Download MATH 103B Homework 2 - Solutions Due April 12, 2013

MATH 103B Homework 2 - Solutions
Due April 12, 2013
Version April 12, 2013
Assigned reading: Chapters 12-13 of Gallian.
Recommended practice questions: Chapter 12 of Gallian, exercises
30, 31, 32, 33
Chapter 13 of Gallian, exercises
1, 2, 5, 6, 9, 15, 18
Assigned questions to hand in:
(1) (Gallian Chapter 12 # 43) Let R Z Z Z and S
or disprove that S is a subring of R.
a, b, c R : a b
c. Prove
Solution: We will prove that S is not a subring of R. Recall that the operations on the
external direct product are componentwise. Consider 1, 0, 1 and 0, 1, 1, both of which
are elements of S. Then multiplication gives
1, 0, 10, 1, 1 0, 0, 1 S
since 0 0 1. Thus, S is not closed under multiplication and hence is not a subring of R.
(2) (Gallian Chapter 12 #44, special case) Suppose that a2
ring. Show that a a for all a in the ring.
a for all elements a of some
Solution: Let a be an element of the ring. Then so is a a. By assumption,
aa
a a2 a aa a
a2 a2 a2 a2
Using cancellation for addition in the ring:
0
a a,
that is,
a
a a a a.
a.
Alternate solution: Let a be an element in the ring. Then so is
a a2 a
a Thm 12.1.3 aa
a2
a. By assumption,
a.
This solution generalizes to prove the bonus once a general version of Theorem 12.1.3. is
proved.
(3) (Gallian Chapter 13 #10) Describe all zero-divisors and units of Z Q Z.
a, b, c Z Q Z, x is a unit if and only if
Solution: We will prove that for x
a, c 1, 1 and b 0, and x is a zero divisor if and only if at least one of a, b, c is 0
(and not all of them are).
First, we prove that 1, b, 1 is a unit (for any choice of b 0) by finding its
multiplicative inverse. Note that the unity of this ring is 1, 1, 1. It is easy to verify that
1, b, 11 1, 1b , 1, 1, b, 11 1, 1b , 1,
1, b, 11 1, 1b , 1,
1, b, 11 1, 1b , 1.
We will show that if x is not of this form, it is not a unit:
Suppose x n, b, c with n 1, 1. Then n is not a unit in Z and so there is no
m Z such that nm 1. But, if y m, d, e is the inverse of x in Z Q Z, then
xy 1, 1, 1 so in particular nm 1. Thus, no such y can exist.
Suppose x a, 0, c. Then for any y n, d, e, the second component of xy is 0 so
xy 1, 1, 1. Thus, x has no multiplicative inverse and is not a unit.
Similar to the first case.
a, 0, 0 for a 0. Then y 0, 1, 1 0, 0, 0 but xy 0, 0, 0.
Consider x
Similarly, if x has at least one zero component then we can find a nonzero element of
Z Q Z whose product with x is 0, 0, 0. Conversely, suppose x is a zero-divisor in
this ring. Then since multiplication is component-wise and Z and Q are each integral
domains, x must have at least one zero component.
(4) (Gallian Chapter 13 #28) Let R be the set of all real-valued functions defined for all real
numbers under function addition and multiplication.
(a) Determine all zero-divisors of R.
(c) Show that every nonzero elements is a zero-divisor or a unit.
Solution:
(a) A function f : R R is a zero-divisor of R if and only if there is some x0 R such
0 if x x0
and notice that f g 0.
that f x 0. In this case, consider g x
1 if x x0
If f has no zeroes then any choice of g such that f g 0 would need f xg x 0
for each x, but f x R and R is a field so no appropriate g x exists.
(c) Suppose f is a nonzero function and is not a zero-divisor. By part (a), that means
that f x 0 for each x R. We will prove that f is a unit. Consider the
1
function g x
, a well-defined real-valued function by our assumption. Then,
f x
f gx
f xg x
Thus, f is a unit.
f x
f x
1 for all x R and g is the multiplicative inverse of f .
(5) (Gallian Chapter 13 # 54) Let R be a ring with m elements. Show that the characteristic
of R divides m.
Solution: Let n be the characteristic of R. That is, it is the smallest positive integer
such that na 0 for each a R. Since R is a finite group (under addition), Lagrange’s
theorem says that the order of each element of R divides m. In particular, for each a R,
0.
ma
By the division theorem, let q
For each a R,
0
ma
Z and 0 r n be such that
m nq r.
nq ra nqa ra
q na ra
2
q 0 ra
0 ra
ra.
Thus, the rth multiple of each ring element vanishes. By definition of the characteristic
and since r n, r cannot be positive. Therefore, r 0 and n divides m.
(6) (Gallian Chapter 13 #32) Let R
ulo 10. Prove that R is a field.
0, 2, 4, 6, 8 under addition and multiplication mod-
Solution: Addition and multiplication modulo 10 are associative and commutative operations, which also satisfy distributivity. Thus, it remains to prove the following:
R is closed under 10 , 10:
10 0 2 4 6 8
10 0 2 4 6 8
0 0 2 4 6 8
0 0 0 0 0 0
2 2 4 6 8 0
2 0 4 8 2 6
4 4 6 8 0 2
4 0 8 6 4 2
6 6 8 0 2 4
6 0 2 4 6 8
8 8 0 2 4 6
8 0 6 2 8 4
R has a unity: 6 acts as the unity of this ring.
Every nonzero element is a unit:
21
41
8,
61
4,
81
6,
2.
(7) (Gallian Chapter 13 #60) In a commutative ring of characteristic 2, prove that the idempotents form a subring. Recall (from question 18) that a is idempotent if a2 a.
Solution: We use the Subring Test.
Nonempty? The additive identity 0 is always idempotent, 0 0 0.
Closure under subtraction? Suppose a2 a and b2 b. Consider
a b2 a ba b
a2 ba a
b b2
Thm 12.1
But, in a ring of characteristic 2, x x
a b
2
ab
a2 ab ab b2
0 for each x, so x
a b
a
b
and we have proved that a b is idempotent.
Closure under multiplication? Suppose a2 a and b2
ab2 Commutativity a2b2
and we have proved that ab is idempotent.
3
ab,
Char 2
a
0b
a b.
x for each x. Thus,
b. Consider