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Let's say that I have a finite field K with characteristic 2. I define @ as a map where @ : K -> K, and x -> $x^2$.

First of all, what are some examples of fields like K? I initially thought it would only be 0 and 1 with addition and multiplication mod 2, but apparently more elements can exist so long as 1+1=0.

Secondly, how would one begin to prove that @ is an automorphism?

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First, the prime field of characteristic $2$ is $\mathbb{F}_2 = ℤ/2ℤ$.

Then, for any irreducible polyonmial $p ∈ \mathbb{F}_2[X]$ of degree $1$ or greater, you can form the factor ring $\mathbb{F}_2 [X]/(p)$ which turns out to be a field. These are examples (and the only examples) of finite fields of characteristic $2$. I don’t know if you already know about this, but otherwise you should make yourself familiar with polynomial rings and a bit of ideal theory.

The argument that such a ring $\mathbb{F}_2 [X]/(p)$ forms a field is as follows:

  • $\mathbb{F}_2 [X]$ is a euclidian ring, hence a principal ideal domain.
  • Therefore, $p ∈ \mathbb{F}_2 [X]$ is irreducible if and only if $(p) ⊂ \mathbb{F}_2 [X]$ is a maximal ideal.
  • A factor ring is a field if and only if the ideal factored out is maximal.

If you want to show that for a finite field $K$ of characteristic $2$, the Frobenius morphism $@\colon K → K$ is an automorphism, it suffices to check five things:

  • It is additive – this follows from the binomial theorem/formula and the fact that $2 = 0$ in $K$.
  • It is multiplicative – because exponentiation in commutative rings always is.
  • It preserves the identity – $@(1) = 1$ which indeed should not be neglected as pointed out by Nishant² in the comments.
  • It is injective – because it is a field homomorphism as just checked and field homomorphisms always have trivial kernel and are hence injective.
  • It is surjective – because it is injective and domain and codomain are finite of the same cardinality.

By the way, Wolfram Alpha knows about the finite fields of characteristic $2$, e.g. of $\mathbb{F}_4$, the field with four elements. You can try out other fields of characteristic $2$, just search for finite fields of any order which is a power of $2$.

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    $\begingroup$ Technically, you'd have to check that 1 is sent to 1 for it to be a field homomorphism. Without that condition, you could send everything to 0. $\endgroup$ – Nishant Apr 3 '14 at 17:50

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