Show that if \(|K| = p^n\), then every element of \(K\) has a unique \(p\)th root in \(K\).
Solution
1. The Frobenius map \(\varphi\colon K \to K\), \(\varphi(x) = x^p\), is an injective ring homomorphism.
The field \(K\) has order \(p^n\), so its prime field is \(\FF_p\) and \(\operatorname{char}K=p\). Hence \(p\) divides \(\binom pk\) for \(0<k<p\), so \((x + y)^p = x^p + y^p\); also \((xy)^p = x^p y^p\) and \(1^p=1\). The kernel of a ring homomorphism out of a field is an ideal not containing \(1\), hence \(0\).
2. \(\varphi\) is bijective.
By step 1, \(\varphi\) is an injective map from the finite set \(K\) to itself, hence surjective.
3. Q.E.D.
By step 2, for every \(a\in K\) there is exactly one \(x\in K\) with \(x^p=\varphi(x)=a\).