# If $R$ is a subring of $\mathbb{Q}$, $m/n\in R$ with $\gcd(m,n)=1$, and $p$ a prime factor of $n$, is $1/p$ an element of $R$?

Let $R$ be any subring (with $1$) of $\mathbb{Q}$. Let $m/n$ be an element of $R$ where $m$ and $n$ are coprime. Let $p$ be a prime divisor of $n$. Could anyone help me show that $1/p$ is an element of $R$?

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HINT. $\$ Scaling Bezout $\rm\ a\ m + b\ n = 1\$ by $\rm\: \dfrac{1}n\$ yields $\rm\:\dfrac{1}n\in R\:,\:$ hence $\rm\ \dfrac{n}p\ \dfrac{1}n = \dfrac{1}p\:\in\: R\:.$
NOTE. $\$ This shows that every ring $\rm\:R$ between $\rm\:\mathbb Z\:$ and $\rm\:\mathbb Q\:$ can be obtained by adjoining inverses of elements of $\rm\:\mathbb Z\:,\:$ i.e. $\rm\:R\:$ is a localization $\rm\:S^{-1} \mathbb Z\:.\:$ The same argument as above shows that this holds true for any domain where such Bezout identities hold. Domains satisfying this propery, i.e. whose subrings of fractions are localizations, are known as QR-domains. They have been quite intensively studied. For example, a QR-domain is necessarily a Prüfer domain, and conversely, a Prüfer domain with torsion class group is a QR-domain. For more see my May 11, 2004 sci.math post.
Ok, so that gives us $am/n+b=1/n$, sorry if i'm being slow here, but how do we know $a,b\in R$ – Freeman Sep 6 '11 at 16:18
@LHS $\:a,b\in \mathbb Z\ \Rightarrow\ a,b\in R\$ since every subring of $\:\mathbb Q\:$ contains $1$ so contains $\;\mathbb Z\:.$ – Bill Dubuque Sep 6 '11 at 16:23
Depends on the definition of subring. The usual definition of ring includes having a multiplicative identity, and of subring as having multiplicative identity. Then clearly $b\in R$. If we do not ask a ring to have a multiplicative identity, the result looked for is false. For example, look at all rationals of the shape $2m/3$, where $m$ ranges over the integers. Thus you need to look carefully at the definition of ring used in your course. – André Nicolas Sep 6 '11 at 16:28