Answers to Exercises; Chapter 4 301- [a(3v2w + w3d) + 2bvw + cw]J;i.
Since 4 $ F, it f o 11 ows that both expressions in square brackets must
vanish, resulting in 0 = f (v- ~4). The third zero of the cubic is -(b/a)-
2v, a member of F.
(b) By (a), we see that g(t) has a zero in F,,-1, hence in Fn-2, etc.
3.10. (b) Using C6 = -C3 - 1, ‘t 1 is straightforward to show that Q(c) is
closed under addition, subtraction and multiplication. As for the inverse of
C3 + C, by the Euclidean algorithm, we obtain
Hence
(9 + t3 + 1) = (t3 - t + l)(P + t) + (t2 - t + 1)(t3 + t) = (t + l)(P -t + 1) + (t - 1)
(t2 - t + 1) = t(t - 1) + 1.1 = (t2 - t + 1) - t[(t3 + t) - (t + l)(P - t + l)]
= (t2 + t + l)(P -t + 1) - t(t3 + t)
= (t2 + t + l)[(P + t3 + 1) - (t3 -t + l)(P + t)] - t(t3 + t)
= (P+t+1)(t6+t3+1)-(t5+t4+t+1)(t3+t).Setting t = C yields
1 = -(C” + c4 + c + l)(C” + C),whence (C3 + C)-’ = -(C” + c4 + C + 1).
Similarly, C-l = -(C” + C2).
(e) There are many ways of verifying the zeros. For example,
f (u2 - 2) = u6 - 6u4 + 9u2 - 1 = (3u - 1)” - 6u4 + 9u2 - 1= -6u(u3 - 3u + 1) = 0.Since the sum of the three zeros is 0, the third zero must be 2 - u - u2.
The reciprocal of any element in Q(u) can be found by using the Euclidean
algorithm as outlined in (b).
4.2. (b) The corresponding real system is 2x - y - 3 = 0, x + 2y + 4 = 0.
4.3. The corresponding real system is px - qy + r = 0, qx + py + s = 0.
The lines are perpendicular and hence intersect in exactly one point.
4.5. (e) The corresponding real system is
(x + l/2)2 - (y + 1)2 = l/4(2x + l)(y + 1) = 0