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.
, 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.
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.
, 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
(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,
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