$\mathbb{R}$ is not the field of fractions of a UFD I need to prove the following.
If $D$ is an UFD and if 
$$\mathbb{R}\cong \operatorname{Frac}(D)$$
Then $\mathbb{R}\cong D$.
I have no idea how to prove it. I tried using the fact that the fraction field is a localisation and that localisation is flat.
 A: Hint: Suppose $p\in D$ is a prime element.  Then, as an element of $\mathbb{R}\cong\operatorname{Frac}(D)$, $p$ has a cube root.  What does this tell you?
Details of how to finish the argument are hidden below.

 It is a contradiction if $p$ has a cube root in $\operatorname{Frac}(D)$, as you can see by considering the factorizations in $D$ of the numerator and denominator of its cube root.  Thus $D$ has no prime elements, which implies every nonzero element of $D$ is a unit.  That is, $D$ is a field, so $D=\operatorname{Frac}(D)\cong\mathbb{R}$.

A: More generally, let $D$ be a UFD and $K=\operatorname{Frac}(D)$.
Then, $K^\times = D^\times \times A$, where $A$ is a free abelian group. 
Moreover, $A$ is trivial iff $D$ is a field and $K=D$.
Otherwise, $A$ and so $K^\times$ admit a nonzero homomorphism to $\mathbb Z$.
Now, $\mathbb R^\times = \{1,-1\} \times \mathbb{R}_{>0}$ is the product of a torsion group by a divisible group and so admits only the zero homomorphism to $\mathbb Z$. See here.
Bottom line: if $\mathbb R= \operatorname{Frac}(D)$, then $D=\mathbb R$.
A: By Eisenstein's criterion, there are irreducible polynomials of any degree in $(\operatorname{Frac} D)[X]$ if $D$ is a UFD but not a field.
