Download Introduction to Ring Theory. Math 228 Homework 5

Introduction to Ring Theory. Math 228
Unless otherwise stated, homework problems are taken from Hungerford, Abstract Algebra,
Second Edition.
Homework 5 - due February 23
3.1.5 d: Which of the following six sets are subrings of M (R)? Which ones have an
identity?
a 0
(d) All matrices of the form
, with a ∈ R.
a 0
Solution: We just need to show that this set is closed under addition, multiplication,
contains the additive identity, and additive inverses.
Since
a 0
+
a 0
a 0
a 0
b 0
b 0
a+b 0
=
a+b 0
b 0
ab 0
=
b 0
ab 0
it is closed under the operations.
0 0
The additive identity from M (R) is
which is an element of the set. The
0
0
a 0
−a 0
additive inverse of
is
which is an element of the set.
a 0
−a 0
Therefore,
the set is a subring of M (R). It has a multiplicative identity, namely,
1 0
, since
1 0
1 0
a 0
a 0
a 0
1 0
=
=
.
1 0
a 0
a 0
a 0
1 0
As an additional observation, it is commutative, even though M (R) is not!
Another way to see that this set is a subring of M (R): Check that it is nonempty,
and that it is closed under substraction and multiplication. Since
a 0
b 0
a−b 0
−
=
a 0
b 0
a−b 0
it is closed under substraction. We already saw that it is closed under multiplication.
0 0
It is nonempty since
is an element of the set.
0 0
3.1.18: Define a new addition ⊕ and multiplication on Z by
a⊕b=a+b−1
a b = a + b − ab,
where the operations on the right-hand side of the equal signs are ordinary addition,
substraction, and multiplication. Prove that, with the new operations ⊕ and , Z is an
integral domain.
Solution: Z is closed under addition and multiplication with these operations, and
additon and multiplication are commutative.
Addition is associative: a ⊕ (b ⊕ c) = a + (b ⊕ c) − 1 = a + b + c − 2 = (a ⊕ b) + c − 1 =
(a ⊕ b) ⊕ c.
Multiplication is associative: a (b c) = a + (b c) − a(b c) = a + b + c − bc −
a(b + c − bc) = a + b − ab + c − (a + b − ab)c = (a b) + c − (a b)c = (a b) c.
Distributive laws: a(b⊕c) = a+(b+c−1)−a(b+c−1) = a+b−ab+a+c−ac−1 =
(a b) ⊕ (a c).
Since is commutative, we do not have to check the other distributive law.
Additive identity is 1: a ⊕ 1 = a + 1 − 1 = a.
Mutiplicative identity is 0: a 0 = a + 0 − a0 = a.
Additive inverse: 2 − a: a ⊕ (2 − a) = a + (2 − a) − 1 = 1.
Therefore, we get a commutative ring with identity. To see that it is an integral
domain, suppose that the product of two numbers gives the Additive identity, i.e.,
suppose that a b = 1.
Then a + b − ab = 1, which leads 0 = 1 − a − b + ab = (1 − a)(1 − b). But this
could only happen iff a = 1 or b = 1, i.e., iff a or b are equal to the Additive identity.
Therefore, Z with these operations is an integral domain.
3.1.24: The addition table and part of the multiplication table for a three-element ring
are given below. Use the distributive laws to complete the multiplication table.
+ r
r r
s s
t t
s
s
t
r
·
r
s
t
t
t
r
s
r
r
r
r
Solution:
We compute s · t = s · (s + s) = s · s + s · s = t + t = s.
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s
r
t
t
r
t · s = (s + s) · s = s · s + s · s = t + t = s.
t · t = (s + s) · t = s · t + s · t = s + s = t.
The complete tables are
+ r
r r
s s
t t
s
s
t
r
·
r
s
t
t
t
r
s
r
r
r
r
s
r
t
s
t
r
s
t
Additional comments: Note that r is the additive identity, while t is the multiplicative
identity. The ring is commutative since the right table is symmetric. Every element
different from r has a multiplicative inverse, so the ring is actually a field.
3.2.2: An element e of a ring R is said to be idempotent if e2 = e.
(a) Find four idempotent elements in the ring M (R).
(b) Find all idempotents in Z12 .
(c) Prove that the only idempotents in an integral domain R are 0R and 1R .
Solution: (a)
1 0
0 1
2
1 0
0 0
2
0 0
0 1
2
0 0
0 0
2
1 0
0 1
1 0
0 0
0 0
0 1
0 0
0 0
=
=
=
=
are all idempotent
(b) In Z12 , we have
02 = 0,
62 = 0,
12 = 1 22 = 4,
72 = 1 82 = 4,
32 = 9,
92 = 9,
42 = 4,
52 = 1,
102 = 4,
112 = 1.
Therefore, the idempotent elements are 0, 1, 4, and 9.
(c) Let e be an idempotent in an integral domain. Then e2 = e, which implies that
e2 − e = 0R , or e(e − 1R ) = 0R . Since we are in an integral domain, we conclude that
either e = 0R or e − 1R = 0R (which implies e = 1R ).
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3.2.5: Let S and T be subrings of a ring R.
(a) Is S ∩ T a subring of R?
(b) Is S ∪ T a subring of R?
Solution: (a) Let a, b ∈ S ∩ T . Then a + b ∈ S and a + b ∈ T since S and T are
(sub)rings. Similarly ab ∈ S and ab ∈ T . Also 0R ∈ S and 0S ∈ T since S and T are
subrings of R. Finally, if a ∈ S ∩ T , then −a ∈ S since a ∈ S and S is a (sub)ring
and the same applies to T . So −a ∈ S ∩ T . By the four step criteria, S ∩ T is a
subring of R.
This can be also proved by using the two step criteria as follows:
0R ∈ S and 0S ∈ T since S and T are subrings of R, 0 ∈ S ∩ T which is then
nonempty. Let a, b ∈ S ∩ T . Then a − b ∈ S and a − b ∈ T since S and T are
(sub)rings. Similarly, ab ∈ S and ab ∈ T . Therefore, both a − b and ab ∈ S ∩ T . and
S ∩ T is a subring of R.
(b) Take R = Z. Let S = {2n|n ∈ Z} (even numbers) and let T = {3n|n ∈ Z}
(multiples of 3). Then we will see that S and T are subrings.
For S the sum of two elements is just another even number (2a + 2b = 2(a + b)) so
it is still in S. The same with the product of two elements ((2a)(2b) = 2(2ab)). For
T , the sum of two elements is a multiple of 3 (3a + 3b = 3(a + b)), which is still in
T . The product of two elements is also a multiple of 3 ((3a)(3b) = 3(3ab)). Also 0 is
both even and multiple of 3, so it is in S and T . The negative of an even number is
even and the negative of a number that is multiple of 3 is still multiple of 3. Then
S and T are subrings.
Now consider S ∪ T . Then 3, 2 ∈ S ∪ T , but 3 + 2 = 5 which is not an element of
S ∪ T since it is neither a multiple of 3 or an even number. Then S ∪ T is not a ring
(and in particular, it is not a subring of R).
3.2.14: Let R and S be nonzero rings (meaning that each of them contains at lest one
nonzero element). Show that R × S contains zero divisors.
Solution: Let r ∈ R and s ∈ S nonzero elements. Take a = (r, 0S ) and b = (0R , s).
Then a and b are nonzero elements in R × S, but ab = (r0R , 0S s) = (0R , 0S ).
3.2.22: Let R and S be rings with identity. What are the units in the ring R × S?
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Solution: We are going to prove that (r, s) ∈ R × S is a unit if and only if r is a
unit in R and s is a unit in S.
Let (r, s) be a unit in R × S. That means that there is (a, b) ∈ R × S such
that (r, s)(a, b) = (a, b)(r, s) = (1R , 1S ). But (r, s)(a, b) = (ra, sb) = (1R , 1S ) and
(a, b)(r, s) = (ar, bs) = (1R , 1S ), which implies that ar = ra = 1R and bs = sb = 1S
and therefore r and s are units in R and S respectively.
On the other hand, if r is a unit in R and s is a unit in S, we have that (r, s)(r−1 , s−1 ) =
(rr−1 , ss−1 ) = (1R , 1S ) = (r−1 r, s−1 s) = (r−1 , s−1 )(r, s).
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