Let \(\FF_q\) be a finite field, where \(q=p^n\). Determine \(\operatorname{Gal}(\FF_q/\FF_p)\) and a generator. How many elements does \(\FF_q\) have, and what is the structure of \(\FF_q^\times\)?
Solution
By definition \(|\FF_q|=q=p^n\). Every element satisfies \[ x^q=x, \] so \(\FF_q\) is the splitting field over \(\FF_p\) of \(x^q-x\).
The Frobenius map \[ \varphi:\FF_q\to\FF_q, \qquad x\mapsto x^p, \] is an \(\FF_p\)-automorphism. Its \(n\)th power is the identity because \(x^{p^n}=x\). If \(0<r<n\), then the fixed field of \(\varphi^r\) has at most \(p^r<q\) elements, so \(\varphi^r\ne1\). Hence \(\varphi\) has order \(n\).
Since \[ [\FF_q:\FF_p]=n, \] and a finite Galois extension has exactly its degree many automorphisms, \[ \operatorname{Gal}(\FF_q/\FF_p) =\langle\varphi\rangle \cong C_n. \]
The multiplicative group \(\FF_q^\times\) has order \(q-1\) and is cyclic. To prove cyclicity, let \(m\) be the exponent of the finite abelian group \(\FF_q^\times\). Every nonzero element is a root of \(x^m-1\), so \[ q-1\le m. \] But always \(m\le q-1\), hence \(m=q-1\). A finite abelian group contains an element whose order equals its exponent, so \(\FF_q^\times\) contains an element of order \(q-1\) and is cyclic.