9.5. A FEW GENERALIZATIONS 241

So when is a field of characteristic p perfect? As observed above, for a field of char-acteristic p,(a+b)p = ap +bp. Also, (ab)p = apbp. It follows that a→ ap is a homomor-phism. This is also one to one because, as mentioned above (a−b)p = ap−bp. Therefore,if ap = bp, it follows that a = b. Therefore, this homomorphism is also one to one.

Let Fp be the collection of ap where a ∈ F. Then clearly Fp is a subfield of F becauseit is the image of a one to one homomorphism. What follows is the condition for a field ofcharacteristic p to be perfect.

Theorem 9.5.11 Let F be a field of characteristic p. Then F is perfect if and only if F= Fp.

Proof: Suppose F= Fp first. Let f (x) be an irreducible polynomial over F. By The-orem 9.5.5, if f ′ (x) and f (x) are relatively prime over F then f (x) has no repeated roots.Suppose then that the two polynomials are not relatively prime. If d (x) divides both f (x)and f ′ (x) with degree of d (x) ≥ 1. Then, since f (x) is irreducible, it follows that d (x) isa multiple of f (x) and so f (x) divides f ′ (x) which is impossible unless f ′ (x) = 0. But iff ′ (x) = 0, then f (x) must be of the form

a0 +a1xp +a2x2p + · · ·+anxnp

since if it had some other nonzero term with exponent not a multiple of p then f ′ (x) couldnot equal zero since you would have something surviving in the expression for the deriva-tive after taking out multiples of p which is like kaxk−1 where a ̸= 0 and k < p. Thus ka ̸= 0.Hence the form of f (x) is as indicated above.

If ak = bpk for some bk ∈ F, then the expression for f (x) is

bp0 +bp

1xp +bp2x2p + · · ·+bp

nxnp =(b0 +b1x+bxx2 + · · ·+bnxn)p

because of the fact noted earlier that a→ ap is a homomorphism. However, this says thatf (x) is not irreducible after all. It follows that there exists ak such that ak /∈ Fp contrary tothe assumption that F= Fp. Hence the greatest common divisor of f ′ (x) and f (x) must be1.

Next consider the other direction. Suppose F ̸= Fp. Then there exists a ∈ F\Fp.Consider the polynomial xp−a. As noted, its derivative equals 0. Therefore, xp−a and itsderivative cannot be relatively prime. In fact, xp−a would divide both. ■

Now suppose F is a finite field. If n ·1 is never equal to 0 then, since the field is finite,k ·1 = m ·1, for some k < m. m > k, and (m− k) ·1 = 0 which is a contradiction. Hence Fis a field of characteristic p for some prime p, by Proposition 9.5.8. The mapping a→ ap

was shown to be a homomorphism which is also one to one. Therefore, Fp is a subfield ofF. It follows that it has characteristic q for some q a prime. However, this requires q = pand so Fp = F. Then the following corollary is obtained from the above theorem.

With this information, here is a convenient version of the fundamental theorem of Ga-lois theory.

Theorem 9.5.12 Let K be a splitting field of any polynomial p(x) ∈ F [x] where F iseither of characteristic 0 or of characteristic p with Fp = F. Let {Li}k

i=0 be the increasingsequence of intermediate fields between F and K. Then each of these is a normal extensionof F and the Galois group G

(L j−1,F

)is a normal subgroup of G(L j,F). In addition to

this,G(L j,F)≃ G(K,F)/G(K,L j)

where the symbol ≃ indicates the two spaces are isomorphic.