Let \(F\) be a finite field of order \(q\) and let \(E/F\) be a field extension. Suppose that an element \(a \in E\) is algebraic over \(F\). Prove that \([F(a) : F]\) is the smallest positive integer \(n\) such that \(a^{q^n} = a\) and that it divides every other such positive integer.
Solution
Set \[ d:=[F(a):F]. \] Because \(a\) is algebraic over \(F\), one has \(d<\infty\).
1. The field \(F(a)\) has exactly \(q^d\) elements, and every element \(x\in F(a)\) satisfies\[ x^{q^d}=x. \]
As a \(d\)-dimensional vector space over the field \(F\) of \(q\) elements, \(F(a)\) has \[ q^d \] elements. Its multiplicative group \(F(a)^\times\) therefore has order \(q^d-1\). By Lagrange’s theorem, every nonzero \(x\in F(a)\) satisfies \[ x^{q^d-1}=1, \] so \[ x^{q^d}=x. \] The equality is also true for \(x=0\).
2. In particular,\[ a^{q^d}=a. \]
This is 1 applied to the element \(a\in F(a)\). Thus \(d\) is one positive integer for which the required Frobenius equality holds.
3. For every positive integer \(n\), the roots of\[ T^{q^n}-T \]in an algebraic closure of \(F\) form a field with exactly \(q^n\) elements.
Let \[ K_n:=\{x:x^{q^n}=x\} \] inside a fixed algebraic closure of \(F\). In characteristic \(p\), where \(q\) is a power of \(p\), the map \[ x\longmapsto x^{q^n} \] is a ring homomorphism. Hence if \(x,y\in K_n\), then \[ (x+y)^{q^n}=x^{q^n}+y^{q^n}=x+y \] and \[ (xy)^{q^n}=x^{q^n}y^{q^n}=xy. \] If \(0\neq x\in K_n\), then \[ (x^{-1})^{q^n}=(x^{q^n})^{-1}=x^{-1}. \] Thus \(K_n\) is a field.
The polynomial \(T^{q^n}-T\) has derivative \[ q^nT^{q^n-1}-1=-1, \] so it has no repeated roots. Its splitting field contains all of its \(q^n\) distinct roots, and those roots are precisely the elements of \(K_n\). Therefore \[ |K_n|=q^n. \]
4. If \(a^{q^n}=a\), then\[ d\mid n. \]
3.Also \(F\subseteq K_n\), because every element \(c\in F\) satisfies \(c^q=c\), hence \(c^{q^n}=c\).Therefore the field generated by \(F\) and \(a\) satisfies\[ F(a)\subseteq K_n. \]By 3,\[ [K_n:F]=n, \]since \(|K_n|=q^n\) and \(|F|=q\).The tower law gives\[ n=[K_n:F]=[K_n:F(a)]\,[F(a):F]. \]Thus\[ d=[F(a):F]\mid n. \]
5. Hence \(d=[F(a):F]\) is the smallest positive integer \(n\) such that \(a^{q^n}=a\), and it divides every other such \(n\).
By 2, \(d\) itself has the required property. By 4, every positive integer with that property is divisible by \(d\). Therefore none is smaller than \(d\), and \(d\) divides all of them.