Rational Numbers form Prime Field

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Theorem

The field of rational numbers $\left({\Q, +, \times}\right)$ is a prime field.


That is, the only subset of $\Q$ which sustains both addition and multiplication are $\Q$ and $\left\{{0}\right\}$, and vacuously $\varnothing$.


Proof

Suppose $F$ is a subfield of $\left({\Q, +, \times}\right)$.

Let $a \in F$.

Then $a^{-1} \in F$ and so $a \times a^{-1} = 1$ and so $1 \in F$.


So, we have that $1 \in F$ and so $-1 \in F$ and thus $1 + \left({-1}\right) = 0 \in F$.

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

Then $n + 1 \in F$ and so by the Principle of Mathematical Induction that all positive integers are in $F$.

But then if $m \in F$ we have that $-m \in F$ and so all negative integers are in $F$.


Next, as we have seen, $m \in F \implies m^{-1} \in F$.

So:

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

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

$\blacksquare$


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