114 Groups and Rings 2008–09
Suggested solutions to exercise set 6
April 8, 2009
28.2 Observe that for any x, y ∈ Dp we have
x > y ⇐⇒ x − (x + y) > y − (x + y) ⇐⇒ −y > −x
(here we are using Theorem 28.2(d) to add −(x + y) to both sides of the
inequality).
Since 0 > c, this gives −c > 0. Now a > b so Theorem 28.2(e) gives
−ac = a(−c) > b(−c) = −bc. So bc > ac.
28.6 Since a > 0 and b > 0, Theorem 28.2(e) gives a > b =⇒ a2 = a·a > ba = ab
and a > b =⇒ ab > b · b = b2 . So a2 > ab and ab > b2 ; by Theorem 28.2(g),
a2 > b2 .
28.7 Let a, b, c ∈ D, an ordered integral domain.
(a) If a > 0 and b > 0 then a ∈ Dp and b ∈ Dp so by the definition of an
ordered integral domain, a + b ∈ Dp , so a + b > 0.
(b) If a > 0 and b > 0 then a ∈ Dp and b ∈ Dp so by the definition of an
ordered integral domain, ab ∈ Dp , so ab > 0.
(c) By the definition of an ordered integral domain, if a ∈ D then exactly
one of −a ∈ Dp ⇐⇒ a < 0, a = 0 or a ∈ Dp ⇐⇒ a > 0 is true.
(d) If a > b then (a + c) − (b + c) = a − b ∈ Dp , so a + c > b + c.
(e) [see the text]
(f) If a 6= 0 then Lemma 28.1 gives a2 ∈ Dp , so a > 0.
(g) If a > b and b > c then a − b ∈ Dp and b − c ∈ Dp , so (a − b) + (b − c) =
a − c ∈ Dp , so a > c.
28.12 This is false. For example, in Z we have 0 > −1 but 02 = 0 < 1 = (−1)2 .
1
28.13 This is true. If a > b then either a > b > 0, a > 0 ≥ b, 0 > a > b or
a = 0 > b. We consider these in turn:
– If a > b > 0 then a2 > b2 (see Exercise 28.6) so
a3 = a · a2 > b · a2 > b · b 2 = b 3
(by Thm 28.2(e), twice). So a3 > b3 .
– If a > 0 ≥ b then −b ≥ 0 so −b3 = (−b)3 ≥ 0, so 0 ≥ b3 , and since
a > 0 we have a3 > 0. So a3 > 0 ≥ b3 , so a3 > b3 .
– If 0 > a > b then −b > −a > 0 so −b3 = (−b)3 > (−a)3 = −a3 (by the
first case) so a3 > b3 .
– If a = 0 > b then −b > 0 so −b3 = (−b)3 > 0 so 0 = a3 > b3 .
29.4 Since D is well-ordered, Lemma 29.1 shows that e is the least element of Dp .
Since a ∈ Dp and e 6= a, this gives e < a. Since a > 0, Theorem 28.2(e) gives
a2 = a · a > e · a = a.
29.7 Suppose that P (n) is a statement that may be true or false for each n ∈ Zp ,
and that P (1) is true and P (n + 1) is true whenever P (n) is true.
Let S = {k ∈ Zp : P (k) is false}.
If S 6= ∅ then, since Z is well-ordered and S ⊆ Zp , there is a least element
m of S. Since P (1) is not false, 1 6∈ S, so m 6= 1. Since m ∈ Zp , this gives
m > 1. So m − 1 ∈ Zp , and m − 1 6∈ S so P (m − 1) is true. But this implies
that P ((m − 1) + 1) = P (m) is true, so m 6∈ S, a contradiction.
So S = ∅, so P (k) is true for every k ∈ Zp .
29.8 If θ : Z → Z is a ring isomorphism then it must send the unit of Z to the
unit of Z. So θ(1) = 1; so for any n ∈ Z, θ(n) = θ(n1) = nθ(1) = n1 = n.
So θ is the identity map.
The map θ : Z → Z, θ(n) = −n is an isomorphism of additive groups which
is not the identity map. It is an isomorphism since θ(n + m) = −(n + m) =
−n + (−m) = θ(n) + θ(m) and θ is clearly one-to-one and onto.
2