Classify finite fields, their subfields, and their field extensions. What are the automorphisms of a finite field?
Solution
1. Every finite field has cardinality \(p^n\) for a unique prime \(p\) and integer \(n\ge1\).
If \(K\) is finite, its characteristic is a prime \(p\). The prime field \(\FF_p\) is therefore a subfield of \(K\), so \(K\) is a finite-dimensional vector space over \(\FF_p\). If \[ [K:\FF_p]=n, \] then \[ |K|=p^n. \]
2. For every prime power \(q=p^n\), there is, up to isomorphism, a unique field \(\FF_q\) with \(q\) elements.
In an algebraic closure of \(\FF_p\), the roots of \[ x^{p^n}-x \] form a field: Frobenius identities show closure under addition, subtraction, multiplication, and inversion. The derivative is \(-1\), so the polynomial has exactly \(p^n\) distinct roots. Thus its root field has \(p^n\) elements.
Conversely, if \(K\) has \(p^n\) elements, then every \(a\in K\) satisfies \[ a^{p^n}=a \] (by Lagrange’s theorem on \(K^\times\), together with \(0\)). Hence \(K\) is a splitting field of \(x^{p^n}-x\) over \(\FF_p\). Splitting fields are unique up to \(\FF_p\)-isomorphism.
3. The subfields of \(\FF_{p^n}\) are exactly\[ \FF_{p^d} \qquad(d\mid n), \]and for each divisor \(d\mid n\) there is exactly one such subfield.
If \(d\mid n\), then \[ p^d-1\mid p^n-1, \] so every root of \(x^{p^d}-x\) is also a root of \(x^{p^n}-x\). Hence the unique field of \(p^d\) elements sits inside \(\FF_{p^n}\).
Conversely, let \(K\subseteq\FF_{p^n}\) be a subfield. Then \(|K|=p^d\) for some \(d\), and \[ [\FF_{p^n}:\FF_p] = [\FF_{p^n}:K]\,[K:\FF_p] \] shows that \(d\mid n\).
Uniqueness follows because any subfield with \(p^d\) elements is exactly the root set of \(x^{p^d}-x\) inside \(\FF_{p^n}\).
4. Every finite extension of \(\FF_q\) of degree \(m\) is isomorphic to \(\FF_{q^m}\).
Write \(q=p^d\). If \(L/\FF_q\) has degree \(m\), then \[ [L:\FF_p]=md, \] so \[ |L|=p^{md}=q^m. \] By 2, there is only one field of that cardinality up to isomorphism.
5. The extension\[ \FF_{q^m}/\FF_q \]is Galois and cyclic of degree \(m\), generated by the \(q\)-Frobenius\[ \operatorname{Fr}_q:x\mapsto x^q. \]
Every element of \(\FF_{q^m}\) is a root of \[ x^{q^m}-x, \] whose derivative is \(-1\). Hence \(\FF_{q^m}\) is the splitting field of a separable polynomial over \(\FF_q\), so the extension is Galois.
The map \(\operatorname{Fr}_q\) fixes \(\FF_q\) and satisfies \[ \operatorname{Fr}_q^m(x)=x^{q^m}=x. \] Its fixed field is exactly \(\FF_q\), because the fixed points are the roots of \(x^q-x\), of which there are exactly \(q\). Therefore its order is \(m\). Since the Galois group also has order \(m\), it is generated by \(\operatorname{Fr}_q\).
6. In particular,\[ \operatorname{Aut}(\FF_{p^n}) =\operatorname{Gal}(\FF_{p^n}/\FF_p) \cong C_n, \]generated by\[ x\mapsto x^p. \]
Every field automorphism fixes the prime field \(\FF_p\). Apply 5 with \(q=p\) and \(m=n\).