Past exam of the options II course of the University of Oxford / 2009 / A / A2

The characteristic of a field is the least positive integer $r$ with $r{1}_K=0$, or zero if none exists. In a finite field a positive such integer exists because the sequence $0,1_K,2\cdot 1_K,\dots$ must repeat.

The prime subfield is the intersection of all subfields of $K$, equivalently the subfield generated by $1_K$. In characteristic $p$ it is the copy of ${\mathbb{F}}_p$ formed by the multiples of $1_K$. Every field automorphism preserves $1_K$ and addition, so it fixes every element of this subfield.

Finally, once the order-$n$ Frobenius automorphism below is established, all its $n$ powers are distinct. Any automorphism fixes $F$ and is determined by its value at the generator $\lambda$. That value must be a root of the degree-$n$ minimal polynomial of $\lambda$, so there are at most $n$ automorphisms. Hence

If the positive characteristic were composite, say $r=uv$ with $1<u,v<r$, then $(u{1}_K)(v{1}_K)=0$ while neither factor is zero, by minimality of $r$. This contradicts a field having no zero divisors. Thus the characteristic is a prime $p$.

Regard $K$ as a vector space over its prime subfield ${\mathbb{F}}_p$. Since $K$ is finite, a maximal linearly independent list terminates and is a finite basis, of some dimension $n\ge 1$. Each field element has a unique coordinate vector in ${\mathbb{F}}_p^n$, so

For a simple field extension, the powers $1,\lambda ,\dots ,{\lambda }^{d-1}$ form a basis when the minimal polynomial has degree $d$: reduction modulo that polynomial spans the extension, and a relation of smaller degree would contradict minimality. Hence $[F(\lambda ):F]=d$. The field is $K$, whose dimension is $n$, giving $d=n$.

Define $\vartheta (x)=x^p$ for every $x\in K$. In characteristic $p$, all intermediate binomial coefficients vanish, so $\vartheta (x+y)=\vartheta (x)+\vartheta (y)$; multiplicativity and preservation of one are immediate. If $x^p=y^p$, then $(x-y{)}^p=0$ and the field property gives $x=y$. Thus this field homomorphism is injective, and finiteness makes it surjective. It is therefore a field automorphism, sending $\lambda$ to ${\lambda }^p$ as required.

Here the iteration index is a positive integer: $1\le m<n$. If ${\vartheta }^m(\lambda )=\lambda$, then ${\vartheta }^m$ fixes $F$ and the generator, so it fixes all of $K$. Every element of $K$ would then solve $T^{p^m}-T=0$. A nonzero polynomial of degree $p^m$ over a field has at most $p^m$ roots, contradicting $|K|=p^n>p^m$. Thus ${\vartheta }^m(\lambda )\ne \lambda$. The printed inequality alone omits positivity of $m$; at $m=0$ the claimed inequality would be false.

The multiplicative group $K^{\times }$ has order $p^n-1$, so Lagrange's theorem for finite groups gives $x^{p^n-1}=1$ for $x\ne 0$. Including zero, $x^{p^n}=x$ for every $x\in K$. Therefore ${\vartheta }^n$ is the identity. No positive smaller power is the identity by part (iii), soFor $n=1$ this says that Frobenius itself is the identity, as it should be.