Power to Characteristic Power of Field is Monomorphism

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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:

$\forall x \in F: \phi_n \left({x}\right) = x^{p^n}$


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


Proof

Proof by induction:

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:

$\phi_n$ is a (field) monomorphism.

$P(0)$ is trivially true:

$\phi_0 \left({x}\right) = x^{p^0} = x^1 = x$

... 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 $P(1)$ is true:

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

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


Induction Hypothesis

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


So this is our induction hypothesis:

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


Then we need to show:

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


Induction Step

This is our induction step:

\(\displaystyle \phi_{k+1} \left({a + b}\right)\) \(=\) \(\displaystyle \left({a + b}\right)^{p^{k+1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({a + b}\right)^{p^k} }\right)^p\)
\(\displaystyle \) \(=\) \(\displaystyle \left({a^{p^k} + b^{p^k} }\right)^p\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \left({a^{p^k} }\right)^p + \left({b^{p^k} }\right)^p\) Basis for the Induction
\(\displaystyle \) \(=\) \(\displaystyle a^{p^{k+1} } + b^{p^{k+1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \phi_{k+1} \left({a}\right) + \phi_{k+1} \left({b}\right)\)


Multiplication is more straightforward:

\(\displaystyle \phi_{k+1} \left({a b}\right)\) \(=\) \(\displaystyle \left({a b}\right)^{p^{k+1} }\)
\(\displaystyle \) \(=\) \(\displaystyle a^{p^{k+1} } b^{p^{k+1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \phi_{k+1} \left({a}\right) \phi_{k+1} \left({b}\right)\)

... and does not rely on the induction process.


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


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

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


So $P \left({k}\right) \implies P \left({k+1}\right)$ 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


Sources