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:


 * $(1): \quad P$ is a subfield of $F$
 * $(2): \quad P \cong \Z_p$.

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 $\left({F, +, \times}\right)$ be a field such that $\operatorname{Char} \left({F}\right) = p$ whose unity is $1_F$.

Let $P$ be a prime subfield of $F$.

From Intersection of Subfields, this has been show to exist.

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


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

Suppose $a, b \in \left[\!\left[{n}\right]\!\right]_p$.

Then $a = n + k_1 p, b = n + k_2 p$

So:

and similarly for $b$, showing that $\phi$ is well-defined.

Let $C_a, C_b \in \Z_p$.

Let $a \in C_a, b \in C_b$ such that $a = a' + k_a p, b = b' + k_b p$.

Then:

Similarly for $\phi \left({C_a}\right) \times \phi \left({C_b}\right)$.

So $\phi$ is a ring homomorphism.

From Homomorphism from Field Either Monomorphism or Zero Homomorphism, it follows that $\phi$ is a ring monomorphism.

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


 * 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 $P_1 = P_2$.