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272 Answers to Exercises and Solutions to Problems

positive. Then, since

(x + u)(z + w) - (y + v)” - r(x + u)(y + v) = x-‘(x + u)(xz - y2 - rxy)

+ u-l(x + U)(UUJ - v2 - ruv) + (2x)-‘(zv - YU)~

and

(XU)(XW + 2yv + ZU) - (xv + yu)2 - r(xu)(xv + yu)

= x2(uw - v2 - ruv) + u”(xz - y2 - rxy),

it follows that f + g and fg are in A along with f and g. If t = 0, then

y = 0 and x > 0, and it is straightforward to check that f + g and fg are

in A along with f and g.

Answers to Exercises

Chapter 3

1.1. (b) Let x be any rational. For any other nonzero rational y, we have

x = y(y-lx).

1.2. Let the coefficients of p(t) and q(t) be ai and bi respectively. If p(t) =

cq(t), then ai = cbi for each i, i.e. clai. The converse is straightforward.

1.3. If at + b = f(t)g(t), the degrees of f(t) and g(t) must be 0 and 1 in

some order.

1.4. If f is a polynomial over R with a nontrivial factorization f = gh over

R, then this is also a factorization over C. Hence an irreducible polynomial

over C is also irreducible over R. An analogous argument applies for R and

Q-

1.5. Suppose, if possible, that t2 + 1 = (at + b)(d + d). Then UC = bd = 1

and ad + bc = 0. Hence, a and c must be nonzero reals with the same sign.

Similarly, b and d are nonzero with the same sign. Hence, ad and bc are

nonzero with the same sign, and so cannot satisfy ad+ bc = 0. Thus, t2 + 1

cannot be factored over R.

1.6. Note that p(t) = f(t)g(t) if and only if p(t - k) = f(t - k)g(t - k).
1.7. (a) t2+c has nonreal zeros and so cannot be factored over R. However,
t2 + c = (t + i&)(t - i J) c is a factorization over C.
(b) t2 - d2 = (t - d)(t + d) is a factorization over 2, Q, R, C.
(c) t2 + c = (t + J-c)(t - J=-) c is a factorization over R, C. Suppose,
if possible, that F c were rational; let it be u/v in lowest terms. Then
-cv2 = u2. Since gcd(u, v) = 1, v must be 1 and so -c is a perfect square, a
contradiction. Hence t2+c does not have rational zeros and so is irreducible
over Q.