Problems and Solutions to Abstract Algebra (Beachy, Blair)

Problems and Solutions for Abstract Algebra

John A. Beachy & William D. Blair

April 24, 2012

Contents

5 Commutative Rings 3

5.1 Commutative Rings; Integral Domains . . . . . . . . . . . . . . . 35.2 Ring Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . 115.3 Ideals and Factor Rings . . . . . . . . . . . . . . . . . . . . . . . 175.4 Quotient Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

6 Fields 32

6.1 Algebraic Elements . . . . . . . . . . . . . . . . . . . . . . . . . . 326.2 Finite and Algebraic Extensions . . . . . . . . . . . . . . . . . . . 366.3 Geometric Constructions . . . . . . . . . . . . . . . . . . . . . . . 386.4 Splitting Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.5 Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456.6 Irreducible Polynomials over Finite Fields . . . . . . . . . . . . . 486.7 Quadratic Reciprocity . . . . . . . . . . . . . . . . . . . . . . . . 53

7 Structure of Groups 56

7.1 Isomorphism Theorems: Automorphisms . . . . . . . . . . . . . . 567.2 Conjugacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

1

CONTENTS 2

Here's some random crap from chapter 4 that I did by accident.

9. Let f(x) = anxn + an−1x

n−1 + · · · a1x+ a0 be a polynomial with rationalcoe�cients. Show that if c is a root of f(x) and c 6= 0, then 1/c is a root ofg(x) = a0x

n + a1xn−1 + · · ·+ an−1x+ an.

Examine g(1/c) =∑nj=0

an−jcj . Now, if this sum is equal to zero, then so

must the sum cn∑nj=0

an−jcj . Now cn

∑nj=0

an−jcj =

∑nj=0 an−jcn−j = f(c) = 0.

We now see that g(1/c) = f(c)/cn. Since f(c) = 0 and c 6= 0, we conclude thatg(1/c) = 0.

10. Let f(x) = anxn+an−1x

n−1 + · · · a1x+a0 be a polynomial with rationalcoe�cients. Show that if c is a root of f(x), and k is a nonzero constant, thenkc is a root of g(x) = bnx

n + bn−1xn−1 + · · ·+ b1x+ b0 where bi = kn−iai.

Examine g(kc) =∑nj=0 k

n−jkjcj = kn∑nj=0 ajc

j . Since f(c) = 0, we know

that∑nj=0 ajc

j = 0. Thus, kn∑nj=0 ajc

j = g(kc) = 0. Thus, kc is a root ofg(x).

Chapter 5

Commutative Rings

5.1 Commutative Rings; Integral Domains

1. Which of the following sets are subrings of the �eld Q of rational num-bers? Assume that m, n are integers with n 6= 0 and (m,n) = 1.

(a){mn |n is odd

}(b)

{mn |n is even

}(c){mn |4 - n

}(d)

{mn |(n, k) = 1

}where k is a �xed positive integer

(a) Since we are very familiar withQ, we know that + is abelian and associa-tive and we know inverses and identity for +. We also know that · distributesover +. Thus, we only need to show closure. Observe that for a, b, c, d ∈ Z,ab + c

d = ad+bcbd . Now, since b and d are both odd (they do not have a factor of

2), there is no way that bd can be even. (Even if there is some cancellation thatoccurs, a factor of 2 can never magically appear in the denominator.)

(b) Examine 12 + 1

2 = 1 = 11 /∈

{mn |n is even

}. Thus, the group property is

not satis�ed, so (b) is not a subring.(c) Examine 1

3 + 27 = 13

28 /∈{mn |4 - n

}. Thus, the group property is not

satis�ed, so (b) is not a subring.(d) Again we only need to show closure. since a

b + cd := ad+bc

bd , and neither bnor d share any factor in common with k, it is clear that (bd, k) = 1. Thus, (d)is a subring.

2. Which of the following sets are subrings of the �eld R of real numbers?(a) A = {m+ n

√2|m,n ∈ Z and n is even}

(b) B = {m+ n√

2|m,n ∈ Z and m is odd}(c) C = {a+ b 3

√2|a, b ∈ Q}

(d) D = {a+ b 3√

2 + c 3√

9|a, b, c ∈ Q}(e) E = {m+ nu|m,n ∈ Z},where u = (1 +

√3)/2

3

CHAPTER 5. COMMUTATIVE RINGS 4

(f) F = {m+ nv|m,n ∈ Z}, where v = (1 +√

5)/2

(a) As in Exercise 1, we will only need to show closure of + (or lack thereof).We see that (a+ b

√2) + (c+ d

√2) = (a+ c) + (b+ d)

√2. The sum of two even

numbers is even, so A is a subring.(b) Examine (3 +

√2) + (3 +

√2) = 6 + 2

√2 /∈ B. Thus, B is not a subring.

(c) Similar to (a), (a + b 3√

2) + (c + d 3√

2) = (a + c) + (b + d) 3√

2. The sumof two rationals is rational, so C is a subring.

(d) Similar to (c), D is a subring.

(e) Examine (1 + u) + (1 + u) = 2 + 2u = 2 + 2 · 1+√3

2 = 3 +√

3 /∈ E. Thus,E is not a subring.

(f) Similarly, F is not a subring.(Note: Sets like A,B,C,D,E, and F can almost be thought of as vector

spaces. For example, D can be thought of as a vector space over Q with basis{1, 3√

2, 3√

9}. Then, one only needs to check that the set from which the co-e�cients come is closed, since one can always write vectors as a unique linearcombination. Sets like A don't technically form a vector space since Z is a ringand not a �eld. However, there is still a concept of basis associated with thisset. Such a set with a basis is called a �free module.�)

3. Consider which of the following conditions on the set of all 2×2 matrices[a bc d

]with rational entries. Which conditions below de�ne a commutative

ring? If the set is a ring, �nd all units.(a) all matrices with d = a, c = 0(b) all matrices with d = a, c = b(c) all matrices with d = a, c = −2b(d) all matrices with d = a, c = −b(e) all matrices with c = 0(f) all matrices with a = 0 and d = 0.

(a) We know that matrix addition is commutative and we know inversesand identity. Furthermore, matrix multiplication distributes over matrix addi-tion. Thus, we only need to check for closure and commutivity of ·. Exam-

ine

[a b0 a

]+

[c d0 c

]=

[a+ c b+ d0 a+ c

]. Also,

[a b0 a

]·[c d0 c

]=[

ac ad+ bc0 ac

]=

[c d0 c

]·[a b0 a

]. Thus, (a) de�nes a commutative ring.

From elementary linear algebra, we see that

[a b0 a

]−1=

[1a

−ba2

0 1a

]which

is de�ned so long as a 6= 0. So, the set of units is the set de�ned by (a) minusany matrix where a = 0.

(b) Examine

[a bb a

]+

[c dd c

]=

[a+ c b+ db+ d a+ c

]. Also,

[a bb a

]·

CHAPTER 5. COMMUTATIVE RINGS 5[c dd c

]=

[ac+ bd ad+ bcad+ bc ac+ bd

]=

[c dd c

]·[a bb a

]. Thus, (b) de�nes a

commutative ring. From elementary linear algebra,

[a bb a

]−1=

[ aa2−b2

−ba2−b2

−ba2−b2

aa2−b2

]which is de�ned as long as a2 − b2 6= 0. Thus, the set of units is the set de�nedby (b) minus those elements where a2 − b2 = 0.

(c) Examine

[a b−2b a

]+

[c d−2d c

]=

[a+ c b+ d−2(b+ d) a+ c

]. Also,[

a b−2b a

]·[c d−2d c

]=

[ac− 2bd ad+ bc−2ad− 2bc ac− 2bd

]=

[c d−2d c

]·[a b−2b a

].

Thus, (c) de�nes a commutative ring. Now,

[a b−2b a

]−1= 1

a2+2b2

[a −b2b a

]which is de�ned as long as a and b are not both equal to 0.

(d) Examine

[a b−b a

]+

[c d−d c

]=

[a+ c b+ d−(b+ d) a+ c

]. Multiplica-

tion is commutative since we can factor out 12 and obtain elements in (c). Thus,

(d) de�nes a commutative ring. We see

[a b−b a

]−1= 1

a2+b2

[a −bb a

]which

again is de�ned as long as a and b are not both 0.

(e) Closure is clear. However,

[3 20 1

]·[

4 50 6

]=

[12 270 6

]6=[

4 50 6

]·[

3 20 1

]=

[12 130 6

]. Thus, (e) does not de�ne a commutative ring.

(f) Closure is clear. However,

[0 12 0

]·[

0 34 0

]=

[4 00 6

]6=[

0 34 0

]·[

0 12 0

]=

[6 00 4

]. Thus, (f) does not de�ne a commutative ring.

4. Let R = {m+ n√

2|m,n ∈ Z}.(a) Show that m+ n

√2 is a unit in R if and only if m2 − 2n2 = ±1.

(b) Show that 1 + 2√

2 has in�nite order in R×.(c) Show that 1 and −1 are the only units that have �nite order in R×.

(a) We see that m + n√

2 is a unit if and only if there is an x ∈ R suchthat x(m + n

√2) = (m + n

√2) = 1. By commutivity of ·, we only need to

consider x(m + n√

2). Now, x(m + n√

2) = 1 if and only if x = 1m+n

√2.

Multiplying numerator and denominator by the conjugate, this implies x =1

m2−2n2 (m − n√

2). For x to be a member of R, we need mm2−2n2 ,

nm2−2n2 ∈ Z.

Thus, x ∈ R if and only if m2 − 2n2 = ±1.

(b) By the binomial theorem, we have that (1 + 2√

2)r =∑rs=0

(rs

)(2√

2)s

which clearly can become arbitrarily large. Thus, o(1 + 2√

2) =∞.(c) First of all, by (a), any unit must have the form ±

√±1 + 2n2 + n

√2.

Then, by the binomial theorem,(±√±1 + 2n2 + n

√2)r

=∑rs=0

(±√±1 + 2n2

)snr−s

√2r−s

.


This can become arbitrarily large unless n = 0 in which case ±√±1 + 2n2 +

n√

2 = ±1.

5. Let R be a subset of an integral domain D. Prove that if R is a ringunder the operations of D, then R is a subring of D.

Closure for + and · is assumed since R is a ring under the operations of D.Since (R,+) is an abelian group, we must have inverses. Thus, if a ∈ R, wemust have −a ∈ R. Finally, suppose that x ∈ R. Then, since 1 · x = x ∈ R, wemust have 1 ∈ R by closure. Thus, R is a subring of D.

6. Let D be a �nite integral domain. Give another proof of Theorem 5.1.9by showing that if d is a nonzero element of D, then d−1 = dk for some positiveinteger k. [Theorem 5.1.9 states that any �nite integral domain is a �eld.]

Let d ∈ D. By closure, d · d ∈ D. Now, D is assumed to be �nite, so forsome m and n, we must have dm = dn. (Otherwise D would have to be in�nite.)Then, dm = dn ⇔ 1 = dn−m by the cancellation property for integral domains.Thus, 1 = d · dn−m−1. Set k = n−m− 1. Then clearly d−1 = dk. Thus, everyelement has an inverse, so D is a �eld.

7. An element a of a commutative ring R is called nilpotent if an = 0 forsome positive integer n. Prove that if u is a unit in R and a is nilpotent, thenu− a is a unit in R.

Consider the element an − 1 = 1. Now, 1 is clearly a root of an − 1, so wecan write an−1 = (a−1) ·p(a) where p(a) is a polynomial in the indeterminanta. Thus, (a− 1) · p(a) = 1, so (a− 1) is a unit. (We see that p(a) ∈ R since itis just linear combos of powers ofa.)

Now, let u ∈ R×. Then, (a − u) = u(au−1 − 1). Since a is nilpotent, so isau−1. Thus, by the previous paragraph, (au−1 − 1) ∈ R×. Since u ∈ R×, wemust have u(au−1 − 1) ∈ R×.

8. Let R be a commutative ring such that a2 = a for all a ∈ R. Show thata+ a = 0 for all a ∈ R.

Notice that (a+1)2 = a2+2a+1. But, a2 = a for all a ∈ R, so (a+1)2 = a+1and a+ 1 = a+ 2a+ 1. Thus, a+ 1 = 3a+ 1 implying 0 = 2a, as desired.

9. Let I be any set and let R be the collection of all subsets of I. De�neaddition and multiplication of subsets A,B ⊆ I as follows:

A+B = (A ∪B) ∩A ∩B and A ·B = A ∩B.

Show that R is a commutative ring under this addition and multiplication.


We simply verify the axioms.(Closure of +) Since A,B ⊆ I, A ∪ B ⊆ I by de�nition of union. Then

the intersection of any set with a set in I is completely contained in I. Thus,A+B ⊆ I.

(Associativity of +) Omitted.(Zero Element) ∅ is the zero element: A+ ∅ = (A− ∅) ∪ (∅ −A) = A ∪ ∅.(Additive Inverses) If A ∈ R, then A + A = (A − A) ∪ (A − A) = ∅. Thus,

every element is its own additive inverse.(Multiplicative identity) I is the multiplicative inverse: A · I = A ∩ I =

I ∩A = A.(Commutivity) Given since ∩ is commutative.(Distributivity of · over +) Omitted.

10. For the ring R de�ned in Exercise 9, write out addition and multiplica-tion tables for the following cases:

(a) I has two elements;(b) I has three elements.

(a) Suppose I = {a, b}. Then 2I = {I, {a}, {b}, ∅}. Here is the additiontable:

Here is the multiplication table:

(b) Suppose I = {a, b, c}. Then 2I = {I, {a, b}, {a, c}, {b, c}, {a}, {b}, {c}, ∅}.Here is the addition table:

Here is the multiplication table:


(b) Omitted

11. A commutative ring R is called a Boolean ring if a2 = a for all a ∈ R.Show that in a Boolean ring the commutative law follows from the other axioms.

Let a, b ∈ R. Then ab = ba⇔ a2b = aba⇔ ab = aba⇔ ab2 = abab⇔ ab =(ab)2. Since R is boolean, the last equality is true, so all the equalities that arederived from it� including ab = ba are true. This completes the proof.


A+B = A ∪B and A ·B = A ∩B.

Is R a commutative ring under this addition and multiplication?

Observe that ∅ is the additive identity of R since A ∪ ∅ = A for all A ⊆ I.However, one can never union a set to any nonempty set and obtain ∅. Thus,unless I = ∅, R is not a ring.

13. Let R be the set of all continuous functions from the set of real numbersto itself.

(a) Show that R is a commutative ring if the formulas (f+g)(x) = f(x)+g(x)and (f · g)(x) = f(x) · g(x) for all x ∈ R are used to de�ne addition andmultiplication of functions.

(b) Which properties in the de�nition of commutative ring fail if the productof two functions is de�ned to be (fg)(x) = f(g(x)) for all x?

(a) (Associativity of +) Given since + is associative.(Additive Identity) The function f ≡ 0 is continuous and (g + f)(x) = g(x)

for all continuous functions g.(Additive Inverses) For a continuous function f , the function −f(x) = (−1) ·

f(x) is its additive inverse.(Commutativity) Given since · is commutative.(Distributivity) (f ·(g+h))(x) = f(x)·[g(x)+h(x)] = f(x)g(x)+f(x)h(x) =

(fg)(x) + (fh)(x).

(b) Commutativity fails and so does distributivity. For example, if f(x) = x2

and g(x) = −x, then (fg)(x) = x2, but (gf) = −x2. Also, if f(x) = x2, g(x) =x3, and h(x) = x4, then f(g + h)(x) = (x3 + x4)2, while f(g(x)) + f(h(x)) =x6 + x8.

14. De�ne new operations on Q by letting a⊕ b = a+ b and a� b = 2ab forall a, b ∈ Q. Show that Q is a commutative ring under these operations.

(Group under +) Given.


(Commutativity of �) a� b = 2ab = 2ba = b� a.(Distributivity of� over⊕) a�(b⊕c) = 2a(b+c) = 2ab+2ac = (a�b)⊕(a�c)

15. De�ne new operations on Z by letting m⊕ n = m+ n− 1 and m� n =m+n−mn, for all m,n ∈ Z. Is Z a commutative ring under these operations?

(Associativity of ⊕) Given by associativity of + and -.(Zero element) 1 is the additive identity since m+ 1− 1 = m for all m ∈ Z.(Additive Inverseses) For allm ∈ Z, −m+2 is such thatm+(−m+2)−1 = 1.(Identity element) 0 is the identity element since m�0 = m+0−m(0) = m.(Commutativity of �) m� n = m+ n−mn = n+m− nm = n�m.(Distributivity of � over ⊕) ` � (m ⊕ n) = ` � (m + n − 1) = ` + m + n −

1− (`m+ `n− `) = (`+m− `m) + (`+ n− `n)− 1 = (`�m)⊕ (`� n).

16. Let R and S be commutative rings. Prove that the set of all orderedpairs (r, s) such that r ∈ R and s ∈ S can be given a ring structure by de�ning

(r1, s1) + (r2, s2) = (r1 + r2, s1 + s2) and (r1, s1) · (r2, s2) = (r1r2, s1s2).

This is called the direct sum of R and S, denoted by R⊕ S.

(Associativity of +) Given by the associativity of vector addition.(Zero Element) The element (0R, 0S) where 0R and 0S are the zero elements

of R and S respectively is the zero element of R⊕ S.(Additive Inverses) If (a, b) ∈ R⊕S, then (−a,−b) is its inverse since (a, b)+

(−a,−b) = (0R, 0S).(Identity Element) The element (1R, 1S) where 1R and 1S are the identity

elements of R and S is the identity element since (a, b) · (1R, 1S) = (a, b).(Commutativity) (a, b) · (c, d) = (ac, bd) = (ca, db) = (c, d) · (a, b).(Distributivity) (a, b)·[(c, d)+(e, f)] = (a, b)·(c+e, d+f) = (ac+ae, bd+bf) =

(a, b) · (c, d) + (a, b) · (e, f).Thus, R⊕ S is a commutative ring.

17. Give addition and multiplication tables for Z2 ⊕ Z2.

Omitted.

18. Generalizing to allow the direct sum of three commutative rings, giveaddition and multiplication tables for Z2 ⊕ Z2 ⊕ Z2.

Omitted.

19. Find all the units of the following rings.(a) Z⊕ Z(b) Z4 ⊕ Z9


(a) From Q16, the identity element is (1, 1). Now, (a, b) · (c, d) = (1, 1) ⇔(ac, bd) = (1, 1). Since a, b, c, d ∈ Z, we can only have ac = 1 and bd = 1 ifa = b = c = d = 1.

(b) From Q16, the identity element is ([1]4, [1]9). Now, ([a]4, [b]9)·([c]4, [d]9) =([1]4, [1]9) ⇔ ([ac]4, [bd]9). Now, we simply need to �nd the units of Z4 andZ9. We know Z×4 = {[1]4, [3]4} and Z9 = {[1]9, [2]9, [4]9, [5]9, [7]9, [8]9}. Thus,(Z4 ⊕ Z9)× = {(x, y)|x ∈ Z4 and y ∈ Z9}.

20. An element e of a ring R is said to be idempotent if e2 = e. Find allidempotent elements of the following rings.

(a) Z8 and Z9

(b) Z10 and Z12

(c) Z⊕ Z(d) Z10 ⊕ Z12

(a) Idempotent elements of Z8: [0]8, [1]8Idempotent elements of Z9: [0]9, [1]9(b) Idempotent elements of Z10: [0]10, [1]10, [5]10, [6]10Idempotent elements of Z12: [0]12, [1]12, [4]12, [9]12(c) Idempotent elements of Z⊕ Z: (0, 0), (1, 1)(d) Idempotent elements of Z10⊕Z12: {(x, y)|x ∈ {idempotent elements of Z10}, y ∈

{idempotent elements of Z12}}

21. Let A be an abelian group, and let R = {(a, b)|a ∈ A and n ∈ Z}.De�ne binary operations + and · on R by (a, n) + (b,m) = (a + b, n + m) and(a, n) · (b,m) = (am+ bn, nm), for all (a, n) and (b,m) in R. Show that R is acommutative ring.

(Group property) Satis�ed since the underlying group structure is the carte-sian product of two groups.

(Identity element) We need an element (x, y) ∈ R such that (x, y) · (a, n) =(xn + ay, n) = (a, n). In order for (x, y) · (a, n) = (a, n), we must have yn = nwhich can only happen if y = 1. Thus, we need xn+ a = a which implies thatx = 0A. Thus, (0A, 1) is the identity element.

(Commutativity) Given.(Distributivity) (a, `)[(b,m)+(c, n)] = (a, `)·(b+c,m+n) = ((a(m+ n) + `(b+ c), `(m+ n)).

(1)(a, `) · (b,m) + (a, `)(c, n) = (am+ `b, `m) + (an+ `c, `n) = (am+ `b+ an+

`c, `m+ `n) = (a(m+ n) + `(b+ c), `(m+ n)). (2)We see that (1) = (2), so distributivity has been shown.

22. Let R be a set that satis�es all the axioms of a commutative ring, withthe exception of the existence of a multiplicative identity element. De�ne binaryoperations + and · on R1 = {(r, n)|r ∈ R,n ∈ Z} by (r, n) + (s,m) = (r +s, n+m) and (r, n) · (s,m) = (rs+ns+mr, nm), for all (r, n) and (s,m) in R1.Show that R1 is a commutative ring with identity (0, 1) and that {(r, 0)|r ∈ R}


satis�es all the conditions of a subring, with the exception that it does not havethe multiplicative identity of R.

(Group property) Satis�ed since the underlying group structure is the carte-sian product of two groups.

(Identity element) Similar to Q21.(Commutativity) Given except for multiplication with the identity element.

Since we showed that (0,1) was the identity element, we know it commutes (byde�nition of identity element).

(Distributivity) (a, `)[(b,m)+(c, n)] = (a, `)·(b+c, n+m) = (a(b+ c) + `(b+ c) + a(n+m), `(n+m)).(1)

(a, `) · (b,m) + (a, `) · (c, n) = (ab + `b + am, `m) + (ac + `c + an, `n) =(ab+`b+am+ac+`c+an, `m+`n) = (a(b+ c) + `(b+ c) + a(m+ n), `(m+ n)).(2)

We see (1) = (2), so distributivity has been shown.

5.2 Ring Homomorphisms

1. Let R be a commutative ring and let D be an integral domain. Letφ : R→ D be a nonzero function such that φ(a+ b) = φ(a) + φ(b) and φ(ab) =φ(a)φ(b), for all a, b ∈ R. Show that φ is a ring homomorphism.

We must simply show that φ(1) = 1. Now, φ(a) = φ(1 · a) = φ(1) · φ(a).Since D is an integral domain, the cancellation property can be used to obtainφ(a) = φ(1) · φ(a)⇔ 1 = φ(1).

2. Let F be a �eld and let φ : F → R be a ring homomorphism. Show thatφ is either zero or one-to-one.

Clearly φ ≡ 0 is a possible mapping.Suppose that φ(a) = φ(b) for some a, b ∈ F . Since F is a �eld, this implies

that φ(ab−1) = 1. This then implies that ab−1 = 1 which implies that a = b.

3. Let F,E be �elds, and let φ : F → E be a ring homomorphism. Showthat if φ is onto, then φ must be an isomorphism.

We only need to show that φ is one-to-one. Suppose that φ(x1) = φ(x2) = y.Then, since E is a �eld, φ(x1) · φ(x2)−1 = 1. Since F is a �eld, we see that thisimplies that φ(x1x

−12 ) = 1. This means that xx−12 = 1, so x1 = x2. Thus, φ is

one-to-one.

4. Show that taking complex conjugates de�nes an automorphism of C. Thatis, for z ∈ C, de�ne φ(z) = z, and show that φ is an automorphism of C.


(1-1) Let z = reiθ. Then φ(reiθ) = re−iθ. This mapping is one-to-one thenby the uniqueness of additive inverses.

(Onto) Let w = reiθ. Then, z = re−iθ gives φ(z) = w.(Preservation) Let z = a+ ib and w = c+ id. Then φ(z + w) = φ[(a+ c) +

i(b+ d)] = (a+ c)− i(b+ d) = (a− ib) + (c− id) = φ(z) + φ(w).Let z = reiθ and w = seiω. Then, φ(zw) = φ(rsei(θ+ω)) = rse−i(θ+ω) =

re−iθse−iω = φ(z)φ(w).(Identity) Satis�ed since C is a �eld. (See Q1.)

5. Show that the identity mapping is the only ring homomorphism from Zinto Z.

Let φ : Z → Z be a ring homomorphism. Then, we must have φ(1) = 1.Since for any n ∈ Z, n = 1 + 1 + · · · + 1 (n terms), we must have φ(n) =φ(1 + 1 + · · ·+ 1) = φ(1) + φ(1) + · · ·+ φ(1) = 1 + 1 + · · ·+ 1 = n.

6. Show that the set of all matrices over Z of the form

[m n2n m

]is a ring

isomorphic to the ring Z[√

2] de�ned in Example 5.1.6.

Let S be the set described in the statement of the problem. De�ne φ :

Z[√

2]→ S via φ(a+ b√

2) =

[a b2b m

].

(1-1) This map is a direct substitution, so it is 1-1.

(Onto). Let

[m n2n m

]∈ S. Then, φ(m+ n

√2) =

[m n2n m

].

(Identity) φ(1) =

[1 00 1

].

(Preservation) φ[(a+b√

2)+(c+d√

2)] = φ[(a+c)+(b+d)√

2] =

[a+ c b+ d2(b+ d) a+ c

]=[

a b2b a

]+

[c d2d c

]= φ(a+ b

√2) + φ(c+ d

√2).

φ[(a+b√

2)(c+d√

2)] = φ[(ac+2bd)+(ad+bc)√

2] =

[ac+ 2bd ad+ bc2(ad+ bc) ac+ 2bd

].

(1)

φ(a+b√

2)φ(c+d√

2) =

[a b2b a

]·[c d2d c

]=

[ac+ 2bd ad+ bc2(ad+ bc) ac+ 2bd

].

(2)We see (1) = (2), so preservation has been shown.

7. De�ne φ : Z[√

2]→ Z[√

2] by φ(m+ n√

2) = m− n√

2, for all m,n ∈ Z.Show that φ is an automorphism of Z[

√2].

This is similar to Q4. We will only show preservation of products. (Theidentity mapping is very easily veri�ed).


φ[(a+ b√

2)(c+ d√

2)] = φ[ac+ 2bd+ (ad+ bc)√

2] = ac+ 2bd− (ad+ bc)√

2.(1)

φ(a+ b√

2)φ(c+ d√

2) = (a− b√

2)(c− d√

2) = ac+ 2bd− (ad+ bc)√

2. (2)(1) = (2), so preservation is shown.

8. Let F be a �eld, and let a ∈ F . De�ne φ : F [x] → F [x] by φ(f(x)) =f(x+ a), for all f(x) ∈ F [x]. Show that φ is an automorphism of F .

(1-1) The transformation f(x+ a) simply takes the graph of f(x) and shiftsit left a units. (This does make sense in an abstract setting since the de�nitionof graph is simply the set of ordered pairs of the form (x, f(x)).) Thus, thismapping of f(x) 7→ f(x+ a) is a direct substitution and thus one-to-one.

(Onto) Let f(x) ∈ F [x]. Then, f(x− a) 7→ f(x).(Identity) Let f ≡ 1. Then, f(x+ a) ≡ 1.(Preservation) Let f(x) = anx

n + an−1xn−1 + · · · + a1x + a0 and g(x) =

bmxm + bm−1x

m−1 + · · ·+ b1x+ b0. Without loss of generality, suppose deg g ≥deg f .

φ[f(x)+g(x)] = φ[bmxm+bm−1x

m−1+· · ·+(an+bn)xn+(an−1+bn−1)xn−1+· · ·+(a1+b1)x+(a0+b0)]= bm(x+a)m+ · · ·+(an+bn)(x+a)n+ · · ·+a0+b0 =(bm(x+ a)m + · · ·+ b0) + (an(x+ a)n + · · ·+ a0) = φ(g(x)) + φ(f(x)).

Let f(x) = (x−r1)(x−r2) · · · (x−rn) and g(x) = (x−s1)(x−s2) · · · (x−sm).φ[f(x)g(x)] = φ[(x−r1) · · · (x−rn)(x−s1) · · · (x−sm)] = (x+a−r1) · · · (x+

a− rn)(x+ a− s1) · · · (x+ a− sm) = φ(f(x))φ(g(x)).

9. Show that the composition of two ring homomorphisms is a ring homo-morphism.

Let φ : S → T and π : R→ S be ring homomorphisms. De�ne φ◦π : R→ Tvia φ ◦ π(x) = φ(π(x)) for all x ∈ R.

(Preservation of +) φ(π(x+ y)) = φ(π(x) + π(y)) = φ(π(x)) + φ(π(y)).(Preservation of ·) φ(π(xy)) = φ(π(x)π(y)) = φ(π(x))φ(π(y)).(Identity) φ(π(1)) = φ(1) = 1.

10. Let R and S be rings, and let φ, θ : R → S be ring homomorphisms.Show that {r ∈ R|φ(r) = θ(r)}is a subring of R.

(Closure under +) Let r, s ∈ R. Then φ(r+ s) = φ(r) +φ(s) = θ(r) + θ(s) =θ(r + s). Thus, (r + s) ∈ R.

(Closure under ·) φ(rs) = φ(r)φ(s) = θ(r)θ(s) = θ(rs). Thus, rs ∈ R.(Additive inverses) Let r ∈ R. Then φ(−r) = −φ(r) = −θ(r) = θ(−r).

Thus, −r ∈ R.(Identity) φ(1) = 1 = θ(1). Thus, 1 ∈ R.

11. Show that the direct sum of two nonzero rings is never an integraldomain.


Let R and S be nonzero rings. Then, we always have (1, 0) · (0, 1) = (0, 0).Neither element is zero, but the product is the zero of R⊕ S.

12. Let R1 and R2 be commutative rings.(a) De�ne π1 : R1⊕R2 → R1 by π1((r1, r2)) = r1, for all (r1, r2) ∈ R1⊕R2

and de�ne π2 : R1 ⊕ R2 → R2 by π2((r1, r2)) = r2 for all (r1, r2) ∈ R1 ⊕ R2.Show that π1 and π2 are ring homomorphisms.

(b) Let R be any ring, and let φ : R→ R1⊕R2 be a function. Show that φ isa ring homomorphism if and only if π1φ and π2φ are both ring homomorphisms.

(a) (Preservation of +) π1[(r, s) + (v, w)] = π1(r + v, s + w) = r + v =π1(r, s) + π1(v, w).

(Preservation of ·) π1[(r, s) + (v, w)] = π1(rv, sw) = rv = π1(r, s)π1(v, w).(Identity) π1(1, 1) = 1.Thus, π1 is a ring homomorphism. Similarly, π2 is a ring homomorphism.

(b) (⇒)Suppose that one of π1φ or π2φ is not a homomorphism. Withoutloss of generality, suppose that π1φ is not a homomorphism. By Q9, if π1and φ are homomorphisms, then π1φ is a homomorphism. By contraposition,if π1φ is not a homomorphism, then one of π1 or φ is not a homomorphism.Part (a) showed that π1 is a homomorphism, so we conclude that φ is not ahomomorphism.

(⇐) Suppose that φ is not a ring homomorphism. Then, at least one ofthree things must occur: (1) φ(r + s) 6= φ(r) + φ(s), (2) φ(rs) 6= φ(r)φ(s), or(3) φ(1) 6= 1.

For (1), suppose that φ(r+ s) = (v, w) while φ(r) = (a, b) and φ(s) = (c, d).Then, π1(φ(v + w)) = π1((a + c, b + d)) = a + c. Similarly, π2(φ(v + w)) =b + d. Now, it is possible for either a + c = v or b + d = w, but not both.For if both held, it would imply that π1(φ(v + w)) = π1(φ(v) + φ(w)) andπ2(φ(v+w)) = π2(φ(v) +φ(w)), which would imply that π1 and π2 are not wellde�ned. Contradiction. Thus, one of π1φ or π2φ is not a homomorphism.

(2) Follows from a similar argument.For (3), if φ(1) 6= (1, 1), then at least one of π1(1, 1) or π2(1, 1) is not equal

to the identity. Thus, one of π1φ or π2φ is not a homomorphism.

13. Find all ring homomorphisms from Z ⊕ Z into Z. That is, �nd allpossible formulas and show why no others are possible.

Let φ : Z⊕Z→ Z be a homomorphism. Then, since the identities must mapto each other, we see that (1, 1) 7→ 1. Then, since n = 1 + 1 + · · ·+ 1 (n terms)for any n ∈ Z, φ(n, n) = n for all n ∈ Z. Thus, we must have φ(m,n) = m orφ(m,n) = n for all (m,n) ∈ Z⊕ Z.

14. Find all ring homomorphisms from Z⊕ Z into Z⊕ Z.


As above, we must have (1, 1)φ7→ (1, 1). By extension, we must have (n, n) 7→

(n, n). Thus, we must have φ(m,n) = (m,n), φ(m,n) = (m,m), or φ(m,n) =(n, n).

15. For the rings Zn and Zk, show that if k|n, then the function φ : Zn → Zkde�ned by φ([x]n) = [x]k, for all [x]n ∈ Zn, is a ring homomorphism. Show thatthis is the only ring homomorphism from Zn to Zk.

(Preservation of +) φ([a]n + [b]n) = φ([a + b]n) = [a + b]k = [a]k + [b]k =φ([a]n) + φ([b]n).

(Preservation of ·) φ([a]n[b]n) = φ([ab]n) = [ab]k = [a]k[b]k = φ([a]n)φ([b]n).Now, as mentioned many times before, if φ : Zn → Zk is a homomorphism,

we must have [1]n 7→ [1]k. Also, since for any r ∈ Z, n = 1 + 1 + · · · + 1 (rterms), we know that φ([r]n) = φ([1]n + [1]n + · · ·+ [1]n) = φ([1]n) + φ([1]n) +· · · + φ([1]n) = [1]k + [1]k + · · · + [1]k = [r]k. Thus φ([x]n) = [x]k is the onlyhomomorphism from Zn to Zk.

16. Are Z9 and Z3 ⊕ Z3 isomorphic as rings?

Let φ : Z9 → Z3 ⊕ Z3. If φ is to be an isomorphism, we need [1]9 7→([1]3, [1]3). However, 3 · [1]9 = [3]9 while 3 · ([1]3, [1]3) = ([0]3, [0]3), so theunderlying additive groups are not isomorphic, so Z9 � Z3 ⊕ Z3.

17. Let S be the subset of Z4 ⊕ Z4 given by {([m]4, [n]4)|m ≡ n(mod 2)}.(a) Show that S is a subring of Z4 ⊕ Z4.(b) Show that S is not isomorphic (as a ring) to any ring of the form Zn,

nor to any direct sum of such rings.

(a) (Closure under +) Let ([a]4, [b]4), ([c]4, [d]4) ∈ S. Then, ([a]4, [b]4) +([c]4, [d]4) = ([a+ c]4, [b+ d]4). Now if [a]2 = [b]2 and [c]2 = [d]2, then [a+ c]2 =[b+ d]2 by simple adding of two valid equations. Thus ([a+ c]4, [b+ d]4) ∈ S.

(Closure under ·) Let ([a]4, [b]4), ([c]4, [d]4) ∈ S. Then, ([a]4, [b]4)·([c]4, [d]4) =([ac]4, [bd]4). Now there are two possibilities: (1) [c]2 = [d]2 = [0]2, or (2)[c]2 = [d]2 = [1]2. So, when we evaluate [ac]2 and [bd]2, in (1) we get [0]2 and[0]2 and in (2) we get [a]2 and [b]2. Thus, ([ac]4, [bd]4) ∈ S.

(Additive inverses) If [a]2 = [b]2, then −[a]2 = −[b]2. Thus, if ([a]4, [b]4) ∈ S,we have −([a]4, [b]4) ∈ S.

(Identity Element) Clearly, [1]2 ⊂ [1]4, so ([1]4, [1]4) ∈ S.Thus, S is a subring of Z4 ⊕ Z4.

(b) We note that the element ([1]4, [1]4) ∈ S has multiplicative order 4.Thus, if S were isomorphic to a direct sum of residue rings, it would have to beisomorphic to Z4 ⊕ Z4. However, ([1]4, [2]4) ∈ Z4 ⊕ Z4 − S. Thus S � Z4.

If S were to be isomorphic to a ring of residues, it would have to be iso-morphic to Z4. However, ([n]4, [n]4) are in S for all n, giving four congruence


classes. In addition, ([1]4, [3]4) ∈ S. Thus, |S| > |Z4|, so S � Z4.

18. De�ne φ : Z → Zm ⊕ Zn by φ(x) = ([x]m, [x]n). Find the kernel andimage of φ. Show that φ is onto if and only if gcd(m,n) = 1.

We see that kerφ = {x ∈ Z : n|x and m|x}. The image is {([x]m, [x]n)|x ∈Z}. If gcd(m,n) = 1, then m and n share no factors, so no integers besides 0and kmn where k ∈ Z can be mapped to ([0]m, [0]n). Since |Zm ⊕ Zn| = mn,the �rst mn integers will be mapped to unique elements of Zm ⊕ Zn. Thus, φis onto.

Conversely, if gcd(m,n) = d 6= 1, then we have m = dr and n = dq for somer, q ∈ Z. Thus, drq 7→ ([0]m, [0]n). Since drq < mn, the mapping will replicateall the previous (drq−1) mappings over and over again without having generatedmn unique elements (the order of Zm ⊕ Zn). Thus, φ is not onto.

19. Let R be the ring given by de�ning new operations on Z by lettingm⊕n = m+n−1 and m�n = m+n−mn. De�ne φ : Z→ R by φ(n) = 1−n.Show that φ is an isomorphism.

(1-1) Given since φ is a linear map.(Onto) Let y ∈ R. Then φ(−y + 1) = 1− (−y + 1) = y. Thus, φ is onto.(Preservation of sums) φ(m+ n) = 1− (m+ n) = 1−m− n (1).φ(m)⊕φ(n) = (1−m)⊕ (1−n) = 1−m+ 1−n− 1 = 1−m−n (2). Now,

(1) = (2), so sums are preserved.(Preservation of products) φ(mn) = 1−mn. (1)φ(m)�φ(n) = (1−m)�(1−n) = 1−m+1−n−(1−m−n+mn) = 1−mn

(2). Again, (1) = (2), so products are preserved.


A+B = (A ∪B) ∩A ∩B and A ·B = A ∩B.

(a) Show that if I has two elements then R ∼= Z2 ⊕ Z2.(b) Show that if I has three elements then R ∼= Z2 ⊕ Z2 ⊕ Z2.

For both of these, one can merely construct multiplication- and additiontables and note that the rings behave in the same way. The multiplication- andaddition tables for R in these two particualar cases were already constructed inQ10 of 5.1.

21. Let R1, R2, ..., Rn be commutative rings. Complete the proof of Propo-sition 5.2.8, to show that R = R1 ⊕R2 ⊕ · · · ⊕Rn is a commutative ring. Thenshow that R× ∼= R×1 ×R

×2 × · · · ×R×n .


(Group Property) Given by Proposition 3.3.4.(Commutativity) Given by the commutativity of R1, ..., Rn.(Identity Element) (1R1

, 1R2, ..., 1Rn) is obviously the identity element.

(Distributivity) (r1, ..., rn)[(s1, ..., sn)+(t1, ..., tn)] = (r1, ..., rn)(s1+t1, ..., sn+tn) = (r1[s1+t1], ..., rn[sn+tn]) = (r1s1+r1t1, ..., rnsn+rntn) = (r1, ..., rn)(s1, ..., sn)+(r1, ..., rn)(t1, ..., tn)

Let u = (u1, ..., un) ∈ R×. We must have uu−1 = 1R = (1R1, ..., 1Rn). Thus,

(u1, ..., un)u−1 = (1R1, ..., 1Rn). We see that because R1, ..., Rn are commutative

rings, this can only happen if ui ∈ R×i and u−1 = (u−11 , ...u−1n ). Thus, R× ∼=R×1 ×R

×2 × · · · ×R×n .

22. Let R be an integral domain. Show that R contains a subring isomorphicto Zp for some prime number p if and only if char(R) = p.

char(R) = p ⇔ p · 1 = 0 for some prime p ⇔ 1 + 1 + · · · + 1 = 0 (p terms)⇔< 1 >∼= Zp.

23. Show that if R is an integral domain with characteristic p > 0, then forall a, b ∈ R we must have (a + b)p = ap + bp. Show by induction that we mustalso have (a+ b)p

n

= apn

+ bpn

for all positive integers n.

By the binomial theorem, (a+b)p =∑pk=0

(pk

)ap−kbk =

∑pk=0

p!k!(p−k)!a

p−kbk.

Clearly, p!k!(p−k)! will always yield a factor of p unless k = 0 or k = p. Thus, all

the terms of the sum are 0 except for the terms where k = 0 or k = p. Thus,(a+ b)p = ap + bp.

We have established a basis for induction. Suppose the statement is true for

all positive integers less than or equal to N . Then, (a+ b)pN+1

= (a+ b)pNp =(

(a+ b)pN)p

=(ap

N

+ bpN)p

=(ap

N)p

+(bpN)p

= apN+1

+ bpN+1

, as desired.

5.3 Ideals and Factor Rings

1. Give a multiplication table for the ring Z2[x]/〈x2 + 1〉.

We see that Z2[x]/〈x2 + 1〉 = {ax + b|a, b ∈ Z2} = {x + 1, x, 1, 0}. We usethe relation x2 + 1 = 0 to obtain the multiplication table.

· 0 1 x x+ 1

0 0 0 0 01 0 1 x x+ 1x 0 x 1 x+ 1

x+ 1 0 x+ 1 x+ 1 0


2. Give a multiplication table for the ring Z2[x]/〈x3 + x2 + x+ 1〉.

We see that Z2[x]/〈x3+x2+x+1〉 = {ax2+bx+c|a, b, c ∈ Z2} = {0, 1, x, x+1, x2, x2 + x, x2 + 1, x2 + x + 1}. We use the relation x3 + x2 + x + 1 = 0 toobtain the multiplication table.

· 0 1 x x+ 1 x2 x2 + x x2 + 1 x2 + x+ 1

0 0 0 0 0 0 0 0 01 1 x x+ 1 x2 x2 + x x2 + 1 x2 + x+ 1x x2 x2 + x x2 + x+ 1 x+ 1 x2 + 1 1

x+ 1 x2 + 1 x+ 1 x2 + 1 0 x2 + xx2 1 x2 + x x2 + 1 x

x2 + x x2 + 1 0 0x2 + 1 0 x2 + 1

x2 + x+ 1 x2

Since the multiplication is commutative, the table is symmetric about thediagonal.

3. Let R be the ring Q[x]/〈x3 + 2x2 − x − 3〉. Describe the elements of Rand give the formulas necessary to describe the product of any two elements.

We see that Q[x]/〈x3 + 2x2 − x− 3〉 = {rx2 + sx+ t|r, s, t ∈ Q}. Using therelations x3 = −2x2 + x + 3 and x4 = x3x = −2x3 + x2 + 3x = 5x2 + x − 6,we can �nd the product of any two elements in R. (The back of the book givessomething less useful.)

4. Give a multiplication table for the ring Z3[x]/〈x2 − 1〉.

We see that Z3[x]/〈x2 − 1〉 = {ax + b|a, b ∈ Z3} = {0, 1, 2, x, x + 1, x +2, 2x, 2x+ 1, 2x+ 2}. It is left to the reader to write out the table. Simply usethe relation x2 = 1 to simplify products.

5. Show that Q[x]/〈x2 − 2〉 ∼= Q[x]/〈x2 + 4x+ 2〉.

De�ne φ : Q[x]/〈x2 − 2〉 → Q[x]/〈x2 + 4x+ 2〉 via φ(f(x)) = f(x+ 2). Wehave seen from Q8 from 5.2 that φ is a homomorphism. We see that φ(x2−2) =x2+4x+2, so φ(〈x2−2〉) = 〈x2+4x+2〉. Thus,Q[x]/〈x2−2〉 ∼= Q[x]/〈x2+4x+2〉by Example 5.3.5.

6. Let R = F [x] and let I be any ideal of R.(a) Prove that there is a unique monic polynomial f(x) with I = 〈f(x)〉.(b) Prove that if I is a maximal ideal of R, then I = 〈p(x)〉 for some monic

irreducible polynomial p(x).


(a) We know that since F [x] is a principal ideal domain, we have thatI = 〈g(x)〉 for some g(x) ∈ F [x]. Suppose g(x) = anx

n + · · · a1x + a0.Then, f(x) = a−1n · g(x) is a unique monic polynomial with the same rootsas g(x). Let h(x) ∈ F [x]. Then, if h(x)/g(x) = q(x) + r(x), then h(x)/f(x) =h(x)/(a−1n g(x)) = (h(x)/g(x)) · an = anq(x) + anr(x). Thus, the map φ :F [x]/〈g(x)〉 → F [x]/〈f(x)〉 de�ned by φ(h(x)) = anh(x) gives a one-to-one,onto map which gives that F [x]/〈g(x)〉 ∼= F [x]/〈f(x)〉. Thus, I = 〈f(x)〉.

(b) From (a), we know that if I is an ideal of F [x], then I = 〈p(x)〉, wherep(x) is monic. Thus, we only need to show that if I is maximal, then p(x) isirreducible. Suppose, then, that p(x) is reducible. Then p(x) = f(x)g(x) forsome f(x) 6= g(x) ∈ F [x]. Clearly, 〈f(x)g(x)〉 ⊂ 〈g(x)〉, 〈f(x)〉. Thus, I is notmaximal.

7. Show that the intersection of two ideals of a commutative ring is againan ideal.

Suppose I and J are ideals of the commutative ring R. Then, let x, y ∈ I∩Jand r ∈ R. Then x± y ∈ I ∩ J since x± y ∈ I and x± y ∈ J (by de�nition ofideal). We also see that r · x ∈ I ∩ J : since x ∈ I and I is an ideal, r · x ∈ I;also, r · x ∈ J since J is an ideal. Thus, I ∩ J is an ideal.

8. Show that if R is a �nite ring, then every prime ideal is maximal.

Let I be a prime ideal of R. Suppose that J is an ideal of R such thatI ⊆ J ⊆ R. Let j ∈ J . Since R is �nite, there must be n > m ∈ Z+ so thatjm = jn. Thus, jm − jn = 0 ∈ I. This implies that jm(1 − jn−m) ∈ I. Now,since I is prime, this implies that either jm ∈ I or (1 − jn−m) ∈ I. If jm ∈ I,then we have jjm−1 ∈ I, implying that j ∈ I or jm−1 ∈ I. If j ∈ I, then I = Jand the problem is solved. Let us suppose that jm−1 ∈ I. Thus, jjm−2 ∈ I.Again, either j ∈ I, or jm−2 ∈ I. Continuing this factorization, we get thatjjm−(m−1) = jj ∈ I which implies j ∈ I. Thus, j ∈ I and I = J .

If (1− jn−m) ∈ I, then jm(1− jn−m) = 0 ∈ I (remembering that jm = jn).Now, jm(1− jn−m)− (1− jn−m) = (jm−1)(1− jn−m) = jm− jn−1 + jn−m =jn−m− 1 ∈ I (by the closure property. Note that jm(1− jn−m)− (1− jn−m) =−(1 − jn−m). Thus, jm − 1 = −1, showing that jm = 0 ∈ I. Then, by theargument in the previous paragraph, j ∈ I. Thus, I = J . [Somehow I think thatsomething is wrong with this paragraph, but I can't �nd any invalid statement.]

9. Find a nonzero prime ideal of Z⊕ Z that is not maximal.

Let I = pZ⊕Z where p is prime. Then, I is a prime ideal since (a, b)·(c, d) ∈ Iimplies that a = kp or b = kp for some k ∈ Z. Thus, I is prime. We will show(Q25) that the only possible prime ideals are those of the form pZ ⊕ Z wherep is prime. However, (Z ⊕ Z)/(pZ ⊕ Z) ∼= Zp which is a �eld, so the ideal


is maximum. Thus, the only nonmaximal primes are 0 ⊕ Z or Z ⊕ 0 sinceZ/(0⊕ Z) ∼= Z which is not a �eld.

10. Let P be a prime ideal of the commutative ring R. Prove that I and Jare ideals of R and I ∩ J ⊆ P , then either I ⊆ P or J ⊆ P .

Supposee that there are x ∈ I, y ∈ J such that x, y /∈ P . Then, since I and Jare ideals, they have the absorption property and so xy ∈ I ∩ J . But I ∩ J ⊆ Pand since P is prime, this implies that x ∈ P or y ∈ P . Contradiction.

11. Let R be a commutative ring, with a ∈ R. The annihilator of a isde�ned by

Ann(a) = {x ∈ R|xa = 0}.Prove that Ann(a) is an ideal of R.

Let x, y ∈ Ann(a). Then, (x ± y) · a = xa ± ya = 0 ± 0 = 0. Thus,x± y ∈ Ann(a). Let r ∈ R and x ∈ Ann(a). Then, (r · x) · a = r · 0 = 0. Thus,rx ∈ Ann(a). Thus, Ann(a) is an ideal.

12. Recall that an element of a commutative ring is said to be nilpotent ifan = 0 for some positive integer n.

(a) Show that the set N of all nilpotent elements of a commutative ring formsan ideal of the ring.

(b) Show that R/N has no nonzero nilpotent elements.(c) Show that N ⊆ P for each prime ideal P of R.

(a) Let x, y ∈ N . Then, xm = yn = 0 for some m,n ∈ Z+. Suppose, without

loss of generality, thatm ≥ n. Now, (x±y)mn =∑mnk=0(−1)k

(mnk

)xmn−kyk =∑m(n−1)

k=0 (−1)k(mnk

)xmn−kyk +

∑mnk=m(n−1)+1(−1)k

(mnk

)xmn−kyk. We see

that both sums are zero since xk = 0 for all k > m and yk = 0 for all k > n.Thus, x± y ∈ N .

Next, suppose r ∈ R and x ∈ N . Then, (r · x)m = rmxm = r · 0 = 0. Thus,rx ∈ N . Thus, N is an ideal.

(b) Let x + N ∈ R/N be such that (x + N)m = xm + N = N for somem ∈ Z+. Since xm + N = N if and only if xm = 0, this implies that R/N hasno nonzero nilpotent elements.

(c) Let x ∈ N . Then, xm = 0 for some m ∈ Z+. Then, 0 ∈ P for all primeideals P , so 0 = x · xm−1 ∈ P , implying that x ∈ P .

13. Let R be a commutative ring with ideals I, J . Let

I + J = {x ∈ R|x = a+ b for some a ∈ I, b ∈ J}.


(a) Show that I + J is an ideal.(b) Determine nZ +mZ in the ring of integers.

(a) Let x, y ∈ I + J . Then, x = a+ b and y = c+ d for a, c ∈ I and b, d ∈ J .Then, x ± y = (a ± c) + (b ± d). Since a ± c ∈ I and b ± d ∈ J , we havex± y ∈ I + J .

Next, let r ∈ R and x ∈ I + J where x = a+ b with a ∈ I and b ∈ J . Then,r · x = r(a + b) = ra + rb. Since I and J are ideals, ra ∈ I and rb ∈ J . Thus,rx ∈ I + J . Thus, I + J is an ideal.

(b) Suppose that gcd(n,m) = d. Then, n = dq1 and m = dq2. Then, letrdq1 + sdq2 ∈ nZ + mZ where r, s ∈ Z. Then, rdq1 ∈ dZ and sdq2 ∈ dZ, sonZ +mZ ⊆ dZ.

Since d = gcd(m,n), we have d = rm + sn for some r, s ∈ Z. Thus, d ∈nZ +mZ.

Thus, mZ + nZ = gcd(m,n)Z.

14. Let R be a commutative ring with ideals I, J . Let

IJ =

{n∑i=1

aibi|ai ∈ I, bi ∈ J, n ∈ Z+

}.

(a) Show that IJ is an ideal contained in I ∩ J .(b) Determine (nZ)(mZ) in the set of integers.

(a) Let x, y ∈ IJ . Then, x± y =∑aibi ±

∑ajbj ∈ IJ . Thus, x± y ∈ IJ .

Since aibi ∈ I ∩ J for all i, we also have x± y ∈ I ∩ J .Next, let r ∈ R and x =

∑aibi ∈ IJ . Then, rx =

∑raibi. Since I and J

are ideals, we have rai ∈ I for all ai, so rx ∈ IJ . On the other hand, rbi ∈ Jfor all bi, so rx ∈ I ∩ J as well.

(b) Suppose that x =∑Ni=1 rinsim where ri, si ∈ Z. Then, x = mn

∑risi.

Clearly x ∈ mnZ. Thus, (mZ)(nZ) ⊆ mnZ.Next, suppose x ∈ mnZ. Then x = kmn for some k ∈ Z. We could write

x = (km)(1 · n) ∈ (mZ)(nZ). Thus, mnZ ⊆ (mZ)(nZ).Thus, (mZ)(nZ) = mnZ.

15. Let M = {f(x, y) ∈ F [x, y]|f(0, 0) = 0} be the maximal ideal of F [x, y]de�ned in Example 5.3.8.

(a) Show that M = {s(x, y)x+ t(x, y)y|s(x, y), t(x, y) ∈ F [x, y]}.(b) Using the de�nition in Exercise 14, �nd M2.

(a) Since [s(x, y)x+t(x, y)y](0,0) = 0, we clearly have {s(x, y)x+t(x, y)y|s, t ∈F [x, y]} ⊆M .


Next, if f(0, 0) = 0, then the constant term of f must be zero. Hence,collect all the terms divisible by x into one set of brackets and everything else(which must be divisible by y since the constant term is zero) into anotherset. Call the �rst expression s(x, y)x and the second expression t(x, y)y. Thus,f(x, y) = s(x, y)x+ t(x, y)y.

(b)M2 =∑

(s1,i(x, y)x+t1,i(x, y)y)(s2,i(x, y)x+t2,i(x, y)y) =∑

[s1,i(x, y)s2,i(x, y)x2+(s1,i(x, y)t2,i(x, y) + t1,i(x, y)s2,i(x, y))xy+ t1,i(x, y)t2,i(x, y)y2]. Not that thereis no constant term. Thus, M2 = M .

16. Let R = {m+n√

2|m,n ∈ Z} and let I = {m+n√

2|m,n ∈ Z and m is even}.(a) Show that I is an ideal of R.(b) Find the well-known commutative ring to which R/I is isomorphic.

(a) Let x = 2a+ b√

2, y = 2c+ d√

2 ∈ I. Then x± y = 2(a± c) + (b± d)√

2.Thus, x± y ∈ I. Then, if x = 2a+ b

√2 ∈ I and r = c+ d

√2 ∈ R, we have rx =

(2a+b√

2)(c+d√

2) = 2ac+4bd+(2ad+bc)√

2 = 2(ac+2bd)+(2ad+bc)√

2 ∈ I.Thus, I is an ideal.

(b) Let us �nd the congruence classes. Anything of the form 2a + b√

2 isin I, so anything of the form (2a+ 1) + b

√2 is not. These are clearly the only

two classes that partition R. De�ne φ : R → Z2 via φ(2a + b√

2) = [0]2 andφ(2a + 1 + b

√2) = [1]2. Then, this is clearly one-to-one and onto. We only

really need to show that [(2a + 1) + b√

2][(2c + 1) + d√

2] = (2e + 1) + f√

2,which will be left to the reader (we have shown everything else). Then we willsee that the multiplication- and addition tables for R are the same as for Z2.Thus, R/I ∼= Z2.

17. Let R be the set of all matrices

[a bc d

]over Q such that a = d and

c = 0.(a) Verify that R is a commutative ring.(b) Let I be the set of all matrices for which a = d = 0. Show that I is an

ideal of R.(c) Use the fundamental homomorphism theorem for rings to show that

R/I ∼= Q.

(a) (Closure of +) Given since Q is closed under +.(Associativity) Given since + is associative in Q.

(Zero element)

[0 00 0

]is the zero element since

[a bc d

]+

[0 00 0

]=[

a bc d

].

(Additive Inverses)

[−a −b−c −d

]is the additive inverse of

[a bc d

].

(Commutativity of +) Given since + is commutative in Q.


(Commutativity of ·)[a b0 a

]·[c d0 c

]=

[ac ad+ bc0 ac

]=

[c d0 c

]·[

a b0 b

].

(Unity element)

[1 00 1

]is the unity element.

(Distributivity) Given because matrix multiplication is distributive.

Thus, R is a commutative ring with 1 =

[1 00 1

].

(b) Let

[0 a0 0

],

[0 b0 0

]∈ I. Then,

[0 a0 0

]±[

0 b0 0

]=

[0 a± b0 0

]∈

I. Now, let

[c d0 c

]∈ R. Then,

[c d0 c

]·[

0 a0 0

]=

[0 ac0 0

]∈ R. Thus,

I is an ideal of R.

(c) De�ne φ : R → Q by φ

(a b0 a

)= a. Clearly, φ(R) = Q. Also,

kerφ =

{[a b0 a

]∈ R

∣∣∣∣ a = 0

}. Clearly, kerφ = I. By the fundamental ho-

momorphism theorem, R/ kerφ ∼= φ(R), so R/I ∼= Q.

18. Let R be a commutative ring with ideals I, J such that I ⊆ J ⊆ R.(a) Show that J/I is an ideal of R/I,(b) Show that the factor ring (R/I)/(J/I) is isomorphic to R/J .(c) Show that J/I is a prime (or maximal) ideal of R/I if and only if J is

a prime (or maximal) ideal of R.

(a) Let a+I, b+I ∈ J/I. Then, (a+I)± (b+I) = (a± b)+I = c+I ∈ J/I.Since R/I = {x + I|x ∈ R}, we know (a + I) ± (b + I) ∈ R/I. Next, supposer ∈ R. Then, r(a+ I) = ra+ I ∈ R/I.

(b) Let π : R/I → R/J be de�ned by π(r+ I) = r+ J . (This is the naturalprojection mapping.) Then, since I ⊆ J ⊆ R, we know that the map is onto,so π(R/I) = R/J . Also, the elements that are mapped to zero (aka J) are theelements of the form j + I, where j ∈ J . These are precisely the elements fromJ/I. Thus, kerπ = J/I. Then, by the fundamental theorem of homomorphisms,we have (R/I)/(J/I) ∼= R/J .

(c) (Prime) (⇐) Suppose that J is a prime ideal of R and that we have(a+ I)(b+ I) = ab+ I ∈ J/I. Since ab+ I ∈ J/I, we have ab ∈ J and we knowthat a ∈ J or b ∈ J . Thus, a + I ∈ J/I or b + I ∈ J/I. Thus, J/I is a primeideal of R/I.

(⇒) Suppose that J is not a prime ideal of R. Then, there are x, y ∈ R suchthat xy ∈ J , yet x /∈ J and y /∈ J . Then, xy + I = (x + I)(y + I) ∈ J/I, yetsince x, y /∈ J , we know that (x+ I), (y + I) /∈ J/I.


(Maximal) By the correspondence theorem, we know that there is a one-to-one correspondence between the ideals of R/I and the ideals of R that containI. We showed in (a) that J/I is an ideal of R/I, so then this ideal correspondsto π−1(J/I) = J .

(⇐) Suppose that J is maximal and that K is an ideal of R such thatK ⊇ J/I. Then, we know that π−1(J/I) = J , and since K ⊇ J/I, π−1(K) ⊇ J .Since J is maximal, we have π−1(K) = J , and so K = J/I.

(⇒) Suppose that J is not maximal and that K ⊇ J/I is an ideal of R/Iwith k ∈ K, but k /∈ J/I. Then, π−1(k) /∈ J and since π−1(K) ⊇ J (sinceK ⊇ J/I), we conclude that π−1(K) ) J . Thus, J is not maximal.

19. Use Exercise 18 together with Proposition 5.3.9 to determine all primeideals and all maximal ideals of Zn.

By Q18, I/nZ is a prime ideal of Z/nZ if and only if I is a prime ideal of Z.Thus, the prime ideals of Z/nZ = Zn are the ideals of the form 〈p〉/nZ where pis a prime number. Since in Z, prime = maximal, we have also found all of themaximal ideals.

20. In the ring Z[i] of Gaussian integers, let 〈p〉 be the ideal generated by aprime number. Show that Z[i]/〈p〉 has p2 elements, and has characteristic p.

We want to show that any element in Z[i]/〈p〉 has the form [a]p + i[b]p. Ifwe can show this, then clearly |Z[i]/〈p〉| = p2 and char(Z[i]/〈p〉) = p.

Now, all multiples of p are in 〈p〉. We know that any number without animaginary part which is not divisible by p is of the form [a]p, which are not in〈p〉, and so form separate equivalence classes. Similarly, pure imaginary numbersform classes of the form i[b]p. Finally, combinations of two classes of these typesare also not in 〈p〉 and so form even more congruence classes. This �combination�completely classi�es all of Z[i] into congruence classes. Thus, any number inZ[i]/〈p〉 has the form [a]p + i[b]p, so our conclusion follows.

21. In the ring Z[i] of Gaussian integers �nd necessary and su�cient condi-tions on integers m and n for the element m+ni to belong to the ideal 〈1 + 2i〉.Use these conditions to determine the ideal 〈1 + 2i〉 ∩ Z of Z.

Recall that 〈1 + 2i〉 = Z[i] · (1 + 2i). Thus, any element in 〈1 + 2i〉 must haveform (a+ ib)(1 + 2i) = a− 2b+ i(2a+ b), where a, b ∈ Z.

Clearly, 〈1 + 2i〉 ∩ Z contains precisely the elements of the form a − 2bwhere 2a + b = 0. This implies that b = −2a, so elements must have the forma− 2(−2a) = a+ 4a = 5a. Thus, 〈1 + 2i〉 ∩ Z = 5Z.

22. In the ring Z[i] of Gaussian integers, show that the ideal 〈5− i〉 is nota prime ideal.


We see that 5−i = (1−i)(3+2i). We need to check that 1−i, 3+2i /∈ 〈5−i〉.We see that 〈5− i〉 has elements of the form (a+ ib)(5− i) = (5a+ b)+ i(5b−a).

If 1 − i ∈ 〈5 − i〉, we would have to have a, b ∈ Z such that 5a + b = 1 and5b − a = −1. This implies that a = 3

13 and b = − 213 . Since neither is in Z, we

conclude that 1− i /∈ 〈5− i〉.Now, if 3 + 2i ∈ 〈5− i〉, we would have to have a, b ∈ Z such that 5a+ b = 3

and 5b − a = 2. This implies that a = b = 12 . Again, we conclude that

3 + 2i /∈ 〈5− i〉.Now, we have found that (1− i)(3 + 2i) = (5− i) ∈ 〈5− i〉, but 1− i, 3 + 2i /∈

〈5− i〉. Thus, 〈5− i〉 is not a prime ideal.

23. Let R be the set of all continuous functions from the set of real numbersinto itself. In Exercise 13 of Section 5.1, we have shown that R is a commutativering if the following formulas

(f + g)(x) = f(x) + g(x) and (f · g)(x) = f(x)g(x)

for all x, are used to de�ne addition and multiplication of functions. Let a bea �xed real number, and let I be the set of all functions f(x) ∈ R such thatf(a) = 0. Show that I is a maximal ideal of R.

(I is an ideal) Let f(x), g(x) ∈ I. Then, (f ± g)(x) = f(x) ± g(x). Wesee (f ± g)(0) = 0 ± 0 = 0. Thus, (f ± g)(x) ∈ I. Let h(x) ∈ R. Then,(h · f)(x) = h(x)f(x). We see that (h · f)(0) = h(0) · 0 = 0. Thus, (h · f)(x) ∈ I.

(I is maximal) We see that I is the set of polynomials with constant termequal to 0. Thus, if we take the factor ring R/I we obtain the polynomials overR which only have constant terms. To see this, note that we obtain an elementof R by taking an element of I and adding a real number. Thus, any elementof R is of the form a + I, where a ∈ R. This ring is isomorphic to R (by thenatural mapping φ(a + I) = a), which is a �eld. Thus, since R/I ∼= R, byProposition 5.3.9, I is maximal.

24. Let I be the smallest ideal of Z[x] that contains both 2 and x. Show thatI is not a principal ideal.

The smallest ideal that contains 2 and x is 〈2〉 ∩ 〈x〉. (We know that 〈a〉is the smallest ideal that contains a. We have also shown that the intersectionof two ideals is an ideal, and we know that the intersection will give us thefewest number of elements.) Now, 〈2〉 = {f(x) ∈ Z[x]|all coe�cients are even}and 〈x〉 = {f(x) ∈ Z[x]|f(x) has no constant term}. Thus, 〈2〉 ∩ 〈x〉 = {f(x) ∈Z[x]|f(x) has even coe�cients with no constant term}. Now, 2 ∈ 〈2〉 ∩ 〈x〉, soif 〈2〉 ∩ 〈x〉 is a principal ideal, we must have f(x) ∈ Z[x] such that 2 ∈ 〈f(x)〉.This would imply that deg f(x) = 0. But then, such an f(x) would be unableto generate x ∈ 〈2〉 ∩ 〈x〉. The reason this fails is because Z does not havemultiplicative inverses, so ax · b 6= x for any a, b ∈ Z.


25. Let R and S be commutative rings, let I be an ideal of R, and let J bean ideal of S.

(a) Show that I ⊕ J as an ideal of R⊕ S.(b) Show that (R⊕ S)/(I ⊕ J) ∼= (R/I)⊕ (S/J).(c) Show that I⊕J is a prime ideal of R⊕S if and only if either I = R and

J is a prime ideal of S, or else I is a prime ideal of R and J = S.(d) Show that if K is any ideal of R ⊕ S, then there exists an ideal I of R

and an ideal J of S such that K = I ⊕ J .

(a) Let (a, b), (c, d) ∈ I⊕J . Then, (a, b)± (c, d) = (a±c, b±d) ∈ I⊕J , sinceI and J are ideals. Next, let (r, s) ∈ R⊕ S. Then, (r, s)(a, b) = (ra, sb) ∈ I ⊕ Jsince I, J are ideals. Thus, I ⊕ J is an ideal of R⊕ S.

(b) De�ne φ : R ⊕ S → (R/I) ⊕ (S/J) via φ(r, s) = (r + I, s + J). Clearly,φ(R ⊕ S) = (R/I) ⊕ (S/J). The elements that are mapped to 0 (aka (I, J))are any element of the form (i, j) where i ∈ I and j ∈ J . Thus, kerφ = I ⊕ J .By the fundamental theorem for homomorphisms, we have (R ⊕ S)/(I ⊕ J) ∼=(R/I)⊕ (S/J).

(c) By Proposition 5.3.9, I ⊕ J is a prime ideal of R ⊕ S if and only if(R⊕S)/(I⊕J) is an integral domain. By (b), (R⊕S)/(I⊕J) ∼= (R/I)⊕(S/J).Thus, I⊕J is prime if and only if (R/I)⊕(S/J) is an integral domain. Now, forany I, J , we can always form the product (i, 1)(1, j) = (i, j) where i ∈ I, j ∈ J .This product will always be such that (i, 1)(1, j) ∈ I⊕J , but (i, 1), (1, j) /∈ I⊕Junless 1 ∈ I (in which case I = R) or 1 ∈ J (in which case J = S).

Without loss of generality, suppose J = S. Then, (R⊕ S)/(I ⊕ S) ∼= R/I ⊕S/S ∼= R/I. Thus, (R⊕S)/(I⊕J) is an integral domain if and only if J = S andR/I is an integral domain. This happens if and only if I is prime (Proposition5.3.9) and J = S.

(d) Suppose that K is an ideal of R ⊕ S. Then, if (a, b), (c, d) ∈ K, and(r, s) ∈ R ⊕ S, we must have (a± c, b± d) ∈ K and (ra, sb) ∈ K. This impliesthat the coordinates are closed under addition and enjoy the absorption propertyseparately. Thus, K = I ⊕ J for ideals I E R and J E S.

26. Let R be the set of all rational numbers m/n such that n is odd.(a) Show that R is a subring of Q.(b) Let 2kR = {m/n ∈ R|m is a multiple of 2k and n is odd}, for any posi-

tive integer k. Show that 2kR is an ideal of R.(c) Show that each proper nonzero ideal of R has the form 2kR, for some

positive integer k.(d) Show that R/2kR is isomorphic to Z2k .(e) Show that 2R is the unique maximal ideal of R.


(a) Let a2m+1 ,

b2n+1 ∈ R. Then,

a2m+1+ b

2n+1 = a(2n+1)+b(2m+1)(2m+1)(2n+1) = a(2n+1)+b(2m+1)

4mn+2(m+n)+1 ∈R. If a

2m+1 ∈ R, then clearly − a2m+1 ∈ R. Thus, R is a subring of Q.

(b) Let m2k

2n+1 ,`2k

2j+1 ∈ I. Then, m2k

2n+1 + `2k

2j+1 = m2k(2j+1)+`2k(2n+1)(2n+1)(2j+1) =

2mj2k+m2k+2`n2k+`2k

(2n+1)(2j+1) = (2mj+m+2`n+`)2k

(2n+1)(2j+1) ∈ 2kR. If a2b+1 ∈ R, then a

2b+1 ·m2k

2n+1 = am2k

4bn+2(b+n)+1 ∈ 2kR. Thus, 2kR is an ideal of R.

(c) Suppose I is an ideal of R that contains the element md where m and

d are odd. Then, by the absorption property, we should be able to obtain anyelement of R. For instance, if we wish to obtain the element a

b ∈ R where b is

odd, we have the product admb ·

md = a

b . Thus, I = R.

(d) We see that 2kR is a principal ideal (as its notation suggests). Now,let us determine the cosets of 〈2k〉. If r ∈ R ∩ 〈2k〉, then r = s2k, wheres ∈ 〈2k〉. Otherwise, by the division algorithm, r = qs2k+r′, where 1 ≤ r < 2k.Based on the value for r′, we �nd the coset for any r ∈ R. Use the mappingφ(r + 2kR) = [r]2k to see that R/2kR ∼= Z2k .

(e) We have found that the only ideals of R are 2kR for some k. Clearly,

2R ⊇ 4R ⊇ · · · ⊇ 2kR ⊇ · · · . (For instance m2k

n ∈ 2R since m2k

n = 21 ·

m2k−1

n .)Thus, 2R is the unique maximal ideal.

5.4 Quotient Fields

Notational convention: The notation[ab

]∼ is equivalent to [a, b]∼. They

both refer to the equivalence relation that builds Q(D) for an integral domainD.

1. Complete the proof of Lemma 5.4.3, to show that multiplication of equiv-alence classes in Q(D) is well-de�ned.

Let ab = p1p2···pkm

p1p2···pkn ∈[mn

]∼ ,

cd = q1q2···q`r

q1q2···q`s ∈[rs

]∼. Then, a

b ·cd = ac

bd =(p1···pk)(q1···qn)mr(p1···pk)(q1···qn)ns ∈

[mrns

]∼ =

[mn

]∼ ·[rs

]∼.

2. Show that the associative and commutative laws hold for addition inQ(D).

(Associative)[ab

]∼ +

([cd

]∼ +

[ef

])=[ab

]∼ +

[cf+eddf

]∼

=[adf+b(cf+ed)

bdf

]∼.

(1) ([ab

]∼ +

[cd

]∼

)+[ef

]∼

=[ad+bcbd

]∼ +

[ef

]∼

=[f(ad+bc)+ebd

bdf

]∼. (2)

We see that (1) = (2), so + is associative.(Commutative)

[ab

]∼+

[cd

]∼ =

[ad+bcbd

]∼ =

[bc+addb

]∼ =

[cd

]∼+

[ab

]∼. Thus,

+ is commutative.


3. Show that the associative and commutative laws hold in Q(D).

(Associative)[ab

]∼ ·

([cd

]∼ ·[ef

]∼

)=[ab

]∼ ·

[cedf

]∼

=[acebdf

]∼

=[acbd

]∼ ·[

ef

]∼

=([ab

]∼ ·[cd

]∼

)·[ef

]∼.

(Commutative)[ab

]∼ ·[cd

]∼ =

[acbd

]∼ =

[cadb

]∼ =

[cd

]∼ ·[ab

]∼.

4. Let φ : D → Q(D) be the mapping φ(D) = [d, 1] de�ned in Theorem5.4.4. Show that φ is an isomorphism if and only if D is a �eld.

(⇒) If φ is an isomorphism, then every element in Q(D) can be written as[d1

]∼. Now, Q(D) is a �eld, so for any

[d1

]∼ ∈ Q(D), we must have

[d′

1

]∼∈

Q(D) such that[d1

]·[d′

1

]=[dd′

1

]=[11

]. Thus, for any

[d1

]∈ Q(D), we have

1 = φ−1[11

]= φ−1

([d1

]·[d′

1

])= φ−1

[d1

]· φ−1

[d′

1

]= d · d′. Thus, any d ∈ D

has an inverse, so D is a �eld.(⇐) If D is a �eld, then D contains all inverses.(1-1) φ is a direct map and is thus one-to-one.

(Onto) Let[xy

]∈ D. Since the numerators and denominators come from D,

a �eld, this expression is equivalent to[xy

]=[x1

]·[y−1

1

]= φ(xy−1).

(Preservation) φ(xy) =[xy1

]=[x1

]·[y1

]= φ(x)φ(y).

φ(x+ y) =[x+y1

]=[x1

]+[y1

]= φ(x) + φ(y).

5. In Theorem 5.4.6, verify that θ̂ is a one-to-one ring homomorphism.

(1-1) Suppose θ̂[ab

]= θ̂

[cd

]. This means that θ(a)θ(b)−1 = θ(c)θ(d)−1.

Now, F is a �eld, so the previous equality is the same as θ(c)−1θ(b)−1 =θ(a)−1θ(d)−1 ⇔ θ−1(cb) = θ−1(ad). Since θ is one-to-one, this implies cb = ad.Thus,

[ab

]−[cd

]=[ad−bcbd

]=[

0bd

]= [0]. Thus,

[ab

]=[cd

].

(Preservation of +) θ̂[ab + c

d

]= θ̂

[ad+bcbd

]= θ(ad + bc)θ(bd)−1 = (θ(ad) +

θ(bc))θ(bd)−1 = (θ(a)θ(d) + θ(b)θ(c)) θ(b)−1θ(d)−1 = θ(a)θ(b)−1 + θ(c)θ(d)−1

= θ̂[ab

]+ θ̂

[cd

].

(Preservation of ·) θ̂[ab ·

cd

]= θ̂

[acbd

]= θ(ac)θ(bd)−1 = θ(a)θ(c)θ(b)−1θ(d)−1 =

θ(a)θ(b)−1θ(c)θ(d)−1 = θ̂[ab

]θ̂[cd

].

6. Let D1 and D2 be integral domains, with quotient �elds Q(D1) andQ(D2), respectively. Let θ : D1 → D2 be a ring homomorphism.

(a) Prove that if θ is one-to-one, then there exists a ring homomorphism

θ̂ : Q(D1)→ Q(D2) such that θ̂([d, 1]) = [θ(d), 1] for all d ∈ D1.(b) Prove that if θ is not one-to-one, then it is impossible to �nd a ring

homomorphism θ̂ : Q(D1)→ Q(D2) that satis�es the conditions of part (a).

(a) (Identity) θ̂[11

]=[θ(1)1

]=[11

]since θ is a homomorphism.


(Preservation of +) θ̂([a1

]+[c1

])= θ̂

[a+c1

]=[θ(a+c)

1

]=[θ(a)+θ(c)

1

]=[

θ(a)1

]+[θ(c)1

]= θ̂(a) + θ̂(c).

(Preservation of ·) θ̂([a1

]·[b1

])= θ̂

[ab1

]=[θ(ab)

1

]=[θ(a)1

]·[θ(b)1

]=

θ̂(a)θ̂(b).

(b) If θ is not one-to one, then there are x 6= y ∈ D1 such that θ(x) = θ(y).

Then, θ̂[x1 + y

1

]=[θ(x+y)

1

]= 2

[θ(x)1

], but since x 6= y, x + y 6= 2x and

θ̂[x+y1

]=[θ(x+y)

1

]6= 2

[θ(x)1

].

7. Determine Q(D) for D = {m+ n√

2|m,n ∈ Z}.

For every a + b√

2 ∈ D, we must introduce the elements[a+b√2

1

]and[

1a+b√2

]. For two elements

[1

a+b√2

]+[

1c+d√2

], we have the equivalent ex-

pression[a−b√2

a2−2b2

]+[c−d√2

c2−2d2

]∈ Q[

√2]. Thus, r+ s ∈ Q[

√2] for all r, s ∈ Q(D).

Also, doing the same sort of trickery, we easily see that rs ∈ Q[√

2] for allrs ∈ Q(D). Thus, Q(D) = Q[

√2].

8. Let p be a prime number and let D = {m/n|m,n ∈ Z and p - n}. Verifythat D is an integral domain and �nd Q(D).

(D is an integral domain) Let mq1p+r1

· nq2p+r2

= mn(q1p+r1)(q2p+r2)

= 0. The

only way this can happen is if one of m or n is 0, in which case mq1p+r1

or nq2p+r2

would be 0. Thus, D is an integral domain.

For the elementp`p

β11 ···p

βkk

pα11 ···p

αnn

, where pi 6= p for all i, the element 1p`

must

be introduced to Q(D). This means that Q(D) must have elements whosedenominator can contain any power of p. Since D already has elements thatcan contain other primes, we simply have Q(D) = Q.

9. Determine Q(D) for D = {m+ ni|n,m ∈ Z} ⊆ C.

Very similar to Q7, Q(D) = Q[i].

10. Considering Z[x] as a subring of Q[x], show that both rings have thesame quotient �elds.

Since nxk ∈ Z[x], surely 1nx

k ∈ Q(Z[x]) for all n, k ∈ Z. Then, sinceQ(Z[x]) is a �eld, this implies that

∑mn x

k ∈ Q(Z[x]) for all m,n, k ∈ Z. Wenow see that Q[x] ⊆ Q(Z[x]). Since Q(Z[x]) is a �eld, it must contain all of theinverses for Q[x]. Thus, Q(Q[x]) ⊆ Q(Z[x]). On the other hand, we have thatZ[x] ⊆ Q[x], so by the same argument about Q(Q[x]) being a �eld, we haveQ(Z[x]) ⊆ Q(Q[x]). Thus, Q(Z[x]) = Q(Q[x]).


11. Show that if P is a prime ideal of D, then DP = {a/b ∈ Q(D)|b /∈ P}is an integral domain with D ⊆ DP ⊆ Q(D).

(DP is an integral domain) Suppose that[ab

]·[cd

]=[acbd

]= [0]. Now, since

P is an ideal, 0 ∈ P , so neither b nor d can equal 0. Also, since D is an integraldomain, we cannot have bd = 0 since b, d 6= 0. Thus, ac = 0 and since D is anintegral domain, either a = 0 (in which case

[ab

]= [0]) or c = 0 (in which case[

cd

]= [0]). Thus, DP is an integral domain.Clearly, DP ⊆ Q(D), since the elements come from Q(D).Let a ∈ D. Then, by de�nition of DP , a ∈ DP . Thus, D ⊆ DP . (It might

be worth pointing out that 1 /∈ P since P is an ideal.)

12. In the ring DP de�ned in Exercise 11, let M = {a/b ∈ DP |a ∈ P}.(a) Show that M is an ideal of DP .(b) Show that DP /M ∼= Q(R/P ), and conclude that M is a maximal ideal

of DP .

(a) Let[p1q1

],[p2q2

]∈ M . Then,

[p1q1

]±[p2q2

]=[p1q2±p2q1

q1q2

]. Since P is

an ideal, p1q2 ± p2q1 ∈ P , and so[p1q1

]±[p2q2

]∈ M . If

[dq′

]∈ DP , then[

dq′

]·[pq

]=[dpq′q

]. Since P is an ideal, dp ∈ P , so

[dq′

]·[pq

]∈ M . Thus, M is

an ideal of DP .(b) I can only assume that there was a misprint in this question as R is not

de�ned. My best guess is that R should be D.

If that is the case, de�ne φ(ab

)=[a+Pb+P

]. We need to show that φ is a

homomorphism. First, if ab ,cd ∈ DP , then φ

(ab + c

d

)= φ

(ad+bcbd

)=[ad+bc+Pbd+P

].

Also, φ(ab

)+φ

(cd

)=[a+Pb+P

]+[c+Pd+P

]=[(d+P )(a+P )+(b+P )(c+P )

(b+P )(d+P )

]=[ad+bc+Pbd+P

].

Thus, + is preserved.

Next, φ(ab ·

cd

)= φ

(acbd

)=[ac+Pbd+P

]=[a+Pb+P

]·[c+Pd+P

]= φ

(ab

)+φ

(cd

). Thus,

· is preserved. Thus, φ is a homomorphism.We see that the elements that get mapped to 0 (which is to say that the

numerator of the function value is P ) are the ones whose numerators are mem-bers of P . This set is precisely M . Thus, by the fundamental theorem forhomomorphisms, we see that DP /M ∼= Q(D/P ).

13. Let R be a commutative ring. A derivation on R is a function ∂ : R→R such that (i) ∂(x+y) = ∂(x)+∂(y) and (ii) ∂(xy) = ∂(x)y+x∂(y). Show thatif ∂ is a derivation on an integral domain D with quotient �eld Q(D), then ∂ canbe extended to a derivation ∂ of Q(D) by de�ning ∂(a/b) = (b∂(a) − a∂(b))/b2

for all a, b ∈ D with b 6= 0.


(i) ∂(ab + c

d

)= ∂

(ad+bcbd

)= bd∂(ad+bc)−(ad+bc)∂(bd)

b2d2 = bd[∂(ad)+∂(bc)]−(ad+bc)[d∂(b)+b∂(d)]b2d2 =

[bda∂(d)+bd2∂(a)+b2d∂(c)+bdc∂(b)]−[ad2∂(b)+adb∂(d)+bcd∂(b)+b2c∂(d)]b2d2 = bd2∂(a)+b2d∂(c)−ad2∂(b)−b2c∂(d)

b2d2 =d2[b∂(a)−a∂(b)]+b2[d∂(c)−c∂(d)]

b2d2 . (1)

Then, ∂(ab

)+∂(cd

)= b∂(a)−a∂(b)

b2 + d∂(c)−c∂(d)d2 = d2[b∂(a)−a∂(b)]+b2[d∂(c)−c∂(d)]

b2d2

(2). Now, (1) = (2), so (i) is satid�ed.

(ii) ∂(ab ·

cd

)= ∂

(acbd

)= bd∂(ac)−ac∂(bd)

b2d2 = bd[a∂(c)+c∂(a)]−ac[b∂(d)+d∂(b)]b2d2 . (1)

Next, ∂(ab

)· cd + a

b · ∂(cd

)= b∂(a)−a∂(b)

b2 · cd + ab ·

d∂(c)−c∂(d)d2 = c[b∂(a)−a∂(b)]

b2d +a[d∂(c)−c∂(d)]

bd2 = cd[b∂(a)−a∂(b)]+ab[d∂(c)−c∂(d)]b2d2 = bd[a∂(c)+c∂(a)]−ac[b∂(d)+d∂(b)

b2d2 . (2)

Now, (1) = (2), so (ii) is satis�ed. Thus, ∂ is a derivation on Q(D).

14. Show that ∂ : Q[x] → Q[x] de�ned by ∂(f(x)) = f ′(x) for all f(x) ∈Q[x] is a derivation. Describe the derivation ∂ de�ned on the quotient �eld ofQ[x].

We already know that ∂(f(x) + g(x)) = f ′(x) + g′(x) and ∂(f(x)g(x)) =g(x)f ′(x) + f(x)g′(x) from calculus. Thus, ∂ is a derivation of Q[x]. We haveshown that Q(Q[x]) = Q(Z[x]), which is just the set of rational functions.

Thus, ∂(f(x)g(x)

)should be de�ned according to the ordinary quotient rule. That

is, ∂(f(x)g(x)

)= g(x)f ′(x)−f(x)g′(x)

(g(x))2 = g(x)∂(f(x))−f(x)∂(g(x))(g(x))2 .

Chapter 6

Fields

6.1 Algebraic Elements

1. Show that the following complex numbers are algebraic over Q.(a)√

2(b)√n, for n ∈ Z+

(c)√

3 +√

5

(d)√

2 +√

3(e) (−1 +

√3i)/2

(f) 3√

2 +√

2

(a) x2 − 2 has√

2 as a root.(b) x2 − n has

√n as a root.

(c) Let x =√

3 +√

5. Then, x −√

3 =√

5. Squaring both sides, we getx2 − 2

√3x+ 3 = 5 which is the same thing as x2 − 2 = 2

√3x. Again, squaring

both sides, we get x4− 4x2 + 4 = 12x2. Thus, the polynomial x4− 16x2 + 4 has√3 +√

5 as a root.

(d) Let x =√

2 +√

3. Then, x2 = 2 +√

3, so x2 − 2 =√

3. Squaring,

x4 − 4x2 + 4 = 3. Thus, x4 − 4x2 + 1 has√

2 +√

3 as a root.

(e) Note that (−1+√

3i)/2 = ei(2π3 ). Thus, if x = ei(

2π3 ), then x3 = e2πi = 1.

Thus, x3 − 1 has the number as a root.(f) Let x = 3

√2 +√

2 = 21/3 + 21/2. Then, x − 21/2 = 21/3. Cubing,x3−3

√2x2 +6x−2

√2 = 2. Thus, x3 +6x−2 = 3

√2x2 +2

√2, so x3 +6x−2 =√

2(3x2 + 2).Squaring, we get x6 + 12x4 − 4x3 + 36x2 − 24x + 4 = 2(9x4 + 12x2 + 4).

Thus, x6 − 6x4 − 4x3 + 12x2 − 24x− 4 has the number as a root.

2. Let F be an extension �eld of K, and let u be a nonzero element of Fthat is algebraic over K. Show that u−1 is also algebraic over K.

32

CHAPTER 6. FIELDS 33

Let f(x) = anxn+· · ·+a1x+a0 ∈ K[x] be such that f(u) = 0. Then, examine

g(x) = a0xn+· · ·+an−1+an. (The coe�cients are in the reverse order.) We will

show that g(u−1) = 0. If∑nk=0 u

−kan−k = 0, then surely un∑nk=0 u

−kan−k =0. This is the same as

∑nk=0 u

n−kan−k which we are given to be equal to 0. Thus,it must be true that our �rst statement,

∑nk=0 u

−kan−k = 0, is true. (Note thatK[x] is an integral domain and so

∑nk=0 u

−kan−k = 0⇔ un∑nk=0 u

−kan−k = 0since u 6= 0.)

3. Suppose that u is algebraic over the �eld K, and that a ∈ K. Show thatu + a is algebraic over K, �nd its minimal polynomial over K, and show thatthe degree of u+ a over K is equal to the degree of u over K.

It is pretty clear that if u ∈ F , we must have K(u) ⊆ F . So, since u + a =u + a · 1 ∈ K(u) (keeping in mind the notion of K(u) being a vector spaceover K with basis {1, u, u2, ...}), we see that u + a ∈ F . Furthermore, sinceK(u) = K(u+a), the degrees of the extension are also equal. However, since wehave to �nd the minimal polynomial anyway, we don't really need the preceedingargument.

Note that f(x + a) gives the graph of f(x) shifted left a units. Thus, iff(x) has a root at u, then f(x+ a) has a root at u− a. We need to prove this,though, since there is no notion of �shifting left� in the abstract setting. Supposef(x) = anx

n + · · ·+ a1x+ a0. Then, set f(x+ a) = g(x) = an(x+ a)n + · · ·+a1(x+a)+a0. (This polynomial is still in K[x] since K[x] is a ring.) We see thatg(u−a) = an(u−a+a)n + · · ·+a1(u−a+a) +a0 = anu

n + · · ·+a1u+a0 = 0.This proves that g(x) is the minimal polynomial, for if there were a polynomialh(x) of smaller degree for which h(u + a) = 0, we would see that h(x − a) hasu as a root. Since f(x) was the minimal polynomial of u, we know that this isimpossible. Thus, g(x) is the minimal polynomial and deg f = deg g. As such,[K(u) : K] = [K(u+ a) : K].

4. Show that√

3 /∈ Q(√

2).

We see that Q(√

2) as a vector space over Q has a basis {1,√

2}. Now,suppose

√3 ∈ Q(

√2). Then

√3 = a

b + cd

√2 for a, b, c, d ∈ Z. This means

that d(√3−a/b)c =

√2. Squaring both sides, we get

a2d2

b2− 2ad2

√3

b +3d2

c2 = 2. Since√3 /∈ Q, we know that this is impossible. Thus,

√3 /∈ Q(

√2).

5. (a) Show that f(x) = x3 + 3x+ 3 is irreducible over Q.(b) Let u be a root of f(x). Express u−1 and (1+u)−1 in the form a+bu+cu2

where a, b, c ∈ Q.

(a) By the rational root theorem, the only possible rational roots are ±3 and±1. By plugging these numbers in, we quickly see that f(x) is irreducible overQ.


(b) InQ(u), we have u(a+bu+cu2) = 1. This gives au+bu2+cu3 = 1. Sinceu3 + 3u+ 3 = 0, we have u3 = −3(u+ 1) and we obtain au+ bu2− 3cu− 3c = 1or (a − 3c)u + bu2 − 3c = 1. Thus, −3c = 1 ⇒ c = − 1

3 . Then, a − 3c = 0 ⇒a− 3 · − 1

3 = 0⇒ a+ 1 = 0⇒ a = −1. Finally, b = 0. Thus, u−1 = −1− 13u

2.InQ(u), we have (1+u)(a+bu+cu2) = 1. Thus, cu3+(b+c)u2+(a+b)u+a =

1. Exploiting the same relation as above, this gives (b+ c)u2 + (a+ b− 3c)u+a−3c = 1. This implies that a = 4, b = −1, c = 1. Thus, (1+u)−1 = 4−u+u2.

6. Show that the intersection of any collections of sub�elds is again a sub-�eld.

Let E be a �eld and {Fi} be a collection of sub�eld. Let x, y ∈⋂Fi. This

means that x and y are in every Fi. Thus, x± y ∈ Fi for all i. Next, if x ∈ Fifor all i, then since Fi are �elds, we must have x

−1 ∈ Fi for all i. Thus,⋂Fi is

a sub�eld of E.

7. Let F = K(u), where u is transcendental over the �eld K. If E is a �eldsuch that K ⊂ E ⊆ F , then show that u is algebraic over E.

To restate what is given, we have K $ E ⊆ K(u). Now, all elements inF = K(u) are of the form a0 + a1u + a2u

2 + · · · where ai ∈ K. The elementsof K must not have u's. Since E 6= K, we must have b1 + b2u+ · · ·+ bnu

n ∈ Ewhere bi ∈ K. But, K(b1 + b2u+ · · ·+ bnu

n) = K(u), so E = F , and obviouslyu would have to be algebraic over E.

8. Let F be an extension �eld of K.(a) Show that F is a vector space over K.(b) Let u ∈ F . Show that u is algebraic over K if and only if the subspace

spanned by {1, u, u2, ...} is a �eld.

(a) Since F is a �eld, it is an abelian group under addition. Thus, i-v (theaxioms for vector spaces are listed on page 458) are satis�ed.

Since K ⊆ F , we know that k · f ∈ F for all k ∈ K and f ∈ F . Thus, (vi)is satis�ed.

Since K ⊆ F and F is a �eld (and thus · is associative), we have k1(k2f) =(k1k2)f for all k1, k2 ∈ K and f ∈ F . Thus, (vii) is satis�ed.

Since K ⊆ F and F is a �eld (and thus + distributes over ·), we have(k1 + k2)f = k1f + k2f for k1, k2 ∈ K and f ∈ F . Thus, (viii) is satis�ed.

For the same reasons, (ix) is satis�ed.Since F is a �eld, we must have 1 ·f = f for all f ∈ F . Thus, (x) is satis�ed.Thus, F is a vector space over K.

(b) (⇒) Suppose that u is algebraic over K. Then, let f(x) = xn + · · · +a1x+ a0 be the minimal polynomial of u. Then, K[x]/〈f(x)〉 ∼= K(u) is a �eld.We know that K(u) is a vector space over K. It is clear that {1, u, ..., un} is abasis for K(u). Thus, span{1, u, ..., un} = K(u).


(⇐) Suppose that span{1, u, u2, ...} is a �eld. There are two possibilities:(1) {1, u, u2, ...} is linearly independent or (2) {1, u, u2, ...} is not linearly inde-pendent.

In the case of (1), since span{1, u, u2, ...} is a �eld, we must have elementsa0 + a1u + · · · + anu

n, b0 + b1u + · · · + bmum ∈ span{1, u, u2, ...} such that

(a0 + a1u+ · · ·+ anun)(b0 + b1u+ · · ·+ bmu

m) = c0 + c1u+ · · ·+ c`u` = 1. This

would involve coe�cients of u terms multiplying to 0. However, the coe�cientscome from K which is a �eld, so multiplication of nonzero elements should neverequal 0. Contradiction. Thus, (1) is not possible.

In the case of (2), then we know that there is k0 +k1u+ · · ·+k`u` = 0 where

ki 6= 0 for some i > 0. Thus, the polynomial k`x` + · · · + k1x + k0 has u as a

root. Thus, u is algebraic over K.

9. Let F be an extension �eld of K. If u ∈ F is transcendental over K,then show that every element of K(u) that is not in K is also transcendentalover K.

Let u = k0 + k1u + · · · + knun ∈ K(u) where ki 6= 0 for some i ≥ 1 (which

is to say the element is from K(u) −K). Suppose that u is algebraic over K.Then, there is some polynomial over K for which u is a root. But, if we takepowers of u, we obtain an expression that has only powers of u. Thus, if wehave a polynomial that has u as a root, there must be a polynomial that has uas a root.

10. Let u and r be positive real numbers, with u 6= 1. It follows from afamous theorem of Gelfand and Schneider that if r is irrational and both u andr are algebraic over Q, then ur must be transcendental over Q. You may usethis result to show that the following numbers are transcendental over Q.

(a) 3√

7√5

(b) 3√

7√5

+ 7

(a) We see that 3√

7√5

=(

3√

7)√5

. Both 3√

7 and√

5 are both well-known

irrational numbers, so 3√

7√5is transcendental over Q.

(b) We already proved that 3√

7√5was transcendental over Q, so by Q9

3√

7√5

+ 7 ∈ Q

(3√

7√5)−Q is also transcendental.

11. Show that there exist irrational numbers a, b ∈ R such that ab is rational.

We see that (2√2)√2 = 22 = 4 ∈ Q. (Note that 2

√2 is transcendental by

Q10 and is thus irrational.)

12. Assuming that π is transcendental over Q, prove that either π + e orπ · e is irrational.


Suppose that π · e = ab for a, b ∈ Z. Then, ab ∈ Q, so a minimal polynomial

for π · e is p(x) = x − cd (π · e). Thus, we have πe ∈ Q(π)(e) = Q(π, e) and

πe ∈ Q. Thus, we have πe ∈ Q, but πe /∈ Q(π). This is imposssible sinceQ ⊆ Q(π). Thus, e is irrational.

6.2 Finite and Algebraic Extensions

Notational convention: If E is an extension �eld of F , we say �E/F is a(nextesion) �eld.� Similarly, if V is a vector space over the �eld F , we say �V/Fis a vector space.� This saves lots of unnecessary typing.

1. Find the degree and a basis for each of the given extensions.(a) Q(

√3) over Q

(b) Q(√

3,√

7) over Q(c) Q(

√3 +√

7) over Q(d) Q(

√2, 3√

2) over Q(e) Q(

√2 + 3√

2) over Q(f) Q(ω) over Q, where ω = (−1 +

√3i)/2

(a) Basis: {1,√

3}. Degree: 2.(b) Basis: {1,

√3,√

7,√

21}. Degree: 4.(c) Basis: {1,

√3,√

7,√

21}. Degree: 4.(d) Basis: {1, 21/2, 21/3, 22/3, 25/6, 21/6}. Degree: 6.(e) Basis: {1, 21/2, 21/3, 22/3, 25/6, 21/6}. Degree: 6.(f) Basis: {1, ei 2π3 , e−i 2π3 }. Degree: 3.

2. Find the degree and a basis for each of the given �eld extensions.(a) Q(

√3,√

21) over Q(√

7)(b) Q(

√3 +√

7) over Q(√

7)(c) Q(

√3,√

7) over Q(√

3 +√

7)

(a) Basis: {1,√

3}. Degree: 2.(b) Basis: {1,

√3}. Degree: 2.

(c) Basis: {1}. Degree: 1.

3. Find the degree of Q( 3√

2, 4√

5) over Q.

[Q( 3√

2, 4√

5) : Q] = [Q( 3√

2) : Q][Q( 3√

2, 4√

5) : Q( 3√

2)] = 3 · 4 = 12.

4. Let F be a �nite extension of K such that [F : K] = p, a prime number.If u ∈ F but u /∈ K, show that F = K(u).

We have that p = [F : K] = [F : K(u)][K(u) : K]. Thus, [F : K(u)] = p or[K(u) : K] = p. Since u /∈ K, we must have [K(u) : K] = p, so [F : K(u)] = 1.Thus, F = K(u).


5. Let f(x) be an irreducible polynomial in K[x]. Show that if F is anextension �eld of K such that deg (f(x)) is relatively prime to [F : K], thenf(x) is irreducible in F [x].

Let u be a root of f(x). Then, deg (f(x)) = q[K : K(u)]. If f(x) is re-ducible over F [x], then u ∈ F , so we have to have K(u) ⊆ F and [F : K] =[F : K(u)][K(u) : K]. However, since gcd (deg (f(x)) , [F : K]) = 1, this isimpossible. Thus, f(x) is not reducible over F [x].

6. Let K ⊆ E ⊆ F be �elds. Prove that if F is algebraic over K, then F isalgebraic over E and E is algebraic over K.

If F/K is algebraic, then clearly, E/K is algebraic since E ⊆ F . Now,suppose that F/K is transcendental. In particular, suppose that there is u ∈ Fwhich is transcendental over E. Then, by Q9 of 6.1, we have that every elementof E(u)−E is transcendental over E. Let u′ ∈ E(u)−E. Then, sinceK ⊆ E, wehave u′ transcendental overK. But, E(u)−E ⊆ F , so we would have an elementof F being transcendental over K. But F/K is algebraic. Contradiction.

7. Let F ⊇ K be �elds, and let R be a ring such that F ⊇ R ⊇ K. If F isan algebraic extension of K, show that R is a �eld. What happens if we do notassume that F is algebraic over K?

Let r ∈ R −K. Since R ⊆ F and F/K is algebraic, there is an irreduciblep(x) ∈ K[x] such that p(r) = 0. Then, since p(x) is irreducible, we obtainK[x]/〈p(x)〉 ∼= K(r) is a �eld. Now, K(r)/K is a vector space, so any elementof K(r) can be written as k0 + k1r + · · ·+ knr

n where ki ∈ K. This expressionis an element of R by the axioms for subring. (Remember that k ∈ K ⇒ k ∈ Rsince K ⊆ R.) Let E =

⋃r∈R−K K(r). We have shown (Q6 from 6.1) that

E is a sub�eld of F . We see that the addition and multiplication of these twoelements of E is again in R. Thus, E ⊆ R. Obviously, we also have R ⊆ E(since we take K ⊆ R and adjoin any missing elements from R). Thus, R = Eand is thus a �eld.

If we do not have F/K algebraic, we cannot construct E, so we cannotconclude that R is a �eld.

8. Determine [Q(√n) : Q] for all n ∈ Z+.

If n is not a square, then a basis for [Q(√n) : Q] is {1,

√n}. Thus, the

degree would be 2. If n = m2, then, n ∈ Q, so the degree would be 1.

9. For any positive integers a, b, show that Q(√a+√b) = Q(

√a,√b).

We know that a basis for Q(√a,√b) is {1,

√a,√b,√ab}. Let us now �nd

a basis for Q(√a +√b). Any element of Q(

√a +√b) will have the form n0 +


n1(√a+√b)+· · ·+nk(

√a+√b)k. Examine (

√a+√b)n =

∑nk=0

(nk

)√ak√

bn−k

.

We see that every element in Q(√a +√b) can have a term involving

√a,√b,√

ab, or none of them at all. Thus, a basis for Q(√a+√b) is {1,

√a,√b,√ab}.

Thus, Q(√a,√b) = Q(

√a+√b).

10. Let F be an extension �eld of K. Let a ∈ F be algebraic over K, andlet t ∈ F be transcendental over K. Show that a+ t is transcendental over K.

Suppose that (a + t) is algebraic over K. Then, there is p(x) ∈ K[x] suchthat p(a+ t) = 0. Thus, there is q(x) = p(x+ a) ∈ K(a)[x] such that t is a rootof q(x). Thus, t is algebraic over K(a). Thus, K(a, t) = K(a)(t) = K(a). Thust is algebraic over K by Proposition 6.2.4. Contradiction.

11. Let F be an algebraic extension of K, and let S be a subset of F suchthat S ⊇ K, S is a vector space over K, and sn ∈ S for all s ∈ S and allpositive integers n. Prove that if char(K) 6= 2, then S is a sub�eld of F .

Let s1, s2 ∈ S. Then, s1 + s2 ∈ S by vector addition. Since sn ∈ S for alls ∈ S and n ∈ Z+, we know that (s1 + s2)2 = s21 + s22 + 2s1s2 ∈ S. Again, byvector addition, we have s21 + s22 + 2s1s2 − s21 − s22 = 2s1s2 ∈ S. Thus, s1s2 ∈ Ssince char(K) 6= 2. Now, distributivity in S follows from the fact that productsare closed and by the fact that S inherits the distributivity of F . Thus, S is aring. Since F/K is algebraic and K ⊆ S ⊆ F , S is a �eld by Q7.

6.3 Geometric Constructions

1. Show that the roots of the polynomial 8x3− 6x− 1 used in Theorem 6.3.9are u1 = cos π9 , u2 = cos 5π

9 , and u3 = cos 7π9 .

If we make the substitution x = cos θ, then we obtain the expression 8 cos3 θ−6 cos θ − 1 = 2(4 cos3 θ − 3 cos θ)− 1 = 2 cos(3θ)− 1. Then, the polynomial has

the roots where 2 cos(3θ) = 1. This happens at θ = (6k+1)π9 or θ = (6k−1)π

9 .Pick any three values of θ to �nd that cos θ is equal to u1, u2, or u3.

2. Use the identity 4 cos3 θ − 3 cos θ − cos(3θ) = 0 to show that the roots ofthe polynomial x3 − 3x+ 1 are u1 = 2 cos 2π

9 , u2 = 2 cos 4π9 , u3 = 2 cos 8π

9 .

Substitute x = 2 cos θ. The polynomial becomes (2−8(1−cos2 θ)) cos θ+1 =2 cos θ − 8 cos θ(1 − cos2 θ) + 1 = −6 cos θ + 8 cos3 θ + 1. This equals 0 when8 cos3 θ − 6 cos θ = −1 ⇔ 4 cos3 θ + −3 cos θ = − 1

2 ⇔ cos(3θ) = − 12 . This

happens at θ = 2(3k+1)π9 orθ = 2(3k−1)π

9 . Hence, the polynomial is equal to zerofor x values u1, u2, and u3.


3. In this exercise we outline how to construct a regular pentagon. Letζ = cos(2π/5) + i sin(2π/5).

(a) Show that ζ is a primitive �fth root of unity.(b) Show that (ζ + ζ−1)2 + (ζ + ζ−1)− 1 = 0.(c) Show that ζ + ζ−1 = (−1 +

√5)/2.

(d) Show that cos(2π/5) = (−1+√

5)/4 and that sin(2π/5) =(√

10 + 2√

5)/4.

(e) Conclude that a regular pentagon is constructible.

(a) By example A.5.3, this is a solution to z5 = 1. Thus, ζ is a primitive�fth root of unity.

(b) Plugging ζ = cos(2θ)+i sin(2θ) where θ = 2π/5, we see that (ζ+ζ−1)2+(ζ+ζ−1)−1 = 4 cos2(θ)+2 cos(θ)−1 where θ = 2π/5. Thus, 4 cos2 θ+2 cos θ−1 = 0 if and only if 4 cos2 θ = 1 − 2 cos θ. Then, 4 cos2 θ = 1 − 2 cos θ ⇔2(cos(2θ) + 1) = 1 − 2 cos θ ⇔ 2 cos(2θ) = −2 cos θ − 1. Now, we can check

to see that this equality is true. In (c), we shall �nd that cos(2π/5) = −1+√5

4 .

Using this, we �nd that LHS= −1−√5

2 =RHS.(c) We see that ζ + ζ−1 = 2 cos(2π/5). In (d), we �nd that cos(2π/5) =

(−1 +√

5)/4, so we must have 2 cos(2π/5) = ζ + ζ−1 = (−1 +√

5)/2.(d) From (b), we have 4 cos2 θ + 2 cos θ − 1 = 0, so if x = cos θ, we have

4x2 + 2x − 1 = 0 which has solutions at x = −1−√5

4 or x = −1+√5

4 . Hence,

cos θ = −1+√5

4 . Since sin2 θ+ cos2 θ = 1, we know that sin2 θ =(−1+

√5

4

)− 1 =

−√5−58 . Thus, sin θ =

√10+2

√5

4 .

(e) We need to �nd

[Q

(−1+

√5

4 ,

√10+2

√5

4

): Q

]. Let us �rst �nd

[Q(−1+

√5

4

): Q].

In Q(−1+

√5

4

), we have a polynomial of the form x− −1+

√5

4 which has −1+√5

4

as a root. Thus, in Q(−1+

√5

4

), we have an x such that x = −1+

√5

4 . Hence,

4x+ 1 =√

5 and 16x2 + 8x+ 1 = 5, so 16x2 + 8x− 4 = 0. Thus, 16x2 + 8x− 4

is the minimum polynomial for −1+√5

4 , so[Q(−1+

√5

4

): Q]

= 2.

Let us now �nd

[Q

(√10+2

√5

4

): Q

(−1+

√5

4

)]. As above, x =

√10+2

√5

4 ⇔

4x =√

10 + 2√

5⇔ 16x2−10−2√

5 = 0. Thus, 16x2−10−2√

5 is the minimum

polynomial of

√10+2

√5

4 over Q(−1−

√5

4

).

Now,

[Q

(−1+

√5

4 ,

√10+2

√5

4

): Q

]=

[Q

(√10+2

√5

4

): Q

(−1+

√5

4

)] [Q(−1+

√5

4

): Q]

=

2 · 2 = 4 = 22. Thus, a regular pentagon is constructible.

4. Prove that a regular heptagon is not constructible.

Needless to say, ζ = cos(2π/7) + i sin(2π/7) = ei2π7 is a primitive seventh

root of unity. As above, x = ei2π7 ⇔ x7 − 1 = 0. Thus, [Q(ζ) : Q] = 7


which is not a power of 2. Hence, a regular heptagon is not constructiblesince [Q(ζ) : Q] = [Q(cos(2π/7), sin(2π/7)) : Q(cos(2π/7))] [Q(cos(2π/7) : Q]and thus [Q(cos(2π/7), sin(2π/7)) : Q(cos(2π/7))] cannot be a power of 2.

6.4 Splitting Fields

1. Determine the splitting �elds in C for the following polynomials (overQ).

(a) x2 − 2(b) x2 + 3(c) x4 + x2 − 6(d) x3 − 5

(a) x2− 2 = 0⇔ x2 = 2⇔ x = ±√

2. Thus, the polynomial has all its rootsin Q(

√2).

(b) x2 + 3 = 0 ⇔ x2 = −3 ⇔ x = ±i√

3. Thus, the polynomial has all itsroots in Q(i

√3).

(c) Let u = x2. Then, u2 + u + 6 = 0 ⇔ (u − 2)(u + 3) = 0 ⇔ u = 2 oru = −3. Thus, x = ±

√2 or x = ±i

√3. Thus, the splitting �eld is Q(

√2, i√

3).

(d) x3 − 5 = 0 ⇔ x3 = 5 ⇔ x = 51/3. Thus, Q(51/3, 51/3ei2π3 , 51/3ei

4π3 ) is

the splitting �eld.

2. Determine the splitting �elds in C for the following polynomials (overQ).

(a) x3 − 1(b) x4 − 1(c) x3 + 3x2 + 3x− 4

(a) x3 − 1 = 0⇔ x3 = 1. Thus, Q(ei2π3 , ei

4π3 ) is the splitting �eld.

(b) x4 − 1 = 0⇔ x4 = 1. Thus, Q(eiπ2 , ei

3π2 ) is the splitting �eld.

(c) By completing the cube, we �nd that x3+3x2+3x−4 = 0⇔ (x+1)3 = 5.

If y = x+1, then the solutions to y3 = 5 are 51/3, 51/3ei2π3 , ei

4π3 . Thus, the roots

of the original polynomial are ω1 = 51/3−1, ω2 = 51/3ei2π3 −1, ω3 = 51/3ei

4π3 −1.

Thus, a splitting �eld for the polynomial would be Q(ω1, ω2, ω3).

3. Determine the splitting �elds over Z2 for the follwing polynomials.(a) x2 + x+ 1(b) x2 + 1(c) x3 + x+ 1(d) x3 + x2 + 1

(a) Following Example 6.4.4, identify Z2 with GL2(Z2). We see that A =[0 11 1

]is such that A2 + A + I = 0. Hence, A is the companion matrix of


x2 + x+ 1. Thus, the set of matrices of the form a0I + a1A where ai ∈ Z2 forman extension �eld of Z2 in which x2 + x+ 1 has a root.

(b) Similarly, let A =

[0 11 0

]. We see that A2+I = 0, so A is the companion

matrix. Thus, {a0I + a1A ∈ GL2(Z2)|a0, a1 ∈ Z2} is a splitting �eld.

(c) Similarly, let A =

0 0 11 0 10 1 0

. We see that A3 + A + I = 0, so A is the

companion matrix. Thus, {a0I + a1A + a2A2 ∈ GL2(Z2)|a0, a1, a2 ∈ Z2} is a

splitting �eld.

(d) Let A =

0 0 11 0 00 1 1

. We see that A3+A2+I = 0, so A is the companion

matrix. Thus, {a0I+a1A+a2A2 ∈ GL2(Z2)|a0, a1, a2 ∈ Z2} is a splitting �eld.

4. Let p be a prime number. Determine the splitting �eld for xp−1 over Q.

We see that xp−1 = 0⇔ xp = 1. Thus, the roots are {1, ei2πp , ei

4πp , ..., ei

2(p−1)πp }.

If p = 2, then the solutions are ±1, so the splitting �eld would be Q. If p 6= 2,

then the splitting �eld would be Q(ei2πp , ..., ei

2(p−1)πp ). (None of the roots except

for 1 are from Q because p is prime and so the fraction in the exponent willnever simplify to an expression in Q).

5. Determine the splitting �eld for xp − x over Zp.

Since p must be prime for Zp to be a �eld, by Fermat's little theorem,xp ≡ x(mod p), so xp − x ≡ 0(mod p) and Zp would be the splitting �eld.

6. Determine the splitting �eld for x9 − x over Z3.

We have that x9 − x = (x3)3 − x ≡ x3 − x(mod 3)Q5≡ 0(mod 3). Thus, the

splitting �eld is Z3.

7. Prove that if F is an extension �eld of K of degree 2, then F is thesplitting �eld over K of some polynomial.

If [F : K] = 2, then F is a vector space over K with some basis {1, u}for some u ∈ F − K. Since [F : K] = 2, we know that u2 can be written ask0 + k1u for some k0, k1 ∈ K. Hence, u2 − k1u− k0 = 0. Thus, the polynomialx2 − k1x− k0 has u as a root and u ∈ F −K. (The other root must also be inF , since we can factor out the root to obtain x2 − k1x− k0 = (x− u)(x− r). Ifr ∈ K, when we foil the expression on the RHS out, we get a polynomial whosecoe�cients do not come from K. Contradiction.) Thus, F is the splitting �eldfor x2 − k1x− k0.


8. Let K be a �eld. For a monic polynomial f(x) = a0 + a1x + · · · +an−1x

n−1+xn in K[x], the following matrix C is called the companion matrix

of f(x):

0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...

......

. . ....

0 0 0 · · · 1−a0 −a1 −a2 · · · −an−1

.

This exercise outlines the proof that f(C) = 0. (That is, a0I + a1C + · · · +an−1C

n−1 +Cn = 0, where I is the n×n identity matrix.) Let v1 = (1, 0, ..., 0),v2 = (0, 1, ..., 0), ... ,vn = (0, 0, ..., 1) be the standard basis vectors for Kn.

(a) Show that viC = vi+1 for i = 1, ..., n− 1 and vnC =∑nj=1−aj−1vj.

(b) Find similar expressions for v1C2, ...,v1C

n−1,v1Cn, and show that v1f(C) =

0.(c) Show that vif(C) = 0, for i = 2, ..., n and conclude that f(C) = 0.

(a) We see that[0 · · · 1 · · · 0

]·

0 1 0 · · · 00 0 1 · · · 00 0 0 · · · 0...

......

. . ....

0 0 0 · · · 1−a0 −a1 −a2 · · · −an−1

will

give a 1×n vector. By matrix multiplication, to obtain the (1, k)th entry of thevector, we take the dot product of the �rst (and only) row of

[0 · · · 1 · · · 0

]with the kth column of C. Thus, the (1, k)th entry will be 0 if the (k, i)th entryof C is 0. The only nonzero entries of C are the superdiagonal and the bottomrow. So the (1, k)th entry of the product will be 1 if k − 1 = i and will be −akif i = n. Hence, viC = vi+1 for i = 1, ..., n− 1 and vnC =

∑nj=1−aj−1vj .

(b) We can write v1C2 = v1CC = v2C = v3. Arguing similarly, we see

that v1Cm = vm+1 for 1 ≤ m ≤ n − 1. Then, v1C

n = v1Cn−1C = vnC =∑n

j=1−aj−1vj .Next, v1f(C) = a0v1I+a1v1C+ · · ·+an−1v1C

n−1 +v1Cn = a0v1 +a1v2 +

· · ·+an−1vn+∑nj=1−aj−1vj . Note that the sum at the end cancels every term

before it. Thus, v1f(C) = 0.

(c) Examine vif(C) = v1Ci−1f(C) = v1

(a0C

i−1 + a1Ci + · · ·+ an−1C

n−1+i + Cn+i)

=v1a0C

i−1+· · ·v1an−1Cn−1+i+v1C

n+i= (v1a0I+· · ·+v1an−1Cn−1+v1C

n)Ci =0 · Ci = 0. Now vif(C) = 0 for all i ≥ 2, so by the axioms for �eld, f(C) = 0.

9. Let K be a �eld, let f(x) = a0+a1x+ · · ·+an−1xn−1+xn ∈ F [x], and letC be the companion matrix of f(x), as de�ned in Exercise 8. Show that the set


R = {b0I + b1C + · · ·+ bn−1Cn−1|bi ∈ F for i = 0, ..., n− 1} is a commutative

ring isomorphic to the ring F [x]/〈f(x)〉.

First of all, I believe that there is a typo in this question, as K is introducedbut not used in any way. I assume that K should have been F .

Example 6.4.4 practically establishes the ring properties, so let us prove theisomorphic claim.

Note that F [x]/〈f(x)〉 can only be a ring if f(x) is irreducible. De�ne φ :F [x]/〈f(x)〉 by φ(b0 + b1x+ · · ·+ bn−1x

n−1) = b0I + b1C + · · ·+ bn−1Cn−1.

(1-1)Recall that by how C was de�ned, b0I + b1C + · · · + bn−1Cn−1 = 0 if

and only if the coe�cients of C terms correspond to the coe�cients of x termsin f(x). Thus, kerφ = 0. Thus, φ is one-to-one.

(Onto)If we wish to obtain a0I + a1C + · · ·+ an−1Cn−1, we simply take the

image φ(a0I + a1x+ · · ·+ an−1xn−1). Thus, φ is onto.

(Preservation of +)Addition in R is clearly the same as for polynomials, soaddition is preserved.

(Preservation of ·) In F [x]/〈f(x)〉, we have the relation xn = −b−1n−1x(b0 +b1x + · · · + bn−2x

n−2). Since b0I + b1C + · · · + bn−1Cn−1 = 0, we have the

relation Cn = −b−1n−1C(b0 + b1C + · · · + bn−2Cn−2). We have similar relations

for xn+1, xn+2, .... Thus, multiplication in R corresponds to multiplication inF [x]/〈f(x)〉.

Thus, F [x]/〈f(x)〉 ∼= R.

10. Strengthen Theorem 6.4.2 by proving under the conditions of the theoremthere exists a splitting �eld F for f(x) over K for which [F : K] is a divisor ofn!.

We need to show that [F : K] = n − k for some 0 ≤ k ≤ 1. We procedeby induction on n. Suppose that n = 1. Then, F = K is a splitting �eld and[F : K] = 1|n!.

Suppose that there exists an F/K such that [F : K]|n! for all n ≤ N .Now, suppose that deg f(x) = N + 1. By Kronecker's Theorem, there is

E/K where r ∈ E is a root of f(x). Hence, f(x) = g(x)(x− r) over K(r) wheredeg g(x) = N . Now, by the inductive hypothesis, there is F/K(r) such thatg(x) splits over F and [F : K(r)] = N −k where 0 ≤ k < n. Since the minimumpolynomial of r is of degree N+1 at most, we have [K(r) : K] ≤ N+1, so we seethat [F : K] = [F : K(r)][K(r) : K] ≤ (N − k)(N + 1). Thus, [F : K]|(N + 1)!,as desired. (Recall that if x, y ≤ z, we have xy|z!.)

11. Let K be a �eld, and let F be an extension �eld of K. Let φ : F → Fbe an automorphism of F such that φ(a) = a, for all a ∈ K. Show that for anypolynomial f(x) ∈ K[x], and any root u ∈ F of f(x), the image φ(u) must be aroot of f(x).

Suppose that f(x) = anxn+· · ·+a1x+a0 has a root at u ∈ F . Then, f(u) = 0

or in other words anun + · · ·+ a1u+ a0 = 0. Then, since φ is a homomorphism,


φ(0) = 0 and we get 0 = φ(anun+· · ·+a1u+a0) = anφ(un)+· · ·+a1φ(u)+a0 =

anφ(u)n + · · ·+ a1φ(u) + a0 (since φ �xes all elements of K). Hence, φ(u) is aroot.

12. Use Exercise 11 to show that there are only two automorphisms of the�eld Q(i): the identity automorphism and, and the one de�ned by φ(a + bi) =a− bi, for all a, b ∈ Q.

We must show that any automorphism of Q(i) must �x Q in order to useQ11. Suppose that φ : Q(i) → Q(i) is an automorphism. Then, φ(m) =φ(1 + 1 + · · ·+ 1) = m ·φ(1) = m. Also, φ

(1n

)= φ(n−1) = [φ(n)]−1 = n−1 = 1

n .

Hence φ(mn

)= φ(m)φ

(1m

)= m

n .The element a+ ib 7→ a+ φ(i)b, so let us simply consider what the possible

mappings of i are. We know that a minimal polynomial for i is found byx = i ⇒ x2 = −1. Thus x2 + 1 is the minimal polynomial of i over Q. Wesee that ±i are the roots to this polynomial. Thus, there are two possiblemappings of i: the identity mapping or mapping to its additive inverse. Thus,a+ ib 7→ a+ ib or a+ ib 7→ a− ib.

13. Use Exercise 11 to show that there are at most four distinct automor-phisms of the �eld Q(

√2,√

3).

As proved in Q12, every element of Q must be �xed by any automorphismcontaining Q. Suppose that φ : Q(

√2,√

3) → Q(√

2,√

3) is an automor-phism. Then, we must have φ(a+ b

√2 + c

√3 + d

√6) = a+ bφ(

√2) + cφ(

√3) +

dφ(√

2)φ(√

3). As in Q12, we only need to consider the mappings of√

2 and√

3.Minimal polynomials for

√2 and

√3 over Q are x2− 2 and x2− 3, respectively.

The roots of these polynomials are ±√

2 and ±√

3, respectively. Thus, by Q11,√2 7→ ±

√2 and

√3 7→ ±

√3. Thus, there are four possible mappings.

14. (a) Show that the splitting �eld of x4 − 2 over Q is Q( 4√

2, i).(b) Show that Q( 4

√2, i) is also the splitting �eld of x4 + 2 over Q.

(a) We see that x4 − 2 = (x+ 21/4i)(x− 21/4i)(x− 21/4)(x+ 21/4). (Thesewere obtained by solving x4 = 2 as discussed in A.5.) Hence, Q(21/4, i) is thesplitting �eld.

(b) The roots of x4 + 2 are 2−1/4 + 2−1/4i, 2−1/4 − 2−1/4i,−2−1/4 + 2−1/4i,and −2−1/4− 2−1/4i. All of these elements are contained in Q( 4

√2, i) and since

each element needs both i and 4√

2, it is the splitting �eld.

15. Use Exercise 11 to show that there are at most eight distinct automor-phisms of the splitting �eld Q( 4

√2, i) of x4 − 2 over Q.

Let φ : Q( 4√

2, i) → Q( 4√

2, i) be an automorphism. Then, φ(a + b(21/4) +c(21/2) + di + e(21/4i) + f(21/2i)) = a + bφ(21/4) + cφ(21/4)φ(21/4) + dφ(i) +


eφ(21/4)φ(i) + fφ(21/4)φ(21/4)φ(i). A minimal polynomial for 21/4 over Q isx4 − 2. By Q11, this means that φ(21/4) can be mapped to any of the foursolutions. Now, x2 + 1 is the minimal polynomial for i and so i 7→ ±1. Thus,there are 4 · 2 = 8 possible automorphisms.

6.5 Finite Fields

1. Give addition and multiplication tables for the �nite �eld GF(23), asdescribed in Example 6.5.3.

Use the relation x8 = x (among others) to construct the multiplication table.Addition is the same as polynomial addition.

Let us examine another way of constructing GF(23). We see that x3 + 1 isirreducible over Z[x], so Z[x]/〈x3+x+1〉 ∼= GF(23). Additionally, we previouslyfound that Z[x]/〈x3+1〉 ∼= {a0I+a1C+a2C

2 ∈ GL3(Z2)|a0, a1, a2 ∈ Z2} where

C =

0 1 00 0 11 0 0

is the companion matrix for x3 + 1. Thus, we can use ordinary

matrix multiplication and addition to �nd the desired tables. This may be aneasier option if a computer is available.

2. Give addition and multiplication tables for the �nite �eld GF(32), and�nd a generator for the gyclic group of nonzero elements under multiplication.

The tables are left to the reader. After constructing the multiplication table,the reader should be able to readily identify the generator. Simply start withthe �rst nonzero element (call it x), look up the product of it with itself. Thentake the product and multiply it by x again. Continue the process until (1) allelements are accounted for or (2) a duplicate is generated. In the case of (2),move on to the next nonzero element and start the process over.

3. Find a generator for the cyclic group of nonzero elements of GF(24).

We see that x4 + x+ 1 is irreducible over Z2[x] and so Z2[x]/〈x4 + x+ 1〉 ∼=GF(24). One can construct the multiplication table of the nonzero elements ofthis �eld and identify the generator. The generator happens to be x+〈x4+x+1〉.

4. Find the splitting �elds over GF(3) for the following polynomials.(a) x4 + 2(b) x4 − 2.

(a) We see that x = 1 is a root, so x4 + 2 = (x− 1)(x3 + x2 + x+ 1). Then,x3 + x2 + x+ 1 has x = 2 as a root, so x4 + 2 = (x− 1)(x− 2)(x2 + 1). This iscompletely factored over Z3, so the splitting �eld is Z3[x]/〈x2 + 1〉 ∼= GF(32).


(b) We see that x4−2 is irreducible over Z3, so the splitting �eld is Z3[x]/〈x4−2〉 ∼= GF(34).

5. Show that x3−x−1 and x3−x+1 are irreducible over GF(3). Constructtheir splitting �elds and explicitly exhibit the isomorphism between these �elds.

Just plug in x = 0, 1, 2 to both polynomials and observe that the output isnever zero. Thus, both polynomials are irreducible. Thus, Z3[x]/〈x3 − x − 1〉is the splitting �eld of x3 − x − 1 and Z3[x]/〈x3 − x + 1〉 is the splitting �eldof x3 − x+ 1. Since both �elds have the same number of elements (namely 27),they must be isomorphic.

De�ne φ : Z3[x]/〈x3−x−1〉 → Z3[x]/〈x3−x+ 1〉 de�ned by φ(f(x) + 〈x3−x− 1〉) = (f(x)− 2) + 〈x3 − x+ 1〉.

(1-1) We have just subtracted 2 from every element, so φ is one-to-one.(Onto) If we wish to obtain g(x) + 〈x3− x+ 1〉, we use the input g(x) + 2 +

〈x3 − x− 1〉. Thus, φ is onto.(Preservation of +) Addition is corresponds to polynomial addition, so + is

clearly preserved.(Preservation of ·) In Z[x]/〈x3−x−1〉, we have the relation x3+〈x3−x−1〉 =

x+1+〈x3−x−1〉. In Z[x]/〈x3−x+1〉, we have the relation x3 +〈x3−x+1〉 =x − 1 + 〈x3 − x + 1〉. Since φ(x + 1 + 〈x3 − x − 1〉) = x − 1 + 〈x3 − x + 1〉,we see that multiplication will correspond. Thus, the two splitting �elds areisomorphic.

6. Show that x3 − x2 + 1 is irreducible over GF(3). Construct its splitting�eld and explicitly exhibit the isomorphism between this �eld and the splitting�eld of x3 − x+ 1 over GF(3).

Just plug in x = 0, 1, 2 and observe that the output is never 0. Thus, thepolynomial is irreducible. The splitting �eld for the polynomial would then beZ3[x]/〈x3 − x2 + 1〉. One can construct multiplication tables for both �elds,identify the generator and map them to each other.

7. Show that if g(x) is irreducible over GF(p) and g(x)|(xpm − x), thendeg(g(x)) is a divisor of m.

We know that xpm − x has splitting �eld GF(pm). Now, if g(x)|(xpm − x),

then g(x) must split over GF(pm) since xpm − x splits over it. Suppose that F

is the splitting �eld of g(x). Then, F is a sub�eld of GF(pm) since g(x) splitsover it. Also, [F : GF(p)] = deg(g(x)) since g(x) is the minimal polynomialover its splitting �eld. By Lemma 6.5.5 and its proof, [GF(pm) : GF(p)] = m.Furthermore, m = [GF(pm) : GF(p)] = [GF(pm) : F ][F : GF(p)] = [GF(pm) :F ] · deg(g(x)). Thus, deg(g(x)) is a divisor of m.

8. Let m,n be positive integers with gcd(m,n) = d. Show that, over any�eld, the greatest common divisor of xm − 1 and xn − 1 is xd − 1.


Let m = qd and n = rd. We see by simple polynomial long division thatxdq− 1 = (xd− 1)(xd(q−1) +xd(q−2) + · · ·+x+ 1) and similarly, xrd− 1 = (xd−1)(xd(r−1) +xd(r−2) + · · ·+x+1). We now examine p1(x) = xd(q−1) + · · ·+x+1and p2(x) = xd(r−1) + · · · + x + 1 to show that these two factors share nodivisors. Assume without loss of generality that q ≥ r. Then, we can writep1(x) = p2(x) + xd(q−1) + xd(q−2) + · · · + xd(r+1) + xdr. We see clearly thatp1(x)

p2(x)=p2(x) + xd(q−1) + · · ·+ xr

p2(x)=p2(x)

p2(x)+xd(q−1) + · · ·+ xr

p2(x). Clearly, the

second term is not perfectly divisible. (If it were, we'd have p2(x) as a factor ofxd(q−1)+· · ·+xr which would imply that p1(x) = p2(x)[1+p3(x)] for some p3(x)of degree less than p2(x). But since p1(x) is of degree q and p2(x) is of degreer such that (q, r) = 1, this is impossible.) Thus, gcd (xm − 1, xn − 1) = xd − 1.

9. If E and F are sub�elds of GF(pn) with pe and pf elements respectively,how many elements does E ∩ F contain? Prove your claim.

Since E and F are sub�elds, they have pe and pf elements where e|n andf |n. Then, E ∩ F is a sub�eld of both E and F , so the number of elementsmust divide both pe and pf . Thus, E ∩ F contains pgcd(e,f) elements.

10. Let p be an odd prime.(a) Show that the set S of squares in GF(pn) contains (pn + 1)/2 elements.(b) Given a ∈ GF(pn), let T = {a− x|x ∈ S}. Show that T ∩ S 6= ∅.(c) Show that every element of GF(pn) is a sum of two squares.(d) What can be said about GF(2n)?

(a) We know that multiplication in GF(pn)× is cyclic with some generatorx. Then x2 generates x2k for all 0 ≤ k < pn. This element generates (pn− 1)/2elements. Let us not forget that 02 = 0 so 0 ∈ S. Thus, |S| = (pn + 1)/2.

(b) Suppose a = 0. Then, 0 ∈ S, so 0− 0 = 0 ∈ T and so T ∩ S 6= ∅. Also,if a = 1, then 1− 1 = 0 ∈ T ∩ S.

Otherwise, we have a = xk0 for some �xed k0 ∈ Z. Then, xk0 − x2` =xk0(1 − x2`−k0). Thus, choose x2` such that x2`−k = 1. (Since GF(pn)× iscyclic, we can always do this.)

(c) It is clear that 0 = 02 + 02 and 1 = 12 + 02 are sums of squares. Now,let xk ∈ GF(pn). Then, by (b), xk − x2` = x2j for some x2` ∈ S and j ∈ Z+.Hence, xk = x2j +x2` = (xj)2 + (x`)2. Thus, any element of GF(pn) is the sumof two squares.

(d) Suppose that y ∈ GF(2n) is such that y = a2 + b2 = (a + b)2. SinceGF(2n) is closed under addition, we have that a + b = c ∈ GF(2n) and thaty = c2. Hence, y is a sum of squares if and only if y is itself a square. (Forinstance, if GF(22) = {0, 1, a, b}, then a2 = b and b2 = a, and so neither a norb can be a sum of squares.)


11. Show that xp − x+ a is irreducible over GF(p) for all nonzero elementsa ∈ GF(p).

We know that xp = x for any x ∈ GF(p). Hence, xp−x+a = x−x+a = a.There are no roots to a constant polynomial, so xp − x + a is irreducible overGF(p).

12. De�ne the function φ : GF(23) → GF(23) by φ(x) = x2, for all x ∈GF(23).

(a) Show that φ is an isomorphism.(b) Choose an irreducible polynomial p(x) to represent GF(23) as Z2/〈p(x)〉,

and give an explicit computation of φ,φ2, and φ3.

(a) (1-1) Here is the multiplication table for GF(23):

Note that every element is a square. Thus, the map is one-to-one.(Onto) Any �nite map that is one-to-one is also onto.(Preservation of +) φ(a+ b) = (a+ b)2 = a2 + b2 = φ(a)+φ(b). (See Lemma

6.5.4.)(Preservation of ·) φ(ab) = (ab)2 = a2b2 = φ(a)φ(b).

(b) Let p(x) = x3 + 1. (Check that this has no roots by using the multipli-cation table above.) Then, Z2/〈p(x)〉 has 8 elements and is thus isomorphic toGF(23). We must have φ(ax2 + bx+ c+ p(x))2 = a2x4 + b2x2 + c2 + p(x). Sincex3 = 1, we know x4 = x, so φ(ax2 + bx + c) = bx2 + a2x + c2. The rest of themaps are similar.

6.6 Irreducible Polynomials over Finite Fields

1. Verify Theorem 6.6.1 in the special case of x16 − x over GF(2), bymultiplying out the appropriate irreducible polynomials from the list given in theanswer to Exercise 12 of section 4.2.


The list of irreducible polynomials is: x, x + 1, x2 + x + 1, x3 + x2 + 1,x3 + x + 1, x4 + x3 + x2 + 1, x4 + x3 + 1, and x4 + x + 1. Now, by Theorem6.6.1, the monic irreducible factors of x2

4 − x should be the monic irreduciblepolynomials of degree 4, 2, and 1. Thus, we must show that x(x+ 1)(x2 + x+1)(x4 + x3 + x2 + x+ 1)(x4 + x3 + 1)(x4 + x+ 1) = x16 − x.

First, x(x+1) = x2+x. Next, (x2+x)(x2+x+1) = x4+2x3+2x2+x = x4+x.Then, (x4 + x)(x4 + x3 + x2 + x + 1) = x8 + x7 + x6 + x3 + x2 + x. Next,(x8 + x7 + x6 + x3 + x2 + x)(x4 + x3 + 1) = x12 + x9 + x8 + x6 + x4 + x3 + x.Finally, (x12 + x9 + x8 + x6 + x4 + x3 + x)(x4 + x+ 1) = x16 + x = x16 − x.

2. Use Theorem 6.6.1 to show that over GF(2) the polynomial x32+x factorsas a product of the terms x, x+ 1, x5 +x2 + 1, x5 +x3 + 1, x5 +x4 +x3 +x+ 1,x5 + x4 + x2 + x+ 1, x5 + x3 + x2 + x+ 1, and x5 + x4 + x3 + x2 + 1.

We have that x32 +x = x25 −x, so by Theorem 6.6.1, the irreducible factors

must be the irreducible polynomials of degree 5 and 1. It is left to the readerto show that the list of factors is complete. To do this, realize that in order forthe polynomial to be irreducible, it must be linear or have an odd number ofterms. Since 5 has no other divisors than 5 and 1, the non-linear polynomialsmust be of degree 5 and have an odd number of terms.

3. Let F be a �eld of characteristic p, with prime sub�eld K = GF(p). Showthat if u ∈ F is a root of the polynomial f(x) ∈ K[x], then up is also a root off(x).

Let f(x) = anxn + · · · + a1x + a0. If f(u) = 0, then 0 = f(u)p = (anu

n +· · · + a1u + a0)p = apn(up)n + · · · + ap1u

p + ap0. (The last equality holds sincechar(F ) = p.) Now, by Fermat's Little Theorem, api ≡ ai(mod p), so the lastexpression in the equality becomes an(up)n + · · ·+ a1(up) + a0 = f(up). Thus,f(up) = 0.

4. Let u be a primitive element of GF(pm) and let M (i)(x) be the minimum

polynomial of ui over GF(p). Show that every element of the form uipk

is alsoa root of M (i)(x).

If u is primitive, then 〈u〉 = GF(pm)•. By de�nition of M (i)(x), we haveM (i)(ui) = 0. By Q3, we know that M (i)

((ui)p

)= M (i)(uip) = 0. Similarly,

since M (i)(uip) = 0, we know that M (i)((uip)p

)= M (i)(uip

2

) = 0. Continuing

this argument, we see that M (i)(uipk

) = 0.

5. Let GF(26) be represented by Z2[x]/〈x6+x+1〉, and let u be any primitiveelement of GF(26). Show that GF(23) = {0, 1, u9, u18, u27, u36, u45, u54}.

I'm not sure why GF(26) would have to have the representation discussedabove. If u is primitive, then 〈u〉 = GF(26)× (which has a multiplicative group


structure). Since GF(23) is a sub�eld of GF(26) (since 3|6), we know thatGF(23)× ≤ GF(26)×. By Cauchy's theorem, any generator of GF(23) musthave order dividing 64. Since |GF(23)| = 8, the only possible element that canbe a generator is u9. Hence GF(23)× = 〈u9〉. Thus, GF(23) = {0, 1, u9, ..., u54}.

6. Let F be a �eld, and let n be a positive integer. An element ζ ∈ F iscalled a primitive nth root of unity if it has order n in the multiplicativegroup F×. Show that no �eld of characteristic p > 0 contains a primitive pthroot of unity.

If F has characteristic p, then |F×| = pk − 1 for some k. Now, by clearly,p is not a divisor of pk − 1. (The division algorithm can be used to show this.)Thus, by Cauchy's Theorem, there cannot exist an element of F× of order p.Hence, there can be no primitive pth root of unity.

7. Let n ∈ Z+, and de�ne τ(n) to be the number of divisors of n.(a) Show that τ is a multiplicative function.(b) Show that if n = pα1

1 pα22 · · · p

αkk , then τ(n) = (α1 +1)(α2 +1) · · · (αk+1).

(c) Show that τ(n) is odd if and only if n is a square.(d) Show that

∑d|n τ(d)µ(n/d) = 1.

(a) Suppose that n has divisors r and s such that (r, s) = 1. Then, r and sshare no factors, so τ(rs) = τ(r)τ(s).

(b) By (a), we have τ(n) = τ(pα11 )τ(pα2

2 ) · · · τ(pαkk ). Then, each pαii has1, pi, p

2i , ..., p

αii as divisors, so τ(pαii ) = αi + 1. Hence, τ(n) = (α1 + 1)(α2 +

1) · · · (αk + 1), as desired.(c) If n is a square, then all of its prime factors are raised to an even power,

so by (b), τ(n) is odd. If τ(n) = τ(pα11 · · · p

αkk ) is odd, then by (b) all of the

factors of (α1 + 1) · · · (αk + 1) must be odd. (For if there were so much as oneαi + 1 that were even, then the expression (α1 + 1) · · · (αk + 1) would be even.)This implies that αi + 1 = 2mi for some mi for all αi. Thus, αi = 2mi + 1, soαi is odd for all i.

(d) We have established that τ is a multiplicative function, so if we de�nef(n) ≡ 1, then τ(n) =

∑d|n f(d) and so by the Mobius Inversion Formula, we

obtain 1 = f(m) =∑n|m µ(m/n)τ(n).

8. Let n ∈ Z+ and de�ne σ(n) =∑d|n d, the sum of positive divisors of n.

(a) Show that σ is a multiplicative function.

(b) Show that if n = pα11 pα2

2 · · · pαkk , then σ(n) =

∏ki=1

((pαi+1i − 1)/(pi − 1)

).

(c) Show that σ(n) is odd if and only if n is a square or two times a square.(d) Show that

∑n|d σ(d)µ(n/d) = n.

(a) Suppose that n has two divisors r and s such that (r, s) = 1. Then, rand s share no divisors, so σ(rs) = σ(r)σ(s). Thus, σ is multiplicative.


(b) By (a), we see that σ(n) =∏ki=1 σ(pαii ). We need to show that σ(pαii ) =

pαi+1i − 1

pi − 1. We do this by induction on αi. For αi = 1, we get σ(pi) = p + 1.

Also,p2i − 1

pi − 1=

(pi + 1)(pi − 1)

pi − 1= pi + 1. Thus, the statement is true for

αi = 1. Suppose it is true for all αi ≤ N . Then, for αi = N + 1, we seethat σ(pN+1

i ) = σ(pNi ) + pN+1i . Thus, by the inductive hypothesis, σ(pN+1

i ) =

pN+1i − 1

pi − 1+pN+1

i =pN+1i − 1 + pN+2 − pN+1

pi − 1=pN+2 − 1

pi − 1, as required. Hence,

σ(n) =∏ki=1

((pαi+1i − 1)/(pi − 1)

).

(c) For any prime number pi, we have τ(pαii ) = αi + 1. Hence, if αi is even,we will have that σ(pαii ) is a sum of an odd number of terms. Thus, σ(pαii ) isodd for all pi if αi is even. Hence, if n is a square, σ(n) = σ(pα1

1 ) · · ·σ(pαkk ) is aproduct of odd numbers and thus odd.

If n is two times a square, then σ(n) is of the form σ(n) = σ(22α1+1)σ(p2α22 ) · · ·σ(p2αkk ).

As previously discovered, σ(p2α22 ) · · ·σ(p2αkk ) is odd. Then, τ(22α1+1) = 2α+ 2.

One of the factors is 1, and the rest are even. Thus, σ(22α1+1) is odd.Conversely, σ(pαii ) is even if αi is odd. (See �rst paragraph of (c)). Hence,

σ(n) = σ(pα11 ) · · ·σ(pαkk ) would be even since one even factor makes the entire

product even.(d) We have shown σ to be multiplicative. Let f(n) = n. Then, σ(n) =∑d|n f(d). Thus, by the Mobius Inversion Formula,

∑n|d σ(d)µ(n/d) = n.

9. A positive integer n is called perfect if it is equal to the sum of its properpositive divisors. Thus, n is perfect if and only if σ(n) = 2n. Prove that n isan even perfect number if and only if n = 2p−1(2p − 1), where p and 2p − 1 areprime numbers.

(⇐) Suppose that p and 2p − 1 are prime. We show that n = 2p−1(2p − 1)is perfect. Examine σ(n) = σ(2p−1)σ(2p − 1). Since 2p − 1 is prime, its onlydivisors are 1 and 2p−1. Thus, σ(n) = σ(2p−1) ·2p = (2p−1)2p from Q8. Thus,σ(n) = 2 · 2p−1(2p − 1) = 2n, as needed.

(⇒) Suppose that p is prime, but 2p− 1 is not. Then σ(n) = σ(2p−1)σ(2p−1) = (2p− 1) · (something greater than 2p). (Since at very least 2p− 1 has threefactors:1, 2p − 1 and p̄, for some prime p̄. If we add these three up we get2p + p̄ > 2p.) Thus, σ(n) > 2n. Thus, n is not perfect.

10. Let D be an integral domain. Show that if f : Z+ → D is a nonzeromultiplicative function, then

∑d|n µ(d)f(d) =

∏p|n(1 − f(p)), for all n ∈ Z+,

where the product is taken over all prime divisors p of n.

Suppose that n has the prime factorization n = pα11 pα2

2 · · · pαkk . We form di-

visors of n by selecting prime factors of n and multiplying them together. Both fand µ are multiplicative, so we see that µ(d)f(d) = µ(qα1

1 qα22 · · · qαmm )f(qα1

1 qα22 · · · qαmm ) =

µ(qα11 )f(qα1

1 ) · · ·µ(qαmm )f(qαmm ). Now, if αi > 1 for any pi prime factor of a di-


visor, then by the de�nition of µ, we have that the entire term must equal zerosince µ(pαii ) = 0. Thus, the only divisors of n that will ever contribute to thesum are those that are products of prime factors.

We now show that∑d|n µ(d)f(d) =

∏p|n(1 − f(p)) by induction on the

number of prime factors of n. Suppose that n = p for some prime p. Then,∑d|n µ(d)f(d) = µ(1)f(1) + µ(p)f(p) = f(1) − f(p) = 1 − f(p). (We see that

f(1) = 1 since we must have f(x) = f(1 · x) = f(1)f(x).) Thus, we haveestablished a basis for induction. Let us suppose that this formula holds for Mprime factors. Now, we need to show that it works for M + 1 factors.

Let n = pα11 · · · p

αM+1

M+1 . Then, n = mpαM+1

M+1 where m = pα11 · · · p

αMM . Then,∑

d|n µ(d)f(d) =∑d|m µ(d)f(d)−f(pM+1)+

∑Mi=1 f(pipM+1)−

∑i 6=j f(pipjpM+1)+

· · · + (−1)M∑i1 6=i2 6=···6=iM f(pi1pi2 · · · piM pM+1). Since f is multiplicative, we

can factor out f(pM+1) and obtain∑d|n µ(d)f(d) =

∑d|m µ(d)f(d)− f(pM+1)(

1−∑Mi=1 f(pi) +

∑i 6=j f(pipj)− · · ·+ (−1)M−1

∑i1 6=i2 6=···iM f(pi1pi2 · · · piM )

).

We notice that the coe�cient of −f(pM+1) on the RHS of the equation is equalto∑d|m µ(d)f(d). Hence,

∑d|n µ(d)f(d) =

∑d|m µ(d)f(d)−f(pM+1)

∑d|m µ(d)f(d).

By the inductive hypothesis, this is equal to∏p|m(1−f(p))−f(pM+1)

∏p|m(1−

f(p)) =(∏

p|m(1− f(p)))

(1−f(pM+1)) =∏p|n(1−f(p)), as desired. (Whew!)

11. Let R be a commutative ring. Let R be the set of all functions f : Z+ →R. For f, g ∈ R, de�ne f + g by ordinary addition of functions: (f + g)(n) =f(n) + g(n), for all n ∈ Z+. De�ne a product * on R as follows:

(f ∗ g)(n) =∑d|n

f(d)g(n/d), for all n ∈ Z+.

The product * is called the convolution product of the functions f and g.De�ne ε : Z+ → R by ε(1) = 1 and ε(n) = 0, for all n > 1.

(a) Show that R is a commutative ring under the operations + and *, withidentity ε.

(b) Show that f ∈ R has a multiplicative inverse if and only if f(1) isinvertible in R.

(c) Show that if f, g ∈ R are multiplicative functions, then so is f ∗ g.(d) Show that if R is an integral domain, then the set of nonzero multiplica-

tive functions in R is a subgroup of R×, the group of units of R.(e) De�ne µR ∈ R as in De�nition 6.6.3, with the understanding that 0 and

1 are (respectively) the additive and multiplicative identities of R. Let ν ∈ R bede�ned by ν(n) = 1, for all n ∈ Z+. Show that µR∗ν = ε, and that µR∗ν∗f = f ,for all f ∈ R.

(a) (Group property of +) Already shown in Ch. 5.(Distributivity) Examine f∗(g+h)(n) = f∗(g(n)+h(n)) =

∑d|n f(d)(g(n/d)+

h(n/d)) =∑d|n[f(d)g(n/d)+f(d)h(n/d)]=

∑d|n f(d)g(n/d)+

∑d|n f(d)h(d/n)=

(f ∗ g)(n) + (f ∗ h)(n).


The other distributive property is shown in a similar way.(Commutativity) We see that (f∗g)(n) =

∑d|n f(d)g(n/d) =

∑d|n f(n/d)g(d) =

(g ∗ f)(n). (Whatever factors aren't in n/d are in d and vice-versa.)(b) Suppose that there is g ∈ R such that g ∗ f = f ∗ g = ε. This happens if

and only if n = 1 implies that 1 = ε(1) = f(1)g(1) = g(1)f(1). This happens ifand only if f(1) is invertible.

(c) Suppose that f and g are multiplicative functions. Suppose that n = pqwhere p and q are prime. Then, (f ∗ g)(pq) = f(1)g(pq) + f(p)g(q) + f(q)g(p) +f(pq)g(1). (1) Next, (f ∗ g)(p)(f ∗ g)(q) = [f(1)g(p) + f(p)g(1)] · [f(1)g(q) +f(q)g(1)] = g(p)g(q) + g(p)f(q) + f(p)g(q) + f(p)f(q). (2) We see that (1) =(2) (keeping in mind that for multiplicative functions f and g, we must havef(1) = g(1) = 1). Hence, f ∗ g is multiplicative.

(d) Suppose that f is multiplicative. Then, as always, f(1) = 1 ∈ R×.Hence, f ∈ R×. By (c), the set of multiplicative functions is closed. Thus, theset of multiplicative functions are a subgroup of R×.

(e) Suppose that n = pα11 · · · p

αkk . Then, (µR ∗ν)(n) = (µR ∗ν)(pα1

1 ) · · · (µR ∗ν)(pαkk ) =

∏kj=1

(∑αki=0 µR(pαij )

). Now, clearly (µR ∗ ν)(1) = 1. Also, it is clear

that if αi > 1 for some i, then (µR ∗ ν)(n) = 0. So, suppose that αi = 1 for all

i. Then, we have (µR ∗ ν)(n) =∏kj=1(1 − 1) = 0. Hence, (µR ∗ ν) = ε. Now

that this has been established, we have that µR ∗ ν ∗ f = ε ∗ f = f (keeping inmind that the ring R is commutative).

12. Let R be a commutative ring. Let f : Z+ → R be any function, and letF : Z+ → R be de�ned by F (n) =

∑d|n f(d), for all n ∈ Z+. Show that if F is

a multiplicative function, then so is f .

Suppose we have (r, s) = 1. Now, by the Mobius Inversion formula, we havef(rs) =

∑n|rs µ(rs/n)F (n) = 1 − F (r) − F (s) + F (rs) = 1 − F (r) − F (s) +

F (r)F (s). (1)Next, f(r)f(s) = (−F (r) + 1)(−F (s) + 1) = F (r)F (s) − F (r) − F (s) + 1.

(2)We see that (1) = (2), so f is multiplicative.

6.7 Quadratic Reciprocity

1. Prove that(abp

)=(ap

)(bp

)for all a, b ∈ Z such that p - a and p - b.

We see that by Euler's Criterion(abp

)≡ (ab)(p−1)/2 ≡ a(p−1)/2b(p−1)/2(mod p).

On the other hand,(ap

)≡ a(p−1)/2(mod p) and

(bp

)≡ b(p−1)/2, so

(ap

)(bp

)≡

a(p−1)/2b(p−1)/2(mod p). Thus,(abp

)=(ap

)(bp

).

2. Compute the following values of the Legendre symbol.


(a)(231997

)(b)

(783997

)(a) From Q1,

(231997

)=(3·7·11997

)=(

3997

) (7

997

) (11997

). Following Example

6.7.1, we �nd that(

3997

)=(9973

)(−1)(997−1)(3−1)/4 ≡ 9972/2(−1)498(mod 3) ≡

1(mod 3). Thus,(

3997

)= 1. Similarly,

(7

997

)= −1 and

(11997

)= −1. Hence,(

231997

)= (1)(−1)(−1) = 1.

(b) Again,(783997

)=(

33

997

) (29997

). We compute, as above, that

(3

997

)= 1.

Also,(

29997

)= −1. Hence,

(783997

)= (1)3(−1) = −1.

3. Is the congruence x2 ≡ 180873(mod 997) solvable?

First of all, 180873 ≡ 416(mod 997). To determine if the congruence is

solvable, we must compute(416997

)=(

2997

)5 ( 13997

). Now, obviously

(2

997

)=

(−1)9972−1

8 = −1. Also,(

13997

)= 1. Hence

(416997

)= −1. Thus, the congruence is

not solvable.

*4. Determine the value of(rp

)for the indicated values of r, where p is an

odd prime subject to the indicated conditions.(a) r = 5, p 6= 5(b) r = 6, p 6= 3(c) r = 7, p 6= 7(d) r = 11, p 6= 11(e) r = 13, p 6= 13

(a) We see that(

5p

)=(p5

)(−1)4(p−1)/4 =

(p5

). Thus,

(5p

)≡ p2(mod 5).

Thus, we have

Value of p modulo 5(

5p

)1 12 -13 -14 1

(b) We see that(

6p

)=(

3p

)(2p

)=(p3

)(−1)

p2−18 . Example 6.7.3 gives the

possible values of(p3

). Using these, we construct the following table.


6p

)1 15 17 -111 -1


(c) We see that(

7p

)=(p7

)(−1)6(p−1)/4. Now,

(p7

)≡ p3(mod 7). If

(p7

)=

1, then p ≡ 1, 2, or 4(mod 7). If(p7

)= −1, then p ≡ 3, 5, or 6(mod 7). In

addition, we have that p ≡ 1 or 3(mod 4). Thus, we create the following table:


7p

)1 -19 -125 -115 123 111 117 -15 -113 -13 119 127 1

[FINISH]5. If a is a quadratic nonresidue of each of the odd primes p and q, is the

congruence x2 ≡ a(mod pq) sovlable?

Suppose that x2 ≡ a(mod pq) has a solution. Then, x2 − a = kpq for somek. This implies that x2 = a + p(kq) = a + q(kp) imlying that x2 ≡ a(mod p)and x2 ≡ a(mod q) have solutions. Contradiction.

6-8. [FINISH]

Chapter 7

Structure of Groups

7.1 Isomorphism Theorems: Automorphisms

See other Scribd document.

7.2 Conjugacy

1. Let H be a subgroup of G. Prove that N(H) is a subgroup of G.

Let a, b ∈ N(H). Then, we have aHa = H and b−1Hb = H. Then,ab−1H(ab−1)−1 = ab−1Hba−1 = a(b−1Hb)a−1 = aHa−1 = H. Thus, N(H) ≤G.

2. Let H be a subgroup of the group G. Prove that the subgroups of Gthat are conjugate to H are in one-to-one correspondence with the left cosets ofN(H) in G.

Let K ∼ H. Then K = aHa−1 for some a ∈ G. De�ne the prescription

aHa−1φ7→ aN(H). Now, if φ(aHa−1) = φ(bHb−1), then we have aN(H) =

bN(H) which implies that ab−1 ∈ N(H). Hence ab−1Hba−1 = H. This isequivalent to aHa−1 = bHb−1. Thus, members of G/N(H) are in a one-to-onecorrespondence with subgroups of G conjugate to H.

3. Let G be a group with subgroups H and K such that H ⊆ K. Show thatH is a normal subgroup of K if and only if K ⊆ N(H).

We see that H E K if and only if aHa−1 = H for all a ∈ K. Since anyelement a such that aHa−1 = H is contained in N(H) by de�nition, we see thatH E K if and only if K ⊆ N(H).

56

CHAPTER 7. STRUCTURE OF GROUPS 57

4. Let p be a prime number, and let C be a cyclic subgroup of order p in Sp.Compute the order of N(C).

The only element of order p in Sp is (1, 2, ..., p). Let K ∼ C. Then, everyelement of K has the same structure as an element in C by Example 7.2.3. Butsince the only element of order p is (1, 2, ..., p), it must be that K = C. Thus,C E Sp. Thus, |C| = p.

5. Let G be a group, let H be a subgroup of G, and let a ∈ G. Show thatthere is a subgroup K of G such that K is conjugate to H and aH = Ka.

We must simply show that aHa−1 ≤ G. Let ah1a−1, ah2a

−1 ∈ aHa−1.Then, (ah1a

−1)(ah2a−1)−1 = (ah1a

−1)(ah−12 a−1) = ah1h2a−1 ∈ aHa−1. Thus,

aHa−1 ≤ G, as desired.

6. Let G be a group, let x, y ∈ G and let n ∈ Z. Show that y is a conjugateof xn if and only if y is the nth power of a conjugate of x.

y = (gxg−1)n ⇔ y = (gxg−1)(gxg−1) · · · (gxg−1)⇔ y = gxng−1 ⇔ y ∼ xn.

7. Find the conjugate subgroups of D4.

First, all normal subgroups are self-conjugate, and since each subgroup of or-der 4 has index 2, the subgroups {e, a2, b, a2b}, {e, a, a2, a3}, and {e, a2, ab, a3b}are all self conjugate.

Next {e, b} ∼ {e, a2} ∼ {e, a2b} (from simple veri�cation) and (ab)a3b(ba3) =a2ba3, so {e, ab} and {e, a3b} are self-conjugate.

8. Find the conjugacy classes of D5.

First,D5 = {e, a, a2, a3, a4, b, ab, a2b, a3b, a4b}. The cyclic subgroup {e, a, a2, a3, a4}has order 5 and is of index 2 and thus normal. Thus, it is its own conjugacyclass. There can be no more subgroups of order 5 (for if there were, there wouldbe more than ten elements). The only possible orders of subgroups would beorder 2. There are �ve such subgroups: {e, ab}, {e, a2b}, {e, a3b}, {e, a4b}, and{e, b}. We multiply {e, ab} on the left and right by the non-identity elements inthe subgroups above and see that all �ve subgroups are conjugate.

9. Describe the conjugacy classes of S5 by listing the types of elements andthe number of each type in each class.


Class Calculation of Cardinality Cardinality

[(1)] The identity mapping is unique 1

[(1,2)]

(52

)10

[(1,2)(3,4)]

(52

)(32

)· 1

215

[(1,2,3)]

(53

)· 2! 20

[(1,2,3)(4,5)] Same 20

[(1,2,3,4)]

(54

)· 3! 30

[(1,2,3,4,5)] 4! 24Total: 120

10. Find the conjugacy classes of A4.

A4 consists of the identity element (1), all 3-cycles (of which there are 8), andthree single products of disjoint transpositions (those being (1,2)(3,4), (1,3)(2,4),and (1,4)(3,2)) . Since disjoint transpositions commute, we know that theproducts of disjoint transpositions cannot be conjugate to any 3-cycle or theidentity. We now �nd the conjugacy classes: (1,3)(2,4)[(1,2)(3,4)](1,3)(2,4) =(1,3)(4,2)(2,1)(4,3)(3,1)(2,4) = (1,2,4)(2,4)(1,3)(1,3,4) = (2,1)(4,3) (1,4)(3,2)[(1,2)(3,4)](1,4)(3,2)= (3,2)(4,1)(1,2)(3,4)(4,1)(3,2) = (3,2)(2,1,4)(1,4,3)(3,2) = (2,4)(1,3) Thus, (1,2)(3,4)∼ (1,3)(2,4) ∼ (1,4)(3,2). Doing a similar analysis on the 3-cycles, we get that(1,2,3) ∼ (4,3,2) ∼ (3,4,1) ∼ (2,1,4) and (1,3,2) ∼ (4,2,3) ∼ (3,1,4) ∼ (2,4,1).

11. Find the conjugacy classes for the quaternion group Q de�ned in Ex-ample 3.3.7.

We see iji−1 = j; kjk−1 = j. Thus, j is its own conjugacy class.Then, jij−1 = −i = kik−1. Thus, i ∼ −i.Then, iki−1 = −k = jkj−1. Thus, k ∼ −k. Then, ±1 are their own

conjugacy classes.

12. Write out the conjugacy class equations for the following groups.(a) A4

(b) S5

(a) Using Q4, clearly |A4| = |Z(A4)|+∑

[A4 : C(x)] = 1 + (3 + 4 + 4) = 12.(b) Using Q9, clearly |S4| = |Z(S5)| +

∑[S5 : C(x)] = 1 + (10 + 15 + 20 +

20 + 30) = 120.

13. Let the dihedral group Dn be given by elements a of order n and b oforder 2, where ba = a−1b. Show that am is conjugate to only itself and a−m,and that amb is conjugate to am+2kb for any integer k.


Suppose that x ∼ am in Dn. We see that x cannot have a factor of b in itsince if a`b = yamy−1, then whether y has a factor of b in it or not, the b's iny and y−1 will cancel. Thus, x must be a power of a. Thus, a` = akbamban−k

for some k. By the fact that ba = a−1b, we see that this implies that a` =akan−man−k = ak+n−m+n−k = a2n−m = a−m. Of course, every element isconjugate to itself as well.

We see that akbambban−k = akbam+n−k = ak+n−m+k = a2k+mb.

14. Show that the Frobenius group F20 (de�ned in Exercise 12 of Section 7.1)is isomorphic to the subgroup of S5 generated by the permutations (1, 2, 3, 4, 5)and (2, 3, 5, 4). Use this fact to help in �nding the conjugacy classes of F20, andits conjugacy class equation.

The fact that 〈(1, 2, 3, 4, 5), (2, 3, 5, 4)〉comes from the work done in Q13 ofSection 7.1, where we found elements a of order 5 and b of order 4 and that anyelement of F20 could be uniquely expressed as a product of these two elements(i.e., they generate F20).

From Q12 of 7.1, Z(F20) = 4. Now,〈(1, 2, 3, 4, 5)〉 = {(1), (1, 2, 3, 4, 5)), (1, 3, 5, 2, 4), (1, 4, 2, 5, 3), (1, 5, 4, 3, 2)} and

〈(2, 3, 5, 4)〉 = {(1), (2, 3, 5, 4), (2, 5)(3, 4), (2, 4, 5, 3)}. So, it is easliy seen that〈(1, 2, 3, 4, 5), (2, 3, 5, 4)〉 = {(1), (1, 2, 3, 4, 5), (1, 3, 5, 2, 4), (1, 4, 2, 5, 3), (1, 5, 4, 3, 2),(2, 3, 5, 4), (2, 5)(3, 4), (2, 4, 5, 3), (1, 3, 2, 5), (1, 5, 3, 4), (1, 2, 4, 3), (1, 4, 5, 2), (1, 5)(2, 4),(1, 4)(2, 3), (1, 3)(4, 5), (1, 2)(3, 5), (1, 4, 3, 5), (1, 2, 5, 4), (1, 5, 2, 3), (1, 3, 4, 2)}.

Thus, the conjugacy classes of F20 are [a], [b], and [b2]. The class equationwould then be |F20| = 4 + 4 + 10 + 2 = 20.

15. Show that if a group G has an element a which has precisely two conju-gates, then G has a nontrivial proper normal subgroup.

Suppose that [a]∼ has precisely two elements. Then, [G : C(a)] = 2 andC(a) E G (since subgroups of index two are normal.)

*16. Show that for each prime p, there exists a nonabelian group of orderp3.

Let us examine the possible centers of G. By Lagrange's theorem and Burn-side's theorem, |Z(G)| = p, or p2. Any group of order p is cyclic (isomorphicto Zp), and the abelian groups of order p2 (recall that all groups of order p2

are abelian) are Zp2 and Zp×Zp. The following conclusions are made from theG/Z(G) theorem.

Z(G) G/Z(G) Conclusion

Zp Zp2 G is abelianZp × Zp ???

Zp2 Zp G is abelian


Thus, we seek to �nd a nonabelian group of order p3 with center Z(G) ∼= Zpand G/Z(G) ∼= Zp × Zp. [FINISH]

17. Let G be a nonabelian group of order p3, for some prime number p.Show that Z(G) must have order p.

The center of any p- group is nontrivial by Burnsides Theorem, so the pos-sibilities for |Z(G)| are p, p2, or p3. However, if |G| = p2, then G would beabelian. Since G is given to be nonabelian (so its center cannot have order p3),we must have |Z(G)| = p.

18. Determine the conjugacy classes of the alternating group A5 and usethis information to show that A5 is a simple group.

We have already found the conjugacy classes for A5 in Q9.

Class Order

[(1)] 1[(1,2)(3,4)] 15[(1,2,3)] 20

[(1,2,3,4,5)] 24

We know that a normal subgroup of A5 (or any group, for that matter) isthe union of conjugacy classes. In the following table, for every row, constructa union of conjugacy classes corresponding to the column headers who have X'sin their cell. The order of this union is given in the right-most column.

[(1)] [(1,2)(3,4)] [(1,2,3)] [(1,2,3,4,5)] Order of union

X 1X X 16X X 21X X 25X X X 36X X X 40X X X 45X X X X 60

Now, by Lagrange's theorem, the order of any group must divide |A5| =60. The only possible unions that have orders dividing 60 are [(1)] = (1) and[(1)] ∪ [(1, 2)(3, 4)] ∪ [(1, 2, 3)] ∪ [(1, 2, 3, 4, 5)] = A5. Hence, A5 has no propernontrivial normal subgroups. Thus, A5 is simple.

Problems and Solutions to Abstract Algebra (Beachy, Blair)

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