# Power to Characteristic Power of Field is Monomorphism

## Theorem

Let $F$ be a field whose characteristic is $p$ where $p \ne 0$.

Let $n \in \Z_{\ge 0}$ be any positive integer.

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

- $\forall x \in F: \map {\phi_n} x = x^{p^n}$

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

## Proof

Proof by induction:

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

- $\phi_n$ is a (field) monomorphism.

$\map P 0$ is trivially true:

- $\map {\phi_0} x = x^{p^0} = x^1 = x$

and we see that $\phi_0$ is the identity automorphism.

This is not the zero homomorphism.

So from Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi_0$ is a ring monomorphism.

### Basis for the Induction

First we need to show that $\map P 1$ is true:

- $\map {\phi_1} x = x^{p^1} = x^p$ is a (field) monomorphism.

This is demonstrated to be a monomorphism in Power to Characteristic of Field is Monomorphism.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

- $\map {\phi_k} x = x^{p^k}$ is a (field) monomorphism.

Then we need to show:

- $\map {\phi_{k + 1} } x = x^{p^{k + 1} }$ is a (field) monomorphism.

### Induction Step

This is our induction step:

\(\ds \map {\phi_{k + 1} } {a + b}\) | \(=\) | \(\ds \paren {a + b}^{p^{k + 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {\paren {a + b}^{p^k} }^p\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {a^{p^k} + b^{p^k} }^p\) | Induction Hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {a^{p^k} }^p + \paren {b^{p^k} }^p\) | Basis for the Induction | |||||||||||

\(\ds \) | \(=\) | \(\ds a^{p^{k + 1} } + b^{p^{k + 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map {\phi_{k + 1} } a + \map {\phi_{k + 1} } b\) |

Multiplication is more straightforward:

\(\ds \map {\phi_{k + 1} } {a b}\) | \(=\) | \(\ds \paren {a b}^{p^{k + 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds a^{p^{k + 1} } b^{p^{k + 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map {\phi_{k + 1} } a \map {\phi_{k + 1} } b\) |

and does not rely on the induction process.

Thus, $\phi_{k + 1}$ is a homomorphism.

$\phi_{k + 1}$ is not the zero homomorphism, since $\map {\phi_{k + 1} } 1 = 1^{p^{k + 1} } = 1 \ne 0$.

So from Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi_{k + 1}$ is a ring monomorphism.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

- $\forall n \in \Z_{\ge 0}: \phi_n$ is a (field) monomorphism.

$\blacksquare$

## Also see

- Prime Power of Sum Modulo Prime, where the same technique is used.

## Sources

- 1964: Iain T. Adamson:
*Introduction to Field Theory*... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Theorem $3.3$ Corollary