Let Fp be the field with p (prime) elements. Also let Fp_bar be its algebraic
closure. Let phi : Fp_bar --> Fp_bar be defi ned by phi(x) = x^p, which we also know satisfi es
phi(x + y) = phi(x) + phi(y):
(i) Show that for any m, phi maps F_p^m onto itself; ie prove that all
finite fi elds are perfect.
(ii) Show that for all m, x ϵ Fp_bar, we have x ϵ Fp if and only if phi(x) = x.
(iii) Show that for all x ϵ Fp_bar, we have x ϵ F_(p^m) iff (phi^m)(x)=x
closure. Let phi : Fp_bar --> Fp_bar be defi ned by phi(x) = x^p, which we also know satisfi es
phi(x + y) = phi(x) + phi(y):
(i) Show that for any m, phi maps F_p^m onto itself; ie prove that all
finite fi elds are perfect.
(ii) Show that for all m, x ϵ Fp_bar, we have x ϵ Fp if and only if phi(x) = x.
(iii) Show that for all x ϵ Fp_bar, we have x ϵ F_(p^m) iff (phi^m)(x)=x
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(i) φ is 1-1, since its kernel is clearly {0}.
Since φ maps F_p^m to itself (all we need is that we have an injection between two finite sets with the same cardinality), φ is automatically onto. Hence, φ is a bijection on F_p^m.
(ii/iii) Note that Fp^m is Galois over Fp, with cyclic Galois group of order m generated by φ. By the Fundamental Theorem of Galois Theory, every subfield of Fp^m corresponds to a subgroup of <φ>. So for every divisor d of m, there is precisely one subfield of Fp^m of degree d over Fp, namely the fixed field of the subgroup generated by φ^d of order m/d, which is isomorphic to Fp^d.
The claim now follows from Fp_bar being the field containing any F_p^d and Fp^d being the fixed field of the subgroup generated by φ^d. (For (ii) let d = 1, and for (iii) let d = m.)
I hope this helps!
Since φ maps F_p^m to itself (all we need is that we have an injection between two finite sets with the same cardinality), φ is automatically onto. Hence, φ is a bijection on F_p^m.
(ii/iii) Note that Fp^m is Galois over Fp, with cyclic Galois group of order m generated by φ. By the Fundamental Theorem of Galois Theory, every subfield of Fp^m corresponds to a subgroup of <φ>. So for every divisor d of m, there is precisely one subfield of Fp^m of degree d over Fp, namely the fixed field of the subgroup generated by φ^d of order m/d, which is isomorphic to Fp^d.
The claim now follows from Fp_bar being the field containing any F_p^d and Fp^d being the fixed field of the subgroup generated by φ^d. (For (ii) let d = 1, and for (iii) let d = m.)
I hope this helps!