Unknown

300 Answers to Exercises and Solutions to Problems

au2-u4+av -v2; d= bv+2uv2 - auv - u3v. Rearranging terms, we find

that

v2-(3u2+a)v+(u4+au2-bu+c)=O

2uv2 + (b - au - u3)v - d = 0

whence (b + ua + 5u3)v - (2u5 + 2au3 - 2bu2 + 2cu + d) = 0.
Eliminating v, we find that

(2u5 + 2au3 - 2bu2 + 2cu + d)2 - (3u2 + a)(2u5 + 2au3 - 2bu2

+ 2cu + d)(5u3 + au + b) + (u4 + au2 - bu - c)(5u3 + au + b)2 = 0,

an equation in u of degree 10 which cannot in general be reduced to an
equation of lower degree.

3.6. a) Suppose, if possible, that fi = a + b&. Then 2 = (a” + 3b2) +
2ab $ 3. Since fi is not rational, ab = 0. But this cannot occur. If 3i13 =
a + b&, then 3 = (a” + 9ab2) + 3b(a2 + b2)fi. Hence b = 0, which is
impossible. Since Q(d) C R, i $Z Q(a).
(b) Suppose i = a + b@. Then -1 = (a2 - 3b2) + 2abi&, so ab = 0.
But either a = 0 or b = 0 is impossible.
(c) d E C but xh $Z Q(i).
(e) (a + b&)-l = (a2 - b2d)-‘(a - b&).
(f) Any field which contains Q and & must contain all the elements of
the form a + bfi, where a, b E Q, hence must contain Q(d). On the other
hand, Q(d) is itself a field, and so the result follows.
(h) It suffices to choose b, c such that b2 - 4c is equal to d multiplied by
a square, say b2 - 4c = 4d. This can be arranged if c + d = 1, b = 2. An
example is t2 $2t + (1 - d).

3.7. (b)

(a + bxh + c& + d&)-l = [(a2 + 2b2 - 3c2 - 6d2)2 - 8(ab - 3cd)2]-1

[(a + b/h) - &(c + dfi)][(a2 + 2b2 - 3c2 - 6d2) - 2h(ab - 3cd)].

3.8. (a)

(a + 21i4b + 21i2c + 23/4d)-’ = [(a” - 4bd + 2~~)~ - 2(2ac - b2 - 2d2)2]-1

[(a + 21j2c) - 21i4(b + 2’/“d)][(a” - 4bd + 2c2) - 21i2(2ac - b2 - 2d2)].

(I+ 2114 + 2112 + p/4)-1 = -(I _ p/4)

(1+ 23’4)-’ = (1 - 4.21i4 + 2.21’2 - 23’4)/(-7).

3.9. (a) Let f(t) = at3 + bt2 + ct + k and suppose that u E F(h) is a zero
of f(t). If u E F, then the result follows. Otherwise, u = v + w&, with
v, w E F, w # 0, and

0 = [a(v” + 3vw2d) + b(v2 + w2d) + cv + k]