TITLE: Is every field the field of fractions for some integral domain?
QUESTION [12 upvotes]: 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 $\operatorname{Quot}(R) \cong \mathbb{F}$?

REPLY [8 votes]: To add to Qiaochu's answer, I'll point out that sometimes there is just one domain satisfying this (as in the case of finite fields) and sometimes there are many, many possible domains.  Take, for example, $$\mathbb{Q} = Frac(\mathbb{Z}) = Frac(\mathbb{Z}[1/2]) = Frac(\mathbb{Z}[m/n])$$ etc.  So the point I'm trying to make is that it's all up in the air.
If this is something that interests you, you might want to read up on commutative algebra, specifically localization, which abstracts the notion of a fraction field.  Eisenbud does a decent job of describing localization commutative algebra book.
Also related is the notion of an "algebraic integer".  The notion behind a ring of algebraic integers is this:
Suppose you add some element to $\mathbb{Q}$, and you end up with a bigger field $K$, (for example, $\mathbb{Q}(\sqrt{2})$).  Then, you look to find a domain $\mathcal{O} \subset \mathbb{Q}$ satisfying some basic properties such as: 

$Frac(\mathcal{O}) =  K$
"Good" factorization (not necessarily unique)
$\mathcal{O}$ is Noetherian

And you end up with a generalization of $\mathcal{O} \subset K$ of  $\mathbb{Z} \subset \mathbb{Q}$