e. fractions with the numerator being in $I^c$ and the denominator $b$ such that $p\not\mid b$.
  3. Prove that all ideals of $\mathbb{Z}$ are of the form $c\mathbb{Z}$ with $c\in \mathbb{Z}$.
  4. Conclude that any ideal $I$ of $R$ is of the form $cR$ for some $c\in \mathbb{Z}$.

Actually steps 1-3 are just so that we can make this conclusion, so you can skip 1-3 altogether by proving 4 directly.

  1. Prove that if $c\in \mathbb{Z}$ is such that $p\not\mid c$, then $c/1$ is a unit (it has an inverse) in $R$. Hence if $p\not\mid c$ then $cR=R$.
  2. Finally let $c=p^n d$ with $p\not\mid d$, then show that $cR=p^nR$.

This shows your proposition...

  • $\begingroup$ Thank you very much! I' ll try to prove it via these steps. $\endgroup$
    – Peter123
    Jul 16, 2016 at 17:14

You're missing the verification that $0\in R$ and that $x\in R$ implies $-x\in R$ (they're easy, but they should be stated).

A classification of the ideals with homomorphisms is certainly interesting and more generalizable. Here's one with elementary methods.

Let $I$ be an ideal in $R$ and assume $I\ne\{0\}$.

There certainly is $k$ such that $p^k/1\in I$: if $a/b\in I$, with $a\ne0$ and $p\nmid b$, let $k$ be the maximum exponent such that $p^k\mid a$, so $p^{k+1}\nmid a$. Thus $a=p^k c$ and $p\nmid c$. Therefore $b/c\in R$ and $$ \frac{p^k}{1}=\frac{a}{b}\frac{b}{c}\in I $$

Now consider the least natural number $n$ such that $p^n/1\in I$. Clearly the ideal generated by $p^n/1$ is contained in $I$. Suppose $x/y\in I$; write again $x=p^m z$, with $p\nmid z$. By construction $m\ge n$ and so $$ \frac{x}{y}=\frac{p^n}{1}\frac{p^{m-n}z}{y} $$ belongs to the ideal generated by $p^n/1$.

  • $\begingroup$ That' s a great solution, thank you. $\endgroup$
    – Peter123
    Jul 18, 2016 at 20:58

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