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 $$P_1 = P_2$$.