# Power to Characteristic of Field is Monomorphism

## Theorem

Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let the characteristic of $F$ be $p$ where $p \ne 0$.

Let $\phi: F \to F$ be the mapping on $F$ defined as:

$\forall x \in F: \map \phi x = x^p$

Then $\phi$ is a (field) monomorphism.

## Proof

Let $a, b \in F$.

First note that:

 $\ds \forall k: 0 < k < p: \,$ $\ds \binom p k$ $\equiv$ $\ds 0$ $\ds \pmod p$ Binomial Coefficient of Prime $\ds \leadsto \ \$ $\ds \binom p k$ $=$ $\ds r p$ for some $r \in \Z$ $\ds \leadsto \ \$ $\ds \binom p k a^k b^{p - k}$ $=$ $\ds r p \cdot a^k b^{p - k}$ $\ds \leadsto \ \$ $\ds \binom p k a^k b^{p - k}$ $=$ $\ds 0_F$ Characteristic of Field by Annihilator: Prime Characteristic $\ds \leadsto \ \$ $\ds \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k}$ $=$ $\ds 0_F$

So:

 $\ds \map \phi {a + b}$ $=$ $\ds \paren {a + b}^p$ $\ds$ $=$ $\ds a^p + \sum_{k \mathop = 1}^{p - 1} \binom p k a^k b^{p - k} + b^p$ Binomial Theorem $\ds$ $=$ $\ds a^p + 0_F + b^p$ $\ds$ $=$ $\ds a^p + b^p$ $\ds$ $=$ $\ds \map \phi a + \map \phi b$

Multiplication is more straightforward:

 $\ds \map \phi {a b}$ $=$ $\ds \paren {a b}^p$ $\ds$ $=$ $\ds a^p b^p$ Power of Product of Commutative Elements in Monoid $\ds$ $=$ $\ds \map \phi a \map \phi b$

Thus, $\phi$ is a (field) homomorphism.

It is clear that $\phi$ is not a zero homomorphism, since:

 $\ds \map \phi {1_F}$ $=$ $\ds 1_F^p$ $\ds$ $=$ $\ds 1_F$ Power of Identity is Identity $\ds$ $\ne$ $\ds 0_F$

Hence, from Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi$ must be a monomorphism.

$\blacksquare$