Let \(K\) and \(L\) be finite fields with \(K \subseteq L\). Prove that \(L\) is Galois over \(K\) and that \(\mathrm{Gal}(L/K)\) is cyclic.
Solution
Write \(q=|K|=p^r\), where \(p\) is the characteristic, and put \(d=[L:K]\). Counting coordinates in a \(K\)-basis gives \(|L|=q^d\).
1. The extension \(L/K\) is Galois.
The group \(L^\times\) has order \(q^d-1\), so every \(a\in L\) satisfies \(a^{q^d}=a\), also for \(a=0\). The factor theorem and comparison of degrees give \[ T^{q^d}-T=\prod_{a\in L}(T-a). \] Consequently \(L\) is the splitting field over \(K\) of \(T^{q^d}-T\): its roots are all the elements of \(L\). The derivative is \(-1\), since \(q^d\) is zero in characteristic \(p\), so the polynomial is separable. A splitting field of a separable polynomial is a Galois extension [@DF04].
2. The automorphism \(\varphi(a)=a^q\) generates\(\operatorname{Gal}(L/K)\) and has order \(d\).
The characteristic-\(p\) binomial identity, iterated \(r\) times, gives \((a+b)^q=a^q+b^q\). Multiplication and the identity are preserved as well, so \(\varphi\) is an injective field homomorphism and is surjective because \(L\) is finite. For \(a\in K\), Lagrange’s theorem in \(K^\times\) gives \(a^q=a\), including zero. Thus \(\varphi\) fixes \(K\).
Step 1 gives \(\varphi^d=1\). For \(0<j<d\), equality \(\varphi^j=1\) would make all \(q^d\) elements of \(L\) roots of the nonzero polynomial \(T^{q^j}-T\), whose degree \(q^j\) is smaller than \(q^d\). The root bound rules this out. Hence the order is exactly \(d\). Since a finite Galois group has order equal to the extension degree [@DF04], these \(d\) powers exhaust \(\operatorname{Gal}(L/K)\). It is therefore cyclic. For \(d=1\), this argument gives the trivial group.