Rational Numbers form Prime Field

Theorem
The field of rational numbers $\struct {\Q, +, \times}$ is a prime field.

That is, the only subset of $\Q$ which sustains both addition and multiplication are $\Q$ and $\set 0$, and vacuously $\O$.

Proof
Let $F$ be a subfield of $\struct {\Q, +, \times}$.

Then $F$ is by definition a field.

Thus by definition:
 * $\struct {F, +}$ is an abelian group

and:
 * $\struct {F^*, \times}$ is an abelian group, where $F^* = F \setminus \set 0$.

and $\times$ is distributive over $+$:
 * $\forall a, b, c \in F: a \times \paren {b + c} = a \times b + a \times c$

Let $a \in F^*$.

Then:

We have $1 \in F$.

Suppose $n \in \Z$ such that $n \in F$.

Then:
 * $n + 1 \in F$

So by the Principle of Mathematical Induction, all positive integers are in $F$.

By in $\struct {F, +}$:
 * $m \in F \implies -m \in F$

So all negative integers are also in $F$.

That is:
 * $\forall m \in \Z: m \in \F$

Recall in $\struct {F^*, \times}$:
 * $m \in F^* \implies m^{-1} \in F^*$

So:
 * $\forall p \in \Z, q \in \Z_{\ne 0}: p \times q^{-1} \in F$

and it follows by the Subfield Test that $F = \Q$.

Also see

 * Number Field has Rational Numbers as Subfield, a direct consequence