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 $R$?
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 $R$?
1 Answers
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.
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0@André The problem explicitly states the rings are "with $1$", so, by convention, ring maps must preserve $1$. Else all kinds of bizarre behavior can occur, e.g. see my remarks in this [Nov 6, 2009 sci.math post.](http://groups.google.com/groups?selm=l2ceiobik1w.fsf@shaggy.csail.mit.edu) – 2011-09-06