Given an integral domain $R$, one can construct its field of fractions (or quotients) $\operatorname{Quot}(R)$ which is of course a field. Does every field arise in this way? That is:
Given a field $\mathbb{F}$, does there exist an integral domain $R$ such that $\operatorname{Quot}(R) \cong \mathbb{F}$?
Best Answer
Yes. $F$. (You can't hope to do better than this in general; consider the finite fields.)
Here's a cute example, though. It turns out that $\mathbb{C}$ is isomorphic to $\overline{ \mathbb{C}(t) }$. From this it follows that $\mathbb{C}$ is isomorphic to the fraction field of the integral closure of $\mathbb{C}[t]$ in the algebraic closure of its fraction field (the analogue of the algebraic integers in this setting).
Here's some geometry. If your field $F$ is finitely generated over some base field $k$, you can think of it as the function field of some variety $X$. Finding a nice domain whose field of fractions is $F$ can be interpreted geometrically as finding an affine variety birational to $X$, which you can do as follows: $F$ is necessarily a finite extension of $k(x_1, ... x_n)$ for some $n$, so we can take the integral closure of $k[x_1, ... x_n]$ in $F$.