MATH 103B Homework 1 - Solutions
Due April 5, 2013
Version April 3, 2013
Assigned reading: Chapter 12 of Gallian.
Recommended practice questions: Chapter 12 of Gallian, exercises
1, 2, 4, 7, 15, 23, 24
Assigned questions to hand in:
(1) (Gallian Chapter 12 # 3) Give an example of a subset of a ring that is a subgroup under
addition but not a subring. Prove that your example works.
Solution (using hint in back of Gallian): In R, consider
n 2 : n Z
A
We will prove that this example is a subgroup under addition but not a subring. We use
the one-step subgroup test to prove that this (nonempty, because includes 0 for example)
subset of Ris a subgroupunder addition. Let x, y A. Then there are m, n Z such
that x m 2 and y n 2. Computing
xy
m 2n 2
m n
2A
because m n Z. We will prove that A isnot a subring of R by showing that it is not
2 A (using n 1). Then
closed under multiplication. Consider x
x2
2 2
2.
But Z A since 2 is not rational. Therefore, since x2
shown that A is not closed under multiplication.
Z, x2 A and we have
Alternate solution: Consider the ring M2 Z of 2 2 matrices with integer entries. Consider the subset
a a
B
: a, b Z .
0 b
Then B M2 Z and is nonempty. (For example, it contains the 0 matrix when we
choose a b 0.) To confirm that B is a subgroup under addition, we use the one-step
subgroup test: let X, Y B so there are a, b, a , b Z such that
a a
0 b
X
Then
½
Y
½
a a
0 b
½
½
½
.
aa aa
B
X Y
0
bb
because a a , b b Z. But, B is not a subring because we will show that it is not
closed under multiplication: for example,
½
½
½
½
½
2 2
0 1
1 1
0 1
2 4
,
0 1
which is not in B since 2 4. Thus, B is a subgroup of M2 Z under addition but is not
a subring.
(2) (Gallian Chapter 12 # 5) Prove Theorem 12.2. Namely: prove that if a ring has a unity,
it is unique and that if a ring element has a multiplicative inverse, it is unique.
Solution: Let R be a ring with identity 0. Suppose x, y
of being a unity. That is, for each r R,
xr
In particular, choosing r
rx
r
ry
r.
y and using the first equation, we have
y
xy
whereas choosing r
yr
and
R both satisfy the properties
x and using the second equation, we have
x.
xy
Thus, y x and so if there is a unity of R, it is unique. We denote such a unity by 1.
Let r R and suppose that x, y R both satisfy the properties of being multiplicative
inverses of r. That is,
rx
We want to show that x
xr
1
ry
and
yr
1.
y. Using the equations from the definitions,
xr
1
yr.
But, r has a multiplication inverse, namely x. So, multiply each side of this equation by
x:
xrx yrx.
By the fact that rx
x
Thus, x
1 and using associativity, we get the chain of equations:
x1
xrx
xrx
yrx
y rx
y1
y.
y and the multiplicative inverse of r is unique.
(3) (Gallian Chapter 12 # 6) Find an integer n that shows that the rings Zn need not have
the following properties (that the ring of integers Z does have)
(a) a2 a implies a 0 or a 1.
(b) ab 0 implies a 0 or b 0.
(c) ab ac and a 0 implies b c.
Is the n you found prime?
Solution: We will use n 6 for all of these. Namely, consider Z6 0, 1, 2, 3, 4, 5.
Recall that to prove that an implication “P implies Q” fails, we need to find an example
where P is true and Q is false.
(a) Consider a 3. Then a2 9 mod 6 3 a but 3 0 and 3 1.
(b) Consider a 2 and b 3. Then ab 2 3 mod 6 6 mod 6 0 but 2 0 and
3 0.
(c) Consider a
2, b
4, c
1. Then ab
2 4 mod 6
8 mod 6
2
21
mod 6 ac and 2 0, but 4 1.
2
(4) (Gallian Chapter 12 # 9) Prove that the intersection of any collection of subrings of a
ring R is a subring of R.
Solution: Let S be the intersection of a collection of subrings a ring R: S
i I Si where
I is some index set and each Si is a subring of R. We use the subring test to prove that
S is a subring of R. Recall that it suffices to prove that S satisfies the following three
properties:
S
a, b S a b S .
a, b S ab S .
We will prove each of these in turn. The following claim is useful:
È
Claim. If S is a subring of R then 0 S
Proof. If S is a subring of R then it is nonempty. Let x S. Then, since S is a ring and
by closure under additive inverses and addition, x x S. But, by definition of the
additive inverse, x x 0. Thus, 0 S.
Using the claim, we see that 0 Si for each i I. Thus, 0 i I Si . To prove the
second two properties, let a, b S be arbitrary elements. We need to prove that a b S
and ab S. By definition of S, a, b Si for each i I. By assumption that each Si is a
ring, a b, ab Si for each i. But, by definition of intersection, this implies that a b S
and ab S. Thus, by the subring test, S is a subring of R.
È
(5) (Gallian Chapter 12 #22) Let R be a commutative ring with unity and let U R denote
the set of units of R. Prove that U R is a group under the multiplication of R.
Solution: We prove that each of the group axioms are satisfied by U R under the multiplication operation in R.
Closure: we need to show that if x, y U R then xy U R as well. By definition
of U R, there are a, b R such that
xa
ax
1
yb
and
by
1
(where 1 is the unity of R). Consider the element ba. By closure of R under multiplication, ba R. Moreover,
and
xyba assoc xyba
x1a
unity
xa
1
baxy assoc baxy
b1y
unity
by
1.
Thus, ba satisfies the properties of being a multiplicative inverse for xy and
xy U R.
Associativity: follows from associativity of multiplication in the ring R.
Identity: we will prove that the unity of R, 1, is the group identity of U R. Namely,
let x U R. Then by definition of unity,
x1
1x
x,
as required for an identity in U R under multiplication.
3
Inverses:
let x U R. We need to find y U R such that xy
yx
1. By
definition of U R there is y R that satisfies this property. It remains to argue that
y is, in fact, in U R. But, this amounts to proving that y itself has a multiplicative
inverse. It does: x.
4