Field of Prime Characteristic has Unique Prime Subfield

Theorem
Let $F$ be a field whose characteristic is $p$.

Then there exists a unique $P \subseteq F$ such that:


 * 1) $P$ is a subfield of $F$;
 * 2) $P \cong \Z_p$.

That is, $P \cong \Z_p$ is a unique minimal subfield of $F$, and all other subfields of $F$ contain $P$.

This field $P$ is called the prime subfield of $F$.

Proof

 * Let $F$ be a field such that $\operatorname{Char} \left({F}\right) = p$.

We can consistently define a mapping $\phi: \Z_p \to F$ by:

$\forall n \in \Z_p: \phi \left({\left[\!\left[{n}\right]\!\right]_p}\right) = \left({n \cdot 1_F}\right)$.

Thus it follows that $P = \operatorname{Im} \left({\phi}\right)$ is a subfield of $F$ such that $P \cong \Z_p$.

Note that $\operatorname{Im} \left({\phi}\right) = \left({0_F, 1 \cdot 1_F, 2 \cdot 1_F, \ldots, \left({n - 1}\right) \cdot 1_F}\right\}$.


 * Let $K$ be a subfield of $F$, and $P = \operatorname{Im} \left({\phi}\right)$ as defined above.

We know that $1_F \in K$.

It follows that $1_F \in K \implies P \subseteq K$.

Thus $K$ contains a subfield $P$ such that $P$ is isomorphic to $\Z_p$.


 * The uniqueness of $P$ follows from the fact that if $P_1$ and $P_2$ are both minimal subfields of $F$, then $P_1 \subseteq P_2$ and $P_2 \subseteq P_1$, thus $P_1 = P_2$.