Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / A / A2

The characteristic is the least positive integer $r$ with $r{1}_K=0$, if it exists, and zero otherwise. The prime subfield is the smallest subfield containing $1_K$; in characteristic $p$ it is $F=\{0,1,\dots ,(p-1)1_K\}\cong {\mathbb{F}}_p$. Every field automorphism fixes $1_K$ and addition, so it fixes every element of this prime subfield.

The additive sequence $0,1,2\cdot 1,\dots$ must repeat because $K$ is finite, so the characteristic is positive. If its least positive value were composite, $r=ab$ with $1<a,b<r$, then $(a{1}_K)(b{1}_K)=r{1}_K=0$ while neither factor would be zero. A field has no zero divisors, giving a contradiction. Hence $\boxed{\mathrm{char}K=p\text{ is prime}}$.

The finite field $K$ is a finite-dimensional vector space over its prime subfield $F$ of size $p$. A basis has some positive length $n$, and every element is uniquely specified by $n$ coefficients in $F$. There are $p$ choices per coefficient, so $\boxed{|K|=p^n}$ and $[K:F]=n$.

Since $K=F(\lambda )$ and the extension is finite, the element $\lambda$ is algebraic. Its minimal polynomial is irreducible, and evaluation gives an isomorphism $F[X]/(m_{\lambda })\cong F[\lambda ]=F(\lambda )=K$. If its degree is $d$, the powers $1,\lambda ,\dots ,{\lambda }^{d-1}$ form a basis. Comparing with $[K:F]=n$ gives $\boxed{\mathrm{deg}{m}_{\lambda }=n}$.

Define $\vartheta (x)=x^p$ on all of $K$. The binomial coefficients between the endpoints are divisible by $p$, so $(x+y{)}^p=x^p+y^p$; also $(xy{)}^p=x^p{y}^p$ and $1^p=1$. Thus this is a field homomorphism. Its kernel is zero, and an injective self-map of the finite set $K$ is surjective. Hence $\boxed{\vartheta \text{ is an automorphism}}$, the Frobenius automorphism, with the required image of $\lambda$. It fixes $F$ by the preceding prime-subfield argument.

Take $0<m<n$, the intended positive-exponent range. If ${\vartheta }^m(\lambda )=\lambda$, then ${\vartheta }^m$ fixes both $F$ and its field generator, so it fixes every element of $K=F(\lambda )$. Every element would then be a root of $X^{p^m}-X$. But that polynomial has degree $p^m<p^n=|K|$, impossible for a nonzero polynomial over a field. Thus $\boxed{{\vartheta }^m(\lambda )\ne \lambda (0<m<n)}$. The source writes only $m<n$; zero must be excluded, since ${\vartheta }^0(\lambda )=\lambda$.

For nonzero $x\in K$, Lagrange's theorem in $K^{\times }$ gives $x^{p^n-1}=1$; zero satisfies $x^{p^n}=x$ too. Thus ${\vartheta }^n$ is the identity, and (iii) excludes every smaller positive order. Hence $\boxed{|⟨\vartheta ⟩|=n}$.

Any field automorphism of $K$ fixes $F$ and is determined by its value on $\lambda$. That value must be a root of the degree-$n$ minimal polynomial, so there are at most $n$ automorphisms. The $n$ powers of $\vartheta$ already attain this bound. Consequently $\boxed{\mathrm{Aut}(K)=⟨\vartheta ⟩\cong C_n}$.