Show that if $R$ is a unique factorization domain with the quotient field $R$ being isomorphic to $\mathbb{R}$, then R is isomorphic to $\mathbb{R}.$

Question

$\mathbf{The \ Problem \ is}:$ Show that if $R$ is a unique factorization domain such that the quotient field of $R$ is isomorphic to $\mathbb{R}$, then R is isomorphic to $\mathbb{R}.$

$\mathbf{My \ approach}:$ Enough to show that $R$ is a field. Suppose not, then $R$ contains a prime(/irreducible) elememt $p.$ Let $Q(R)$ be the quotient field of $R$ and $\phi:Q(R)\to \mathbb{R}$ is a field isomorphism. Now, we can localize $R$ at the prime ideal $\langle p\rangle.$ Then $R_p$ is a subring of $Q(R),$ then I was thinking to compute $\phi(R_p).$ Also, we can consider the all possible $n$-th roots of $r:=\phi(\frac{p}{1})(\neq 0)$ for $r>0$ and all odd radicals of $r$ for $r<0.$ For $r>0$ and $n>1,$ we get a sequence $\bigl\{\frac{a_n}{b_n}\bigr\}_n\in Q(R)$ and without loss of generality, we can assume that $gcd(a_n,b_n)=1$ such that $\phi\bigl({\bigl(\frac{a_n}{b_n}\bigr)}^n\bigr)=r=\phi(\frac{p}{1})$ which implies $a_n^n=pb_n^n\implies p\mid a_n$ as $gcd(a_n,b_n)=1$ and also $a_n\mid p\implies p\sim a_n$(associates). So, $b_n^n\sim p^{n-1}\implies p\mid b_n$ (as $n>1$), contradicting $gcd(a_n,b_n)=1.$ Does this seem correct ?

Any hint is really appreciated.

Answer 1

That looks correct but with a lot of redundancy. I think you need neither the localization nor that sequence you construct: the term with $n=2$ suffices.

That is, first identify $R$ with a subring of $\mathbb R$ (its image under $\phi$). Now assume w.l.o.g. $0 < p \in R$ is irreducible. Since $\sqrt{p} \in \mathbb R$, there are $a,b \in R$ with $gcd(a,b)=1$ and $\frac{a}{b}=\sqrt{p}$ i.e. $a^2=pb^2$, so that (just like in the classical proof of irrationality of square roots of rational primes) $p$ divides $a$, but then it also divides $b$, contradiction to $gcd(a,b)=1$.

Answer 2

UFDs are integrally closed so the Lemma $\Rightarrow R\:\!$ has no $\,\overbrace{{\rm primes}\ p}^{>\ 0\,\ \rm wlog\!}\:$ by $\,\color{#c00}{p = \sqrt p^{\,2}},\,$ so $R\:\!$ is a field (else $R$ has a nonunit, which has a prime factor $p,\,$ by $R$ a UFD).

Lemma $ $ Suppose that a domain $R\:\!$ is integrally-closed in its ordered fraction field $\,K.\,$ If further $K$ is $\rm\color{#c00}{closed}$ under square roots of positive elements then so $\rm\color{#c00}{too}$ is $R$.

Proof $\,\ 0 < r\in R\subset K\Rightarrow \sqrt r\in K,\,$ so $\, \sqrt r\in R\,$ by $R$ integrally closed.