Introduction to Abstract Algebra, Solutions Manual, 4th Edition

6.4 Finite Fields

1.

(a) In , 2² = 4, 2³ = 8, 2⁴ = 5, 2⁵ = − 1, so o(2) = 10. Thus 2 is a primitive element.

(c) since x³ + x + 1 is irreducible in . In this case GF(8)^∗ has order 7, so every nonzero element except 1 is primitive by Lagrange's theorem.

3. Both p and q have no root in , so they are irreducible. Hence both rings are fields of order 2³ and so are isomorphic by Theorem 4.

4. (a)

5. First x⁴ + x + 1 is irreducible over (it has no root, x² + x + 1 is the only irreducible quadratic, and (x² + x + 1)² = x⁴ + x² + 1). Hence

Now t³ ≠ 1 and t⁵ = t(t + 1) = t + t² ≠ 1, so o(t) = 15. Thus t is primitive. Since 16 = 2⁴ the subfields of GF(2⁴) are GF(2), GF(2²) and GF(2⁴). Clearly and GF(2⁴) = GF(16). Finally o(t⁵) = 3 so by the discussion following Corollary 2 of Theorem 7,