12
$\begingroup$

Out of the blue, I asked myself the following question:

Is an infinite field always isomorphic to the fraction field of an integral domain which is itself not a field?

Please note that the above setup avoid answering by a field is its own fraction field.

I think this question is natural as it is kind of a converse to the following:

Proposition. The rings which can be embedded into some field are exactly the integral domains.

As fraction fields have no characteristic properties that I am aware of, I am lost on how to tackle this problem. Any enlightenment will be greatly appreciated!

Edit 1. My question only involves infinite field as a finite integral domain is itself a field and the result failed to be true in this case. - pointed out by carmichael561 in the comments below.

Edit 2. The result holds for field of characteristic zero. - link provided by Arthur. As for now, the interesting question is the following:

Is an infinite field of prime characteristic isomorphic to a non-trivial fraction field?

$\endgroup$
  • 3
    $\begingroup$ Since a finite integral domain is a field, the answer would seem to be no for finite fields. $\endgroup$ – carmichael561 May 31 '17 at 21:05
  • 1
    $\begingroup$ What integral domain would you suggest to obtain $\Bbb R$? (I'm not saying there is none, but I don't think there is a natural or even constructive one) $\endgroup$ – Hagen von Eitzen May 31 '17 at 21:07
  • 1
    $\begingroup$ @HagenvonEitzen Actually, $\mathbb{R}$ is the genesis of my thoughts. I failed to see it as a non-trivial fraction field and also failed to prove there is none. $\endgroup$ – C. Falcon May 31 '17 at 21:12
  • 1
    $\begingroup$ The answer to this question (it was the top of related questions) answers it for characteristic $0$, it seems. A transcendence basis as used in the answer does require the axiom of choice. $\endgroup$ – Arthur May 31 '17 at 21:22
  • 1
    $\begingroup$ mathoverflow.net/questions/47103/… $\endgroup$ – Jorge Fernández Hidalgo May 31 '17 at 21:38
4
$\begingroup$

A field $K$ is the fraction field of a proper subring iff $K$ is not algebraic over $\mathbb{F}_p$ for some $p$. First, if $K$ is algebraic over $\mathbb{F}_p$, then every subring of $K$ is a field (since the subring generated by any single element is finite and a finite domain is a field), so $K$ cannot be the fraction field of a proper subring.

Conversely, if $K$ is not algebraic over $\mathbb{F}_p$ for any $p$, let $B$ be a transcendence basis for $K$ over the prime field and let $R$ be the subring of $K$ generated by $B$. Note that $R$ is not a field: if $K$ has characteristic $0$, this is because $R$ is a polynomial ring over $\mathbb{Z}$, and if $K$ has characteristic $p$, this is because $R$ is a polynomial ring over $\mathbb{F}_p$ in at least one variable since $B$ is nonempty by hypothesis. Let $S$ be the integral closure of $R$ in $K$. Since $K$ is algebraic over $R$, $K$ is the field of fractions of $S$. Since $S$ is integral over $R$ and $R$ is not a field, $S$ is not a field either.

$\endgroup$
  • $\begingroup$ I am grateful you gave me a detailed answer, I was able to follow it. Thank you! $\endgroup$ – C. Falcon May 31 '17 at 21:51

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.