114 Groups and Rings 2008–09
Suggested solutions to exercise set 7
April 14, 2009
30.8 If Z6 can be embedded in a field F , then F contains a subring R ⊆ F with
R ≈ Z6 . Since [2] [3] = [0] in Z6 , R contains zero divisors (the elements
identified with [2] and [3] by the isomorphism of R and Z6 ). So F contains
zero divisors; but F is a field, so an integral domain, so cannot contain any
zero divisors. This contradiction shows that Z6 cannot be embedded in a
field.
30.9 Let D be a field with unit e. Let θ : D → FD be the map θ(a) = [a, e]. We
have seen in the proof of Lemma 30.4 that θ preserves the ring operations,
and is one-to-one. We claim that θ is onto: for if [x, y] ∈ FD for some
(x, y) ∈ D × D0 , then y is a non-zero element of the field D, so y −1 ∈ D.
Now
xe = (xy −1 )y,
so (x, y) ∼ (xy −1 , e) ⇐⇒ [x, y] = [xy −1 , e].
So [x, y] = [xy −1 , e] = θ(xy −1 ) ∈ θ(D). So θ is onto and D ≈ θ(D) = FD .
31.2 Recall from section 28 that for a, b, c ∈ F ,
a > b, c > 0 =⇒ ac > bc and a > b, c < 0 =⇒ ac < bc.
If a < 0 then a−1 < 0; for otherwise, a < 0 and a−1 > 0, so e = aa−1 < 0,
which is false (see page 138). So if a < 0 and b < 0 then a−1 < 0 and b−1 < 0
so a−1 b−1 > 0.
So if 0 > b > a then c = a−1 b−1 > 0, so 0 > bc > ac, that is, 0 > a−1 > b−1 .
31.4 If a is rational, a 6= 0 and b is irrational then ab is a real number. Suppose
that ab is rational. Since the rational numbers form a field, a−1 is rational,
and the product of the two rational numbers a−1 and ab is also rational. So
b = a−1 ab is rational, a contradiction. So ab must be irrational.
1
31.8 Let x = 0.035. We have
99x = 100x − x = 3.535 − 0.035 = 3.5 =
so x =
7
.
2×99
7
2
So
1.935 = 1.9 + 0.035 =
19
7
1916
+
=
.
10 2 × 99
990
31.23 This is true (provided a 6= 0). If a is irrational and a−1 is rational, then
aa−1 = 1 is irrational by exercise 31.4, which is false since 1 ∈ Q. So a−1
must be irrational. [If a = 0 then the question is moot since a−1 is not
defined.]
√
√
√ √
√
√
31.25 2 and 3 are both irrational, as is 6 = 2 3. On the other hand, 2
and √12 are both irrational, but their product is 1, which is rational.
32.6 Recall Fermat’s Little Theorem: if p is prime then for any x ∈ Zp we have
xp = x in Zp , so xp − x + [1] = [1] 6= [0]. So the polynomial equation
xp − x + [1] = [0] has no solution in Zp and has degree p − 1 > 1. So Zp is
not algebraically closed.
32.17 Let θ : C → C be an isomorphism with θ(a) = a for each a ∈ R. Since
θ(−1) = −1, we have
θ(i)2 = θ(i2 ) = θ(−1) = −1.
Let θ(i) = x + yi where x, y ∈ R. We have
(x + yi)2 = −1 ⇐⇒ x2 − y 2 + 2xyi = −1 ⇐⇒ x2 − y 2 = −1 and 2xy = 0,
so xy = 0, so x = 0 or y = 0. If y = 0 then x2 = −1, which has no real
solutions (as x2 > 0 for all x ∈ R with x 6= 0 by Lemma 28.1). So we must
have x = 0 and −y 2 = −1 ⇐⇒ y 2 = 1 ⇐⇒ y = ±1. So either θ(i) = i or
θ(i) = −i.
For any a, b ∈ R we have
θ(a + bi) = θ(a) + θ(bi) = a + θ(b)θ(i) = a + bθ(i).
So if θ(i) = i then θ(a + bi) = a + bi for all a, b ∈ R; and if θ(i) = −i then
θ(a + bi) = a + b(−i) = a − bi for all a, b ∈ R.
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