Assignment 13a
3, 4, 6, 7, 11, 21, 24, 43, 46, 56, 57
3. Show that a commutative ring with the cancelation property (under multiplication) has no zerodivisors.
Let a 6= 0, and suppose that ab = 0 for some b in the ring. Since we are in a ring (Theorem
12.1), a0 = 0. So we have ab = a0, and by the cancelation property we see that b = 0. Thus
there are no zero-divisors in the ring.
4. List all zero-divisors in Z20 . Can you see a relationship between the zero-divisors of Z20 and the
units of Z20 ?
The zero-divisors are elements a 6= 0 that satisfy ab = 0 for some b 6= 0. So we need a to have a
common factor with n = 20, but not be equal to 20 (≡ 0). Thus the zero-divisors are all non-zero
elements that are not relatively prime to 20: 2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18.
Here are specific examples for each of these zero-divisors:
2 · 10 = 20 ≡ 0
4 · 5 = 20 ≡ 0
6 · 10 = 60 ≡ 0
8 · 5 = 40 ≡ 0
12 · 5 = 60 ≡ 0
14 · 10 = 140 ≡ 0
15 · 4 = 60 ≡ 0
16 · 5 = 80 ≡ 0
18 · 10 = 180 ≡ 0
The zero-divisors in Z20 are all the non-zero elements that are not units.
(U (n) = {1, 3, 7, 9, 11, 13, 17, 19}).
6. Find a nonzero element in a ring that is neither a zero-divisor nor a unit.
There are lots of options, but the easiest that comes to mind for me is 2 ∈ Z. The integers have
no zero-divisors, and 2 is not a unit in the integers. To be more careful about this, suppose k ∈ Z
and consider 2k. If k ≥ 1 then 2k ≥ 2. If k ≤ 0 then 2k ≤ 0. Thus 2k is never equal to 1 if k ∈ Z.
7. Let R be a finite commutative ring with unity. Prove that every nonzero element of R is either
a zero-divisor or a unit. What happens if we drop the “finite” condition on R?
Proof A: This will follow the proof of Theorem 13.2 with a few modifications. Let a ∈ R be
non-zero and suppose that a is not a zero-divisor. First I will prove the cancelation property just
for a. If ab = ac, then ab − ac = 0 and a(b − c) = 0. Since a is not a zero-divisor, then b − c = 0
so b = c. (This is the same as the proof for Theorem 13.1, except there it was for every a ∈ R,
but here we just have a special a that we know is not a zero-divisor.)
Consider the set {an | n ∈ N} = {a, a2 , a3 , . . . }. Since R is finite, we must have ai = aj for some
i, j with i > j. Then since we have the cancelation property for a and we have ai−j aj = 1 · aj
(remember we have unity), then cancelation gives us ai−j = 1. If a = 1 then a is clearly a unit
since 1−1 = 1. If a 6= 1 then i − j > 1 so we can factor out one copy of a to get ai−j−1 a1 = 1.
Thus the element ai−j−1 is the multiplicative inverse of a, so a is a unit.
Thus every nonzero element of this ring that is not a zero-divisor is a unit. In other words, every
nonzero element is either a zero-divisor or a unit.
If we drop the finite condition then the result does not hold true. For example, Z is a commutative ring with unity, but 2 is neither a zero-divisor nor a unit.
Proof B: You can also follow the outline in the back of the book for this problem. The main
thing to realize is that the finite condition is important in the second line of their hint. This
comes from the fact that if R is a finite set and f : R → R is a function, then f is 1-1 if and only
if f is onto (see problem 12 in chapter 5). Thus if f (r) = sr, we see that the condition S 6= R is
the same as f is not onto, so that f is not 1-1, which is the fact that they use.
√
√
11. Let d be an integer. Prove that Z[ d] = {a + b d | a, b ∈ Z} is an integral domain.
√
First note that R = Z[ d] is a subset of the complex numbers. Thus we use the subring test to
show that R is a ring.
√
√
√
d
∈
R.
Let
(a
+
b
d),
(f
+
g
d) ∈ R, so a, b, f, g ∈ Z.
Note that R√is nonempty
since
0
=
0
+
0
√
√
)+(b−g)
d
and
since
(a−f
),
(b−g)
∈
Z so R is closed under
Then (a+b d)−(f +g √d) = (a−f
√
√
√
√ 2
√
subtraction. Also (a + b d)(f + g d) = af + ag d + bf d + bg( d) = (af + bgd) + (ag + bf ) d
and since (af + bgd), (ag + bf ) ∈ Z, we see that R is closed under multiplication. Thus by the
subring test R is a ring.
√
R has unity since 1 = 1 + 0 d ∈ R. And since R ⊂ C, we know that R has no zero-divisors.
Thus R is an integral domain.
21. Let R be the ring of real-valued continuous functions on [−1, 1]. Show that R has zero-divisors.
(
(
0
if x < 12
x + 12 if x < − 12
and g(x) =
Let f, g ∈ R be defined by f (x) =
x − 12 if x ≥ 21
0
if x ≥ − 21
Note that both f, g are continuous functions and that neither is the zero function z(x) = 0 which
is the additive identity 0 in the ring R. Consider f g. We see that (f g)(x) = f (x)g(x), and if
x < 21 then f (x) = 0, and if x > − 12 then g(x) = 0 so either way (f g)(x) = 0 and thus f g = z
and R has zero-divisors.
24. Find a zero-divisor in Z5 [i] = {a + bi | a, b ∈ Z5 }.
Suppose (a + bi)(c + di) = 0. Then (ac − bd) + i(ad + bc) = 0 so ac − bd = 0 and ad + bc = 0.
Since we are just asked to find a zero-divisor (instead of trying to characterize all zero-divisors),
we’ll just plug some numbers in and see if we can make it work. Remember that we are working
mod 5.
If a = 2, b = 3 then we need 2c − 3d = 0 and 2d + 3c = 0. Adding the equations gives 5c − d = 0
and since 5 = 0 then −d = 0 so d = 0. Plugging this in we get 2c = 0, but in Z5 the only solution
to this is c = 0 which makes c + di = 0 so we haven’t found a zero-divisor. So we’ll continue
trying other options.
If a = 2, b = 4 then we have 2c − 4d = 0 and 2d + 4c = 0. Then 6c − 2d = 0 or c − 2d = 0 so
c = 2d. So let’s pick c = 2, d = 1. Then (2 + 4i)(2 + i) = 4 + 2i + 8i + 4i2 = 4 + 10i − 4 = 10i = 0.
Since neither element equals 0, then we see that (2 + 4i) is a zero-divisor. And so is (2 + i).
Note: As far as a complete answer goes, only the last line or two is necessary. I just wanted to
include some scratch work so you could see how to go about it to figure out the answer (not just
showing that the two elements multiply to 0).
43. The nonzero elements of Z3 [i] form an Abelian group of order 8 under multiplication. Is it isomorphic to Z8 , Z4 ⊕ Z2 , or Z2 ⊕ Z2 ⊕ Z2 ?
Recall that these are all Abelian groups of order 8 (up to isomorphism) by the Fundamental
Theorem of Finite Abelian Groups. We will check the orders of elements and compare. By
looking at the table on page 258, we see that Z3 [i] has only 2 elements of order ≤ 2, namely 1
and 2 (These elements squared are the identity 1). The group Z2 ⊕ Z2 ⊕ Z2 has more elements
of order 2, such as (1, 0, 0), (1, 1, 0), (1, 1, 1) to name a few. So our group is not Z2 ⊕ Z2 ⊕ Z2 .
The group Z4 ⊕ Z2 also has more elements of order 2 such as (2, 0), (0, 1), (2, 1), so that’s not our
group either.
Therefore the group of units of Z3 [i] is isomorphic to Z8 .
46. Suppose that a and b belong to a commutative ring and ab is a zero-divisor. Show that either a
or b is a zero-divisor.
Since ab is a zero-divisor, we know that ab 6= 0 so neither a nor b equals 0. Also, there exists an
element c 6= 0 such that (ab)c = 0. If bc = 0, then b is a zero-divisor since neither b nor c is 0. If
bc 6= 0, then since a(bc) = 0 we see that a is a zero-divisor.
Thus either a or b is a zero-divisor. (or both, of course)
56. Find all solutions of x2 − x + 2 = 0 over Z3 [i].
Since Z3 [i] is a field, it is an integral domain (see the first paragraph of page 257) so if we can
factor the polynomial we can find all the solutions that way. Unfortunately it looks like it won’t
factor in an obvious way, so we’ll just plug in different values and see when we get 0. Remember
we are working mod 3.
x
0
1
2
i
1+i
2+i
2i
1 + 2i
2 + 2i
x2 − x + 2
0−0+2=2
1−1+2=2
4−2+2=1
−1 − i + 2 = 1 − i
(1 + i)2 − (1 + i) + 2 = 1 + 2i + i2 − 1 − i + 2 = 1 + i
(2 + i)2 − (2 + i) + 2 = 4 + 4i + i2 − 2 − i + 2 = 0 + 0i = 0
(2i)2 − (2i) + 2 = −4 − 2i + 2 = 1 + i
(1 + 2i)2 − (1 + 2i) + 2 = 1 + 4i + 4i2 − 1 − 2i + 2 = 1 + 2i
(2 + 2i)2 − (2 + 2i) + 2 = 4 + 8i + 4i2 − 2 − 2i + 2 = 0 + 0i = 0
So the solutions are x = 2 + i and x = 2 + 2i.
Apparently this polynomial factors (over Z3 [i]) as (x − (2 + i))(x − (2 + 2i)) = x2 − (2 + 2i)x −
(2 + i)x + (2 + i)(2 + 2i) = x2 − x(4 + 3i) + 4 + 6i + 2i2 = x2 − x + 2.
57. Consider the equation x2 − 5x + 6 = 0.
(a) How many solutions does this equation have in Z7 ?
Since 7 is prime, Z7 is a field and therefore an integral domain, so we can find all solutions
by factoring. x2 − 5x + 6 = (x − 2)(x − 3) so that there are 2 solutions, namely x = 2, 3.
(b) Find all solutions of this equation in Z8 .
Since 8 is not prime, Z8 is not an integral domain so we cannot just factor to find all
solutions. We do know that x = 2, 3 will still be solutions but we must check all values to
see if there are others.
x x2 − 5x + 6
0 0−0+6=6
1 1−5+6=2
2 4 − 10 + 6 = 0
3 9 − 15 + 6 = 0
4 16 − 20 + 6 = 2
5 25 − 25 + 6 = 6
6 36 − 30 + 6 = 12 = 4
7 49 − 35 + 6 = 20 = 4
So the solutions in Z8 are x = 2, 3.
(c) Find all solutions of this equation in Z12 .
x x2 − 5x + 6
0 0−0+6=6
1 1−5+6=2
2 4 − 10 + 6 = 0
3 9 − 15 + 6 = 0
4 16 − 20 + 6 = 2
5 25 − 25 + 6 = 6
6 36 − 30 + 6 = 12 = 0
7 49 − 35 + 6 = 20 = 8
8 64 − 40 + 6 = 30 = 6
9 81 − 45 + 6 = 42 = 6
10 100 − 50 + 6 = 56 = 6
11 121 − 55 + 6 = 72 = 0
So the solutions in Z12 are x = 2, 3, 6, 11.
(d) Find all solutions of this equation in Z14 .
x x2 − 5x + 6
0 0−0+6=6
1 1−5+6=2
2 4 − 10 + 6 = 0
3 9 − 15 + 6 = 0
4 16 − 20 + 6 = 2
5 25 − 25 + 6 = 6
6 36 − 30 + 6 = 12
7 49 − 35 + 6 = 20 = 6
8 64 − 40 + 6 = 30 = 2
9 81 − 45 + 6 = 42 = 0
10 100 − 50 + 6 = 56 = 0
11 121 − 55 + 6 = 72 = 2
12 144 − 60 + 6 = 90 = 6
13 169 − 65 + 6 = 110 = 12
So the solutions in Z12 are x = 2, 3, 9, 10.