Rational Numbers form Prime Field

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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:

\(\ds a\) \(\in\) \(\ds F^*\)
\(\ds \leadsto \ \ \) \(\ds a^{-1}\) \(\in\) \(\ds F^*\) Group Axiom $\text G 3$: Existence of Inverse Element in $\struct {F^*, \times}$
\(\ds \leadsto \ \ \) \(\ds a^{-1} \times a = 1\) \(\in\) \(\ds F^*\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(\in\) \(\ds F\)
\(\ds \leadsto \ \ \) \(\ds -1\) \(\in\) \(\ds F\) Group Axiom $\text G 3$: Existence of Inverse Element in $\struct {F, +}$
\(\ds \leadsto \ \ \) \(\ds 1 + \paren {-1} = 0\) \(\in\) \(\ds F\)


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 Group Axiom $\text G 3$: Existence of Inverse Element 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 Group Axiom $\text G 3$: Existence of Inverse Element 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 $F = \Q$.

$\blacksquare$


Also see


Sources

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