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Answers and Solutions to Selected Homework Problems From
Section 2.5
S. F. Ellermeyer
5. Since gcd (2; 4) 6= 1, then 2 is a zero divisor (and not a unit) in Z4 .
In fact, we see that 2 2 = 0 in Z4 . Thus 2x = 2y in Z4 does not
imply that x = y. In particular, we can write the equation 2x = 2y
as 2 (x + ( y)) = 0 and we know that any choice of x and y such that
x + ( y) = 2 will make this equation be true. Thus suppose that x = 3
and y = 1 (meaning that x + ( y) = 2 is true). Then 2x = 2 3 = 2
and 2y = 2 1 = 2 so 2x = 2y, but x 6= y.
In Z5 , all non–zero elements (including 2) are units. In particular,
3 2 = 1 in Z5 . Thus if we have 2x = 2y, we can multiply both sides
of this equation by 3 to obtain 3 (2x) = 2 (2y) which gives (3 2) x =
(3 2) y which gives 1 x = 1 y which gives x = y.
By similar reasoning, since 2 is a zero divisor in Z20 , we see that 2x = 2y
does not imply that x = y. For example, suppose that x = 6 and
y = 16. Then 2x = 2 6 = 12 and 2y = 2 16 = 12 so 2x = 2y but
x 6= y.
In Z15 , 2 is a unit and 8 2 = 1. Thus 2x = 2y does imply that x = y.
6. In the ring M2;2 (R) (which is a non–commutative ring) suppose we
take
1 0
0 5
A=
and B =
.
0
2
0
2
Then
(A + B)2 =
1
0
5
4
1
0
5
4
1
0
0
2
0
0
=
1
0
15
16
and
A2 + 2AB + B 2
=
1
0
0
2
=
1 0
0 4
=
1 0
0 16
1
0
+
0
2
0 10
0 8
+2
+
0
0
10
4
1
5
2
+
0
0
5
2
0
0
5
2
which shows that (A + B)2 6= A2 +2AB+B 2 . Using the ring properties,
we see that if x and y are any two members of a ring then
(x + y)2 = (x + y) (x + y)
= (x + y) x + (x + y) y
= x2 + yx + xy + y 2 .
(Note that yx need not be equal to xy so we can’t replace the yx + xy
with 2xy.)
Returning to our example with matrices, note that
A2 + BA + AB + B 2
=
1 0
0 4
+
0
0
5
2
=
1 0
0 4
+
0
0
10
4
=
1
0
1
0
0
2
+
0 5
0 4
+
+
1
0
0
2
0
0
0
0
5
2
+
0
0
10
4
10
4
15
16
which shows that (A + B)2 = A2 + BA + AB + B 2 .
11. Let us try plugging each member of Z5 into the expression x2 + x + 4:
02 + 0 + 4 = 4 6= 0
12 + 1 + 4 = 1 6= 0
22 + 2 + 4 = 0
33 + 3 + 4 = 1 6= 0
42 + 4 + 4 = 4 6= 0
so we see that x = 2 does satisfy the equation x2 + x + 4 = 0 in Z5 .
2
Now we will try this in Z7 :
02 + 0 + 4 = 4 6= 0
12 + 1 + 4 = 6 6= 0
22 + 2 + 4 = 3 6= 0
32 + 3 + 4 = 2 6= 0
42 + 4 + 4 = 3 6= 0
52 + 5 + 4 = 6 6= 0
62 + 6 + 4 = 4 6= 0
showing that x2 + x + 4 = 0 does not have any solutions in Z7 .
12. The units of M2;2 (R) are those matrices, A 2 M2;2 (R) for which there
exists a matrix, B, such that AB = BA = I. These are precisely the
matrices that are invertible. In Linear Algebra, there are many criteria
that can be used to determine whether or not a matrix is invertible. One
such criteria is that a matrix, A, is invertible if and only if det (A) 6= 0.
p
13. Suppose that x = a + b 5 where a and b are rational numbers and
suppose that x 6= 0. (This implies that either a 6= 0 or b 6= 0.) If b = 0,
then x = a 6= 0 and it is obvious that the multiplicative inverse of
p x is
1=a (which is also a rational number and also a member of Q 5 ).
p
Thus, let us suppose that b 6= 0. If a = 0, then x = b 5 and we can
see that the multiplicative inverse of x is
p
1
1p
p =
52Q
5 .
5b
b 5
We are left to consider the case that b 6= 0 and a 6= 0. In this case, the
multiplicative inverse of x is
p
1
a b 5
1
p
p =
p
a+b 5
a+b 5 a b 5
p
a b 5
= 2
a
5b2
p
a
b
= 2
+
5.
a
5b2
a2 5b2
3
The above computations are legitimate as long as we are not dividing
by 0 in any of these computations. To see that we are not dividing
by 0, note that, since we are
p assuming that a 6= 0pand b 6= 0, then
it must be true
p that a + b 5 6= 0 because if a + bp 5 = 0, then we
would have 5 = a=b which is impossible because 5 is an irrational
number and
p a=b is a rational number. By similar reasoning, we see
that a b 5 6= 0.
14. We can see that Z3 [i] has exactly nine members. They are 0, i, 2i, 1,
1 + i, 1 + 2i, 2, 2 + i, and 2 + 2i.
To make sure that Z3 [i] is closed under addition and multiplication,
we take two members of Z3 [i], x = a + bi and y = c + di (where a; b;
c; and d are elements of Z3 ) and observe that
x + y = (a + bi) + (c + di) = (a + c) + (b + d) i 2 Z3 [i]
and
xy = (a + bi) (c + di)
= ac + adi + bci + bdi2
= (ac bd) + (ad + bc) i 2 Z3 [i] .
The additive identity of Z3 [i] is 0 and the multiplicative identity is 1.
It is clear that addition and multiplication in Z3 [i] are both associative
and commutative. Also, if x = a + bi; y = c + di, and z = e + f i are
members of Z3 [i], then
x (y + z) = (a + bi) ((c + e) + (d + f ) i)
= a (c + e) + a (d + f ) i + b (c + e) i + b (d + f ) i2
= (ac + ae bd bf ) + (ad + af + bc + be) i
= (ac bd) + (ad + bc) i + (ae bf ) + (af + be) i
= (a + bi) (c + di) + (a + bi) (e + f i)
= xy + xz
which shows that the distributive law holds. Thus Z3 [i] is a commutative ring with unity. To see that every non–zero member of Z3 [i] is a
unit, we observe that
i 2i = 1i 2i = (1 2) i2 =
4
2=1
so i and 2i are multiplicative inverses of each other. Likewise
1 1=1
(1 + i) (2 + i) = 1
(1 + 2i) (2 + 2i) = 1
2 2 = 1.
This shows that Z3 [i] is a …eld.
Since the multiplicative inverse of 2+i is 1+i, we can solve the equation
(2 + i) x = 1 + 2i
by multiplying both sides of this equation by 1 + i to obtain
(1 + i) (2 + i) x = (1 + i) (1 + 2i)
which gives
x = 2.
To see that this solution is correct, observe that
(2 + i) (2) = 1 + 2i.
15. It is clear that Z5 [i] is a commutative ring with unity. We ask "Does
Z5 [i] have any zero divisors?" The answer is yes because
(2 + i) (3 + i) = 0
in Z5 [i]. Therefore Z5 [i] is not an integral domain.
16. In an integral domain, the equation x2 = x can be written as x x +
( 1 x) = 0 or as (x + ( 1)) x = 0. Since an integral domain has no
zero divisors, it must be true that either x + ( 1) = 0 or that x = 0.
The solution of x + ( 1) = 0 is x = 1. Thus x = 0 and x = 1 are the
only two solutions of x2 = x. Note: If 1 = 0, which is really not ruled
out by the de…nitions given in the textbook, then x2 = x has only one
solution. However, this can happen only in the case of a one element
…eld.
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