Past exam of the options II course of the University of Oxford / 2010 / A (Introduction to Fields)

The extension $K/F$ is a vector space over $F$. Since $K$ is finite, a maximal linearly independent subset is finite and spans it; let its size be $d$. Relative to this basis, every element of $K$ has a unique list of $d$ coefficients in $F$. There are $q$ choices for each coefficient, soThe inclusion $F\subseteq K$ makes $d\ge 1$.

Write $Q=q^m$ and take a splitting field $L$ of $X^Q-X$ over $F$. Its formal derivative is $-1$, since the characteristic divides $Q$, so it has $Q$ distinct roots in $L$. Let $K$ be their set. The Frobenius endomorphism identity givesThus $K$ contains $0,1$ and is closed under addition, negatives and multiplication. If $u\ne 0$ belongs to $K$, then $(u^{-1}{)}^Q=(u^Q{)}^{-1}=u^{-1}$, so it is closed under inverses too. Hence $K$ is a field with exactly $Q$ elements. Every $c\in F$ satisfies $c^q=c$ by Lagrange's theorem for finite groups applied to $F^{\times }$, together with the zero case; iterating gives $c^Q=c$. Therefore

By the allowed finite multiplicative-subgroup theorem, the group $K^{\times }$ is cyclic. Let $z$ generate it. Every nonzero element of $K$ is a power of $z$, hence belongs to $F(z)$; zero belongs as well. Since the reverse inclusion is automatic,Thus a finite extension of a finite field is a simple field extension.

For $\alpha \ne 0$, Lagrange's theorem for finite groups in $K^{\times }$, whose order is $q^m-1$, gives ${\alpha }^{q^m-1}=1$. The zero element satisfies the equation too, so every $\alpha \in K$ has ${\alpha }^{q^m}=\alpha$.

Now let $f\in F[X]$ be irreducible of degree $m$. The quotient $F[X]/(f)$ is a field with $q^m$ elements, by its basis $1,\overline{X},\dots ,{\overline{X}}^{m-1}$. Its element $\overline{X}$ therefore satisfies ${\overline{X}}^{q^m}-\overline{X}=0$. This means precisely that $X^{q^m}-X$ lies in the ideal $(f)$, giving