# Introduction to Ring Theory. Math 228 Homework 5 - due ... mlalin/math22810/hwk228-5.pdfIntroduction to Ring Theory. Math 228 Unless otherwise stated, homework problems are taken from Hungerford, Abstract Algebra, Second Edition. Homework 5

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• 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 anidentity?

(d) All matrices of the form

(a 0a 0

), with a R.

Solution: We just need to show that this set is closed under addition, multiplication,contains the additive identity, and additive inverses.

Since (a 0a 0

)+

(b 0b 0

)=

(a + b 0a + b 0

)(

a 0a 0

) (b 0b 0

)=

(ab 0ab 0

)it is closed under the operations.

The additive identity from M(R) is(

0 00 0

)which is an element of the set. The

(a 0a 0

)is

(a 0a 0

)which is an element of the set.

Therefore, the set is a subring of M(R). It has a multiplicative identity, namely,(1 01 0

), since

(1 01 0

) (a 0a 0

)=

(a 0a 0

)=

(a 0a 0

) (1 01 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 0a 0

)

(b 0b 0

)=

(a b 0a b 0

)it is closed under substraction. We already saw that it is closed under multiplication.

It is nonempty since

(0 00 0

)is an element of the set.

• 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 anintegral domain.

Solution: Z is closed under addition and multiplication with these operations, andadditon and multiplication are commutative.

Addition is associative: a (b c) = a+(b c)1 = a+ b+ c2 = (a b)+ c1 =(a b) c.Multiplication is associative: a (b c) = a + (b c) a(b c) = a + b + c bca(b + c bc) = a + b ab + c (a + b ab)c = (a b) + c (a b)c = (a b) c.Distributive laws: a(bc) = a+(b+c1)a(b+c1) = a+bab+a+cac1 =(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 integraldomain, 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 thiscould 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 ringare given below. Use the distributive laws to complete the multiplication table.

+ r s tr r s ts s t rt t r s

r s tr r r rs r tt r

Solution:

We compute s t = s (s + s) = s s + s s = t + t = s.

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• 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 s tr r s ts s t rt t r s

r s tr r r rs r t st r s t

Additional comments: Note that r is the additive identity, while t is the multiplicativeidentity. The ring is commutative since the right table is symmetric. Every elementdifferent 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 00 1

)2=

(1 00 1

)(

1 00 0

)2=

(1 00 0

)(

0 00 1

)2=

(0 00 1

)(

0 00 0

)2=

(0 00 0

)are all idempotent

(b) In Z12, we have

02 = 0, 12 = 1 22 = 4, 32 = 9, 42 = 4, 52 = 1,

62 = 0, 72 = 1 82 = 4, 92 = 9, 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 thate2 e = 0R, or e(e 1R) = 0R. Since we are in an integral domain, we conclude thateither 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 aresubrings of R. Finally, if a S T , then a S since a S and S is a (sub)ringand the same applies to T . So a S T . By the four step criteria, S T is asubring 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 thennonempty. 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 . andS 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)) soit is still in S. The same with the product of two elements ((2a)(2b) = 2(2ab)). ForT , the sum of two elements is a multiple of 3 (3a + 3b = 3(a + b)), which is still inT . The product of two elements is also a multiple of 3 ((3a)(3b) = 3(3ab)). Also 0 isboth even and multiple of 3, so it is in S and T . The negative of an even number iseven and the negative of a number that is multiple of 3 is still multiple of 3. ThenS and T are subrings.

Now consider S T . Then 3, 2 S T , but 3 + 2 = 5 which is not an element ofS 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 onenonzero 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, 0Ss) = (0R, 0S).

3.2.22: Let R and S be rings with identity. What are the units in the ring R S?

Page 4

• Solution: We are going to prove that (r, s) R S is a unit if and only if r is aunit 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 suchthat (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 = 1Sand 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)(r1, s1) =(rr1, ss1) = (1R, 1S) = (r

1r, s1s) = (r1, s1)(r, s).

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