Download Homework 2 January 19, 2006 Math 522 Direction: This homework

Homework 2
January 19, 2006
Math 522
Direction: This homework is due on January 26, 2006. In order to receive full credit, answer
each problem completely and must show all work.
1. What is the set of the units (that is, the multiplicative inverses) of the ring (Z10 , +, ·)? Call
this set U (10). Show that U (10) is a group under multiplication modulo 10.
Answer: The elements of Z10 that have multiplicative inverse are: 1, 3, 7, 9. Hence we have
U (10) = {1, 3, 7, 9}. This set U (10) forms a group under multiplication modulo 10 since
(a) identity in U (10) is 1,
(b) each element of U (10) has an inverse: 1−1 = 1,
3−1 = 7,
7−1 = 3,
9−1 = 1,
(c) the binary operation multiplication mudulo 10 is associative.
2. Let x and y be two elements of the commutative ring with unity (R, +, ·). By induction
n X
n n−k k
show that the binomial theorem (x + y)n =
x
y holds on this commutative ring
k
k=0
(R, +, ·).
Answer: Let n ∈ N be a natural number. The binomial theorem says that
(x + y)n =
n X
n
k=0
k
xn−k y k =
n n
n n−1
n n
x +
x
y + ··· +
y .
0
1
n
We want to prove the above statement by induction on n. Since
(x + y)n+1
= (x + y)(x + y)n
n n
n n−1
n n
= (x + y)
x +
x
y + ··· +
y
0
1
n
n n+1
n n
n
n n
n n−1 2
n n+1
=
x
+
x y + ··· +
xy n +
x y+
x
y + ··· +
y
0
1
n
0
1
n
n
n
n
n
n n+1
n n+1
x
+
+
xn y + · · · +
+
xn+1 y +
y
=
0
1
n−1
n
n
0
n + 1 n+1
n+1 n
n + 1 n+1
n + 1 n+1
=
x
+
x y + ··· +
x
y+
y
,
0
1
n
n+1
the formula holds for all positive integers n. The last line of the above formula follows from the
facts
n
n+1
n
n+1
n
n
n+1
=
,
=
, and
+
=
0
0
n
n+1
k−1
k
k
for all k = 1, 2, 3, ..., n.
Remark: If the ring is not commutative, then xy 6= yx, and we can not have such a binomial
theorem for noncommutative rings.
3. An element x of a ring (R, +, ·) is said to be a nilpotent element if there exists a positive
integer n such that xn = 0. Show that if x is a nilpotent element of the ring (R, +, ·) with
unity 1, then 1 − x and 1 + x both have multiplicative inverse. [Hint: Use the factorization
1 − xn = (1 − x) 1 + x + x2 + · · · + xn−1 for 1 − x and a similar formula for x + 1.]
Answer: Since x is a nilpotent element of the ring (R, +, ·), we have
xn = 0
for some positive integer n. We want to show that 1 − x has a multiplicative inverse. From the
identity 1 − xn = (1 − x) 1 + x + x2 + · · · + xn−1 , using xn = 0 we get
1 = (1 − x) 1 + x + x2 + · · · + xn−1 .
Therefore (1 − x) has a multiplicative inverse and it is given by 1 + x + x2 + · · · + xn−1 . Similarly
from the identity
1 + (−1)n−1 xn = (1 + x) 1 − x + x2 − x3 + · · · + (−1)n−1 xn−1 ,
we get 1 = (1 + x) 1 − x + x2 − x3 + · · · + (−1)n−1 xn−1 , since xn = 0. Therefore, (1 + x)
has an inverse.
4. Using binomial theorem show that if x and y are two nilpotent elements in the commutative
ring with unity (R, +, ·), then x + y is also a nilpotent element in the ring (R, +, ·).
Answer: Since x and y are nilpotent elements, therefore there exist positive integers m and n
such that xm = 0 and y n = 0. We want to show that x + y is nilpotent. That is, we want to
find a positive integer k such that (x + y)k = 0. We guess this k to be m + n and compute
(x + y)m+n
m + n m+n
m + n m+n−1
m + n m+n
=
x
+
x
y + ··· +
y
0
1
m+n
m+n m n
m + n m n−1
m+n m n
=
x x +
x x
y + ··· +
y y
0
1
m+n
= 0 + 0 + ··· + 0
=0
Hence x + y is nilpotent.
5. Find two zero-divisors in the ring (M2 (Z), +, ·) of two-by-two matrices with integer entries.
Answer: Two zero-divisors in the ring (M2 (Z), +, ·) are
1 1
0 0
and
0 −1
0
1
since
1
0
1
0
0 −1
0
1
=
0
0
0
0
0
0
0
0
or
0 −1
0
1
1
0
1
0
=
.
Remark: Note that (M2 (Z), +, ·) is a noncommutative ring. If X and Y are two elements
in M2 (Z), then XY 6= Y X. I have selected the above matrices in such a way that no matter
which way you multiply them you will get a zero matrix.
6. Let Z be the set of integers. Let P(Z) be the power set of Z, that is the collection of all
subsets of Z. Let ⊕ and be two binary operations on P(Z) defined as X ⊕Y = (X ∪Y )\(X ∩Y )
and X Y = X ∩ Y . Show that the algebraic system (P(Z), ⊕, ) is a ring. What is the
characteristic of this ring?
Answer: Let Z be the set of integers and P(Z) be the set of all subsets of Z. We want to
show (P(Z), ⊕, ) is a ring. That is (P(Z), ⊕) is an abelian group, the binary operation is
associative, and distributive over ⊕ on the set P(Z).
(i) We show P(Z) is closed under ⊕. Let X and Y in P(Z). Then X ⊕ Y = (X ∪ Y ) \ (X ∩ Y ) ∈
P(Z). Hence X ⊕ Y belongs to P(Z) whenever X and Y are in P(Z). Thus P(Z) is closed
under ⊕.
(ii) Next we show ∅ is the additive identity. This is true because
X ⊕ ∅ = (X ∪ ∅) \ (X ∩ ∅) = X \ ∅ = X.
Similarly ∅ ⊕ X = X. Hence ∅ serves as the identity in P(Z).
(iii) X is the inverse of itself in (P(Z), ⊕) since
X ⊕ X = (X ∪ X) \ (X ∩ X) = X \ X = ∅.
Hence each set in (P(Z)) has an inverse with respect to ⊕.
(iv) Next we show ⊕ is associative. Let X, Y , W be three sets in P(Z). We have to show that
(X ⊕ Y ) ⊕ W = X ⊕ (Y ⊕ W ).
(You have to do a lot of book keeping for this part. I leave that to you.)
(iv) Next we show ⊕ is commutative. Since for any three sets X, Y , W in P(Z)
X ⊕ Y = (X ∪ Y ) \ (X ∩ Y ) = (Y ∪ X) \ (Y ∩ X) = Y ⊕ X
the binary operation ⊕ is commutative.
Hence we have shown that (P(Z), ⊕) is an abelian group.
Next we show that the binary operation is associative. Let X, Y , W be three sets in
the set P(Z). Since
(X Y ) W = (X ∩ Y ) W = X ∩ Y ∩ W = X ∩ (Y ∩ W ) = X (Y ∩ W ) = X (Y W )
the binary operation is associative.
Finally, we show the binary operation is distributive over ⊕. Since for any three sets X,
Y , W in P(Z), we have
X [Y ⊕ W ] = X ∩ [(Y ∪ W ) \ (Y ∩ W )]
= X ∩ (Y ∪ W ) \ X ∩ (Y ∩ W )
= (X ∩ Y ) ∪ (X ∩ W ) \ (X ∩ Y ) ∩ (X ∩ W )
= (X ∩ Y ) ⊕ (X ∩ W )
= (X Y ) ⊕ (X W ),
the binary operation is left distributive over ⊕ on the set P(Z). Similarly, it can be show
that the binary operation is right distributive over ⊕ on the set P(Z). Hence (P(Z), ⊕, ) is
a ring.
The characteristic of this ring is 2 since for any X ∈ P(Z), we see that X ⊕ X = ∅ which
is 2X = ∅.
7. Give an example of a finite field (F, +, ·) that have two nonzero elements a and b such that
a2 + b2 = 0.
Answer: Consider the field (Z5 , +, ·). Let a = 3 and b = 4. Then a2 + b2 = 9 + 16 =
25 mod 5 = 0.
8. Let x and y be any two elements in the integral domain (Z7 , +, ·) whose characteristic is 7.
Prove that (x + y)7 = x7 + y 7 .
Answer: Using the binomial theorem, we see that
7 7
7 6
7 5 2
7
7 7
(x + y)7 =
x +
x y+
x y + ··· +
xy 6 +
y
0
1
2
6
7
= x7 + 7x6 y + (3)7x5 y 2 + · · · + 7xy 6 + y 7
= x7 + 0 + 0 + · · · + 0 + y 7
= x7 + y 7 .
9. The algebraic structure (Z3 ⊕ Z6 , +, ·), where the binary operations + and · are defined as
(a, b) + (c, d) = (a + c, b + d) and (a, b) · (c, d) = (ac, bd), forms a ring (similar to the external
direct product of groups you have seen in chapter seven). What is the characteristic of this ring
(Z3 ⊕ Z6 , +, ·)?
Answer: Char(Z3 ⊕ Z6 ) = lcm(3, 6) = 6.
10. Prove that the ring of Gaussian integers (Z[i], +, ·), where Z[i] = { a + bi | a, b ∈ Z }, has no
zero-divisors.
Answer: Suppose (Z[i], +, ·) has a zero-divisor, say a + bi. Since a + bi is a zero divisor, there
exists a nonzero element c + di such that
(a + bi)(c + di) = 0
which is
ac − bd + (bc + ad)i = 0.
Therefore
ac − bd = 0
bc + ad = 0
which is
a −b
b
a
c
0
=
.
d
0
a −b
Since det
6= 0, solving the above equation, we get c = 0 and d = 0 and this is a
b
a
contradiction to the fact that c + di is a nonzero element.