Prime Power Mapping on Galois Field is Automorphism

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Theorem

Let $\GF$ be a Galois field whose zero is $0_\GF$ and whose characteristic is $p$.

Let $\sigma: \GF \to \GF$ be defined as:

$\forall x \in \GF: \map \sigma x = x^p$


Then $\sigma$ is an automorphism of $\GF$.


Proof

Let $x, y \in \GF$.

Then:

\(\ds \map \sigma {x y}\) \(=\) \(\ds \paren {x y}^p\) Definition of $\sigma$
\(\ds \) \(=\) \(\ds x^p y^p\) Power of Product of Commutative Elements in Group
\(\ds \) \(=\) \(\ds \map \sigma x \map \sigma y\) Definition of $\sigma$


and:

\(\ds \map \sigma {x + y}\) \(=\) \(\ds \paren {x + y}^p\) Definition of $\sigma$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^p \dbinom p k x^k y^{p - k}\) Binomial Theorem
\(\ds \) \(\equiv\) \(\ds x^p + y^p\) \(\ds \pmod p\) Power of Sum Modulo Prime
\(\ds \) \(=\) \(\ds \map \sigma x + \map \sigma y\) Definition of $\sigma$

Thus it has been demonstrated that $\sigma$ is a homomorphism.


Then we have:

\(\ds \map \ker \sigma\) \(=\) \(\ds \set {x \in \GF: \map \sigma x = 0_\GF}\) Definition of Kernel of Ring Homomorphism
\(\ds \) \(=\) \(\ds \set {x \in \GF: x^p = 0_\GF}\) Definition of $\sigma$
\(\ds \) \(=\) \(\ds \set {x \in \GF: x = 0_\GF}\) Congruence of Powers
\(\ds \) \(=\) \(\ds \set {0_\GF}\) Congruence of Powers

From Kernel is Trivial iff Monomorphism, $\sigma$ is a monomorphism.

That is, $\sigma$ is an injection.

Then from Injection from Finite Set to Itself is Surjection, $\sigma$ is a surjection.


Thus $\sigma$ is a bijective homomorphism to itself.

The result follows by definition of automorphism.

$\blacksquare$


Sources

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