Abstract Algebra Answers

7/25/2019 Abstract Algebra Answers

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MATH 4442 & MATH 6442: MODERN ALGEBRA II

HOMEWORK SETS AND EXAMS

Yongwei Yao

2016 SPRING SEMESTER

GEORGIA STATE UNIVERSITY

Contents

HW Set #01, Solutions 1

HW Set #02, Solutions 3HW Set #03, Solutions 5HW Set #04, Solutions 7Midterm I, Review 9Midterm I, Solutions 10HW Set #05, Solutions 11HW Set #06, Solutions 13HW Set #07, Solutions 15HW Set #08, Solutions 17Midterm II, Review 19Midterm II, Solutions 20

HW Set #09, Solutions 21HW Set #10, Solutions 23Extra Credit Set, Solutionsnot really 25

Note. There are four (4) problems in each homework set. Math 6442 students need to doall 4 problems while Math 4442 students need to do any three (3) problems out the four.If a Math 4442 student submits all 4 problems, then one of the lowest score(s) is dropped.There is a bonus point for a Math 4442 student doing all 4 problems correctly.

Problems for extra credits are available; seethe last page of this file.There are three (3) PDF files for the homework sets and exams, one with the problems

only, one with hints, and one with solutions. Links are available below.

PROBLEMS HINTS SOLUTIONS

(R, +, ) . . .ab = 0 a= 0 b= 0 . . .R/ Ker() =Im() . . . I, m I R . . .IJ P I P J R. . .I= (a) in PID
http://./4(6)442probs-16Sp.pdfhttp://./4(6)442hints-16Sp.pdfhttp://./4(6)442hints-16Sp.pdfhttp://./4(6)442probs-16Sp.pdf


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Math 4442/6442 (Spring 2016) Homework Set #01(Due 01/19) Solutions

Problem 1.1. LetRbe a ring and a, b R.(1) Expand (a+b)3.(2) Further assume that R is commutative. Expand (a+b)3.

Solution. (1) We expand (a+b)3 as follows

(a+b)3 = (a+b)(a+b)(a+b) = (a+b)((a+b)(a+b))

= (a+b)(a(a+b) +b(a+b)) = (a+b)(aa+ab+ba+bb)

=a(aa+ab+ba+bb) +b(aa+ab+ba+bb)

=aaa+aab+aba+abb+baa+bab+bba+bbb

=a3 +a2b+aba+ab2 +ba2 +bab+b2a+b3.

(2) IfR is commutative, then

(a+b)3 =a3 +a2b+aba+ab2 +ba2 +bab+b2a+b3

=a3 +a2b+ba2 +aba+ab2 +b2a+b3

=a3 +a2b+a2b+a2b+ab2 +ab2 +ab2 +b3 =a3 + 3a2b+ 3ab2 +b3.

This completes the problem.

Problem 1.2. Let R ={a+b 32 + c 34 | a, b, cZ}. Consider (R, +,), with the usualaddition and multiplication.

(1) Determine whether (R, +) is an abelian group. No need to justify.(2) Determine whether R is closed under multiplication. Justify your claim.(3) Determine whether (R, +,) is a ring. No need to justify.(4) If (R, +,) is a ring, determine whether it is a commutative ring.(5) If (R, +,) is a ring, determine whether it has unity.

Solution. (1) Yes, (R, +) is an abelian group. (This is easy to verify.)(2) We claim thatRis closed under multiplication. Let x, y R, so thatx= a+b 32+c 34

and y = a+b 3

2 +c 3

4 witha, b, c, a, b, c Z. Then we havexy= (a+b

3

2 +c 3

4)(a+b 3

2 +c 3

4)

=aa+ab 3

2 +ac 3

4 +ba 3

2 +bb 3

4 +bc 3

8 +ca 3

4 +cb 3

8 +cc 3

16

=aa+ab 3

2 +ac 3

4 +ba 3

2 +bb 3

4 + 2bc+ca 3

4 + 2cb+ 2cc 3

2

=aa+ 2bc+ 2cb+ab 3

2 +ba 3

2 + 2cc 3

2 +ac 3

4 +bb 3

4 +ca 3

4

= (aa+ 2bc+ 2cb) + (ab+ba+ 2cc) 3

2 + (ac+bb+ca) 3

4 R

sinceaa+ 2bc+ 2cb Z,ab+ ba+ 2cc Z, andac+ bb+ ca Z. This shows that R isclosed under multiplication.(3) Yes, (R, +,) is a ring. By (2), R is closed under multiplication. Also, note that

associativity and distribution hold for both addition and multiplication in C, they clearlyhold inR. Finally, (R, +) is an abelian group, as noted in (1). Therefore, (R, +,) is a ringunder the usual addition and multiplication.

(4) Yes,R is a commutative ring. Indeed,xy = yxfor allx, y C, hence for allx, y R.(5) Yes, R has unity. Indeed, 1 = 1 + 0 3

2 + 0 3

4 R and 1x= x = x1 for all x R.

1


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Problem 1.3. Let S= {a+b 312 +c 318 | a, b, c Z}. Consider (S,+,), with the usualaddition and multiplication.

(1) Determine whether (S,+) is an abelian group. No need to justify.(2) Determine whether S is closed under multiplication. Justify your claim.(3) Determine whether (S,+,) is a ring. No need to justify.(4) If (S,+,

) is a ring, determine whether it is a commutative ring.

(5) If (S,+,) is a ring, determine whether it has unity.Solution. (1) Yes, (S,+) is an abelian group. (This is easy to verify.)

(2) We claim that Sis closed under multiplication. Let x, y S, so that x = a + b 312 +c 3

18 and y = a+b 3

12 +c 3

18 with a, b, c, a, b, c Z. Then we havexy= (a+b

3

12 +c 3

18)(a+b 3

12 +c 3

18)

=aa+ab 3

12 +ac 3

18 +ba 3

12 +bb 3

144 +bc 3

216 +ca 3

18 +cb 3

216 +cc 3

324

=aa+ab 3

12 +ac 3

18 +ba 3

12 + 2bb 3

18 + 6bc+ca 3

18 + 6cb+ 3cc 3

12

=aa+ 6bc+ 6cb+ab 3

12 +ba 3

12 + 3cc 3

12 +ac 3

18 + 2bb 3

18 +ca 3

18

= (aa+ 6bc+ 6cb) + (ab+ba+ 3cc) 312 + (ac+ 2bb+ca) 318 Ssinceaa+ 6bc+ 6cb Z, ab+ba+ 3cc Z, andac+ 2bb+ca Z. This shows that Sis closed under multiplication.

(3) Yes, (S,+,) is a ring. By (2), S is closed under multiplication. Also, note thatassociativity and distribution hold for both addition and multiplication in C, they clearlyhold in S. Finally, (S, +) is an abelian group, as noted in (1). Therefore, (S,+,) is a ringunder the usual addition and multiplication.

(4) Yes,Sis a commutative ring. Indeed,xy = yxfor allx, y C, hence for all x, y S.(5) Yes,Shas unity. Indeed, 1 = 1 + 0 3

1 2 + 0 3

18 Sand 1x= x = x1 for all x S.

Problem 1.4. Consider R =

{ 0 0 |, R}, which is a ring under the usual matrixaddition and matrix multiplication.

(1) Find concrete a, b R such that a = 0R, b = 0R andab= 0R.(2) Find concrete r, s, t R such that rt= st,t = 0R andr =s.

Solution. (1) Consider the following elements in R:

a=

1 00 0

and b=

0 01 0

.

It is clear that a = 0R,b = 0R and ab = 0R.(2) Consider the following elements in R:

r=

13 00 0

, s=

0 028 0

and t=

0 036 0

.

It is clear that rt= st, t

= 0R and r

=s.


(R, +, ) . . .ab = 0 a= 0 b= 0 . . .R/ Ker() =Im() . . . I, m I R . . .IJ P I P J R. . .I= (a) in PID2


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Problem 2.1. Let= 12 (1 +5) C andR = {a+b | a, b Z} C.

(1) Determine whetherRis a subring ofC (under the usual addition and multiplication).(2) IfR is a ring, isR an integral domain? with unity? Justify your answers.

Solution. (1) Since is a root of 2x

2

2x+ 3 (i.e., 22

2+ 3 = 0), we see2 = 3/2.

Or, computing directly, we see

2 = ( 12 (1 +5))2 = = 1 + 12

5 = 32 + .(We are going to show2 / R rigorously.) Suppose 2 R. Then2 =a+b, meaning

32

+ = a+b

for some a, b Z. This implies(1 b)= a+ 3

2.

Note that a + 32= 0, since a Z. Thus 1 b = 0. Therefore we obtain=

a+ 32

1 b Q,

which is a contradiction. This proves 2 / R.Since Rbut 2 / R, we see that R is not closed under multiplication. HenceR is not

a subring ofC. (SoRis not a ring under usual addition and multiplication.)(2) AsR is not a ring, it is not an integral domain (and the question involving unity does

not apply).

Problem 2.2. Let M2(R) = a bc d | a, b, c, d R, which is a ring with unity under

the usual matrix addition and matrix multiplication. Let R = a 0

0 d | a, d R andS= a 00 0 | a R. Answer the following questions (with no justifications needed):

(1) Is R a subring ofM2(R)? IsRcommutative? with unity?(2) Find the unity ofR (if it exists) explicitly. Does it agree with the unity ofM2(R)?(3) Is Sa subring ofM2(R)? IsScommutative? with unity?(4) Find the unity ofS(if it exists) explicitly. Does it agree with the unity ofM2(R)?

Solution. Skipping the details/justifications, we answer the questions as follows:(1) Yes, R is a subring ofM2(R) andR is commutative with unity 1R =

1 00 1

.

(2) The unity ofR is 1R=

1 00 1

, which agrees with the unity ofM2(R).

(3) Yes, S is a subring ofM2(R), commutative, and with unity 1S= 1 00 0 .(4) The unity ofS is 1S=

1 00 0

, which does not agree with the unity ofM2(R).

Problem 2.3. LetR be a ring with unity which is denoted by 1R, andSbe a subring ofRwith unity 1S= 1R. (Recall that U(R) denotes the set of all units ofR.)

(1) Prove that U(R) is closed under multiplication (i.e., under the multiplication ofR).(2) Prove U(S) U(R).(3) Is U(R) a group under multiplication? There is no need to justify.(4) Prove or disprove: U(S) = U(R) S.

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Proof/Solution. (1) Let u, v U(R). Then there exist u1, v1 Rsuch thatuu1 = 1R= u

1u and vv1 = 1R=v1v.

Considerv1u1 R. By associativity and the meaning of 1R, we see(uv)(v1u1) =u(vv1)u1 =u1Ru

1 =uu1 = 1R

(v1

u1

)(uv) =v1

(u1

u)v= v1

1Rv= v1

v= 1R,which verifies uv U(R) with (uv)1 =v1u1. Thus U(R) is closed under multiplication.

(2) Let u U(S). Then there exists x Ssuch that ux= 1S=xu. Now, since u, x Rand 1S= 1R, we see u U(R). This shows U(S) U(R). (Thus U(S) U(R) S.)

(3) Yes, U(R) a group under multiplication. (This should be easy to prove.)(4) We disprove it with a counterexample. Let S=Z and R= Q. Then S is a subring

ofR with 1S= 1 = 1R. However,

U(Z) = {1,1} and U(Q) = Q \ {0}.In particular, we see U(Z) = U(Q) Z. (For example, 13 U(Q) Zbut 13 / U(Z).)Problem 2.4 (Math 6442). Let R be a ring such that x2 =x for all x

R. (Such ring is

called a Boolean ring.) We prove that R is commutative via the following steps.(1) Prove that a = afor all a R.(2) Prove that ab+ba= 0 for all a, b R.(3) Prove that ab = ba for alla, b R.

Proof. (1) Let a R. By considering (a+a)2, we seea+a= (a+a)2 =a2 +a2 +a2 +a2 =a+a+a+a.

In short, a+a= a+a+a+a. By cancellation, we get 0R= a+a, hence a = a.Alternatively, we can verifya = aas follows:

a= a2 = (a)2 = a,which relies on the property (x)(y) =xy for allx, y R.

(2) Let a, b R. By considering (a+b)2, we seea+b= (a+b)2 =a2 +ab+ba+b2 =a+ab+ba+a.

In short, we have a+b= a+ab+ba+a. By cancellation, we get 0R = ab+ba.(3) Let a, b R. From (1), we see

ab= ab.From (2), we get 0R = ab+ba, which implies

ba= ab.Combining the above, we see

ab= ab= ba.In short, ab = ba for all a, b R. ThusR is commutative.




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Problem 3.1. In R = Z100 = {0,1, . . . , 98, 99}, consider x= 17 and y = 15.(1) Find the inverse element ofx in Z100 explicitly, ifx U(R).(2) Find a non-zero element z Z100 such that yz= 0R andzy= 0R.

Solution. As gcd(17, 100) = 1 and gcd(15, 100) = 5, we see 17 is a unit in Z100 and 15 is azero-divisor inZ100.

(1) To find the inverse of 17 in Z100, it suffices to findm, n Z such that 17m +100n= 1.We accomplish this via repeated division algorithm (a.k.a. Euclidean algorithm):

100 = 17 5 + 15,17 = 15 1 + 2,15 = 2 7 + 1,

2 = 1 2 + 0.Then, reading the above equations backward, we get

gcd(17, 100) = 1 = 15

2

7

= 15 (17 15 1)7= 17 7 + 15 8= 17 7 + (100 17 5)8= 17 47 + 100 8 = 17(47) + 100(8).

Thus 17(47) = 1 800, which implies 17(47) 1 mod 100. Hence17 47 = 47 17 = 1.

In other words,

171

= 47 = 53 Z100.(2) Let z= 60 Z100 for example. It is easy to see yz= 0 =yz.

Problem 3.2. LetR = Z8[x] be the polynomial ring over Z8 = {0, . . . , 7} on the variablex.Considerf=f(x) = 1 + 2x+ 3x2 andg = g(x) = 4 + 5x in Z8[x].

(1) Compute f+g and f g.(2) Determine deg(f), deg(g), deg(f+g) and deg(f g).(3) Find concrete f1, f2 R with deg(f1) = 2 = deg(f2) such that deg(f1+f2) = 1.(4) Find concrete g1, g2 Rwith deg(g1) = 1 = deg(g2) such that deg(g1g2) = 1.

Solution. (1) By routine computation (details omitted), we see

f+g = 5 + 7x+ 3x2

f g = 4 + 5x+ 6x2 + 7x3.

(2) We see deg(f) = 2, deg(g) = 1, deg(f+ g) = 2 and deg(f g) = 3.(3) Let f1 = 1x+ 3x

2 and f2= 5x2, both of degree 2. It is clear that

deg(f1+f2) = deg(1x) = 1.

(4) Let g1= 1 + 2x and g2 = 4x, both of degree 1. It is clear that

deg(g1g2) = deg(4x) = 1.

Now the solution is complete.5


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Problem 3.3. LetS= Z[

2] = {a+b2 | a, b Z} andR= Q[2] = {+2 |, Q}.(It is clear that both Sand R are commutative rings with 1S= 1 = 1R and S is a subringofR, which you may assume without proof.) Consider x= 3 + 2

2 and y = 3 + 4

2.

(1) Determine whether x is an invertible element ofS. If so, find its inverse in S.(2) Determine whether y is an invertible element ofR. If so, find its inverse in R.

Solution. (1) Consider 3 22 S, which is in S. By direct computation, we have3 + 2

2

3 2

2

=

3 2

2

3 + 2

2

= 32 222 = 9 8 = 1.This shows x = 3 + 2

2 U(S) and x1 = (3 + 22)1 = 3 22.

(2) We have (3 + 4

2)(3 42) = 32 (42)2 = 9 32 = 23. Consider123

3 4

2

= 323 + 423

2 R.It is straightforward to see 3

23+ 4

23

2

3 + 4

2

=

3 + 4

2 3

23+ 4

23

2

= (3+4

2)(342)23 = 1.

This shows that y= 3 + 4

2 U(R) with y1 = 323 + 423

2 R.Problem 3.4 (Math 6442). Let R = Q[2] ={+2 |, Q}, as in Problem 3.3 .Letz= a+b

2 = 0 with a, b Q.

(1) Prove (a+b

2)(a b2) = 0.(2) Prove a +b

2 U(R).

(3) Prove or disprove: R is a field.

Proof. (1) We have (a + b

2)(a b2) =a2 (b2)2 =a22b2. To prove by contradiction,suppose a2 2b2 = 0 so that a2 = 2b2. Consider the two possible cases:

Ifb = 0, then a2 = 0 and hence a = 0, which implies a+b2 = 0, a contradiction. Ifb = 0, then 2 = a2

b2 and hence

2 {a

b,a

b} Q, a contradiction.

Therefore (a+b2)(a b2) =a2

2b2

= 0, as required.(2) In part (1), we have concluded (a+b2)(a b2) =a2 2b2 Q \ {0}. Considerw= 1a22b2

a b

2

= aa22b2 + ba22b2

2 R.

It is clear thatwz= zw=

a+b

2

aa22b2 +

ba22b2

2

= 1,

which showsz= a+b

2 U(R).(3) WeprovethatR is a field as follows: Clearly,R is a non-zero commutative ring with

unity. Then by part (2), every non-zero element ofR is invertible. Thus R is a field.




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Problem 4.1. Consider 1 2x Z[x] Z[[x]] and 1 2x Z4[x], whereZ4 = {0,1, 2, 3}.(1) Is 1 2x invertible inZ[x]? Explain why or why not.(2) True or false: 1 2x U(Z[[x]]) with (1 2x)1 = 1 + 2x + 4x2 + 8x3 + Z[[x]].(3) Is 1

2x invertible inZ4[x]? If so, find its inverse explicitly.

Solution. (1) We claim that 12xis not invertible in Z[x]. Indeed, for every 0 =g(x) Z[x],we have

deg

(1 2x)g(x)= deg(1 2x) + deg(g(x)) deg(1 2x)> deg(1).Hence (1 2x)g(x) = 1 for all g(x) Z[x], which verifies 1 2x / U(Z[x]).

(2) True. It is straightforward to verify (details skipped)

(1 2x)

i=0

2ixi

= 1 =

i=0

2ixi

(1 2x).

(3) Yes, 1 2x U(Z4[x]). Indeed, it is straightforward to see(1 2x)(1 2x) = 1.

Thus 1 2x U(Z4[x]) with (1 2x)1 = 1 2x= 1 + 2x Z4[x].Problem 4.2. LetRbe a ring, and a R a (fixed) element ofR. Define Sas follows:

S= {r R | ar= ra}.Prove that Sis a subring ofR.

Proof. It is clear that S R. To show that Sis not empty, note thata0R = 0R= 0Ra.

Thus 0R

Sby the definition/construction ofS. (We also seea

S, asaa= aa.)Next, let x, y S, which simply means

ax= xa and ay= ya.

Then we see

a(x y) =ax ay=xa ya = (x y)a,

and moreover

a(xy) = (ax)y= (xa)y

=x(ay) =x(ya) = (xy)a.

In short, for x, y S, we havea(x y) = (x y)a and a(xy) = (xy)a,

which impliesx y S and xy S

by the definition/construction ofS. By the subring criterion, we conclude that Sis a subringofR. The proof is now complete.

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Problem 4.3. Leta(x) = 4x4 + 4x3 + 1x2 + 7x + 8 Z9[x] andb(x) =x2 + 2x + 3 Z9[x].Findq(x), r(x) Z9[x] satisfying

a(x) =b(x)q(x) +r(x)

such that r(x) = 0 or deg(r(x))


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Math 4442/6442 (Spring 2016) Midterm Exam I (02/16) Review Problems

Rings, subrings, fields: Problems1.1,1.2,1.3, 1.42.1, 2.2,3.4,4.2,4.4.

Zero-divisors, non-zero-divisors: Problems1.4,3.1,4.4.

Commutative rings, unity: Problems1.1,1.2,1.3,2.1, 2.2,2.4.

Invertible elements: Problems2.3,3.1,3.3,3.4,4.1,4.4.

Polynomial rings, power series rings: Problems3.2,4.1,4.3.

Subring criterion: Problems2.1, 2.2,4.2.

Lecture notes and textbooks: All we have covered,

Note: The above list is not intended to be complete. The problems inthe actual test may vary in difficulty as well as in content. Going over,understanding, and digesting the problems listed above will definitely help.However, simply memorizing the solutions of the problems may not help youas much.

You are strongly encouraged to practice more problems (than the oneslisted above) on your own.

9


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Math 4442/6442 (Spring 2016) Midterm Exam I(02/16) Solutions

Solutions

have been withdrawn

from the site




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Problem 5.1. Consider the ring Z under the usual addition and multiplication, as well asthe ring M2(R) =

a bc d

| a, b, c, d R under matrix addition and matrix multiplication.Determine whether each of the functions f , g andhis a ring homomorphism.

(1) Define f:Z

M2(R) byf(m) = m 00 m for all m

Z.

(2) Define g :Z M2(R) byg(m) = m mm m for all m Z.(3) Define h :Z Zby h(m) =m3 for all m Z.

Solution. (1) We claim that f is a ring homomorphism. Indeed, for allm, n Z,f(m+n) =

m+n 0

0 m+n

=

m 00 m

+

n 00 n

= f(m) +f(n),

f(mn) =

mn 00 mn

=

m 00 m

n 00 n

= f(m)f(n).

(2) We claim that g is not a ring homomorphism. For example, g(1 2)= g(1)g(2) asfollows

g(1 2) =g(2) =

2 22 2

,

g(1)g(2) = 1 1

1 1 2 2

2 2

= 4 4

4 4

.(3) We claim that h is not a ring homomorphism. For example, h(1+2) =h(1) + h(2) as

follows

h(1 + 2) =h(3) = 33 = 27,

h(1) +h(2) = 13 + 23 = 1 + 8 = 9.

Problem 5.2. Consider the ringZ and the ring Z6 ={0, . . . , 4,5}. Define f: ZZ6 byf(m) = 4m Z6 for all m Z.

(1) Determine whether fis a ring homomorphism.(2) Determine Ker(f) explicitly, iffis a ring homomorphism.

(3) Determine Im(f) explicitly, iffis a ring homomorphism.Solution. (1) We claim that f is a ring homomorphism. Indeed, for allm, n Z,

f(m+n) = 4(m+n) = 4m+ 4n= 4m+ 4n= f(m) +f(n),

f(mn) = 4mn= 4mn+ 12mn= 16mn= (4m)(4n) = 4m 4n= f(m)f(n).

(2) It is routine to see/verify Ker(f) = {3k | k Z}. Indeed, for all m Z,m Ker(f) f(m) = 0 Z6

4m= 0 Z6 6 | 4m 3 | 2m 3 | m.

In other word, Ker(f) consists of (precisely all) multiples of 3.(3) Finally, it is routine to verify Im(f) = {0, 2, 4}. Indeed, we have

f(3k) = 12k= 0 Z6f(3k+ 1) = 12k+ 4 = 4 Z6f(3k+ 2) = 12k+ 8 = 2 Z6,

for allk Z.11


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Problem 5.3. LetAand R be rings and f: A R a ring homomorphism.(1) Give a concrete example ofA, R and a ring homomorphism f: A R such that A

is commutative and R is not commutative.(2) Prove that ifA is commutative andfis surjective (i.e., onto) thenR is commutative.

Solution/Proof. (1) One such example is f: Z M2(R) defined by f(m) = m 00 m for all

m Z, as inProblem 5.1 (1).(2) Assume thatAis commutative andfis surjective (i.e., onto). Letr, s Rbe arbitraryelements (ofR). As f is onto, there exist a, bA such that f(a) =r and f(b) =s. SinceAis commutative, we have ab = ba. Hence

rs= f(a)f(b) =f(ab) =f(ba) =f(b)f(a) =sr.

In short, rs= sr for all r, s R, showing that R is commutative. Problem 5.4 (Math 6442). Let A and R be rings and f: A R a ring homomorphism.

(1) Give a concrete example ofA, R and a ring homomorphism g : AR such that Ahas unity and R does not have unity.

(2) Prove that ifA has unity 1A and f is surjective (i.e., onto) then R has unity.

Solution/Proof. (1) Let A be any ring with unity (e.g., A =Z or A ={0}) and R be anyring without unity (e.g., R = 2Z = {2k | k Z}). Define/construct f: A R by

f(a) = 0 for all a A.Sincefis clearly a ring homomorphism, we have an example as desired.

(2) Assume that A has unity 1A and f is surjective (i.e., onto). Note thatf(1A) R.We are going to show that f(1A) is the unity ofR. To this end, let r R be an arbitraryelement (ofR). Since fis onto, there exists a Asuch that f(a) =r. Consequently,

f(1A)r= f(1A)f(a) =f(1Aa) =f(a) =r,

rf(1A) =f(a)f(1A) =f(a1A) =f(a) =r.

As f(1A)r = r = rf(1A) for all r R, we conclude that f(1A) is the unity of R. Thiscompletes the proof that R has unity.




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Problem 6.1. Consider the ringZ3 = {[0], [1], [2]} and the ringZ6 = {0, . . . , 4, 5}. Defineg :Z3 Z6 byg([0]) = 0, g([1]) = 4 andg([2]) = 2.

(1) Determine whether g is a ring homomorphism.(2) Determine Ker(g) explicitly, ifg is a ring homomorphism.

(3) Determine Im(g) explicitly, ifg is a ring homomorphism.Solution. (1) We claim that g is a ring homomorphism. One can verifyg(x+y) =g(x)+g(y)for allx, y Z3 by exhaustion as follows (with details skipped)

g([0] + [0]) =g([0]) +g([0]), g([0] + [1]) =g([0]) +g([1]), g([0] + [2]) =g([0]) +g([2]),

g([1] + [0]) =g([1]) +g([0]), g([1] + [1]) =g([1]) +g([1]), g([1] + [2]) =g([1]) +g([2]),

g([2] + [0]) =g([2]) +g([0]), g([2] + [1]) =g([2]) +g([1]), g([2] + [2]) =g([2]) +g([2]).

Similarly, one can verifyg(xy) =g(x)g(y) for all x, y Z3 by exhaustion as followsg([0] [0]) =g([0])g([0]), g([0] [1]) =g([0])g([1]), g([0] [2]) =g([0])g([2]),

g([1] [0]) =g([1])g([0]), g([1] [1]) =g([1])g([1]), g([1] [2]) =g([1])g([2]),g([2] [0]) =g([2])g([0]), g([2] [1]) =g([2])g([1]), g([2] [2]) =g([2])g([2]).

(In fact, this ring homomorphismg is closed related to the ring homomorphismfas definedinProblem 5.2 .)

(2) It is straightforward to see Ker(g) = {[0]}.(3) It is straightforward to see Im(g) = {0, 2, 4}.

Problem 6.2. LetR be a ring and z R a (fixed) element. Define I= {x R | zx = 0R}.Prove that I is a right ideal ofR.

Proof. It is clear that Iis a subset ofR.First of all, from z0R = 0R, we see 0R

Iand hence I

= .

Next, let x, y I andr R; so that x, y R andzx= 0R= zy. Consequentlyz(x y) =zx zy = 0R 0R= 0R,

which implies x y I. Moreover,z(xr) = (zx)r= 0Rr= 0R,

which implies xr I.Therefore, by the right ideal criterion, I is a right ideal ofR.

Problem 6.3. LetR be a ring and let z Rbe a (fixed) element. Define J= {az| a R}.Prove that Jis a left ideal ofR.

Proof. It is clear that Jis a subset ofR.First of all, we have 0R = 0Rz J, which implies J= .Next, let x, y J andr R; so that x = azandy = bzfor somea, b R. Consequently

x y= az bz= (a b)z with a b R;rx= r(az) = (ra)z with ra R.

Thusx y J andrx J.Therefore, by the left ideal criterion, Jis a left ideal ofR.

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Problem 6.4. LetRbe a ring and let I , Jbe ideals (i.e., two-sided ideals) ofR.

(1) Prove that I+Jis an ideal ofR. Recall that I+J= {a+b | a I, b J}.(2) Prove that I Jis an ideal ofR.

Proof. It is clear that bothI+J and I Jare subsets ofR.(1) Since 0R Iand 0R J, we see 0R = 0R+ 0R I+ J. Next, let z1, z2 I+ J and

r R. Thus zi = ai+bi withai I and bi J. Consequentlyz1 z2= (a1 a2) + (b1 b2) I+J,

rz1 = ra1+rb1 I+J,z1r= a1r+b1r I+J.

This proves that I+Jis an ideal ofR, as required.(2) As I and Jare ideals, we have 0R I and 0R J. Hence 0R I J. Next, let

x, y I J andr R. Thus x, y I andx, y J. Consequently, we seex y I and x y J,

xr I and xr J,

rx I and rx J.Therefore

x y I J, xr I J and rx I J.By the ideal criterion, I Jis an ideal ofR.




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Problem 7.1. LetRbe a ring and let I , Jbe ideals ofR.

(1) Prove I I+ J andJ I+J.(2) Prove I J I J. (RecallI J= {

ni=1aibi| n N, ai I, bi J}.)

(3) Prove or disprove: I

Jis (always) an ideal ofR.

Proof/Solution. (1) Indeed, for all a I andb J, we havea= a+ 0R I+J and b= 0R+b I+J.

ThusI I+ J andJ I+ J.(2) We prove the statement IJ I J as follows: Let xI J. Then we can write x as

follows:

x=n

i=1

aibi= a1b1+ +anbn

withn N, ai I, bi J. Since bothI and Jare ideals ofR, we see aibi I and aibi Jfor alli= 1, . . . , n. Consequently

x=n

i=1

aibi I and x=n

i=1

aibi J.

So x I J. In summary, every x IJ satisfiesx I J, which proves I J I J. (3) We disprove the statement that I J is (always) an ideal ofR as follows: Consider

the ring R = Z; let I= (2) andJ= (3), both being ideals ofZ. It is clear that

2 I J, 3 I J but 2 + 3 = 5 / I J.This verifies that I J is not(always) an ideal ofR.Problem 7.2. LetR be a ring and Ian ideal ofR. (Thus there is the quotient ring R/I.)

Prove that R/I is commutative if ab ba I for alla, b R.Proof. Assume ab ba I for all a, b R. (We need to show that the quotient ringR/I iscommutative.) Let , R/Ibe arbitrary elements in R/I. (We need to show=.)By the construction ofR/I, there exist a, b R such that

= a+I and =b+I.

By the operations ofR/I, we see

= (a+I)(b+I) =ab+I

= (b+I)(a+I) =ba+I.

Sinceab

ba

I, we get

ab+I=ba+I

by the criterion r+I=s+I r s I. Combining the above, we derive= (a+I)(b+I) =ab+I

=ba+I= (b+I)(a+I) =.

In short, = for all , R/I. This completes the proof that R/I is commutative ifab ba I for alla, b R.

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Problem 7.3. Consider rings Z and Z12 = {[0], . . . , [11]}. Define functionsf1, f2 :Z Z12byf1(m) = [8m] and f2(m) = [9m] for all m Z.

(1) Determine which of the above (i.e., f1, f2) is a ring homomorphism; rename it f.(2) Find Ker(f) and Im(f) explicitly. (Again,f is the one that is a homomorphism.)(3) By applying the Fundamental Theorem of Homomorphisms, show that there is a ring

isomorphismg :Z4

=

Im(f).(4) Describe g explicitly in terms ofg(0) = [?], g(1) = [?], g(2) = [?], and g(3) = [?].Solution. (1) We claim that f1 is not a ring homomorphism andf2 is a ring homomorphism.Indeed, we have f1(1 1) =f1(1) = [8] = [64] = [8][8] = f1(1)f1(1). For f2, we have

f2(m+n) =[9(m+n)] = [9m+ 9n] = [9m] + [9n] =f2(m) +f2(n) and

f2(mn) = [9mn] = [9mn] + [72mn] = [81mn] = [(9m)(9n)] = [9m][9n] =f2(m)f2(n)

for allm, n Z, showing that f2 is a ring homomorphism. (So from now on, f2= f.)(2) We have Ker(f) = (4) = {4k | k Z} Z and Im(f) = {[0], [3], [6], [9]} Z12.(3) & (4) By the Fundamental Theorem of Homomorphisms, there is a ring isomorphism

g :Z/ Ker(f) =

Im(f) defined by g(m+ Ker(f)) =f(m). Since Ker(f) = (4), we see that

Z/ Ker(f) = Z/(4) = Z4. Putting things together, we see g :Z4 = Im(f) withg(0) =f(0) = [0], g(1) =f(1) = [9], g(2) =f(2) = [6], and g(3) =f(3) = [3].

Problem 7.4. LetRbe a division ring.

(1) Prove that there are exactly two distinct left ideals ofR. (What are they?)(2) How many right ideals does Rpossess? What are they? No justification is needed.(3) How many ideals does R possess? What are they? No justification is needed.

Proof/Solution. Note that{0R} and R are distinct ideals of the division ring R (since R isnot the zero ring). (We are going to show that there are no (left or right) ideals other than{0R} andR in any division ringR.)

(1) It suffices to show that every non-zero left ideal is necessarily R. To this end, let Ibeany non-zero left ideal ofR, so that there exists x Iwithx = 0R. AsR is a division ring,we see x is invertible so thatx1 R. Thus 1R= x1x Isince I is a left ideal. Now that1R I, we consequently see

r= r1R I for all r R.(Or, r=r1R =r(x

1x) = (rx1)x I for all r R.) This shows R Iand hence I=R.So there are precisely two distinct left ideals (namely{0R} and R) in a division ring R.

(2) There are two distinct right ideals, namely{0R} and R, precisely. The proof is verysimilar to the proof in (1) above. (Indeed, let I be any non-zero right ideal of R, so thatthere isx I withx = 0R. Thenx is invertible; hence for allr R, we see

r= 1Rr= (xx1

)r= x(x1

r) I.This showsR I. ThereforeI=R.)

(3) There are precisely two distinct ideals, namely{0R}andR, by the work above.




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Problem 8.1. Let R be a ring and I an ideal ofR. Prove that the quotient ring R/I iscommutative only if ab ba Ifor all a, b R. (Compare withProblem 7.2 .)Proof. Assume that the quotient ring R/I is commutative. (We need to showab ba I

for alla, b

R.) Let a, b

R be arbitrary elements in R. (We need to showab

ba

I.)

By the construction ofR/I, we have

a+I R/I and b+I R/I.By the operations ofR/I, we see

(a+I)(b+I) =ab+I and (b+I)(a+I) =ba+I.

SinceR/I is commutative, we have

(a+I)(b+I) = (b+I)(a+I).

Combining the above, we get

ab+I= (a+I)(b+I)

= (b+I)(a+I) =ba+I.

Now that ab+I=ba+I, we seeab ba I

by the criterion r+ I = s+ I r s I. This completes the proof that R/I iscommutative only ifab ba I for alla, b R. Problem 8.2. Define a function :R[x]C by(f(x)) =f(i) for all f(x)R[x]. (Forexample, we have (2x13 3x3 +5) = 2i13 3 i3 + 5 = 5 + (2 + 3)i C.)

(1) Compute (1 2x+ 3x2),(4 5x) and(1 2x+ 3x2)(4 5x) directly.(2) True or false: (1 2x+ 3x2)(4 5x) =(1 2x+ 3x

2)(4 5x).(3) True or false: is a ring homomorphism. No justification is necessary.(4) Find a concrete f(x) R[x] such that deg(f(x)) = 3 and(f(x)) = 0.(5) Find a concrete g(x) R[x] such that (g(x)) = 13 45i.

Solution. (1) It is routine to carry out the computation (details skipped)

(1 2x+ 3x2) = 2 2i(4 5x) = 4 5i

(1 2x+ 3x2)(4 5x)= (4 13x+ 22x2 15x3) = 18 + 2i.(2) True. Indeed, it is routine to see

(1 2x+ 3x2)(4 5x) = (2 2i)(4 5i)= 18 + 2i= (1 2x+ 3x2)(4 5x).

(3) As covered in class, is a ring homomorphism(4) Let f(x) =x3 +x= (x2 + 1)x. Then deg(f(x)) = 3 and moreover

(f(x)) =(x3 +x) =i3 +i= i+i= 0.(5) Let g(x) = 13 45x. It is clear that

(g(x)) =(13 45x) = 13 45i.17


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Problem 8.3. Consider : R[x] C by(f(x)) = f(i) for all f(x) R[x], which is thesame as inProblem 8.2 above.

(1) Prove Ker() = (x2 + 1), the ideal generated byx2 + 1 inR[x].(2) Prove that is onto. (So what is Im()?)(3) ProveR[x]/(x2 + 1) = C.

Proof. (1) For alla(x) (x2

+ 1), we seea(x) =b(x)(x2

+ 1) for someb(x) R[x] and hence(a(x)) =a(i) =b(i)(i2 + 1) =b(i)0 = 0,

which implies a(x) Ker(). Hence (x2 + 1) Ker().Conversely, let f(x) Ker(); so that f(i) = (f(x)) = 0. (We are going to show

f(x) (x2 + 1).) As x2 + 1 is monic, there exists q(x), r(x) R[x] such thatf(x) =q(x)(x2 + 1) +r(x) with r(x) = 0 or deg(r(x)) 1.

The above condition onr(x) allows us to write r(x) =r0+r1x withr0, r1 R. Then0 =f(i) =q(i)(i2 + 1) +r(i) =q(i)0 +r(i) =r(i) =r0+r1i.

Fromr0+r1i= 0, we seer0 = r1 = 0 and hencer(x) = 0. Thusf(x) =q(x)(x2+1) (x2+1).

This shows Ker() (x2

+ 1). All in all, we see Ker() = (x2

+ 1).(2) For a+bi C with a, b R, there exists a +bx R[x] such that (a+bx) =a+bi.This proves that is onto. (Hence Im() = C.)

(3) By the Fundamental Theorem of Homomorphisms, R[x]/ Ker()= Im(). Now, inlight of Ker() = (x2 + 1) and Im() = C, we seeR[x]/(x2 + 1) = C. Problem 8.4. Let R be a ring, z R a (fixed) element ofR, and Ja (fixed) left ideal ofR. Define I= {r R | rz J}. Prove that Iis a left ideal ofR.Proof. It is clear that I R. First, since 0Rz= 0R J, we see 0R Iand hence I= .

Next, let a, b I(so that a, b Rwithaz J andbz J) and let r R. Then(a

b)z= az

bz

J

because az J, bz J andJ is a left ideal. Thus a b I. Also, we have(ra)z= r(az) J

because az J,r Rand J is a left ideal. Thus ra I. By the criterion for left ideals, weprove that Iis a left ideal ofR, as required.




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Math 4442/6442 (Spring 2016) Midterm Exam II (03/29) Review Problems

Materials covered earlier: Midterm I;Homework Sets1,2,3, 4.

Ring homomorphism, kernel, image: Problems5.1,5.2,6.1,6.2,7.3,8.2,8.3,etcetera.

Proofs/examples involving ring homomorphism: Problems5.3,5.4,6.2,8.4.

Left/right/2-sided ideal, criteria: Problems6.3,6.4,7.1,7.4,8.4.

Quotient ring: Problems7.2,8.1.

Fundamental theorem of ring homomorphisms: Problems7.3,8.3.

Lecture notes and textbooks: All we have covered.

Note: The above list is not intended to be complete. The problems inthe actual test may vary in difficulty as well as in content. Going over,understanding, and digesting the problems listed above will definitely help.However, simply memorizing the solutions of the problems may not help youas much.

You are strongly encouraged to practice more problems (than the oneslisted above) on your own.

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Math 4442/6442 (Spring 2016) Midterm Exam II (03/29) Solutions

Solutions

have been withdrawn

from the site




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Problem 9.1. Consider the ring Z, which is commutative with unity. LetI1 = (0), I2 =(9),I3= (23) andI4= (28) (be principal ideals generated by 0, 9, 23 and 28 respectively).

(1) List all Ii that are prime ideals.(2) List all Ii that are maximal ideals.

(3) For each Ii that is not a prime ideal, find ai, bi Z \ Ii such that aibi Ii.(4) For each Ii that is not a maximal ideal, find a maximal ideal Mi such that Ii Mi.

Solution. (1) Among I1, . . . , I 4, the prime ideals are I1 = (0) and I3= (23).(2) AmongI1, . . . , I 4, the only maximal ideal is I3= (23).(3) We provide elements as required in the following

For I2, there exist 3, 3 Z \ I2 such that 3 3 = 9 I2. For I4, there exist 4, 7 Z \ I4 such that 4 7 = 28 I4. (Or, we have 2, 14 Z \ I4

and 2 14 = 28 I4.)(4) We may choose M1 = (2), M2 = (3) and M4 = (7). It should be straightforward to

see that Mi are maximal ideals and Ii Mi for alli= 1, 2,4.Problem 9.2. Consider the ideal I= (35,119) inZ.

(1) Find a Zsuch that I= (a).(2) Write a as a linear combination of 35 and 119.

Solution. To solve the problem systematically (rather than ad hoc), we apply the EuclideanAlgorithm.

(1) To find gcd(35, 119), we apply the Euclidean Algorithm as follows

119 = 3 35 + 14,35 = 2 14 + 7,14 = 2 7 + 0.

Thusa= gcd(35, 119) = 7. That is, I= (35, 119) = (7).(2) To find a linear combination, we have

a= 7 = 35 2 14= 35 2(119 3 35) = 7 35 2 119.

In summary, we have a = 7 = 7 35 2 119.Problem 9.3. Consider the ideal J= (35, 119, 79) ofZ.

(1) Find b Z such that J= (b).(2) Write b as a linear combination of 35, 119 and 79.

Solution. To solve the problem systematically (rather than ad hoc), we apply the Euclidean

Algorithm.(1) We have gcd(35, 119,79) = gcd(gcd(35, 119), 79) = gcd(7, 79). Then

79 = 11 7 + 2,7 = 3 2 + 1,2 = 2 1 + 0

Thus b = gcd(35, 119, 79) = gcd(7, 79) = 1. That is, J = (35, 119, 79) = (1). (HenceJ= Z.)

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(2) To find a linear combination, we have

b= 1 = 7 3 2= 7 3(79 11 7)= 34 7 3 799.2

= 34(7 35 2 119) 3 79 = 238 35 68 119 3 79.In summary, we have b = 1 = 238 35 68 119 3 79.Problem 9.4 (Math 6442). Let R be a commutative ring with unity and P an ideal ofR.Prove that P is a prime ideal ofR if R/Pis an integral domain.

Proof. Assume thatR/P is an integral domain. (We need to prove thatPis a prime ideal.)To this end, let a, b Rsuch that ab P. Hence ab+P = 0R+P. Thus we see

(a+P)(b+P) =ab+P = 0R+P = 0R/P R/P.Now, asR/P is an integral domain and (a+P)(b+P) = 0R/P, we must have

a+P = 0R+P or b+P = 0R+P,

which readily impliesa P or b P.

In summary,ab P = a P or b P. This establishes that Pis a prime ideal ofR.




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Problem 10.1. Consider an ideal I= (x2 +x+ 1, x3 + 2x) inZ3[x], whereZ3= {0, 1, 2}.(1) Find themonic polynomial f(x) Z3[x] such that I= (f(x)).(2) Express f(x) as a linear combination ofx2 +x+ 1 andx3 + 2x (withinZ3[x]).

Solution. (1) We find gcd(x2 + x + 1, x3 + 2x) in Z3[x] by the Euclidean Algorithm as follows

x3 + 2x= (x 1)(x2 +x+ 1) + (2x+ 1),x2 +x+ 1 = (2x+ 1)(2x+ 1) + 0.

The monic polynomial associated to 2x+ 1 is (2)1(2x+ 1) = 2(2x+ 1) = x+ 2. Thusf(x) = gcd(x2 +x+ 1, x3 + 2x) =x+ 2 and (x2 +x+ 1, x3 + 2x) = (x+ 2).

(2) From (1) above, we see

2x+ 1 =x3 + 2x (x 1)(x2 +x+ 1).Therefore

f(x) =x+ 2 = 2(2x+ 1)

= 2

(x3 + 2x) (x 1)(x2 +x+ 1)= 2(x3 + 2x) (2x 2)(x2 +x+ 1)= 2(x3 + 2x) + (x+ 2)(x2 +x+ 1).

In summary,f(x) =x+ 2 = 2(x3 + 2x) + (x+ 2)(x2 +x+ 1), as required.

Problem 10.2. In the ringR[x], consider M= (x), the ideal generated by x.

(1) True or false: Mconsists of all polynomials ofR[x] whose constant terms are 0.(2) Prove that M is a maximal ideal ofR[x].(3) Multiple choice: The quotient ringR[x]/(x) is isomorphic to Q R C

Solution. Note that M= (x) =

xp(x) |p(x) R

[x]

consists of all multiples ofx.(1) True. Indeed, a polynomial is a multiple ofx if and only if its constant term is 0.(2) As 1 / Mbut 1R[x], we see M R[x]. (It remains to verify that, for every ideal

I ofR[x], ifM I then I =R[x].) Let Ibe any such ideal (so that M I). Then thereexists a polynomial f(x) Ibut f(x) / M. To be concrete, write

f(x) =anxn + +a1x+a0 with n 0 and ai R for i = 0,1, . . . , n.

Note that a0= 0 (since otherwise f(x) (x)) and anxn + +a1x (x) I. Thusa0 = (anx

n + +a1x+a0) (anxn + +a1x) I.Now, since a0 I and a0 U(R[x]) (as a0 R \ {0}), we see I = R[x]. (Indeed, for all

p(x)

R[x], we have p(x) =a0(a

10 p(x))

I since a0

Iand I is an ideal ofR[x].)

(3) We claim that R[x]/(x) is isomorphic to R. (To justify the claim, we define a ringhomomorphism :R[x] R by

(p(x)) = p(0) for all p(x) R[x].Note that Ker() = (x) and Im() =R (details omitted, similar toProblem 4.4 ). By theFundamental Theorem of Homomorphisms, we see R[x]/(x)= R.) (Now we can use (3)to prove (2) as follows: From (3), we see thatR[x]/(x) is a field, which then implies thatM= (x) is a maximal ideal ofR[x].)

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Problem 10.3. Let R be a PID (i.e., a commutative integral domain with unity in whichall ideals are principal), and a, b R. Prove that there existsd R such that d| a, d| band d = ar+bsfor some r, s R. (Here d | ameans a = dt for some t R.)Proof. Consider the ideal (a, b). SinceR is a PID, there existsd Rsuch that (a, b) = (d).(It suffices to show d | a, d | bandd = ar+bs for some r, s R.)

Asa, b

(a, b), we see a, b

(d) ={

dz|

z

R}

. This simply means

a= dx and b= dy

for some x, y R. Therefore d | aand d | b.Finally, asd (d), we see d (a, b) = {ar+bs | r, s R}. This immediately implies

d= ar+bs

for some r, s R. Now the proof is complete. Problem 10.4 (Math 6442). LetR be a commutative ring with unity and Pan ideal ofR.Prove that P is a prime ideal ofR only if R/Pis an integral domain.

Proof. Assume thatP is prime. (We need to prove thatR/Pis an integral domain.) To this

end, let a+P and b +P in R/P such that(a+P)(b+P) = 0R/P.

This translates toab+P = 0R+P.

which impliesab P.

Now, since P is a prime ideal, we see

a P or b P,which implies

a+P = 0R/P or b+P = 0R/P.

In summary, if (a + P)(b + P) = 0R/P, thena + P = 0R/P or b + P = 0R/P. This establishesthat R/P is an integral domain. The proof is now complete.




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Math 4442/6442 (Spring 2016) Extra Credit Set Solutions

Each problem carries 3 points. To claim the eatra points for a problem, you must solve itcompletely and correctly. You may attempt a problem for as many times as you wishby 04/19. Feel free to use the results we have covered.

The points you get here will be added to your total score from homework assignments.

Each represents a correct solution submitted.

Problem E-1. Let R be a ring, and Si subrings ofR (for i = 1,2) such that R = S1 S2.ProveR = S1 or R = S2.

Proof. Keep working on this.

Problem E-2. Let R be a ring with d, e R such that d is not a right zero-divisor andex= x for allx R. Prove that R has unity. Proof. Keep working on this.

Problem E-3. LetR be afinitering such that x2 =x for allx R(thusR is commutativebyProblem 2.4 ). Prove that R has unity.


Problem E-4. Let R be a ring with unity and x, y R. Assume thatxy = 1R and x isnot a left zero-divisor. Prove x U(R). Proof. Keep working on this.

Problem E-5. LetRbe an integral domain,e R, andIa non-zero left ideal ofR. Assumethat ex = x for allx I. Prove that Rhas unity. Proof. Keep working on this.

Problem E-6. Let R be a commutative ring with unity and let ai R. Provei=0aix

i U(R[[x]]) a0 U(R).Proof. Keep working on this.

Problem E-7. Consider the ring M2(R) of all 2 2 real matrices over R.(1) Determine whether the zero ideal{0M2(R)}is maximal in M2(R).(2) Determine whether the quotient ring M2(R)/{0M2(R)} is a division ring.

(The same questions can be asked for Mn(F) with Fa field and 2 n N.)Proof. Keep working on this.

Problem E-8. For any field F, prove that F[[x]] is a PID.


Problem E-9. Let R be a commutative ring with unity, a, b R, and ma maximal ideal.Prove that, ifab m2 then either a m2 or b m.Proof. Keep working on this.


(R, +, ) . . .ab = 0 a= 0 b= 0 . . .R/ Ker() =Im() . . . I, m I R . . .IJ P I P J R. . .I= (a) in PID

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