# Fields $k$ with subrings whose quotient field is $k$

Let $k$ be a given field, any field. Is there a subring $A$ of $k$ such that $k$ is the quotient field of $A$?

Let's restrict ourselves to fields I know anything about; subfields of $\mathbb{C}$, finite fields ($\mathbb{Z}/(p)$ and $(\mathbb{Z}/(p))[x]/(f)$ where $f$ is irreducible of degree $n$), and function fields of any characteristic.

Without thinking too much about it, it seems that such a subring exists if $k$ is a number field, if $k$ is a function field of any characteristic, but not so for finite fields. Is it perhaps the case that infinite fields always have such a subring? If so there would for example exist a subring $A \subset \mathbb{C}$ for which that would be true. I can't imagine what ring that would be (the theory of transcendental extensions might hold the key to answering that, but I don't know any of that stuff).

Anyway that's all I have to contribute, I'm just curious, so if anyone could list (types of) fields with such a subring, no motivation necessary, I'd appreciate it.

• The field $k$ is its own field of fractions, so perhaps you want to require that $A$ be a proper subring? In this case, the answer is no, as you seem to surmise, because the finite field $\mathbf{F}_p$ has no subrings other than itself. – Keenan Kidwell Jul 8 '13 at 1:30

Let $k$ have characteristic zero. If $k$ is algebraic over $\mathbf{Q}$, then you can take the ring of algebraic integers of $k$. Otherwise, choose a transcendence basis $S$ for $k$ over $\mathbf{Q}$. Then $k$ is an algebraic extension of $\mathbf{Q}(S)$. Now let $A$ be the integral closure in $k$ of $\mathbf{Q}[S]$. Then the field of fractions of $A$ will be $k$. Note that since $S\neq\emptyset$, $\mathbf{Q}[S]$ is not a field, so the integral closure $A$ can't be a field either, which means that $A$ is not equal to $k$.
The same argument works for a field of characteristic $p$ as long as it has positive transcendence degree over its prime field.
The only other case is where $k$ is algebraic over $\mathbf{F}_p$. Then you are out of luck. Indeed, if $A$ is any subring of $k$, then $A$ is integral over $\mathbf{F}_p$, hence is itself a field. So the field of fractions of any subring of $k$ is itself.