Introduction to Abstract Algebra, Solutions Manual, 4th Edition

3.2 Integral Domains and Fields

1.

a. 1, − 4.

c. 0, 1.

3. If e² = e in a domain, then e(1 − e) = 0 so e = 0 or e = 1. If aⁿ = 0, n ≥ 1, then a = 0. For if a ≠ 0 then aaⁿ⁻¹ = 0 gives aⁿ⁻¹ = 0, . . ., and eventually a = 0, a contradiction.

5. Let . Then A² = 0 but A ≠ 0. So M_n(R) is not a domain.

7. If ab = 0 then (ba)² = b(ab)a = 0, so ba = 0 by hypothesis.

9. In , 1² + 2² = 0; in , let a² + b² = 0. If either a = 0 or b = 0, the other is 0 (x² = 0 ⇒ x = 0 in a field). If a ≠ 0, b ≠ 0 then a, b {1, 2}. But 1² + 1² ≠ 0, 1² + 2² = 2 ≠ 0, 2² + 2² = 3 ≠ 0.

11. The group F^∗ = F  {0} has order q − 1 so a^q−1 = 1 for all a ≠ 0 (by Lagrange's theorem). Thus a^q = a if a ≠ 0; this also holds if a = 0.

13. Since |F| = p is prime, (F, +) is cyclic and is generated by 1 (or any nonzero element) by Lagrange's theorem. Hence the map given by is an isomorphism of additive groups. It is a ring isomorphism ...