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Example: GF(64)

The field GF(64) has several interesting properties that smaller fields do not share: it has two subfields such that neither is contained in the other; not all generators (elements with minimal polynomial of degree 6 over GF(2)) are primitive elements; and the primitive elements are not all conjugate under the Galois group.

The order of this field being 2, and the divisors of 6 being 1, 2, 3, 6, the subfields of GF(64) are GF(2), GF(2) = GF(4), GF(2) = GF(8), and GF(64) itself. As 2 and 3 are coprime, the intersection of GF(4) and GF(8) in GF(64) is the prime field GF(2).

The union of GF(4) and GF(8) has thus 10 elements. The remaining 54 elements of GF(64) generate GF(64) in the sense that no other subfield contains any of them. It follows that they are roots of irreducible polynomials of degree 6 over GF(2). This implies that, over GF(2), there are exactly 9 = 54 / 6 irreducible monic polynomials of degree 6. This may be verified by factoring X − X over GF(2).

The elements of GF(64) are primitive n th roots of unity for some n dividing 63. As the 3rd and the 7th roots of unity belong to GF(4) and GF(8), respectively, the 54 generators are primitive n th roots of unity for some n in {9, 21, 63}. Euler's totient function shows that there are 6 primitive 9 th roots of unity, 12 primitive 21 st roots of unity, and 36 primitive 63 rd roots of unity. Summing these numbers, one finds again 54 elements.

By factoring the cyclotomic polynomials over GF(2), one finds that:

This shows that the best choice to construct GF(64) is to define it as GF(2) / (X + X + 1). In fact, this generator is a primitive element, and this polynomial is the irreducible polynomial that produces the easiest Euclidean division.