Solutions
Math 310, Homework #7
Theorem HW7.1. Prove proposition 68. It states: Let R be a ring. We have the following
properties:
1. The ring R only has one additive identity. That is, if 00 ∈ R with 00 + b = b + 00 = b for every
b ∈ R, then 00 = 0.
2. The ring R only has one multiplicative identity. That is, if 10 ∈ R with 10 b = b10 = b for every
b ∈ R, then 10 = 1.
3. For every a ∈ R, there is only one additive inverse of a. That is, if b ∈ R with a+b = b+a = 0,
then b = −a.
4. For every a ∈ R, the additive inverse of the additive inverse of a is a itself. That is,
−(−a) = a.
5. For every a ∈ R, then the additive inverse of a is the product of the additive inverse of 1 and
a. That is, (−1)a = −a. In particular, (−1)2 = 1.
6. More generally, for any a, b ∈ R, then (−a)b = −ab = a(−b). Note: There are two things
to prove here!
7. For every a ∈ R, we have a · 0 = 0 · a = 0.
8. The cancellation law for addition holds. That is, if a, b, c ∈ R with a + b = a + c, then b = c.
9. The additive inverse of a sum is the sum of the additive inverses. That is, for a, b ∈ R, we
have −(a + b) = −a + −b.
10. For a, b, c ∈ R, if ab = ba = 1 and ac = 1, then b = c.
Proof.
1. Since 00 is an additive identity, we have 0 = 0 + 00 . Since 0 is an additive identity, we
have 0 + 00 = 00 . By transitivity of equality, we have 0 = 00 .
2. Since 10 is a multiplicative identity, we have 1 = 1 · 10 . Since 1 is a multiplicative identity,
we have 1 · 10 = 10 . By transitivity of equality, we have 1 = 10 .
3. Suppose b and b0 are both additive inverses of a. Then,
b = b+0
= b + (a + b0 )
= (b + a) + b0
= 0 + b0
= b0
by additive identity
as b is an additive inverse of a
by associativity of addition
as b is an additive inverse of a
by additive identity
0
so that b = b0 as desired.
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Solutions
Math 310, Homework #7
4. We have that −a + a = a + (−a) = 0 as −a is the additive inverse of a. This means that
a is an additive inverse of −a. Since −(−a) is the additive inverse of −a, we have that
−a + (−(−a)) = −(−a) + (−a) = 0. By the uniqueness of additive inverses (see #3), we
have that a = −(−a) as desired.
5. By the uniqueness of additive inverses, it suffices to show that (−1)a + a = a + (−1)a = 0.
We have the first equality by the commutativity of addition, so we show the second holds.
We have,
a + (−1)a = 1 · a + (−1) · a
= (1 + (−1)) · a
= 0·a
=0
by multiplicative identity
by distributivity
by additive inverses
by # 7
so that both equalities hold and (−1)a = −a as they are both additive inverses of a. Moreover, (−1)2 = (−1)(−1) so letting a = −1, we see that (−1)2 is the additive inverse of −1,
which is 1 by part # 4.
6. For the first equality, we have
(−a)b = ((−1)a)b
= (−1)(ab)
= −ab
by # 5
by associativity of multiplication
by # 5.
For the second equality, we show that a(−b) is an additive inverse of ab so that −ab = a(−b)
by part # 3. We have, ab + a(−b) = a(−b) + ab = a(−b + b) = a · 0 = 0 by commutativity of
addition, distributivity, and # 7. Thus, a(−b) = −ab by the uniqueness of additive inverses
(see # 3).
7. We have a · 0 = a · (0 + 0) = a · 0 + a · 0 by the distributive property and the fact that 0 is an
additive identity. Adding −(a · 0) to both sides gives
0 = −(a · 0) + a · 0 = −(a · 0) + a · 0 + a · 0 = a · 0
so that a · 0 = 0. To see that 0 · a = 0, we repeat the argument. We have 0 · a = (0 + 0) · a =
0 · a + 0 · a by the distributive property and the fact that 0 is an additive identity. Adding
−(0 · a) to both sides gives
0 = −(0 · a) + 0 · a = −(0 · a) + 0 · a + 0 · a = 0 · a
so that 0 · a = 0 as well.
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Solutions
Math 310, Homework #7
8. Say a + b = a + c. Then, we have
b = 0+b
= (−a + a) + b
= −a + (a + b)
= −a + (a + c)
= (−a + a) + c
= 0+c
=c
by additive identity
by additive inverse of a
by associativity of addition
by assumption
by associativity of addition
by additive inverse of a
by additive identity
so that b = c as desired.
9. We have −(a + b) + (a + b) = (a + b) + −(a + b) as −(a + b) is an additive inverse of a + b.
We also have
(a + b) + (−a + −b) = (−a + −b) + (a + b)
= −a + (−b + a) + b
= −a + (a + −b) + b
= (−a + a) + (−b + b)
= 0+0
=0
by commutativity of addition
by associativity of addition
by commutativity of addition
by associativity of addition
by additive inverses of a and b
by additive identity
so that −a + −b is also an additive inverse of a + b. By the uniqueness of additive inverses
(see # 3), we have that −a + −b = −(a + b). Here’s an alternate proof:
We have
−(a + b) = (−1)(a + b)
= (−1)a + (−1)b
= −a + −b
by # 5
by distributivity
by # 5
so that −(a + b) = −a + −b as desired.
10. We have
b = b·1
= b · (ac)
= (ba) · c
= 1·c
=c
by multiplicative identity
by assumption that ac = 1
by associativity of multiplication
by assumption that ba = 1
by multiplicative identity
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Solutions
Math 310, Homework #7
so that b = c as desired.
Theorem HW 7.2.
Let R be a ring. An element e ∈ R is called an idempotent if e2 = e. The trivial idempotents are 0
and 1.
(a) Give an example of a ring with non-trivial idempotents.
(b) Prove that if e is an idempotent, so is 1 − e.
(c) Prove that if R has no zero divisors, then the only idempotents of R are the trivial ones.
(a) There
many possible examples here. The one I had in mind is M2 (Z). There, the
are 1 0
element
is an idempotent since
0 0
1 0
1 0
1 0
·
=
.
0 0
0 0
0 0
1 0
However
is neither the additive identity nor the multiplicative identity of M2 (Z).
0 0
Proof.
(b) Suppose that e2 = e in a ring R. Then,
(1 − e)2 = (1 − e) · (1 − e) = 1 · (1 − e) − e · (1 − e)
by distributivity
= 1 − e − e + (−e)2
= 1 − e − e + e2
by distributivity
see below for why (−a)2 = a2 for any a ∈ R
by assumption that e2 = e
by additive inverses and identity
= 1−e−e+e
= 1−e
so that (1 − e)2 = 1 − e and 1 − e is also an idempotent. It remains to show that (−a)2 = a2
in any ring. We have
(−a)2 = (−a)(−a) = ((−1)a)(−a)
= (−1)(a(−a))
= (−1)(−aa)
= −(−aa)
by Proposition 68, part # 5
by associativity of multiplication
by Proposition 68, part # 6
by Proposition 68, part # 5
= aa = a2
so that (−a)2 = a2 as desired, thus completing the proof.
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by Proposition 68, part # 4
Solutions
Math 310, Homework #7
(c) Suppose that R has no zero divisors, and let e ∈ R be an idempotent. Then, e(1 − e) =
e − e2 = e − e = 0 so that e = 0 or 1 − e = 0 and e = 1. Thus, the only idempotents in a ring
with no zero divisors are 0 and 1.
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