Describe the Galois group of \(\mathbb{F}_{p^n}\) over \(\mathbb{F}_p\), where \(p\) is a prime and \(n \geq 1\).
Find fields \(K\) and \(F\) such that \(K\) is a splitting field over \(F\), but \(K\) is NOT separable over \(F\).
Solution
1. In part (a), the group is cyclic of order \(n\), generated by theFrobenius automorphism \(\varphi(x)=x^p\).
Write \(E=\mathbb F_{p^n}\). In characteristic \(p\), the binomial theorem gives \((x+y)^p=x^p+y^p\), and \((xy)^p=x^py^p\). Thus \(\varphi\) is a field homomorphism fixing \(\mathbb F_p\). It is injective and hence, since \(E\) is finite, surjective.
The multiplicative group \(E^\times\) has order \(p^n-1\). Lagrange’s theorem [@DF04] gives \(x^{p^n}=x\) for every nonzero \(x\), and the same equality holds at zero. Hence \(\varphi^n=1\). For \(0<d<n\), the equality \(\varphi^d=1\) would make all \(p^n\) elements of \(E\) roots of the nonzero polynomial \(X^{p^d}-X\), whose degree is \(p^d<p^n\). This is impossible, so \(\varphi\) has order exactly \(n\).
Since a vector space of dimension \(m\) over \(\mathbb F_p\) has \(p^m\) elements, \([E:\mathbb F_p]=n\). A finite extension of degree \(n\) has at most \(n\) embeddings into an algebraic closure: choose a finite tower obtained by adjoining generators, and note that at a step of degree \(d\) each embedding of the preceding field has at most \(d\) extensions, because the image of the new generator must be a root of the image of its minimal polynomial. Multiplication of these bounds along the tower gives the degree \(n\). Consequently the \(n\) distinct powers of \(\varphi\) exhaust all \(\mathbb F_p\)-automorphisms of \(E\).
Moreover, \(X^{p^n}-X\) splits over \(E\), has derivative \(-1\), and its roots are exactly all the elements of \(E\). Therefore \(E\) is its splitting field and \(E/\mathbb F_p\) is Galois [@DF04]. Explicitly, \[ \operatorname{Gal}(E/\mathbb F_p) =\{x\mapsto x^{p^j}:0\leq j<n\} \cong\mathbb Z/n\mathbb Z. \] For \(n=1\) this is the trivial group.
2. In part (b), take \(K=\mathbb F_p(u)\) and\(F=\mathbb F_p(u^p)\), where \(u\) is an indeterminate.
Put \(t=u^p\in F\). The element \(u\) satisfies \(X^p-t\), so \([F(u):F]\leq p\). To obtain equality, suppose \[ \sum_{j=0}^{p-1}a_j(u^p)u^j=0, \qquad a_j(T)\in\mathbb F_p(T). \] Multiplication by a common nonzero polynomial denominator gives \(\sum_{j=0}^{p-1}P_j(u^p)u^j=0\) with \(P_j\in\mathbb F_p[T]\). The powers of \(u\) occurring in the \(j\)th summand are all congruent to \(j\) modulo \(p\). Distinct summands therefore have disjoint monomial supports. Since \(u\) is transcendental, every coefficient vanishes, so every \(P_j\), and hence every \(a_j\), is zero. Thus \(1,u,\ldots,u^{p-1}\) are linearly independent over \(F\), and \([K:F]=p\).
It follows that the degree-\(p\) polynomial \(X^p-t\) is the minimal polynomial of \(u\) over \(F\). In \(K[X]\) it factors as \[ X^p-t=(X-u)^p. \] Its only root is \(u\), and that root generates \(K\) over \(F\). Therefore \(K\) is its splitting field. Its minimal polynomial has a repeated root, so \(u\) is not separable over \(F\) and \(K/F\) is not separable. In fact every \(r(u)\in\mathbb F_p(u)\) satisfies \(r(u)^p=r(u^p)\in F\), so the extension is purely inseparable.