13
$\begingroup$

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 $\operatorname{Quot}(R) \cong \mathbb{F}$?

$\endgroup$
  • 7
    $\begingroup$ Same question (with the same title!) on MathOverflow. $\endgroup$ – user26857 Jun 9 '13 at 7:42
18
$\begingroup$

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 $F$.

$\endgroup$
8
$\begingroup$

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:

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.