0

Let $R$ be an integral domain with field of fractions $K$. For a nonzero element $x \in R$, let $R[1/x]$ denote the set of elements of $K$ of the form $r/x^{n}$, where $r \in R$ and $n \in \mathbb{N}$. Show that if $a$ and $b$ are nonzero elements of $R$ with $R = (a, b)$, then $R[1/a] \cap R[1/b] = R$.

$R \subset R[1/a] \cap R[1/b]$ clearly. Take $y \in R[1/a] \cap R[1/b]$ so that $y = r/a^{n} = s/b^{m}$ for some $r, s \in R, n, m \in \mathbb{N}$. Then $rb^{m} = sa^{n}$. So $(a) \cap (b) \neq 0$. I want to say something like $a, b$ are coprime, since $(a, b) = (1)$, whence $ax + by = 1$ for some $x, y \in R$. But $R$ is not necessarily a UFD. I'm having trouble getting past this! I can manipulate my equations, combine them etc, but it gets me nowhere. I suspect I am missing something silly.

Freddie
  • 1,759

1 Answers1

2

The common Bezout-based proof of Euclid's Lemma shows $\,a\mid bc\,\Rightarrow\,a\mid c,\,$ i.e. $$ax+by=1,\,a\mid bc\,\Rightarrow\, a\mid axc,byc\,\Rightarrow\, a\mid (ax+by)c = c$$

Thus by induction $\,a^n\mid r b^m\,\Rightarrow\,a^n\mid r\,\Rightarrow\, y = r/a^n \in R$

Remark $ $ More conceptually, call $\,d\in R\,$ a denominator of the fraction $\,y\,$ if $\,dy \in R.\,$ One easily checks that the set of all denominators of $y$ is an ideal $\cal D.\,$ It contains $\,a^n,\,b^m$ so it contains $(a^n)+(b^m) = (1)\,$ [by here], so $1$ is a denominator of $y,\,$ i.e. $\,1\cdot y\in R.$

For further discussion see denominator ideals and order ideals.

Bill Dubuque
  • 272,048
  • Nice argument! I think you might mean $byc$ in the first display line. – Freddie Jun 11 '19 at 01:30
  • Oh wait, I see that $a \mid rb^{m} \implies a \mid r$ follows by induction, but what about $a^{n} \mid b^{m}r \implies a^{n} \mid r$ ? EDIT: Nevermind that was silly. Thanks again. – Freddie Jun 11 '19 at 01:34