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
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 89 \gamma$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $9$: Rings: Exercise $15$