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EXERCISES

201

EXERCISES

For Exercises 9.1 to 9.6 calculate the quotients and remainders.

9.1. Divide 3

x

4

+

4x3

x2

+

5x

1

by 2x2

+

x

2+

1 in Q[x].

6

 

4

 

3

 

 

 

by

 

3

 

 

 

[

].

9.2. Divide x

7

+ x

6

4

4

 

+ 5

x

 

 

x

3

+ 2

x

+ 1 in

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

 

 

 

 

R x

 

9.3. Divide x

 

5+ x

 

4+ x

 

+3

x +21

 

by x

3

+ x + 1 in Z2[x].

9.4. Divide 2x

 

+ x

+ 2x

 

+ x

+ 2

by x

 

+ 2x + 2 in Z3[x].

9.5. Divide 17

+ 11i

 

 

 

 

 

 

 

 

by 3 + 4i

in Z[i].

 

 

9.6. Divide 20

+ 8i

 

 

 

 

 

 

 

 

 

by 7 2i

in Z[i].

 

 

For Exercises 9.7 to 9.13, find the greatest common divisors of the elements a, b in the given euclidean ring, and find elements s, t in the ring so that as + bt = gcd(a, b).

9.7.a = 33, b = 42 in Z.

9.8.a = 2891, b = 1589 in Z.

9.9.a = 2x3 4x2 8x + 1, b = 2x3 5x2 5x + 2 Q[x].

9.10.a = x6 x3 16x2 + 12x 2, b = x5 2x2 16x + 8 Q[x].

9.11.a = x4 + x + 1, b = x3 + x2 + x Z3[x].

9.12.a = x4 + 2, b = x3 + 3 Z5[x].

9.13.a = 4 i, b = 1 + i Z[i].

For Exercises 9.14 to 9.17, find one solution to each equation with x, y Z.

9.14.

15x + 36y = 3.

9.15.

24x + 29y = 1.

9.16.

24x + 29y = 6.

9.17.

11x + 31y = 1.

For Exercises 9.18 to 9.21, find the inverse to the element in the given field.

9.18.

[4] in Z7.

9.19.

[24]

in

Z29.

9.20.

[35] in Z101.

9.21.

[11]

in

Z31.

Find all integral solutions to the equations in Exercises 9.22 to 9.24.

9.22.

27x + 15y = 13.

9.23. 12x + 20y = 14.

9.24.

28x + 20y = 16.

 

Factor the polynomials in Exercises 9.25 to 9.36 into irreducible factors in the given ring.

9.25.

x5 1

in Q[x].

9.26.

x5

+ 1 in Z2[x].

9.27.

x4 + 1

in Z5[x].

9.28.

2x3

+ x2

+ 4x + 2 in Q[x].

9.29.

x4

9x + 3

in Q[x].

9.30.

2x3

+ x2

+ 4x + 2 in C[x].

9.31.

x3

4x + 1

in Q[x].

9.32.

x4

+ 3x3

+ 9x 9 in Q[x].

9.33.

x8

16

in C[x].

9.34.

x8

16

R[x].

9.35.

x8

16

in Q[x].

9.36.

x8

16

Z17[x].

9.37. Find all irreducible polynomials of degree 5 over Z2.

202

9 POLYNOMIAL AND EUCLIDEAN RINGS

9.38.Find an irreducible polynomial of degree 2 over Z5.

9.39.Find an irreducible polynomial of degree 3 over Z7.

9.40.Find the kernel and image ofthe ring morphism ψ: R[x] C defined by

=p(i), where i = −1.

9.41. Find the kernel andimage of the ring morphism ψ: R[x] C defined by

ψ(p(x)) = p(1 + 3i).

In Exercises 9.42 to 9.47, are the polynomials irreducible in the given ring? Give reasons.

9.42.x3 + x2 + x + 1 in Q[x].

9.43.3x8 4x6 + 8x5 10x + 6 in Q[x].

9.44.

x4

+ x2 6 in Q[x].

9.45.

4x3 + 3x2 + x + 1 in Z5[x].

9.46.

x5

+ 15 in Q[x].

9.47.

x4 2x3 + x2 + 1 in R[x].

9.48.Is Z[x] a euclidean ring when δ(f (x)) = degf (x) for any nonzero polynomial? Is Z[x] a euclidean ring with any other definition of δ(f (x))?

9.49. Can you define a division algorithm in R[x, y]? How would you divide x3 + 3xy + y + 4 by xy + y3 + 2?

9.50. Let Lp be the set of all linear functions f : Zp Zp of the form f (x) = ax + b, where a = 0 in Zp . Show that (Lp , Ž ) is a group of order p(p 1) under composition.

9.51. If p is a prime, prove that (x a)|(xp1 1) in Zp [x] for all nonzero a

in Zp . Hence prove that

 

 

 

 

 

 

 

 

 

 

 

 

 

xp1 1 = (x 1)(x 2) · · · (x p + 1)

in Zp [x].

 

9.52. (Wilson’s theorem) Prove that (n 1)! ≡ −1

mod

n if and

only

if n

is prime.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

9.53. Prove that 2/5 is irrational.

 

+

 

 

 

 

 

 

9.54. Find a polynomial in Q[x] with

 

 

as

a root. Then

prove

that

2

3

 

+

 

is irrational.

 

 

 

 

 

 

 

 

 

2

3

 

 

 

 

 

 

 

 

 

9.55.Is 5 irreducible in Z[i]?

9.56.Show that Z[5] = {a + b5|a, b Z} does not have the unique factorization property.

9.57.Prove that a gaussian integer is irreducible if and only if it is an invertible element times one of the following gaussian integers:

(1)any prime p in Z with p 3 mod 4.

(2)1 + i.

(3)a + bi, where a is positive and even, and a2 + b2 = p, for some prime p in Z such that p 1 mod 4.

9.58.If r/s is a rational root, in its lowest terms, of a polynomial p(x) with integral coefficients, show that p(x) = (sx r)g(x) for some polynomial g(x) with integral coefficients.

EXERCISES

203

9.59.Prove that r/s, in its lowest terms, cannot be a root of the integral polynomial p(x) unless (s r)|p(1). This can be used to shorten the list of possible rational roots of an integral polynomial.

9.60.Let m = m1m2 · · · mr and Mi = m/mi . If gcd(mi , mj ) = 1 for i = j , each of the congruences Mi y 1 mod mi has a solution y bi mod mi . Prove that the solution to the simultaneous congruences

x a1 mod m1, x a2 mod m2, . . . , x ar mod mr

r

Mi bi ai mod m.

is x i=1

For Exercises 9.61 to 9.64, solve the simultaneous congruences.

9.61. x 5 mod 7

 

 

 

9.62. x

41 mod 65

x 4 mod 6.

 

 

 

 

x

35 mod 72.

9.63. x 0 mod 2

 

 

 

9.64. x

9 mod 12

x 1 mod 3

 

 

 

 

x

3 mod 13

x 2 mod 5.

 

 

 

 

x 6 mod 25.

 

 

320

461

5264

72

 

is nonzero.

9.65. Prove that det

702

1008

967

44

 

 

2333

46

127

 

 

 

91

 

 

 

 

 

 

 

 

 

164216 1862 469

9.66.Solve the following simultaneous equations:

26x 141y = −697

55x 112y = 202

(a) in Z2, (b) in Z3, and (c) in Z5. Then use the Chinese remainder theorem to solve them in Z assuming they have a pair of integral solutions between

0 and 29.

 

 

 

 

 

 

 

676

117

522

is positive and less than 100.

9.67. The value of det

375

65

290

 

 

825

143

639

 

Find its value without using a calculator. (If you get tired of doing arithmetic, calculate its value mod 10 and mod 11 and then use the Chinese remainder theorem.)

9.68.The polynomial x3 + 5x Z6[x] has six roots. Does this contradict Theorem 9.6?

9.69.If R is an integral domain and R[x] is euclidean, show that R must be a field.

9.70.Assume that R is a euclidean domain in which δ(a + b) max{δ(a), δ(b)} whenever a, b, and a + b are all nonzero. Show that the quotient and remainder in the division algorithm are uniquely determined.

10

QUOTIENT RINGS

In this chapter we define a quotient ring in a way similar to our definition of a quotient group. The analogue of a normal subgroup is called an ideal, and a quotient ring consists of the set of cosets of the ring by one of its ideals. As in groups, we have a morphism theorem connecting morphisms, ideals, and quotient rings. We discover under what conditions quotient rings are fields. This will enable us to fulfill our long-range goal of extending the number systems by defining new fields using quotient rings of some familiar rings.

IDEALS AND QUOTIENT RINGS

If (R, +, ·) is any ring and (S, +) is any subgroup of the abelian group (R, +), then the quotient group (R/S, +) has already been defined. However, R/S does not have a ring structure induced on it by R unless S is a special kind of subgroup called an ideal.

A nonempty subset I of a ring R is called an ideal of R if the following conditions are satisfied for all x, y I and r R:

(i)x y I .

(ii)x · r and r · x I .

Condition (i) implies that (I, +) is a subgroup of (R, +). In any ring R, R itself is an ideal, and {0} is an ideal.

Proposition 10.1. Let a be an element of a commutative ring R. The set

{ar|r R} of all multiples of a is an ideal of R called the principal ideal generated by a. This ideal is denoted by (a).

Proof. Let ar, as (a) and t R. Then ar as = a(r s) (a) and (ar)t = a(rt) (a). Hence (a) is an ideal of R.

Modern Algebra with Applications, Second Edition, by William J. Gilbert and W. Keith Nicholson ISBN 0-471-41451-4 Copyright 2004 John Wiley & Sons, Inc.

IDEALS AND QUOTIENT RINGS

205

For example, (n) = nZ, consisting of all integer multiples of n, is the principal ideal generated by n in Z.

The set of all polynomials in Q[x] that contain x2 2 as a factor is the principal ideal (x2 2) = {(x2 2) · p(x)|p(x) Q[x]} generated by x2 2 in Q[x]. The set of all real polynomials that have zero constant term is the principal ideal (x) = {x · p(x)|p(x) R[x]} generated by x in R[x]. It is also the set of real polynomials with 0 as a root.

The set of all real polynomials, in two variables x and y, that have a zero constant term is an ideal of R[x, y]. However, this ideal is not principal (see Exercise 10.30).

However, every ideal is principal in many commutative rings; these are called principal ideal rings.

Theorem 10.2. A euclidean ring is a principal ideal ring.

Proof. Let I be any ideal of the euclidean ring R. If I = {0}, then I = (0), the principal ideal generated by 0. Otherwise, I contains nonzero elements. Let b be a nonzero element of I for which δ(b) is minimal. If a is any other element in I , then, by the division algorithm, there exist q, r R such that

a = q · b + r where r = 0 or δ(r) < δ(b).

Now r = a q · b I . Since b is

a nonzero element of I for which δ(b) is

minimal, it follows that r must be

zero and a = q · b. Therefore, a (b) and

I (b).

Conversely, any element of (b) is of the form q · b for some q R, so q · b I . Therefore, I (b), which proves that I = (b). Hence R is a principal ideal

ring.

 

Corollary 10.3. Z is a principal ideal ring, so is F [x], if F is a field.

 

Proof. This follows because Z and F [x] are euclidean rings.

 

Proposition 10.4. Let I be ideal of the ring R. If I contains the identity 1, then I is the entire ring R.

Proof. Let 1 I and r R. Then r = r · 1 I , so I = R.

Let I be any ideal in a ring R. Then (I, +) is a normal subgroup of (R, +), and we denote the coset of I in R that contains r by I + r. Hence

I + r = {i + r R|i I }.

The cosets of I in R are the equivalence classes under the congruence relation modulo I . We have

r1 r2 modI if and only if r1 r2 I.

206

10 QUOTIENT RINGS

By Theorem 4.18, the set of cosets R/I = {I + r|r R} is an abelian group under the operation defined by

(I + r1) + (I + r2) = I + (r1 + r2).

In fact, we get a ring structure in R/I .

Theorem 10.5. Let I be an ideal in the ring R. Then the set of cosets forms a ring (R/I, +, ·) under the operations defined by

(I + r1) + (I + r2) = I + (r1 + r2)

and

(I + r1)(I + r2) = I + (r1r2).

This ring (R/I, +, ·) is called the quotient ring (or factor ring) of R by I .

Proof. As mentioned above, (R/I, +) is an abelian group; thus we only have to verify the axioms related to multiplication.

We first show that multiplication is well defined on cosets. Let I + r1 = I + r1

and I + r2 = I + r2, so that r1 r1 = i1 I

and r2 r2 = i2 I . Then

r r

(i

r

)(i

r

)

=

i i

r i

i r

r r .

1 2 =

1 +

1

2 +

2

 

1 2

+ 1 2 +

1 2 +

1 2

Now, since I is an ideal, i1i2, r1i2

and i1r2 I . Hence r1r2 r1r2 I , so

I + r1r2 = I + r1r2, which shows that multiplication is well defined on R/I .

Multiplication is

associative and

distributive over

addition. If r1, r2, r3

R, then

(I + r1){(I + r2)(I + r3)} = (I + r1)(I + r2r3) = I + r1(r2r3) = I + (r1r2)r3

= (I + r1r2)(I + r3) = {(I + r1)(I + r2)}(I + r3).

Also,

(I + r1){(I + r2) + (I + r3)} = (I + r1){I + (r2 + r3)} = I + r1(r2 + r3)

=I + (r1r2 + r1r3) = (I + r1r2) + (I + r1r3)

={(I + r1)(I + r2)} + {(I + r1)(I + r3)}.

The other distributive law can be proved similarly. The multiplicative identity is I + 1. Hence (R/I, +, ·) is a ring.