# Is an infinite field always isomorphic to a non-trivial fraction field?

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?

• Since a finite integral domain is a field, the answer would seem to be no for finite fields. – carmichael561 May 31 '17 at 21:05
• 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) – Hagen von Eitzen May 31 '17 at 21:07
• @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. – C. Falcon May 31 '17 at 21:12
• 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. – Arthur May 31 '17 at 21:22
• mathoverflow.net/questions/47103/… – Jorge Fernández Hidalgo May 31 '17 at 21:38

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.