We will show that for any power pk of p, there is a unique finite field of order pk, which is unique up to isomorphism1 . For convenience, write q = pk. Fix an algebraic closure Fp of Fp. Look at the set
It follows from the above exercise that Fq is a field (why?).
Since the elements of S are roots of the polynomial Xq - X which has degree q, there can be no more than q of them.
The derivative of the polynomial Xq -X is the constant polynomial -1. Therefore, all its roots in Fp are distinct. This means that S has exactly q elements. Therefore there exists a subfield of order q in Fp. In particular there exists a finite field of order q.
On the other hand, in any field of order q, the multiplicative group of non-zero elements in the field has order q - 1. Therefore, each element of the field satisfies xq-1 = 1, or xq = x. Thus any subfield of Fq of order q must be equal to S.
Now any field of order q must have characteristic p, hence is an algebraic extension of Fp. Therefore, it is isomorphic to some subfield of Fp. We have seen that only such field is Fq. It follows that every field of order q is isomorphic to Fq. We have proved the following theorem:
We present the proof of the following theorem straight out of Serre’s book [Ser73].
Proof. If d is an integer ≥ 1, then let φ(d) denote the number of integers x with 1 ≤ x ≤ d such that (x,d) = 1. In other words, the image of x in Z∕dZ is a generator of Z∕dZ. The function φ(d) is called the Euler totient function.
Proof. If d∣n, let Cd denote the unique subgroup of order d in Z∕nZ, and Φd denote the generators of Cd. Then Z∕nZ is the disjoint union of the Φd. Φd had φ(d) elements. Adding up cardinalities, n = ∑ d∣nφ(d). □
Proof. Let d∣n. If there exists x H of order d, the subgroup
In general, if E is an extension of a field F, then every element x E can be thought of as an F-linear endomorphism of the F-vector space E, when it acts on E by multiplication. The trace of this map is denoted tr E∕F(x). The function trE∕F : E → F is called the trace function of E over F. Likewise, the determinant of multiplication by x is denoted NE∕F (x). The function NE∕F : E → F is called the norm map of E over F.
Since Fq2 is a quadratic extension of Fq, its Galois group is cyclic of order 2. Clearly, the map F : xxp is an automorphism of Fq2 that fixes Fq. Therefore, it must be the non-trivial element in the Galois group of Fq2 over Fq. F is called the Frobenius automorphism. In analogy with complex conjugation, we write F(x) = x for each x Fq2.
Let N and tr denote the norm and trace maps of Fq2 over Fq respectively. Then
Let ψ0 : Fq → C* be a non-trivial additive character. Such a character is completely determined by its value at 1, which can be any pth root of unity different from 1. Then ψ : Fq → C* defined by ψ(x) = ψ(trFq∕Fp(x)) is a non-trivial additive character of Fq.
Proof. The map x′ψx′ is clearly an injective homomorphism. By Proposition 1.2, it must also be onto. □