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:

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


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


Sources