Solutions to Modern Algebra (John R. Durbin, 5E)

Jason Rosendalejason.rosendale@gmail.com

December 21, 2011

This work was done as an undergraduate student: if you really don’t understand something in one of theseproofs, it is very possible that it doesn’t make sense because it’s wrong. Any questions or corrections can bedirected to jason.rosendale@gmail.com.

1 Mappings and Operations

1.26 Let α : S → S be a mapping that is one-to-one but not onto. Because the mapping is not onto, we cannotdirectly define an inverse mapping α−1 because it would be undefined for some elements in S. However,we can define a second function β : S → S as:

β(s) =

{α−1(s) iff s ∈ α(S)s iff s 6∈ α(S)

By showing that β is onto but not one-to-one, the “only if” half of the proof will be demonstrated. First,proof that the function β is onto:

α is well-defined assumed→ (∀a ∈ S)(∃b ∈ S)(α(a) = b) definition of well-defined→ (∀a ∈ S)(∃b ∈ S)(β(b) = a) definition of β→ β is onto definition of onto

Proof that the function β is not one-to-one:

α is not onto assumed→ (∃b ∈ S)(∀a ∈ S)(α(a) 6= b) definition of onto→ (∃b ∈ S)(∀a ∈ S)(α(a) 6= b ∧ b 6∈ α(S)) definition of image

Because α is one-to-one, we know that α(b) has an image in S. And because b 6∈ α(S), we knowthat α(b) 6= b. So:

→ (∃b, c ∈ S)(∀a ∈ S)(α(a) 6= b ∧ b 6∈ α(S) ∧ α(b) = c ∧ c 6= b)→ (∃b, c ∈ S)(∀a ∈ S)(β(b) = b ∧ β(c) = b ∧ c 6= b) definition of β→ β is not one-to-one definition of one-to-one

Now, assume that α is some function that is onto but not one-to-one. Because the mapping is not one-to-one, we can’t directly define an inverse mapping because it would be overdefined for some elements of S.However, we can define a function γ : S → S as:

γ(s) = min{r ∈ S : α(r) = s}

This function maps each element s onto the smallest element r such that α(r) = s.

The function γ is one-to-one. If we assume that γ(s1) = γ(s2), then by the definition of γ this means thatmin{r1 ∈ S : α(r1) = s1} = min{r2 ∈ S : α(r2) = s2}. This equality just means that the two minimum

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elements are the same: r1 = r2. So there is some r such that α(r) = s1 and α(r) = s2. Because α iswell-defined, this shows that s1 = s2. Therefore γ is one-to-one.

The function γ is not onto. Because α is not one-to-one, there is some r, s ∈ S such that α(r) = α(s) eventhough r 6= s. But the definition of γ causes there to be no element that maps to the larger of the twoelements r, s. This larger element is not in the image of γ(S), and therefore γ is not onto.

This shows that a function that is one-to-one but not onto can be used to construct a function that is ontobut not one-to-one, and vice-versa.

1.27 x ∈ α(A ∪B) assumed↔ (∃s ∈ A ∪B)(α(s) = x) definition of the domain of α↔ (∃s1 ∈ A)(α(s1) = x) ∨ (∃s2 ∈ B)(α(s2) = x) definition of set unions↔ x ∈ α(A) ∨ x ∈ α(B) definition of the domain of α↔ x ∈ α(A) ∪ α(B) definition of set unions

1.28 x ∈ α(A ∩B) assumed↔ (∃s ∈ A ∩B)(α(s) = x) definition of the domain of α→ (∃s1 ∈ A)(α(s1) = x) ∧ (∃s2 ∈ B)(α(s2) = x) definition of set intersection

Note that the previous step can not be justified in the other direction without assuming that s2 = s1.

↔ x ∈ α(A) ∧ x ∈ α(B) definition of the domain of α↔ x ∈ α(A) ∩ α(B)

1.29 As noted in the previous problem, the proof is bidirectional if we can assume that α(s1) = x ∧ α(s2) = ximplies that s1 = s2.

1.30 Let R represent the infinite subset of S. If R is infinite, then there is a mapping α : R → R that isone-to-one but not onto. Define a new function β : S → S to be:

β(s) =

{α(s) iff s ∈ Rs iff s 6∈ R

This function is one-to-one. Assume that a 6= b:

case i) If a ∈ R and b ∈ R, then β(a) = α(a) and β(b) = α(b). And because α is one-to-one, the fact thata 6= b implies that α(a) 6= α(b) and therefore β(a) 6= β(b).

case ii) If a 6∈ R and b 6∈ R, then β(a) = a and β(b) = b. So the fact that a 6= b implies that β(a) 6= β(b).

case iii) If a ∈ R and b 6 inR, then β(a) = α(a) and β(b) = b. And the ranges of these functions meansthat α(a) ∈ R and b 6∈ R. So α(a) 6= b, and therefore β(a) 6= b.

case iv) Similar to case (iii)

So in all cases, a 6= b → β(a) 6= β(b). By contrapositive, β is one-to-one. It can also be shown that β isnot onto: because α is not onto, there is some r ∈ R that is not in the domain of α(R). But the image ofβ is (α(R) ∪ R̃): so this r is not in the image of β.

Therefore β : S → S is a function that is one-to-one but not onto: therefore, S is an infinite set.

2 Composition

2.20 Let α : A→ B be an invertible function with an inverse β : B → A. Then, by the definition of invertible,α ◦ β = ιB and β ◦ α = ιA. But this is exactly the definition for a function β : B → A that is invertiblewith an inverse α : A→ B.

2.23 The domains and codomains of β and γ are equal by definition. We need only show that β(t) = γ(t) forevery t ∈ T . Proof by contradiction:

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(∃t ∈ T )(β(t) 6= γ(t)) hypothesis of contradiction→ (∃t ∈ T )(β(t) 6= γ(t)) ∧ (∀t1 ∈ T )(∃s ∈ S)(α(s) = t1) definition of onto→ (∃t ∈ T )(β(t) 6= γ(t)) ∧ (∃s ∈ S)(α(s) = t) choose t1 = t→ (∃t ∈ T )(β(α(s)) 6= γ(α(s))) algebraic replacement→ (∃t ∈ T )(β ◦ α(s) 6= γ ◦ α(s)) definition of ◦→ β ◦ α 6= γ ◦ α definition of mapping equality

And the last statement is false, since we are told that β ◦ α = γ ◦ α.

2.24 The domains and codomains of β and γ are equal by definition. We need only show that β(s) = γ(s) forevery s ∈ S.

α ◦ β = α ◦ γ given→ (∀s ∈ S)(α ◦ β(s) = α ◦ γ(s)) definition of mapping equality→ (∀s ∈ S)(α(β(s)) = α(γ(s))) definition of composition→ (∀s ∈ S)(β(s) = γ(s)) α is one-to-one→ β = γ definition of mapping equality

2.27 (a) If both functions are invertible, they are both one-to-one and onto (theorem 2.2). Therefore, by 2.1,their composition is both one-to-one and onto. Therefore, by theorem 2.2, their composition is invertible.

2.27 (b) The truth of this statement follows directly from theorem 2.1 parts (b) and (d).

9 Equivalence, Congruence, Divisibility

9.5a reflexivity(x, y) ∈ R× R assumed→ (x, y) ∈ R× R ∧ y ∈ R definition of cartesian product→ (x, y) ∈ R× R ∧ y = y reflexivity of =→ (x, y) ∼ (x, y) definition of ∼

symmetry(x1, y1) ∼ (x2, y2) assumed→ (x1, y1), (x2, y2) ∈ R× R ∧ y1 = y2 definition of ∼→ (x1, y1), (x2, y2) ∈ R× R ∧ y2 = y1 symmetry of =→ (x2, y2) ∼ (x1, y1) definition of ∼

transitivity(x1, y1) ∼ (x2, y2) ∧ (x2, y2) ∼ (x3, y3) assumed→ (x1, y1), (x2, y2), (x3, y3) ∈ R× R ∧ y1 = y2 ∧ y1 = y3 definition of ∼→ (x1, y1), (x3, y3) ∈ R× R ∧ y1 = y3 transitivity of =→ (x1, y1) ∼ (x3, y3) definition of ∼

9.5b Each equivalence class represents a horizontal line of the form y = c for some constant c ∈ R.

9.5c {(x, y) ∈ R× R : x = 0}

9.6 It is not transitive: (1, 2) ∼ (1, 3) and (1, 3) ∼ (3, 3), but (1, 2) 6∼ (3, 3).

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9.7 reflexivitya ∈ R assumed→ a = a reflexivity of =→ |a| = |a| definition of absolute value

symmetry|a| = |b| assumed→ |a| = b ∨ |a| = −b definition of absolute value→ (a = b ∨ −a = b) ∨ (a = −b ∨ −a = −b) definition of absolute value→ (b = a ∨ b = −a) ∨ (−b = a ∨ −b = −a) symmetry of =→ (b = |a|) ∨ (−b = |a|) definition of absolute value→ |b| = |a| definition of absolute value

transitivity|a| = |b| ∧ |b| = |c| assumed→ (|a| = b ∨ |a| = −b) ∧ (b = |c| ∨ −b = |c|) definition of absolute value→ (|a| = b = |c|) ∨ (|a| = −b = |c|) ∨ (|a| = b ∧ −b = |c|) ∨ (|a| = −b ∧ b = |c|) logical distributivity→ (|a| = b = |c|) ∨ (|a| = −b = |c|) ∨ (|a| = b ∧ b = −|c|) ∨ (−|a| = b ∧ b = |c|)→ (|a| = |c|) ∨ (|a| = |c|) ∨ (|a| = −|c|) ∨ (−|a| = |c|) transitivity of =

It cannot be the case that |a| = −|c| or −|a| = |c| unless a = 0 = c,so:→ (|a| = |c|)One set of equivalence class representatives is {x ∈ R : x ≥ 0}.

9.8 reflexivitya ∈ N→ a = a reflexivity of =→ a = 1a multiplicative identity of N→ a = 100a x0 = 1 by definition→ a ∼ a definition of ∼

symmetrya ∼ b assumed→ (∃n ∈ Z)(a = 10nb) definition of ∼→ (∃n ∈ Z)(10−na = b) existence of multiplicative inverses in R→ (∃n ∈ Z)(a10−n = b) commutativity of multiplication in R→ b ∼ a definition of ∼

transitivitya ∼ b ∧ b ∼ c assumed→ (∃m,n ∈ Z)(a = 10nb ∧ b = 10mc definition of ∼→ (∃m,n ∈ Z)(a = 10n10mc algebraic replacement→ (∃m,n ∈ Z)(a = 10n+mc→ a ∼ c definition of ∼One set of equivalence class representatives is {x : x 6= 0(mod 10)}.

9.9 The first equivalence relation satisfies symmetry and transitivity, but not reflexivity: 0 6∼ 0. The secondrelation satisfies reflexivity and symmetry, but not transitivity: 1 ∼ 0 and 0 ∼ −1, but 1 6∼ −1.

9.10 One set of equivalence classes would be the set of lines {y = tan(θ)x : −π2 < θ < π2 } ∪ {x = 0}.

9.11 One set of equivalence classes would be {θ : −π2 ≤ θ ≤ π2 }. The range of the inverse sine function is

usually restricted to this set of equivalence classes in order to guarantee that the function is well-definedand one-to-one.

9.12 One set of equivalence classes would be {θ : 0 ≤ θ ≤ π}.

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9.15 reflexivityx ∈ R assumed→ x− x = 0 algebra→ x ∼ x definition of ∼

symmetryx ∼ y assumed→ (∃n ∈ Z)(x− y = n) definition of ∼→ (∃n ∈ Z)(−1(x− y) = −n) algebra→ (∃n ∈ Z)(y − x = −n) algebra→ y ∼ x definition of ∼

transitivityx ∼ y ∧ y ∼ z assumed→ (∃m,n ∈ Z)(x− y = m ∧ y − z = n) definition of ∼→ (∃m,n ∈ Z)(x− y = m ∧ y = n+ z) algebra→ (∃m,n ∈ Z)(x− (n+ z) = m) algebraic replacement→ (∃m,n ∈ Z)(x− z = m+ n) algebra→ x ∼ z definition of ∼

Each equivalence class is a set of real numbers that are equivalent to one another modulus 1 (i.e., they haveidentical digits following the decimal point). One set of equivalence class representatives is the half-openinterval (0, 1].

9.16 reflexivity(x, y) ∈ R× R assumed→ x ∈ R definition of cartesian product→ x− x ∈ Z reflexivity of the relation in 9.15→ (x, y) ∼ (x, y) definition of ∼

symmetry(a, b) ∼ (x, y) assumed→ a− x ∈ Z definition of ∼→ x− a ∈ Z symmetry of the relation in 9.15→ (x, y) ∼ (a, b)

transitivity(a, b) ∼ (m,n) ∧ (m,n) ∼ (x, y) assumed→ a−m ∈ Z ∧m− x ∈ Z definition of ∼→ a− x ∈ Z transitivity of the relation in 9.15→ (a, b) ∼ (x, y) definition of ∼

(0, 0) belongs to the equivalence class consisting of all points on vertical lines of the form x = c, c ∈ Z.One set of equivalence class representatives is {(x, 0) : 0 < x ≤ 1}.

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9.17 reflexivity(x, y) ∈ R× R assumed→ x ∈ R ∧ y ∈ R definition of cartesian product→ x− x =∈ Z ∧ y − y ∈ R reflexivity of the relation in 9.15→ (x, y) ∼ (x, y) definition of ∼

symmetry(a, b) ∼ (x, y) assumed→ a− x ∈ Z ∧ b− y ∈ Z definition of ∼→ x− a ∈ Z ∧ y − b ∈ Z symmetry of the relation in 9.15→ (x, y) ∼ (a, b) definition of ∼

transitivity(a, b) ∼ (m,n) ∧ (m,n) ∼ (x, y) assumed→ a−m ∈ Z ∧m− x ∈ Z ∧ b− n ∈ Z ∧ n− y ∈ Z definition of ∼→ a− x ∈ Z ∧ b− y ∈ Z transitivity of the relation in 9.15→ (a, b) ∼ (x, y) definition of ∼

(0, 0) belongs to the equivalence class of intersections of vertical lines of the form x = c, c ∈ Z andhorizontal lines of the form y = d, d ∈ Z. This can be thought of as the set of all points at which thecoordinate lines intersect on a piece of graph paper. One set of equivalence class representatives would be{(x, y) : 0 < x ≤ 1, 0 < y ≤ 1}. This can be thought of as the set of all points in one unit square on apiece of graph paper.

9.18 The mapping α : S → S defined as α(x) = x is a bijective function for any set S. From theorem 2.2,this means that S ∼ S: so ∼ is reflexive. If S ∼ T , then the definition of invertibility guarantees thatT ∼ S: so ∼ is symmetric. If S ∼ T and T ∼ U , then from theorem 2.1 parts (a) and (c), S ∼ U : so ∼ istransitive.

9.19 a) Let a be any element of S. For every b ∈ S, there is some permutation group (a b) ∈ Sn. So a ∼ bfor every b. This shows that there is only one equivalence class in this case: that equivalence classcontains every element of S.

b) The equivalence classes are [1] = {1}, [2] = [3] = {2, 3}, [4] = {4}. One set of equivalence classrepresentatives is {1, 2, 4}.

c) The equivalence classes are [1] = [2] = [3] = {1, 2, 3}, [4] = {4}, [5] = {5}. One set of equivalence classrepresentatives is {1, 4, 5}.

9.20 The properties of ∼ are directly inherited from the properties of =, which is trivially an equivalencerelation. One set of equivalence class representatives is the set {0 + a1x + a2x

2 + . . . + anxn : ai ∈ R} of

polynomials with a zero constant term.

9.21 a) . . . (∃a ∈ S)(a 6∼ a)

b) . . . (∃a, b ∈ S)(a ∼ b ∧ b 6∼ a)

c) . . . (∃a, b, c ∈ S)(a ∼ b ∧ b ∼ c ∧ a 6∼ c)

9.22 This proof is assuming that there is some x, y such that x ∼ y. If S consisted of only the element x, thisset would be trivially symmetric and transitive. But it might not be the case that x ∼ x (for instance, if∼ were defined as “is greater than”).

10 The Division Algorithm

10.9 This equivalence relation is equivalent to congruence modulo 10.

10.10 This is equivalent to congruence modulo 10k.

10.13 a|b ∧ b|c→ (∃m,n ∈ Z)(am = b ∧ bn = c) definition of divisibility→ (∃m,n ∈ Z)(amn = c) algebraic replacement→ (a|c) definition of divisibility

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10.14 a|b ∧ b|a→ (∃m,n ∈ Z)(am = b ∧ bn = a) definition of divisibility→ (∃m,n ∈ Z)(am = b ∧ bn = a ∧ amn = a) algebraic replacement→ (∃m,n ∈ Z)(am = b ∧ bn = a ∧mn = 1) uniqueness of identity element→ (∃m,n ∈ Z)(am = b ∧ bn = a ∧ [(m = 1 = n) ∨ (m = −1 = n)]) given→ (∃m,n ∈ Z)([a(1) = b ∧ b(1) = a] ∨ [a(−1) = b ∧ b(−1) = a]) algebraic replacement→ (a = b) ∨ (a = −b)→ a = ±b

10.15 a ≡ b(mod n) ∧ c ≡ b(mod n) given(∃u, v ∈ Z)([a = b+ un] ∧ [c = d+ vn]) definition of ≡n(∃u, v ∈ Z)(a+ c = b+ d+ n(u+ v)) add both sidesa+ c ≡ b+ d(mod n) definition of ≡n

10.16 a ≡ b(mod n) ∧ c ≡ b(mod n) given(∃u, v ∈ Z)([a = b+ un] ∧ [c = d+ vn]) definition of ≡n(∃u, v ∈ Z)(ac = bd+ bvn+ dun+ n2uv) multiply both sides(∃u, v ∈ Z)(ac = bd+ nbv + ndu+ n2uv) commutativity of multiplication on R(∃u, v ∈ Z)(ac = bd+ n(bv + du+ nuv)) distributivity of Rac ≡ bd(mod n) definition of ≡n

10.17 Let a = 2, b = 4, n = 4. Then 22 ≡4 0 ≡4 42, but 2 ≡4 2 6= 0 ≡4 4.

10.18 a ≡ b(mod n)→ (∃u ∈ Z)(a = b+ un) definition of ≡n→ (∃u ∈ Z)(a2 = b2 + 2bun+ u2n2) square both sides→ (∃u ∈ Z)(a2 = b2 + n2bu+ n2u2) commutativity of multiplication on R→ (∃u ∈ Z)(a2 = b2 + n(2bu+ nu2)) distributivity of R→ a2 ≡ b2(mod n)

10.19 a ≡ b(mod n) ∧ n|a→ (∃u, k ∈ Z)([a = b+ un] ∧ [nk = a]) definitions of ≡n and divisibility→ (∃u, k ∈ Z)(nk = b+ un) algebraic replacement→ (∃u, k ∈ Z)(n(k − u) = b) algebra→ n|b definition of divisibility

10.20 m|n ∧ a ≡ b(mod n)→ (∃u, k ∈ Z)(mk = n ∧ a = b+ un) definitions of ≡n and divisibility→ (∃u, k ∈ Z)(a = b+ umk) algebraic replacement→ (∃u, k ∈ Z)(a = b+m(uk)) commutativity of multiplication on R→ a ≡ b(mod m) definition of ≡m

10.21 a ≡ b(mod n)↔ n|(a− b) definition of equivalence modulo n↔ (∃u ∈ Z)(a− b = un) definition of divisibility↔ (∃u ∈ Z)(a = b+ un) algebra

The second part of the problem is true because a− b = un implies both a = un+ b and b = un+ a.

10.22 a, b, n ∈ Z ∧ n > 0→ (∃q, r ∈ Z)(a = nq + r) ∧ (∃s, t ∈ Z)(b = ns+ t) division algorithm→ (∃q, r, s, t ∈ Z)(a− b = n(q − s) + (r − t) algebra→ (∃q, r, s, t ∈ Z)(a− (r − t) = b+ n(q − s)) algebra→ (∃q, r, s, t ∈ Z)(a+ (t− r) = b+ n(q − s)) algebra→ (∃t, r ∈ Z)(a+ (t− r) ≡ b(mod n)) definition of ≡n→ (∃x ∈ Z)(a+ x ≡ b(mod n)) (t− r) ∈ Z

10.23 Let a = 0, b = 1, n = 10. There is no x such that 0x = 1(mod 10).

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10.24 a is an odd integer→ (∃k ∈ Z)(a = 2k + 1) definition of odd→ (∃k ∈ Z)(a2 = 4k2 + 4k + 1) algebra→ (∃k ∈ Z)(a2 = 4(k)(k + 1) + 1) algebra

→ (∃k ∈ Z)(a2 = 8( (k)(k+1)2 ) + 1) (k)(k+1)

2 ∈ Z because (k)(k + 1) is even→ a2 ≡ 1(mod 8) definition of ≡n

10.25 (a) 2 (b) 3 (c) 4 (d) k+1

10.26 (a) 2 (b) 3 (c) k+1

10.27 (a) The set of negative integers has no least element.

(b) { 1x : x ∈ N} has no least element.

10.28 i) (∃a, b ∈ Z) ∧ b > 0→ (∃!q, r ∈ Z)(a = bq + r ∧ 0 ≤ r < b) division algorithm

ii) (∃a, b ∈ Z) ∧ b < 0→ (∃!q, r ∈ Z)(a = (−b)q + r ∧ 0 ≤ r < −b) division algorithm→ (∃!q, r ∈ Z)(a = b(−q) + r ∧ 0 ≤ r < −b) algebra

So, in either case, (∃!q, r ∈ Z)(a = bq + r, 0 ≤ r < |b|).

10.29a Let S = {x : 10x ≡ 1(mod 9). Proof by induction that S = N:

S is nonempty:10 = 1 + 9→ 101 = 1 + 9(1)→ 101 ≡ 1(mod 9)→ 1 ∈ S definition of S

induction:n ∈ S hypothesis of induction→ 10n ≡ 1(mod 9) definition of membership in S→ (∃u ∈ Z)(10n = 1 + 9u) definition of ≡n→ (∃u ∈ Z)(10n+1 = 10 + 90u) algebra→ (∃u ∈ Z)(10n+1 = 1 + 9 + 90u) algebra→ (∃u ∈ Z)(10n+1 = 1 + 9(1 + 10u)) algebra→ 10n+1 ≡ 1(mod 9) definition of ≡n→ n+ 1 ∈ S definition of membership in S

Therefore S = N, which means 10n ≡ 1(mod 9) for all positive integers.

10.29b Each base-10 integer k can be expressed in expanded decimal form as

k = a0(100) + a1(101) + . . .+ an(10n)

where ai is ith digit to the left of the decimal point. From the conclusion in part (a) of this problem, weconclude that there exist u0, u1, . . . , un ∈ Z such that:

k = a0(1 + 9u0) + a1(1 + 9u1) + . . .+ an(1 + 9un)→ k = (a0 + a1 + . . .+ an) + 9(a0u0 + a1u1 + . . .+ anun)→ k = (a0 + a1 + . . .+ an)(mod 9)

11 Integers Modulo n

11.7 The fact that it’s a group is given, so we need only prove commutativity:

[a]⊕ [b] = [a+ b] definition of ⊕ operation= [b+ a] commutativity of addition on Z= [b]⊕ [a] definition of ⊕ operation

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11.8 [a]� [b] = [ab] definition of � operation= [ba] commutativity of multiplication on Z= [b]� [a] definition of � operation

11.9 [a]� ([b]⊕ [c]) = [a]� ([b+ c]) definition of � operation= [a(b+ c)] definition of ⊕ operation= [ab+ ac] distributivity of Z= [ab]⊕ [ac] definition of ⊕ operation= ([a]� [b])⊕ ([a]� [c]) definition of � operation

11.10 [a] + [−a] = [a+ (−a)] definition of ⊕ operation= [0] property of negatives in Z

11.11 Proof by contradiction:

(∃a ∈ Z)([a]� [0] = [1]) assumed→ (∃a ∈ Z)([a0] = [1]) definition of � operation→ ([0] = [1]) property of 0 in Z

The conclusion is only true for Z1: in any other case, our assumption must be false.

11.12 Associativity:

[a]� ([b]� [c]) = [a]� ([bc]) definition of � operation= [a(bc)] definition of � operation= [(ab)c] associativity of multiplication on Z= [ab]� [c] definition of � operation= ([a]� [b])� [c] definition of � operation

Commutativity:

[a]� [b] = [ab] definition of � operation= [ba] commutativity of multiplication in Z= [b]� [a] definition of � operation

Existence of an identity element:

[a]� [1]= [a1] definition of � operation= [a] multiplicative identity in Z= [1a] multiplicative identity in Z= [1]� [a] definition of � operation

11.13 Associativity, commutativity, nonemptiness, and the existence of an identity element were shown inproblem 11.12. We need only show closure under the operation �, which can be done with a Cayley table:

� [1] [2][1] [1] [2][2] [2] [1]

11.14 Associativity, commutativity, nonemptiness, and the existence of an identity element were shown inproblem 11.12. But Z#

4 is not closed, since [2]� [2] = [0], and [0] /∈ Z#4 .

11.15 The problem doesn’t specify which operation it wants us to prove this for, but it’s only a group under⊕. The subgroup inherits associativity from Z6 and its nonemptiness is given. We need to show closureunder ⊕, the existence of a unique identity element, and the existence of an inverse for each element. Thefollowing Cayley table shows that all of these requirements are met:

⊕ [0] [2] [4][0] [0] [2] [4][2] [2] [4] [0][4] [4] [0] [2]

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11.16 As in the previous problem, we need only to show closure, the existence of a unique identity element,and the existence of an inverse for each element.

⊕ [0] [3] [6] [9][0] [0] [3] [6] [9][3] [3] [6] [9] [0][6] [6] [9] [0] [3][9] [9] [0] [3] [6]

11.17(a) Proof that there is no more than 1 nonidentity element that is its own inverse:

[a] is its own inverse assumed→ ([a]⊕ [a] ≡n [0]) definition of inverse→ ([a+ a] ≡n [0]) definition of ⊕ operation→ (∃u ∈ Z)(a+ a = 0 + un) definition of modular equivalence→ (∃u ∈ Z)(2a = un) algebra→ (∃u ∈ Z)(a = un2 ) algebra→ (∃u, k ∈ Z)((a = un2 ) ∧ (u = 2k ∨ u = 2k + 1)) all integers are either even or odd→ (∃u, k ∈ Z)((a = un2 ∧ u = 2k) ∨ (a = un2 ∧ u = 2k + 1)) logical distributivity→ (∃k ∈ Z)((a = 2k n2 ) ∨ (a = (2k + 1)n2 ) algebraic replacement→ (∃k ∈ Z)((a = kn) ∨ (a = kn+ n

2 ) algebra→ ([a] ≡n [0]) ∨ ([a] ≡n [n2 ]) definition of modular equivalence

So if [a] is its own inverse, there are only two possible values it may take. One is the additive identity, andthe other is [n2 ]. And it’s only possible for [a] = [n2 ] if n

2 is an integer, which is only true if n is even. Butthis only proves that there is at most one nonidentity element of Z2k that is its own inverse. To show thatthere is always at least one if n is even:

(n is even) assumed→ n

2 ∈ Z definition of even→ [n2 ] ∈ Zn definition of Zn→ [n2 + n

2 ] = [n] ≡n [0]→ [n2 ]⊕ [n2 ] ≡n [0] definition of ⊕ operation

11.17(b) (see previous proof)

11.17(c) [x] = [0]⊕ [1]⊕ . . .⊕ [n− 1] assumed→ [x] = [0 + 1 + . . .+ n− 1] definition of ⊕ operation

→ [x] = [n(n−1)2 ] rules for series summation

→ (∃u ∈ Z)(x = nu+ n(n−1)2 ) definition of modular equivalence

→ (∃u ∈ Z)(x = n(u+ 12n−

12 )) algebra

If n is even, this last statement implies:(∃u, k ∈ Z)(x = n(u+ 1

2n−12 ) ∧ n = 2k)

→ (∃u, k ∈ Z)(x = n(u+ k − 12 ))

→ (∃u, k ∈ Z)(x = n(u+ k − 1 + 12 ))

→ (∃u, k ∈ Z)(x = n(u+ k − 1) + n2 )

→ [x] ≡n [n2 ]

Otherwise, if n is odd, the same statement implies:(∃u, k ∈ Z)(x = n(u+ 1

2n−12 ) ∧ n = 2k + 1)

→ (∃u, k ∈ Z)(x = n(u+ k)→ [x] ≡n [0]

So [x] is equivalent mod n to either [0] or [n2 ].

11.17(d) The modulus of the sum of the series is equal to the modulus of the median (centermost) element ofthe series if n is odd, and [0] otherwise.

11.18 [0] ∼ [1] and [−1] ∼ [−1]. But [0] � [−1] = [−1] and [1] � [−1] = [0], so � is not well-defined.

10

12 Greatest Common Divisors

12.12 Theorem 12.2 proves that there is at least one choice of m,n such that (a, b) = am+ bn. And for everyx ∈ Z, (a(m + xb) + b(n − xa)) = am + bn. So there are an infinite number of choices for m,n such that(a, b) = am+ bn.

12.13 x = gcd(a, b) assume gcd(a, b) is defined→ x|a ∧ x|b ∧ (y|a ∧ y|b→ y|x) definition of gcd→ (∃pi ∈ Z)(xp1 = a ∧ xp2 = b ∧ (yp3 = a ∧ yp4 = b→ yp5 = x)) definition of divisibility→ (∃pi ∈ Z)(xcp1 = ac ∧ xcp2 = bc ∧ (ycp3 = ac ∧ ycp4 = bc→ ycp5 = xc)) algebra→ xc|ac ∧ xc|bc ∧ (yc|ac ∧ yc|bc→ yc|xc) definition of divisibility→ xc = gcd(ac, bc) definition of gcd→ gcd(a, b)c = gcd(ac, bc) (x = gcd(a, b))

12.14 d = gcd(a, b) assumed→ (∃m,n ∈ Z)(d = am+ bn) theorem 12.2→ (∃m,n ∈ Z)(1 = a

dm+ bdn) the fractions are defined because d|a, d|b.

→ 1 = gcd(ad ,bd ) corollary of theorem 12.2

1 = gcd(ad ,bd ) assumed

→ (∃m,n ∈ Z)(1 = adm+ b

dn) corollary of theorem 12.2→ (∃m,n ∈ Z)(d = am+ bn) algebra→ (∃m,n ∈ Z)(d = am+ bn) ∧ gcd(a, b)|a ∧ gcd(a, b)|b property of gcd→ (∃m,n, r, s ∈ Z)(d = am+ bn ∧ a = gcd(a, b)r ∧ b = gcd(a, b)s) definition of divisibility→ (∃m,n, r, s ∈ Z)(d = gcd(a, b)rm+ gcd(a, b)sn) algebraic replacement→ (∃m,n, r, s ∈ Z)(d = gcd(a, b)(rm+ sn)) distributivity→ gcd(a, b)|d definition of divisibility→ gcd(a, b)|d ∧ d|a ∧ d|b given in the problem description→ gcd(a, b)|d ∧ d| gcd(a, b) d|a ∧ d|b→ d| gcd(a, b)→ d = gcd(a, b)

12.15 x = gcd(a, p)→ x|a ∧ x|p partial definition of gcd→ x|a ∧ (x = 1 ∨ x = p) definition of p prime→ (x|a ∧ x = 1) ∨ (x|a ∧ x = p) logical distributivity→ (1|a ∧ x = 1) ∨ (p|a ∧ x = p) algebraic replacement→ (1|a ∧ x = 1) p 6 | a was given→ x = 1→ gcd(a, p) = 1 definition of x from first step

12.16 gcd(a, c) = 1 ∧ gcd(b, c) = 1→ (∃m,n ∈ Z)(1 = ma+ nc) ∧ (∃r, s ∈ Z)(1 = rb+ sc) theorem 12.2→ (∃m,n, r, s ∈ Z)(1 = (ma+ nc)(rb+ sc)) multiply both equations→ (∃m,n, r, s ∈ Z)(1 = ab(mr) + c(ams+ bnr + cns)) algebra→ gcd(ab, c) = 1 theorem 12.2

12.17 gcd(ab, c) = 1 assumption→ (∃m,n ∈ Z)(1 = (ab)m+ (c)n) theorem 12.2→ (∃m,n ∈ Z)((1 = (ab)m+ (c)n) ∧ (1 = (ab)m+ (c)n)) p→ p ∧ p→ (∃m,n ∈ Z)((1 = (a)bm+ (c)n) ∧ (1 = (b)am+ (c)n)) associativity and commutativity→ gcd(a, c) = 1 ∧ gcd(b, c) = 1 theorem 12.2

12.18 c|ab ∧ gcd(a, c) = d assumed→ (∃r ∈ Z)(cr = ab) ∧ (∃s, t ∈ Z)(d = as+ ct) definition of divisibility and theorem 12.2→ (∃r, s, t ∈ Z)(cr = ab ∧ bd = abs+ bct) algebra→ (∃r, s, t ∈ Z)(bd = crs+ bct) algebraic replacement→ (∃r, s, t ∈ Z)(bd = c(rs+ bt)) algebra→ c|bd definition of divisibility

11

12.19 gcd(a, b) = gcd(bq + r, b) definition of a= gcd(bq + r + (−q)b, b) gcd(a, b) = gcd(a+ nb, b), from the Euclidean algorithm= gcd(r, b) algebra

12.20 gcd(a, b) = 1 ∧ c|a assumed→ (∃m,n, p ∈ Z)(1 = ma+ nb ∧ cp = a) theorem 12.2 and gcd→ (∃m,n, p ∈ Z)(1 = m(cp) + nb) algebraic replacement→ (∃m,n, p ∈ Z)(1 = c(mp) + b(n)) commutativity and associativity→ gcd(c, b) = 1 theorem 12.2

12.21 gcd(a,m) = 1 assumed→ (∃x, s ∈ Z)(1 = ax+ms) theorem 12.2→ (∃x, s ∈ Z)(ax = (−s)m+ 1) algebra→ (∃x ∈ Z)(ax ≡ 1(mod m)) definition of mod m

12.22 n− n−n(modp)p

12.23 Let D = {x ∈ N : x is a common divisor of a, b, and c}. We know that this set is nonempty, since 1 ∈ D.We also know that it is finite, since x must be less than the smallest of a, b, or c. Because D is a finite,nonempty set of positive integers, we know that it must have a greatest element. So we know that a, b, andc have a greatest common divisor. To prove that this divisor can be expressed as a linear combination:

x = gcd(a, b, c) assumed→ (x|a ∧ x|b ∧ x|c) ∧ ((y|a ∧ y|b ∧ y|c)→ y|x) definition of gcd→ ((x|a ∧ x|b) ∧ x|c) ∧ (((y|a ∧ y|b) ∧ y|c)→ y|x) associativity→ ((x| gcd(a, b)) ∧ x|c) ∧ ((y| gcd(a, b)) ∧ y|c)→ y|x definition of gcd→ x = gcd(gcd(a, b), c) definition of gcd→ (∃m,n ∈ Z)(x = gcd(a, b)m+ cn) Theorem 12.2→ (∃m,n, r, s ∈ Z)(x = (ar + bs)m+ cn) Theorem 12.2→ (∃m,n, r, s ∈ Z)(x = arm+ bsm+ cn) algebra

12.24 (a) One of a, b is nonzero, so one of 0 + a, 0− a, b+ 0, or −b+ 0 must be a positive integer.

(b) We know that S has a least element by the well-ordering principle. We know that this least elementcan be express as d = am+ bn because that’s the definition of membership in S.

(c) We know that a, b ∈ S because a = 1a+ 0b and b = 0a+ 1b. So if we prove that d divides everythingin S, we will of course have proven that it divides a and b.

(d) We can assume that at least one such k exists because S is nonempty. We can assume the existenceof the integers q and r from the division algorithm of chapter 10. The remaining steps of part (d) arebasic algebra.

(e) r = 0 because we assumed that 0 ≤ r < d and then showed that it was a member of S. But we’vealready assumed that d is the smallest positive element of S, so r cannot be a smaller positive element:it must be 0. So the fact that each element k ∈ S can be written k = dq + r actually tells us thatk = dq + 0, or that d|k for all k ∈ S.

(f) Let d = gcd(a, b):c|a ∧ c|b ∧ (d = gcd(a, b)) assumed→ (∃m,n, r, s ∈ Z)(cn = a ∧ cm = b ∧ d = ar + bs) definition of divisibility, Theorem 12.2→ (∃m,n, r, s ∈ Z)(d = cnr + cms) algebraic replacment→ (∃m,n, r, s ∈ Z)(d = c(nr +ms)) algebra→ c|d definition of divisibility

13 Factorization, Euler’s Phi Function

In all of the following problems, exponents of 0 are allowed in the standard form. For example:

12

7 = pe11 pe22 · · · p

ekk → 7 = 20305071110 . . .

6 = pf11 pf22 · · · p

fkk → 6 = 21315070110 . . .

7× 6 = pe1+f11 pe2+f22 · · · pek+fkk → 42 = 21315071110 . . .

Because each pi always represents the ith prime, this makes it easier to compare or multiply two standardform numbers.

13.5 m|n↔ (∃u ∈ Z)(mu = n) definition of divisibility↔ (∃u ∈ Z)(ps11 p

s22 · · · p

skk )u = (pt11 p

t22 · · · p

tkk ) definitions of m and n

↔ (∃u ∈ Z)u = pt1−s11 pt2−s22 · · · pt2−s2k algebra↔ (∀i ∈ N)(si ≤ ti) standard form cannot have negative exponents

13.15 n is odd assumed→ n = 20pe22 · · · p

ekk FToA, p1 = 2

→ 2n = 21pe22 · · · pekk algebra

→ φ(2n) = φ(21pe22 · · · pekk ) φ is well-defined

→ φ(2n) = φ(21)φ(pe22 · · · pekk ) theorem 13.3

→ φ(2n) = 2(1− 1/2)φ(pe22 · · · pekk ) theorem 13.1

→ φ(2n) = φ(pe22 · · · pekk ) algebra

→ φ(2n) = φ(1pe22 · · · pekk ) property of the identity element

→ φ(2n) = φ(20pe22 · · · pekk ) x0 = 1 by definition

→ φ(2n) = φ(n) definition of n from first step

13.16 n is even assumed(∃u ∈ Z)(n = 2u) definition of even→ n = 2(pe11 p

e22 · · · p

ekk ) FToA

→ 2n = 22(pe11 pe22 · · · p

ekk ) algebra

→ 2n = (pe1+21 pe22 · · · p

ekk ) p1 = 2

→ φ(2n) = φ(pe1+21 pe22 · · · p

ekk ) φ is well-defined

→ φ(2n) = φ(pe1+21 )φ(pe22 · · · p

ekk ) theorem 13.3

→ φ(2n) = pe1+21 (1− 1/2)φ(pe22 · · · p

ekk ) theorem 13.1

→ φ(2n) = 2pe11 φ(pe22 · · · pekk ) algebra

→ φ(2n) = 2φ(pe11 pe22 · · · p

ekk ) theorem 13.3

→ φ(2n) = 2φ(n) definition of n from first step

13.17 gcd(a, b) = 1 ∧ a|m ∧ b|m assumed→ (∃r)(gcd(a, b) = 1 ∧ ar = m ∧ b|m) definition of divisibility→ (∃r)(gcd(a, b) = 1 ∧ ar = m ∧ b|ar) algebraic replacement→ (∃r)(gcd(a, b) = 1 ∧ ar = m ∧ b|r) lemma 13.1→ (∃r, s)(ar = m ∧ bs = r) definition of divisibility→ (∃r, s)(abs = m) algebraic replacement→ ab|m definition of divisibility

13.18(a) Proof by contradiction:

(n = pe11 pe22 · · · p

ekk ) ∧ (∃i ∈ N)(ei ≥ 2) assumption

→ (n = p2i (pe11 p

e22 · · · p

ei−2i · · · pekk ) ∧ (∃i ∈ N)(ei ≥ 2) algebra

→ p2i |n definition of divisibility→ n is not square free definition of square free

So, by contrapositive, if n is square free then there is no ei ≥ 2. To prove that this holds in the otherdirection, we use another proof by contradiction:

13

n is not square free assumption→ (∃x ∈ Z)(x2|n) definition of square free→ (∃u, x ∈ Z)(n = x2u) definition of divisibility→ (∃u, x ∈ Z)(n = x2(pe11 p

e22 · · · p

ekk )) FToA

→ (∃u, x ∈ Z)(n = x2(pe11 pe22 · · · p

ekk ) ∧ [(∀pi)(x 6= pi) ∨ (∃pi)(x = pi)]) excluded middle

→ (∃u, x ∈ Z)(n = (x2pe11 pe22 · · · p

ekk ) ∨ n = (pe11 p

e22 · · · p

ei+2i · · · pekk )) FToA

→ n = (pe11 pe22 · · · p

ekk ) ∧ (∃i ∈ N)(ei ≥ 2)

13.18(b) (∃x ∈ Z)(n = x2) assumption↔ n = (pe11 p

e22 · · · p

ekk )2 FToA

↔ n = (p2e11 p2e22 · · · p2ekk ) algebra↔ each exponent of n, in standard form, is even

13.18(c) Let n = (pe11 pe22 · · · p

ekk ). Let q be the product of every element in the set {pi : ei is odd} (note that if

p5k is an element of n, it is only p1k that is an element of q). Each element of q is raised only to the firstpower, so it is square-free from 13.18(a). Also, n

q has only even exponents, so it is a perfect square from

13.18(b). And since n = q nq , this means n is the product of a perfect square and a square-free integer.

13.19√n is rational assumption→ (∃q, r ∈ N)

√n = q

r definition of rational

→ (∃q, r ∈ N)n = q2

r2 algebra→ (∃q, r ∈ N)r2n = q2 algebra

→ (p2r11 p2r22 · · · p2rkk )(pn11 pn2

2 · · · pnk

k ) = (p2q11 p2q22 · · · p2qkk ) FToA, definition of squares

→ (p2r1+n11 p2r2+n2

2 · · · p2rk+nk

k ) = (p2q11 p2q22 · · · p2qkk ) algebra→ (∀i ∈ N)(2ri + ni = 2qi) properties of exponents→ (∀i ∈ N)(ni = 2(qi − ri)) algebra→ (∀i ∈ N)(2|ni) definition of divisibility→ (∀i ∈ N)(ni is even) definition of even→ pn1

1 pn22 · · · p

nk

k is a perfect square 13.18(b)→ n is a perfect square definition of n

13.20 A general proof for all x√n:

x√n is rational assumption→ (∃q, r ∈ N) x

√n = q

r definition of rational

→ (∃q, r ∈ N)n = qx

rx algebra→ (∃q, r ∈ N)rxn = qx algebra→ (pxr11 pxr22 · · · pxrkk )(pn1

1 pn22 · · · p

nk

k ) = (pxq11 pxq22 · · · pxqkk ) FToA, definition of squares→ (pxr1+n1

1 pxr2+n22 · · · pxrk+nk

k ) = (pxq11 pxq22 · · · pxqkk ) algebra→ (∀i ∈ N)(xri + ni = xqi) properties of exponents→ (∀i ∈ N)(ni = x(qi − ri)) algebra→ (∀i ∈ N)(x|ni) definition of divisibility→ (∃m ∈ N)(pn1

1 pn22 · · · p

nk

k = mx) 13.18(b)→ (∃m ∈ N)(n = mx) definition of n

This shows that 3√n is rational only if n is a perfect cube. 2 is not a perfect cube, so 3

√2 is not rational.

13.21 (see problem 13.20)

14 Elementary Properties

14.11(b) b = be= b(aa−1)= (ba)a−1

= (ca)a−1 ba = ca was given= c(aa−1)= ce= c

14

14.12 xa = b→ (xa)a−1 = ba−1

→ x(aa−1) = ba−1

→ xe = ba−1

→ x = ba−1

14.13 axb = c→ a−1axb = a−1c→ a−1axbb−1 = a−1cb−1

→ (a−1a)x(bb−1) = a−1cb−1

→ exe = a−1cb−1

→ x = a−1cb−1

14.14(a) o(a) = m ∧ o(b) = n assumed→ am = e ∧ bn = e definition of o(x)→ amn = en ∧ bmn = em algebra→ amnbmn = emn algebra→ (ab)mn = emn commutativity, from G being abelian→ (ab)mn = e

14.14(b) (ab)mn = e assumed→ o(ab)|mn from theorem 14.3→ o(ab)|o(a)o(b) from definition of o(a) = m, o(b) = n

14.14(c) In Z4, o([2]) = 2, o([2])o([2]) = 4, and o([2]⊕ [2]) = 1.

14.15 Let a, b be permutation groups with a = (1 2), b = (1 3). Both of these groups are order 2. But(ab)4 = (132)4 = (132).

14.16(a) LetG be an abelian group, and let S be the elements of finite order inG. To prove that S is a subgroup:

e is in S:e1 = e→ o(e) = 1 definition of o(x)→ e ∈ S definition of membership in S

S is closed under its operation:a ∈ S ∧ b ∈ S assumed→ (∃m,n ∈ N)(o(a) = m ∧ o(b) = n) definition of membership in S→ (∃m,n ∈ N)(am = e ∧ bn = e) definition of o(x)→ (∃m,n ∈ N)(amn = en ∧ bmn = em) algebra→ (∃m,n ∈ N)(amn = e ∧ bmn = e) property of the unity element→ (∃m,n ∈ N)(amnbmn = e) algebra→ (∃m,n ∈ N)((ab)mn = e) from the commutativity of abelian groups→ (∃m,n ∈ N)(o(ab)|mn) theorem 14.3→ (∃m,n ∈ N)(o(ab) ≤ mn) property of divisibility→ o(ab) is finite→ ab ∈ S definition of membership in S

15

existence of inverses in S:a ∈ S assumed→ (∃m ∈ N)(o(a) = m) definition of membership in S→ (∃m ∈ N)(am = e) definition of o(a)→ (∃m ∈ N)(am = e ∧ em = e) property of the identity element→ (∃m ∈ N)(am = e ∧ (aa−1)m = e) definition of inverses→ (∃m ∈ N)(am = e ∧ am(a−1)m = e) commutativity of abelian groups→ (∃m ∈ N)(e(a−1)m = e) algebraic substitution→ (∃m ∈ N)((a−1)m = e)→ (∃m ∈ N)(o(a−1)|m) theorem 14.3→ (∃m ∈ N)(o(a−1) ≤ m) property of divisibility→ o(a−1) is finite→ a−1 ∈ S

14.16(b) The only positive rational with a finite order is 1, with o(1) = 1.

14.18(a) ab = ba assumed↔ (ab)(ab)−1 = ba(ab)−1

↔ e = ba(ab)−1 property of inverses↔ (ba)−1e = (ba)−1ba(ab)−1

↔ (ba)−1 = (ab)−1 property of inverses↔ a−1b−1 = b−1a−1 theorem 14.1

14.18(b) ab = ba assumed→ a(ab) = a(ba)→ a(ab)b = a(ba)b→ (aa)(bb) = (ab)(ab) associativity→ a2b2 = (ab)2

14.19 〈m〉 ⊆ 〈n〉 assumed↔ (x ∈ 〈m〉 → x ∈ 〈n〉) definition of subset↔ (∃j ∈ N)(x = mj)→ (∃k ∈ N)(x = nk) definition of membership in 〈m〉 or 〈n〉↔ (∀j ∈ N)(∃k ∈ N)(mj = nk) quantification of the conditional

When j = 1, this means that m = nk. So m must be an integral power of n. This means that anecessary condition is that (∃i ∈ N)(m = ni). Continuing the proof:

↔ (∀j ∈ N)(∃k ∈ N)(mj = nk) ∧ (∃i ∈ N)(m = ni)↔ (∀j ∈ N)(∃i, k ∈ N)(nij = nk) algebraic substitution

And this last condition can always be made true by choosing k = ij. So the condition that m = ni

is both necessary and sufficient for 〈m〉 ⊆ 〈n〉.

14.22(a) a = cb−1

↔ a(bb−1) = cb−1 property of inverses↔ ab = c right cancellation↔ ab = (aa−1)c property of inverses↔ b = a−1c left cancellation

14.22(b) (1 2 3)(1 2) = (1 3) ab = c(1 2)(1 3) 6= (1 2 3) b−1c 6= a

14.23 Let a represent the nonidentity element of the group.

a2 = e assumed↔ aa = e definition of a2

↔ aa = aa−1 property of inverses↔ a = a−1 left cancellation

16

14.24 Assume G is a group with 2n elements. From the definition of a group, each a ∈ G must have a uniqueinverse. Because the identity element is its own inverse, this means that the other 2n − 1 elements musthave a unique inverse among themselves. Since a maximum of n − 1 pairs can be formed from 2n − 1elements, at least one element must be its own inverse by the pigeonhole principle.

14.25 (∀a, b ∈ G)((ab)−1 = a−1b−1) assumed↔ (∀a, b ∈ G)((ab)−1 = (ba)−1) theorem 14.1↔ (∀a, b ∈ G)((ab)−1(ab) = (ba)−1(ab))↔ (∀a, b ∈ G)(e = (ba)−1(ab))↔ (∀a, b ∈ G)((ba)e = (ba)(ba)−1(ab))↔ (∀a, b ∈ G)(ba = ab)↔ G is abelian definition of abelian

14.28 λ is onto G:b ∈ G assumed→ b ∈ G ∧ a ∈ G definition of a→ b ∈ G ∧ a−1 ∈ G existence of inverses in groups→ a−1b ∈ G G is closed under its operation→ λ(a−1b) = a(a−1b) = b definition of λ→ b ∈ Rng(λ)

λ is one-to-one:λ(x) = λ(y) assumed→ ax = ay definition of λ→ x = y left cancellation

14.29 (∀a, b ∈ G)(o(a) = 2 ∧ o(b) = 2) assumed→ (∀a, b ∈ G)(o(a) = 2 ∧ o(b) = 2 ∧ ab ∈ G) G is closed→ (∀a, b ∈ G)(o(a) = 2 ∧ o(b) = 2 ∧ (o(ab) = 1 ∨ o(ab) = 2)) ab either is or isn’t the identity→ (∀a, b ∈ G)(a2 = e ∧ b2 = e ∧ (ab)2 = e) in either case, (ab)2 = e→ (∀a, b ∈ G)(a2b2 = e = (ab)2) algebra→ (∀a, b ∈ G)(aabb = abab)→ (∀a, b ∈ G)(abb = bab) left cancellation→ (∀a, b ∈ G)(ab = ba) right cancellation→ G is abelian defintion of abelian

14.30 We will prove the seemingly unrelated claim that o(a−1) ≤ o(a) and then show that this proves thato(a) = o(a−1).

Lemma: o(a−1) ≤ o(a):o(a) = m assumed→ am = e definition of o(x)→ am = e ∧ a0 = e definition of a0

→ am = e ∧ am−m = e additive inverse→ am = e ∧ ama−m = e additive inverse→ ea−m = e algebraic substitution→ a−m = e algebraic substitution→ (a−1)m = e definition of negative exponents→ o(a−1)|m theorem 14.3→ o(a−1) ≤ m property of divisibility→ o(a−1) ≤ o(a) definition of m

This shows not only that o(a−1) ≤ o(a), but also that o((a−1)−1) ≤ o(a−1). So o(a−1) ≤ o(a) ∧ o(a) ≤o(a−1), which means that o(a) = o(a−1).

17

14.31 o(a−1ba) = m assumed→ (a−1ba)m = e definition of o(x)↔ a−1(baa−1)m−1ba = e associativity

(If the justification for the previous step is unclear, see problem 14.32 for a clearer example.)↔ a−1(b)m−1ba = e property of inverses↔ aa−1(b)m−1baa−1 = aea−1 left and right multiplication↔ (b)m−1b = aa−1 cancellation of inverses↔ bm = e property of inverses→ o(b) ≤ m

And since m = o(a−1ba), this last step tell us that o(b) ≤ o(a−1ba). The steps of this proof can be followedin reverse to justify the claim that (o(b) = n) → (o(a−1ba) ≤ o(b)). And, since o(b) ≤ o(a−1ba) ≤ o(b),this mean that o(b) = o(a−1ba).

14.32 o(ab) = m assumed→ (ab)m = e definition of o(x)→ a(ba)m−1b = e associativity→ a−1a(ba)m−1b = a−1e left multiplication→ (ba)m−1b = a−1

→ (ba)m−1ba = a−1a right multiplication→ (ba)m = e→ o(ba)|m theorem 14.3→ o(ba) ≤ m property of divisibility→ o(ba) ≤ o(ab) definition of m

And of course, taking o(ba) = n and following these same steps would prove that o(ab) ≤ o(ba), soo(ab) = o(ba).

14.33 Every group G is a subgroup of itself, so if the claim were true then every group G would have a subgroupwith |G| elements. In other words, every group would be a cyclic group. So every noncyclic group is acounterexample. One specific example is S3, which is a subgroup of itself and has order 6 but has noelements of order 6.

14.34 If e is the only element of G, then 〈e〉 = {e} = G and therefore G is cyclic. Otherwise, let a be anonidentity element of G. Then 〈a〉 6= {e}, since a1 ∈ 〈a〉 and a1 6= e. And the only other subgroup of Gis G itself, so 〈a〉 = G and therefore G is cyclic.

14.35 G is finite given→ (∃k ∈ N)(|G| = k) definition of finite→ (∀a ∈ G)(∃k ∈ N)(|〈a〉| ≤ k) every subset of a finite set is finite→ (∀a ∈ G)(∃k ∈ N)(o(a) ≤ k) corollary to theorem 14.3→ (∀a ∈ G)(∃k,m ∈ N)(am = e ∧m ≤ k) definition of o(x)→ (∀a ∈ G)(∃m ∈ N)(am = e) we don’t really care about the size of m→ (∀a ∈ G)(∃m ∈ N)(aam−1 = e)→ (∀a ∈ G)(∃m ∈ N)(am−1 = a−1) uniqueness of inverse, or left multiplication→ (∀a ∈ G)(∃b ∈ G)(b = a−1) change of variable

So part (c) of theorem 7.1 is only necessary when analyzing an infinite group like N which fulfills require-ments (a) and (b), but not part (c).

14.36 Let x =lcm[r, s]. The fact that α, β are disjoint means that they commute: αβ = βα.

18

α is an r-cycle ∧ β is an s-cycle assumed→ αr = e ∧ βs = e definition of a cycle (p38)→ αr = e ∧ βs = e ∧ r|x ∧ s|x x =lcm[r, s]→ (∃m,n ∈ N)(αr = e ∧ βs = e ∧ rm = x ∧ sn = x) definition of divisibility→ (∃m,n ∈ N)(αrm = em ∧ βsn = en ∧ rm = x ∧ sn = x)→ αx = e ∧ βx = e algebraic replacement→ αxβx = e algebra→ (αβ)x = e α and β are commutative→ o(αβ)|x theorem 14.3→ o(αβ) ≤ x property of divisibility→ o(αβ) ≤ lcm[r, s] definition of x

So we’ve shown here that o(αβ) is no greater than the least common multiple of [r, s]. Now we provethat it is a multiple of [r, s]:

o(αβ) = y assumed→ (αβ)y = e definition of o(x)→ αyβy = e commutativity of α, β→ αy = e ∧ βy = e disjointness of α, β→ o(α)|y ∧ o(β)|y theorem 14.3→ o(α)|o(αβ) ∧ o(β)|o(αβ) algebraic replacement of o(αβ) = y→ o(αβ) is a multiple of both o(α) and o(β) definition of divisibility→ o(αβ) is a multiple of both r and s definition of r, s

Since o(αβ) is a common multiple of (r, s) and it’s less than or equal to the least such multiple, it mustbe equal to it. o(αβ) =lcm[r, s].

15 Generators, Direct Products

15.14 〈∅〉, by definition, is the intersection of all subgroups of G that contain ∅. A subgroup H contains ∅ iff∅ ⊆ H. And this is true for all sets. So 〈∅〉 represents the intersection of all subgroups of G. And since{e} is a subgroup of G and every subgroup of G must contain {e}, 〈∅〉 = {e}.

15.15 The necessary and sufficient condition is that S is a subgroup of G. Proof:

If S is a group, rather than just an arbitrary set, then the smallest group that contains S is equal to Sitself: i.e., 〈S〉 = S. So S being a group is a sufficient condition for 〈S〉 = S. If S is not a group, then atleast one of the following three things must be true:

i) S is lacking the identity element of G. But 〈S〉 must contain the identity element, since it is a union ofsubgroups of G each of which contains the identity element. In this case, S 6= 〈S〉.

ii) S does not contain the inverse for at least one element of S. But 〈S〉 is a union of subgroups of Gthat contain every element of S, so each subgroup must also contain the inverse of each element ofS. Therefore 〈S〉 must contain the inverse of each element of S. In this case, S 6= 〈S〉.

iii) S is not closed under the operation of G: there is some a, b ∈ S such that ab 6∈ S. But 〈S〉 is aunion of subgroups of G that are closed under the operation of G, so it cannot be the case thata, b ∈ 〈S〉 ∧ ab 6∈ 〈S〉. So S 6= 〈S〉.

In each of these three cases, we see that 〈S〉 6= S. So S being a group is a necessary condition for 〈S〉 = S.

15.16 Let (a, x), (b, y), and (c, z) be elements of A×B.

((a, x)(b, y))(c, z)= (ab, xy)(c, z) definition of direct product operation= ((ab)c, (xy)z) definition of direct product operation= (a(bc), x(yz)) associativity of groups A and B= (a, x)(bc, yz) definition of direct product operation= (a, x)((b, y)(c, z)) definition of direct product operation

19

15.17 existence of an identity element:eA ∈ A ∧ eB ∈ {eB} necessary property of groups→ (eA, eB) ∈ A× {eB} definition of cartesian product

(a, eB)(eA, eB)= (aeA, eBeB)= (a, eB)= (eAa, eBeB)= (eA, eB)(a, eB)

completeness of inverses(a, eB) ∈ A× {eB} assumed→ a ∈ A ∧ eB ∈ {eB} definition of cartesian product→ a−1 ∈ A ∧ eB ∈ {eB} existence of inverse elements→ (a−1, eB) ∈ A× {eB} definition of cartesian product

(a, eB)(a−1, eB)= (aa−1, eBeB)= (eA, eB)= (a−1a, eBeB)= (a−1, eB)(a, eB)

closed under the operation((a, eB) ∈ A× {eB}) ∧ ((x, eB) ∈ A× {eB}) assumed→ (a ∈ A ∧ eB ∈ {eB}) ∧ (x ∈ A ∧ eB ∈ {eB}) definition of cartesian product→ ax ∈ A ∧ eBeB ∈ {eB} closure property of A→ (ax, eBeB) ∈ A× {eB} definition of cartesian product→ (a, eB)(x, eB) ∈ A× {eB} direct product operation

15.18 A×B is abelian↔ (∀a, x ∈ A, b, y ∈ B)(a, b)(x, y) = (x, y)(a, b) definition of abelian↔ (∀a, x ∈ A, b, y ∈ B)(ax, by) = (xa, yb) direct product operation↔ (∀a, x ∈ A, b, y ∈ B)(ax = xa) ∧ (by = yb) definition of ordered pair equality↔ A is abelian ∧B is abelian definition of abelian

15.19 〈[1]2〉 = Z2 and 〈[1]4〉 = Z4, so both of these groups are cyclic. But there is no element of A × B thatcan generate both ([0], [2]) and ([1], [2]) So A×B is not a cyclic group.

15.20 existence of an identity element:eA ∈ A ∧ eB ∈ B necessary property of groups→ (eA, eB) ∈ A×B definition of cartesian product

(a, b)(eA, eB)= (aeA, beB)= (a, b)= (eAa, eBb)= (eA, eB)(a, b)

completeness of inverses(a, b) ∈ A×B assumed→ a ∈ A ∧ b ∈ B definition of cartesian product→ a−1 ∈ A ∧ b−1 ∈ B existence of inverse elements→ (a−1, b−1) ∈ A×B definition of cartesian product

(a, b)(a−1, b−1)= (aa−1, bb−1) direct product operation= (eA, eB) property of inverses= (a−1a, b−1b) property of inverses= (a−1, b−1)(a, b) direct product operation

20

closed under the operation((a, b) ∈ A×B) ∧ ((x, y) ∈ A×B) assumed→ (a ∈ A ∧ b ∈ B) ∧ (x ∈ A ∧ y ∈ B) definition of cartesian product→ ax ∈ A ∧ by ∈ B closure property of A and B→ (ax, by) ∈ A×B definition of cartesian product→ (a, b)(x, y) ∈ A×B direct product operation

At this point, the only operation defined for the group Z has been addition, which causes the notation ofproperties like “x|y” to differ from their notations under multiplicative groups. The following table may makethe next several problems clearer:

expression notation in multiplicative groups: notation in additive groups:x|a ↔ (∃m ∈ Z)(mx = a) (∃m ∈ Z)(xm = a)x ∈ 〈a〉 ↔ (∃m ∈ Z)(am = x) (∃m ∈ Z)(am = x)x ∈ 〈a, b〉 ↔ (∃m,n ∈ Z)(x = am+ bn) (∃m,n ∈ Z)(x = ambn)x = gcd(a, b) → x|a ∧ x|b same, but note different meaning of |x = gcd(a, b) → (∃m,n ∈ Z)(x = am+ bn) (∃m,n ∈ Z)(x = ambn)x = (am)n ↔ x = amn x = am

n

15.23 x ∈ 〈a, b〉 assumed→ (∃m,n ∈ Z)(x = ambn) definition of x ∈ 〈a, b〉→ (∃m,n ∈ Z)(x = ambn ∧ d|a ∧ d|b) definition of d and the gcd→ (∃m,n, r, s ∈ Z)(x = ambn ∧ dr = a ∧ ds = b) def. of divisibility in additive group→ (∃m,n, r, s ∈ Z)(x = (dr)m(ds)n) algebraic replacement→ (∃m,n, r, s ∈ Z)(x = dr

m

dsn

) algebra→ (∃m,n, r, s ∈ Z)(x = dr

msn) algebra→ x ∈ 〈d〉 definition of 〈d〉

x ∈ 〈d〉 assumed→ (∃r ∈ Z)(x = dr) definition of x ∈ 〈a〉→ (∃r ∈ Z)(x = (gcd(a, b))r) definition of d→ (∃m,n, r ∈ Z)(x = (ambn)r) definition of gcd→ (∃m,n, r ∈ Z)(x = am

r

bnr

) algebra→ x ∈ 〈a, b〉 definition of x ∈ 〈a, b〉

This shows that 〈a, b〉 ⊆ 〈d〉 and 〈d〉 ⊆ 〈a, b〉, so 〈a, b〉 = 〈d〉.

15.24 x ∈ 〈m〉 assumed→ (∃r ∈ Z)(x = mr) definition of x ∈ 〈a〉→ (∃r ∈ Z)(x = mr ∧ a|m ∧ b|m) definition of m and the lcm→ (∃r, s, t ∈ Z)(x = mr ∧ as = m ∧ bt = m) definition of divisibility→ (∃r, s, t ∈ Z)(x = (as)r ∧ x = (bt)r) algebraic replacement→ (∃r, s, t ∈ Z)(x = as

r ∧ x = btr

) algebra→ x ∈ 〈a〉 ∧ x ∈ 〈b〉 definition of 〈a〉, 〈b〉→ x ∈ 〈a〉 ∩ 〈b〉 definition of intersection

x ∈ 〈a〉 ∩ 〈b〉 assumed→ x ∈ 〈a〉 ∧ x ∈ 〈b〉 definition of intersection→ (∃r, s ∈ N)(x = ar ∧ x = bs)→ a|x ∧ b|x definition of divisibility→ m|x definition of m and the lcm→ (∃r ∈ N)(mr = x) definition of divisibility→ x ∈ 〈m〉 definition of x ∈ 〈a〉

This shows that 〈a〉 ∩ 〈b〉 ⊆ 〈m〉 and 〈m〉 ⊆ 〈a〉 ∩ 〈b〉, so 〈a〉 ∩ 〈b〉 = 〈m〉.

21

15.25 1 = gcd(a, n) assumed↔ (∃r, s ∈ N)(1 = arns) corollary of theorem 12.2↔ (∃r, s ∈ N)(1 = ar(mod n)) definition of modular equivalence↔ (∃r ∈ N)([1]n = [ar]n) change of notation↔ (∃r ∈ N)([1]n = [a]rn) [aaa] . . . = [a][a][a] . . .↔ [1]n ∈ 〈[a]n〉↔ 〈[a]〉 = Zn [1] is cyclic in all Zn

15.26 x ∈ 〈[a]〉→ (∃r ∈ N)(x = [a]r) definition of x ∈ 〈[a]〉→ (∃r ∈ N)(x = [a]r ∧ d|a) definition of d and gcd→ (∃r, s, t ∈ N)(x = [a]r ∧ ds = a) definition of divisibility→ (∃r, s, t ∈ N)(x = [ds]r) algebraic replacement→ (∃r, s, t ∈ N)(x = [d]s

r

) algebra→ x ∈ 〈[d]〉 definition of x ∈ 〈[a]〉

x ∈ 〈[d]〉 assumed→ (∃r ∈ N)(x = [d]r) definition of x ∈ 〈[a]〉→ (∃r, s, t ∈ N)(x = [d]r ∧ d = asnt) theorem 12.2→ (∃r, s, t ∈ N)(x = [asnt]r) algebraic replacement→ (∃r, s, t ∈ N)(x = [a]s

r

[n]tr

) algebra→ (∃r, s, t ∈ N)(x = [a]s

r

[0]tr

) [n]n = [0]→ (∃r, s, t ∈ N)(x = [a]s

r

) [0] is the identity in Z→ x ∈ 〈[a]〉 definition of x ∈ 〈[a]〉

This shows that 〈[a]〉 ⊆ 〈[d]〉 and 〈[d]〉 ⊆ 〈[a]〉, so 〈[a]〉 = 〈[d]〉.

15.27 gcd(a, n) = gcd(b, n) assumed→ (∃x ∈ Z)(gcd(a, n) = x = gcd(b, n))→ (∃x ∈ Z)(〈[a]〉 = 〈[x]〉 = 〈[b]〉) exercise 15.26→ (〈[a]〉 = 〈[b]〉) transitivity of =

For the second half of the proof, let x = gcd(a, n) and let y = gcd(b, n).

〈[a]〉 = 〈[b]〉 assumed→ 〈[a]〉 = 〈[x]〉 = 〈[y]〉 = 〈[b]〉 exercise 15.26→ 〈[b]〉 = 〈[x]〉 commutativity of =→ [b] ∈ 〈[x]〉 choice of specific element of 〈[b]〉→ (∃r ∈ Z)([b] = [xr]) definition of membership in 〈[x]〉→ (∃r, s ∈ Z)(b = nsxr) definition of modular equivalence

And we know that x|n because x = gcd(a, n), so :→ (∃r, s, t ∈ Z)(b = nsxr ∧ xt = n) definition of divisibility→ (∃r, s, t ∈ Z)(b = (xt)sxr) algebraic replacement→ (∃r, s, t ∈ Z)(b = xt

sr) algebra→ x|b definition of divisibility→ x|b ∧ x|n x = gcd(a, n)→ x| gcd(b, n) property of the gcd→ gcd(a, n)| gcd(b, n) definition of x

We know that gcd(a, n)| gcd(b, n). If we use the same steps with 〈[a]〉 = 〈[y]〉 instead of 〈[b]〉 = 〈[x]〉,it can also be shown that gcd(b, n)| gcd(a, n). In Z, this means that gcd(a, n) = gcd(b, n).

22

15.28 existence of an identity elementeA ∈ A→ (eA, eA) ∈ A×A(a, a)(eA) = (aeA, aeA) = (a, a) = (eAa, eAa) = (eA, a)(eA, a)

existence of inverse elements(a, a) ∈ A× → a ∈ A→ a−1 ∈ A→ (a−1, a−1) ∈ A×A(a, a)(a−1, a−1) = (aa−1, aa−1) = (eA, eA) = (a−1a, a−1a) = (a−1, a−1)(a, a)

closure(a, a), (b, b) ∈ A×A assumed→ a ∈ A ∧ b ∈ A definition of cartesian product→ ab ∈ A closure property of A→ (ab, ab) ∈ A×A definition of cartesian product→ (a, a)(b, b) ∈ A×A operation of direct products

The diagonal subgroup of A represents the set of points in the cartesian plane along the line y = x.

15.29 a) Every group must have an identity element. If the a subgroup H of Z has only one element, thenthat element must be the additive identity. and 〈0〉 = {0} = H.

b) If H has more than one element, it must contain some nonzero element a. And, because H is a group,it must also contain −a. One of these two elements must be positive.

c) If a = bq + r, then a− bq = r And since both a and b are elements of H, then r ∈ H.

d) b is the least positive element of H and r ∈ H, so 0 ≤ r < b implies r = 0.

e) By the division algorithm, all a ∈ H can be expressed as a = bq+ r with 0 ≤ r < b. If r is always zero,this means that all a ∈ H are multiples of b. This means that H = 〈b〉.

16 Cosets

16.9 a ∈ Hb (c)↔ a ∈ {hb : h ∈ H} definition of Hb↔ (∃h ∈ H)(a = hb) (b)↔ (∃h ∈ H)(ab−1 = h)↔ ab−1 ∈ H (a)↔ a ∼ b definition of ∼↔ a ∼ b ∧ b ∼ a transitivity of ∼↔ (∀x ∈ H)((x ∼ a→ x ∼ b) ∧ (x ∼ b→ x ∼ a))

This last step is justified by the fact that ∼ is an equivalence relation. Because ∼ must be transitive,a ∼ b is both a necessary and sufficient condition for (x ∼ a → x ∼ b). Similarly, b ∼ a is both anecessary and sufficient condition for (x ∼ b→ x ∼ a).

↔ (∀x ∈ H)(x ∼ a↔ x ∼ b)↔ (∀x ∈ H)(xa−1 ∈ H ↔ xb−1 ∈ H) definition of ∼↔ (∀x ∈ H)(x ∈ Ha↔ x ∈ Hb) already shown that (a)↔ (c)↔ (Ha ⊆ Hb ∧Hb ⊆ Ha) definition of subset↔ (Ha = Hb) definition of equality, (d)

23

16.10 reflexivea ∈ G assumed→ a ∈ G ∧ a−1 ∈ G inverse property of groups→ a−1a ∈ G closure property of groups→ a ∼ a

symmetrica ∼ b assumed→ a−1b ∈ H definition of ∼→ (a−1b)−1 ∈ H closure property of groups→ b−1a ∈ H→ b ∼ a definition of ∼

transitivea ∼ b ∧ b ∼ c assumed→ a−1b ∈ H ∧ b−1c ∈ H definition of ∼→ (a−1b)(b−1c) ∈ H closure property of groups→ a−1c ∈ H→ a ∼ c definition of ∼

16.11 b ∈ aH (c)↔ b ∈ {ah : h ∈ H} definition of aH↔ (∃h ∈ H)(b = ah) (b)↔ (∃h ∈ H)(a−1b = h)↔ a−1b ∈ H (a)↔ a ∼ b definition of ∼↔ a ∼ b ∧ b ∼ a transitivity of ∼↔ (∀x ∈ H)((x ∼ a→ x ∼ b) ∧ (x ∼ b→ x ∼ a)) a ∼ b is necessary and sufficient for (x ∼ a→ x ∼ b).

This last step is justified by the fact that ∼ is an equivalence relation. Because ∼ must be transitive,a ∼ b is both a necessary and sufficient condition for (x ∼ a → x ∼ b). Similarly, b ∼ a is both anecessary and sufficient condition for (x ∼ b→ x ∼ a).

↔ (∀x ∈ H)(x ∼ a↔ x ∼ b)↔ (∀x ∈ H)(a ∼ x↔ b ∼ x) transitivity of ∼↔ (∀x ∈ H)(a−1x ∈ H ↔ b−1x ∈ H) definition of ∼↔ (∀x ∈ H)(x ∈ aH ↔ x ∈ bH) already shown that (a)↔ (c)↔ (aH ⊆ bH ∧ bH ⊆ aH) definition of subset↔ (aH = bH) definition of equality, (d)

16.12 x ∈ Ha assumed↔ (∃h ∈ H)(x = ha) definition of Ha↔ (∃h ∈ H)(x = ah) commutativity of abelian groups↔ x ∈ aH definition of aH

16.16 Let H(x, y) be an arbitrary right coset of A× {eb} in A×B.

(m,n) ∈ H(x, y) ∩ {eA} ×B→ (m,n) ∈ {(a1, eB)(x, y) : (a1, eB) ∈ H} ∧ (m,n) ∈ {eA ×B} definition of H(x, y)→ (m,n) ∈ {(a1x, eBy) : (a1, eB) ∈ H} ∧ (m,n) ∈ {eA ×B} direct product operation→ ∃(a1 ∈ A, b ∈ B)((m,n) = (a1x, eBy) ∧ (m,n) = (eA, b))→ ∃(a1 ∈ A, b ∈ B)(m = a1x ∧m = eA ∧ n = eBy ∧ n = b)→ m = eA ∧ n = eBy→ (m,n) = (eA, eBy)

16.19 Let S be a subset of G. Let H,K be unique subgroups of G.

24

H 6= K assumed→ (H 6⊆ K) ∨ (K 6⊆ H) definition of set inequality→ (∃h ∈ H)(h /∈ K) ∨ (∃k ∈ K)(k /∈ H) definition of set membership→ (∃h ∈ H)(ha /∈ Ka) ∨ (∃k ∈ K)(ka /∈ Ha)→ (Ha 6⊆ Ka) ∨ (Ka 6⊆ Ha) definition of set inequality→ Ha 6= Ka

So if H and K are unique subgroups, they have unique right cosets. Therefore S cannot be the right cosetof two unique subgroups.

16.20 If HaK and HbK are not disjoint, then there is some x that is a member of both groups. So thereexists elements of H and K such that h1ak1 = x = h2bk2. A bit of algebraic manipulation shows that thisimplies a = (h−11 h2)b(k2k

−11 ) and b = (h−12 h1)a(k1k

−12 ). And this shows that these two sets are identical,

since y ∈ HaK means that y = h3(h−11 h2)b(k2k−11 )k3 which in turn implies that y ∈ HbK. And y ∈ HbK

means that y = h3(h−12 h1)a(k1k−12 )k3, which in turn implies that y ∈ HaK.

16 Lagrange’s Theorem

Construct the lattice thingies for 11 and 17.

17.1

|S3| = 3! = 6 and |〈(1 2)〉| = |{(1 2), (1)}| = 2, so [S3 : 〈(1 2)〉] = 6/2 = 3.

17.2

|S4| = 4! = 24 and |〈(1 2 3)〉| = |{(1 2 3), (1 3 2), (1)}| = 3, so [S4 : 〈(1 2 3)〉] = 24/3 = 8.

17.3

|Z10| = 10 and |〈[2]〉| = |{[2], [4], [6], [8], [0]}| = 5, so [Z10 : 〈[2]〉] = 10/5 = 2.

17.7

From Lagrange’s theorem we know that both 4 and 10 divide |G|. From theorem 12.3, this means that |G|is divisible by 20, the least common multiple of (4,10). We’re also given that |G| < 50. There are only twomultiples of 20 that are less than 50, so we know that

|G| ∈ {20, 40}

17.9

We’re told that |H| = 6 and that [G : H] > 4. From this we can infer that

[G : H] > 4|G|/|H| > 4 (definition of the index [G : H])|G|/6 > 4 (we’re told that |H| = 6)|G| > 24

We are also given an upper bound for |G|, so we know that 24 < |G| < 50. We can further restrict thepossible values of |G| with Lagrange’s theorem: the fact that |H| = 6 tells us that |G| is a multiple of 6. Theset of all multiples of 6 within this interval is

|G| ∈ {30, 36, 42, 48}

17.11

The subgroups are {[0]}, {[0], [2], [4]}, {[0], [3]}, and Z6.

25

17.13

The subgroups are {(0, 0)}, {(1, 0), (0, 0)}, {(0, 1), (0, 0)}, {(1, 1), (0, 0)}, and {(1, 1), (1, 0), (0, 1), (0, 0)}.

17.15

The subgroups are

{(0, 0)} {(0, 0), (1, 0), (0, 1), (1, 1), (0, 2), (1, 2), (0, 3), (1, 3)}{(0, 0), (0, 1), (0, 2), (0, 3)} {(0, 0), (1, 0)}{(0, 0), (0, 2)} {(0, 0), (1, 0), (0, 2), (1, 2)}{(0, 0), (1, 1), (0, 2), (1, 3)} {(0, 0), (1, 0), (0, 2), (1, 2)}

17.17

The subgroups are

〈[0]〉 = {[0]}〈[1]〉 = Z36

〈[2]〉 = {[0], [2], [4], [6], [8], [10], [12], [14], [16], [18], [20], [22], [24], [26], [28], [30], [32], [34]}〈[3]〉 = {[0], [3], [6], [9], [12], [15], [18], [21], [24], [27], [30], [33]}〈[4]〉 = {[0], [4], [8], [12], [16], [20], [24], [28], [32]}〈[6]〉 = {[0], [6], [12], [18], [24], [30]}〈[9]〉 = {[0], [9], [18], [27]}〈[12]〉 = {[0], [12], [24]}〈[18]〉 = {[0], [18]}

17.20

From the definition of the index and the fact that |M ×N | = |M | · |N |, we know that

[G×H : A×B] =|G×H||A×B|

=|G| · |H||A| · |B|

From Lagrange’s theorem we know that |G| = |A| · [G : A] and |H| = |B| · [H : B] and therefore this becomes

|A| · [G : A] · |B| · [H : B]

|A| · |B|= [G : A] · [H : B]

17.24

Let a be an arbitrary element of G other than the identity element. We are justified in assuming this elementexists: if G contained only the identity element, then 〈e〉 = G. But we are told that G is not cyclic, so this can’tbe the case. Therefore we are justified in choosing a to be any non-identity element of G.

We now consider the generated subgroup 〈a〉. By Lagrange’s theorem, the only possible sizes for subgroupsof G are 1, p, and p2. We can rule out the possibility of |〈a〉| = 1, because this would force 〈a〉 = {e} and we’vealready assumed that a was not the identity element. We can also rule out the possibility of |〈a〉| = p2, becausethis would imply that 〈a〉 = G and we’re told that G is not cyclic. So we are forced to conclude that |〈a〉| = p.

Corollary 2 of theorem 14.3 tells us that o(a) = |〈a〉|, so it must be the case that o(a) = p. From this, wecan directly show that ap = e by appealing to the definition of “the order of an element a”: the definition (p78)tells us that p is the smallest integer such that ap = e. We have shown that ap = e for any arbitrary nonidentityelement a.

To complete the proof we need only to show that ap = e when a is the identity element, which is triviallytrue.

17.27

Proof by contradiction: assume that the intersection A ∩ B contains a non-identity element c and consider thegenerated subgroup 〈c〉. Because c is an element of both A and B, Lagrange’s theorem tells us that |〈c〉| divides

26

both |A| and |B|. From the properties of the GCD, we know then that |〈c〉| must also divide gcd(|A|, |B|) = 1.The only group of size one is {e}, so the fact that o(c)|1 proves that 〈c〉 = {e} and therefore that c = e.

We assumed that we could choose a nonidentity element of A∩B and proved that this element must be theidentity: by contradiction, it must not be possible to choose a nonidentity element of A ∩B.

17.34

We’re told that H is a subgroup of G, so by Lagrange’s theorem we know that |G| = |H|[G : H] or, equivalently,

that [G : H] = |G||H| . We’re told that K ⊆ H and that K is a group, so K is a subgroup of H. By Lagrange’s

theorem we know that [H : K] = |H||K| . We can now use these equivalencies to show that

[G : H][H : K] =|G||H||H||K|

=|G||K|

= [G : K]

which is what we were asked to prove.

18 Isomorphisms

18.1 Let θ : (Z,+)→ ({2m : m ∈ Z}, ∗) be defined as θ(x) = 2x.θ(x+ y) = 2x+y

= 2x2y

= θ(x) ∗ θ(y)

The proof that θ is a bijection is trivial.

18.2 Let θ : (Z× Z,+)→ ({2m3n : m,n ∈ Z}, ∗) be defined as θ((x, y)) = 2m3n.

θ((a, b) + (x, y)) = θ((a+ x, b+ y))= 2a+x3b+y

= 2a3b2x3y

= θ(a, b) ∗ θ(x, y)

The proof that θ is a bijection is trivial.

18.3 * a b c da a b c db b c d ac c d a bd d a b c

The isomorphism for this group is θ : ({a, b, c, d}, ∗)→ (Z4,+) defined as:θ(a) = [0], θ(b) = [1], θ(c) = [2], θ(d) = [3]

18.4 * a b c da a b c db b a d cc c d a bd d c b a

The isomorphism for this group is θ : ({a, b, c, d}, ∗)→ (Z2 × Z2,+) defined as:θ(a) = ([0], [0]), θ(b) = ([0], [1]), θ(c) = ([1][0]), θ(d) = ([1], [1])

18.7 The proof of theorem 18.1 does not assume that the function G→ H is one-to-one.

18.8 Let G = ({a, b, c}, ∗) and let H = ({a, b, c},#). An isomorphic function can be defined as θ : G→ H withθ(a) = b, θ(b) = c, and θ(c) = a.

18.10 Let (θ : G×H → H ×G) be defined as θ((g, h)) = (h, g).

27

θ((g, h)(i, j)) = θ((g ∗ i, h#j)) definition of direct product operation= (h#j, g ∗ i) definition of θ= (h, g)(j, i) definition of the direct product operation= θ(g, h)θ(i, j) definition of θ

The proof that θ is a bijection is trivial.

18.11 θ is one-to-oneθ(x1) = θ(x2) assumed→ ex1 = ex2 definition of θ→ ex1−x2 = 1 division→ ex1−x2 = e0

→ x1 − x2 = 0→ x1 = x2

θ is ontox ∈ R+ assumed→ ln(x) ∈ R ln : R+ → R→ θ(ln(x)) = eln(x) = x→ (∃k ∈ R)(θ(k) = x)

θ is homomorphicθ(x+ y) = ex+y

= exey

= θ(x)θ(y)

18.12 ⊕ [0] [1] [2] [3][0] [0] [1] [2] [3][1] [1] [2] [3] [0][2] [2] [3] [0] [1][3] [3] [0] [1] [2]

� [1] [2] [3] [4][1] [1] [2] [3] [4][2] [2] [4] [1] [3][3] [3] [1] [4] [2][4] [4] [3] [2] [1]

An isomorphic function θ : (Z4,⊕)→ (Z#5 ,�) can be defined as θ([0]) = [1], θ([1]) = [2], θ([2]) = [4], θ([3]) =[3].

18.14 Let θ : (Zmn,⊕)→ (Zm × Zn,×) be defined as θ([a]mn) = ([a]m, [a]n).

θ is homomorphicθ([a]mn ⊕ [b]mn)= θ([a+ b]mn) definition of ⊕= ([a+ b]m, [a+ b]n) definition of θ= ([a]m ⊕ [b]m, [a]n ⊕ [b]n) definition of ⊕= ([a]m, [a]n)× ([b]m, [b]n) definition of direct product operation= θ([a]mn)× θ([bmn]) definition of θ

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θ is one-to-oneθ([a]mn) = θ([b]mn) assumed→ ([a]m, [a]n) = ([b]m, [b]n)) definition of θ→ ([a]m = [b]m ∧ [a]n = [b]n) def. of ordered pair equality→ (∃r, s ∈ Z)(a = mr + b ∧ a = ns+ b) definition of modular equality→ (∃r, s ∈ Z)(a− b = mr ∧ a− b = ns) algebra→ m|(a− b) ∧ n|(a− b) definition of divisibility→lcm(m,n)|(a− b) theorem 12.1, definition of lcm→ mn|(a− b) lcm of two primes is their product→ (∃r ∈ Z)((a− b) = mnr) definition of divisibility→ (∃r ∈ Z)(a = mnr + b) algebra→ [a]mn = [b]mn definition of modular equality

θ is ontogcd(m,n) = 1 given→ (∃r, s ∈ Z)(1 = mr + ns) Theorem 12.2 corollary→ (∀a, b ∈ Z)(∃r, s ∈ Z)((a− b) = mr(a− b) + ns(a− b)) algebra→ (∀a, b ∈ Z)(∃r, s ∈ Z)((a− b) = m(ra− rb) + n(sa− sb)) algebra→ (∀a, b ∈ Z)(∃r, s ∈ Z)(a+m(rb− ra) = b+ n(sa− sb)) algebra→ (∀a, b ∈ Z)(∃r, s, z ∈ Z)((z = a+m(rb− ra)) ∧ (z = b+ n(sa− sb))) algebra→ (∀a, b ∈ Z)(∃z ∈ Z)([z]m = [a]m ∧ [z]n = [b]n) definition of modular equivalence→ (∀a, b ∈ Z)(∃z ∈ Z)(θ([z]) = ([a]m, [b]n)) definition of θ→ (∀([a]m, [b]n) ∈ Zm × Zn)(∃z ∈ Z)(θ([z]) = ([a]m, [b]n)) change of variable→ θ is onto Zm × Zn definition of onto

18.15 existence of an identity elementeG ∈ G ∧ eH ∈ H necessary property of groups→ eG ∈ G ∧ eH ∈ H ∧ eH ∈ B property of subgroups, lemma 7.1→ θ(eG) = eH ∧ eH ∈ B theorem 18.2→ θ(eG) ∈ B algebraic replacement→ eG ∈ A definition of A

existence of inversesa ∈ A assumed→ (a ∈ G ∧ θ(a) ∈ B)) definition of A→ (a−1 ∈ G ∧ (θ(a))−1 ∈ B) closure property of groups→ (a−1 ∈ G ∧ (θ(a−1)) ∈ B) theorem 18.2→ a−1 ∈ A definition of A

closurea ∈ A ∧ b ∈ A→ (a ∈ G ∧ θ(a) ∈ B) ∧ (b ∈ G ∧ θ(b) ∈ B) definition of A→ (a ∈ G ∧ b ∈ G) ∧ (θ(a) ∈ B ∧ θ(b) ∈ B) rearrangement of terms→ (a ∗ b ∈ G) ∧ (θ(a)#θ(b) ∈ B) closure property of groups→ a ∗ b ∈ G ∧ θ(a ∗ b) ∈ B homomorphism→ a ∗ b ∈ A definition of A

18.16 This problem refers back to several examples from previous chapters. Here are the definitions of all thegroups involves in this problem:

4.2 The function αa,b : R→ R is defined as αa,b(x) = ax+ b for some given value of a and b.

5.8 The set A is defined as A = {αa,b : a, b ∈ R}: the set of all possible α functions. This set forms agroup under the operation of function composition.

5.10 GL(2,R) is defined as the set of all 2 × 2 matrices with nonzero determinants. This set is a groupunder the operation of matrix multiplication.

29

18.16 G is defined to be the set of all matrices of the form

[a b0 1

]with a 6= 0. This is a subgroup of

GL(2,R).

Now, let the function θ : G → A be defined as θ

([a b0 1

])= αa,b. The proofs that this function are

one-to-one, onto, and isomorphic are straightforward but incredibly tedious to typeset, so to hell with that.

19 More on Isomorphism

19.11 From theorem 19.3, any group of order 59 is isomorphic to Z59. Because Z59 is cyclic, any groupisomorphic to it is cyclic. So there are no noncyclic groups of order 59.

19.12 39 is the product of distinct primes. Therefore, from the corollary of theorem 19.3 at the bottom of p99,every group of order 39 is isomorphic to Z39. Because Z39 is cyclic, from theorem 19.1(c) this means thatany group of order 39 is cyclic.

19.17 If G is cyclic, then (∃a ∈ G)(G = 〈a〉 = {a0, a1, . . . , an−1}). Define the function θ : {a0, a1, . . . , an−1} →Zn as θ(ax = [x]). The proof that this function is a bijection is trivial. To prove that this function preservesoperations:

θ(ax · ay) = θ(ax+y) = [x+ y] = [x]⊕ [y] = θ(ax)⊕ θ(ay)

19.18 If G is an infinite cyclic group, then (∃a ∈ G)(G = 〈a〉 = {an : n ∈ Z}). Define the function θ : {an : n ∈Z} → Z as θ(ax = x). The proof that this function is a bijection is trivial. To prove that this functionpreserves operations:

θ(ax · ay) = θ(ax+y) = x+ y = θ(ax) + θ(ay)

19.19 Let θ represent the isomorphic function from G→ H. G is cyclic, so let g represent the element for which〈g〉 = G.

G is cyclic assumed→ (∀a ∈ G)(∃k ∈ Z)(a = gk)→ (∀a ∈ G)(∃k ∈ Z)(θ(a) = θ(gk)) θ is well-defined→ (∀a ∈ G)(∃k ∈ Z)(θ(a) = θ(g)k) theorem 18.2(c)→ (∀θ(a) ∈ θ(G))(∃k ∈ Z)(θ(a) = θ(g)k) change of variable→ (∀θ(a) ∈ H)(∃k ∈ Z)(θ(a) = θ(g)k) θ is onto H→ H is cyclic definition of cyclic

19.20 For any subgroup S ofG, we know that θ(S) is a subgroup ofH from 18.2(d). Because θ is an isomorphism,S ≈ θ(S) and therefore o(S) = o(θ(S)).

19.21 G has a an element a of order n assumed→ 〈a〉 is a cyclic subgroup of G of order n definition of o(a)→ θ(〈a〉) is isomorphic to 〈a〉 18.2(e)→ θ(〈a〉) is a cyclic group of order n 19.1(a), 19.1(c)→ θ(〈a〉) has an element of order n definition of cyclic

19.22 (∀x ∈ G)(x = x−1)(∀x ∈ G)(xx = eG) algebra(∀x ∈ G)(θ(xx) = θ(eG)) θ is one-to-one(∀x ∈ G)(θ(xx) = eH) 18.2(a)(∀x ∈ G)(θ(x)θ(x) = eH) isomorphism property of θ(∀θ(x) ∈ H)(θ(x)θ(x) = eH) θ is onto(∀y ∈ H)(yy = eH) change of variables(∀y ∈ H)(y = y−1) uniqueness of inverses

19.23 The truth of this claim is a direct consequence of exercise 19.21 combined with the bijection property ofθ.

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19.25 Let θ, λ be arbitrary isomorophisms from G→ G. Let AutG = {f : f is an isomorphism G→ G}.

existence of an identity elementLet ig be the identity function iG(x) = x. Then for all λ ∈ AutG:

(λ ◦ i)(x) = λ(i(x)) = λ(x) = i(λ(x)) = (i ◦ λ)(x)

existence of an inverse functionBecause the members λ ∈ AutG are isomorphisms, they are all bijective functions and therefore theyall have a well-defined inverse function λ−1.

(λ ◦ λ−1)(x) = λ(λ−1(x)) = x = λ−1(λ(x)) = (λ−1 ◦ λ)(x)

closure

(λ ◦ θ)(x+ y) = (λ(θ(x+ y))) = λ(θ(x) ∗ θ(y)) = λ(θ(x)) • λ(θ(y)) = (λ ◦ θ)(x) • (λ ◦ θ)(y)

19.26 We know that this function is one-to-one and onto because, by the properties of groups, each a ∈ G hasone unique inverse in G. So we need only prove that θ is operation preserving:

θ(xy) = (xy)−1 = y−1x−1 = x−1y−1(from abelianism) = θ(x)θ(y)

31

19.27 lemma[a] is a generator for Zn→ 〈[a]〉 = Zn definition of a generator→ {[a]k : k ∈ N} = Zn definition of 〈[a]〉→ Zn ⊆ {[a]k : k ∈ N}→ Zn ⊆ {[ak] : k ∈ N}→ (∀[x] ∈ Zn)(∃k ∈ N)([x] = [ak]) definition of subset→ each [x] ∈ Zn can be expressed as [ak] for some k ∈ N conclusion for people who skipped to the

end

θ is one-to-oneθ([x]) = θ([y]) assumption→ (∃r, s ∈ N)([x] = [ar] ∧ [y] = [as] ∧ θ([x]) = θ([y])) lemma→ (∃r, s ∈ N)([x] = [ar] ∧ [y] = [as] ∧ θ([ar]) = θ([as])) algebraic replacement→ (∃r, s ∈ N)([x] = [ar] ∧ [y] = [as] ∧ ([ara] = [asa])) definition of θ→ (∃r, s ∈ N, v ∈ Z)([x] = [ar] ∧ [y] = [as] ∧ (ar+1 = nv + as+1)) definition of modular equivalence→ (∃r, s ∈ N, v ∈ Z)([x] = [ar] ∧ [y] = [as] ∧ (ar = n(va−1) + as)) multiply by a−1

→ (∃r, s ∈ N)([x] = [ar] ∧ [y] = [as] ∧ [ar] = [as]) definition of modular equivalence→ [x] = [y] algebraic replacement

θ is onto (short, convincing, but possibly invalid proof)(∀[y] ∈ Zn)(θ([ya−1] = [y]))

θ is onto (longer, baffling, but valid proof)[y] ∈Rng(θ) assumed→ [y] ∈ Zn definition of Rng(θ)→ [y] ∈ Zn ∧ [e] ∈ Zn identity element of Zn→ [y] ∈ Zn ∧ θ([e]) ∈ Zn definition of Rng(θ)→ [y] ∈ Zn ∧ [a] ∈ Zn definition of θ→ [y] ∈ Zn ∧ [a−1] ∈ Zn inverse property of group Zn→ [ya−1] ∈ Zn closure property of group Zn→ [ya−1] ∈ Zn ∧ θ([ya−1]) = [y] definition of θ→ [ya−1] ∈Dom(θ) ∧ θ([ya−1]) = [y] definition of Dom(θ)→ (∃[x] ∈Dom(θ))(θ([x]) = [y]) quantification→ θ is onto definition of onto

θ is isomorphicθ([x]⊕ [y]) = θ([x+ y]) = [(x+ y)a] = [xa+ ya]θ([x])⊕ θ([y]) = [xa]⊕ [ya] = [xa+ ya]

19.28 This can be proven by the process of elimination. There are only four possible functions from Z2 → Z2.Only two of the functions are bijections, and only one of those bijections maps the identity element ontoitself.

21 Homomorphisms of Groups

21.7 Let θ : G→ H be a homomorphism.

32

a, b ∈ θ(G) assumed→ a ∈ θ(G) ∧ b ∈ θ(G) ∧ ab ∈ θ(G) closure property: θ(G) is a group by 18.2(d)→ (∃x, y ∈ G)(θ(x) = a ∧ θ(y) = b ∧ ab ∈ θ(G)) definition of a ∈ θ(G)→ (∃x, y ∈ G)(θ(x) = a ∧ θ(y) = b ∧ θ(x)θ(y) ∈ θ(G)) algebraic replacement→ (∃x, y ∈ G)(θ(x) = a ∧ θ(y) = b ∧ θ(xy) ∈ θ(G)) homomorphism property→ (∃x, y ∈ G)(θ(x) = a ∧ θ(y) = b ∧ θ(xy) = θ(yx)) abelian property of G→ (∃x, y ∈ G)(θ(x) = a ∧ θ(y) = b ∧ θ(x)θ(y) = θ(y)θ(x)) homomorphism property→ (∃x, y ∈ G)(θ(x) = a ∧ θ(y) = b ∧ ab = ba) algebraic replacement→ ab = ba

21.8 Let G = 〈g〉 = {g0, g1, . . .}.

y ∈ θ(G) assumed→ (∃x ∈ G)(θ(x) = y) definition of y ∈ θ→ (∃x ∈ G)(x ∈ G ∧ θ(x) = y)→ (∃x ∈ G)(x ∈ {gn : n ∈ Z} ∧ θ(x) = y)→ (∃x ∈ G,n ∈ Z)(x = gn ∧ θ(x) = y) definition of x ∈ 〈x〉→ (∃n ∈ Z)(θ(gn) = y) algebraic replacement→ (∃n ∈ Z)(θ(g)n = y) 18.2(c)→ y ∈ 〈θ(g)〉 definition of 〈θ(g)〉

21.9 existence of identityeG ∈ G true from group properties→ eG ∈ A true from subgroup properties→ θ(eG) ∈ θ(A) definition of images→ eH ∈ θ(A) 18.2(a)

existence of inversex ∈ θ(A) assumed→ (∃a ∈ A)(θ(a) = x) definition of θ(A)→ (∃a ∈ A)(a ∈ A ∧ θ(a) = x)→ (∃a ∈ A)(a−1 ∈ A ∧ θ(a) = x) existence of inverses→ (∃a ∈ A)(a−1 ∈ A ∧ θ(a)−1 = x−1) existence of inverses→ (∃a ∈ A)(a−1 ∈ A ∧ θ(a−1) = x−1) 18.2(c)→ (∃a−1 ∈ A)(θ(a−1) = x−1) change of variable→ x−1 ∈ θ(A) definition of θ(A)

closedx ∈ θ(A) ∧ y ∈ θ(A) assumed→ (∃a, b ∈ A)(θ(a) = x ∧ θ(b) = y) definition of θ(A)→ (∃a, b ∈ A)(θ(a) = x ∧ θ(b) = y ∧ ab ∈ A) closure property of A→ (∃a, b ∈ A)(θ(a)θ(b) = xy ∧ ab ∈ A) algebra→ (∃a, b ∈ A)(θ(ab) = xy ∧ ab ∈ A) homomorphism θ→ (∃ab ∈ A)(θ(ab) = xy) change of variable→ xy ∈ θ(A) definition of θ(A)

33

21.10 existence of an identity elementeH ∈ B true by properties of subgroups→ eH ∈ B ∧ θ(eG) = eH 18.2(a)→ eH ∈ B ∧ θ(eG) = eH ∧ eG ∈ G property of groups→ θ(eG) ∈ B ∧ eG ∈ G algebraic replacement→ eG ∈ θ−1(B) definition of θ−1

existence of an inversea ∈ θ−1(B) assumed→ a ∈ G ∧ θ(a) ∈ B definition of θ−1

→ a−1 ∈ G ∧ θ(a)−1 ∈ B inverse properties of groups G and B→ a−1 ∈ G ∧ θ(a−1) ∈ B 18.2(c)→ a−1 ∈ θ−1(B) definition of θ

closure a ∈ θ−1(B) ∧ b ∈ θ−1(B) assumeda ∈ G ∧ θ(a) ∈ B ∧ b ∈ G ∧ θ(b) ∈ B definition of θ−1

ab ∈ G ∧ θ(a) ∈ B ∧ θ(b) ∈ B closure property of Gab ∈ G ∧ θ(a)θ(b) ∈ B closure property of Bab ∈ G ∧ θ(ab) ∈ B homomorphism→ ab ∈ θ−1(B) definition of θ−1

21.11 (β ◦ α)(xy)= β(α(xy)) definition of composition= β(α(x) • α(y)) homomorphism of α= β(α(x)) ∗ β(α(y)) homomorphism of β= (β ◦ α)(x) ∗ (β ◦ α)(y) definition of composition

21.12a x ∈ ker(α) assumed→ α(x) = eH definition of the kernel→ β(α(x)) = β(eH) β is well-defined→ (β ◦ α)(x) = β(eH) definition of composition→ (β ◦ α)(x) = eK 18.2(a)→ x ∈ ker(β ◦ α) definition of kernel

21.12b Let G = N, H = {2n : n ∈ N}, and K = {0}. Define the function α : G → H to be α(x) = 2x anddefine the function β : H → K to be β(x) = 0. The kernel of α is {0} while the kernel of β ◦ α is N.

21.13 θ(a+ b) = [k(a+ b)] = [ka+ kb] = [ka]⊕ [kb] = θ(a)⊕ θ(b)

21.17 o(a) = n assumed→ an = eG theorem 14.3(a)→ θ(an) = θ(eG) θ is well-defined→ θ(an) = eH thoerem 18.2(a)→ θ(a)n = eH thoerem 18.2(c)→ o(θ(a))|n theorem 14.3(b)→ o(θ(a))|o(a) definition of n

21.18 θ([0]) = θ([3]) = (1)θ([1]) = θ([4]) = (1 2 3)θ([2]) = θ([5]) = (1 3 2)ker = {[0], [3]}

21.19 False. Let G = S3. G is obviously a normal subgroup of itself, but (1 3 2)(1 2)(1 2 3) 6= (1 2).

34

21.20 (∀n ∈ N, g ∈ G)(g−1ng ∈ N) assumed→ (∀n ∈ N, g ∈ G)((g−1ng)−1 ∈ N) existence on inverses in group N→ (∀n ∈ N, g ∈ G)((gn−1g−1) ∈ N)

And because every element in a group has a unique inverse, the set of all n−1 ∈ N is identical tothe set of all n ∈ N . So we can change the variables:

→ (∀n ∈ N, g ∈ G)((gng−1) ∈ N)

The same justifications can be easily used to show the converse.

21.23 Let h be an arbitrary element of H and let m be an arbitrary element of θ(N). Proof that hmh−1 mustthen be in θ(N):

h ∈ H ∧m ∈ θ(N) assumed→ (∃g ∈ G)(θ(g) = h) ∧ (∃n ∈ N)(θ(n) = m) def. of images, since G is onto H→ (∃g ∈ G,n ∈ N)(θ(g) = h ∧ θ(n) = m ∧ gng−1 ∈ N) normalcy of N→ (∃g ∈ G,n ∈ N)(θ(g) = h ∧ θ(n) = m ∧ θ(gng−1) ∈ θ(N)) θ is well-defined→ (∃g ∈ G,n ∈ N)(θ(g)θ(n)θ(g−1) ∈ θ(N)) def of images→ (∃g ∈ G,n ∈ N)(θ(g)θ(n)θ(g)−1 ∈ θ(N)) 18.2(c)→ hmh−1 ∈ θ(N) algebraic replacement

21.24 (a, b) ∈ A×B ∧ (e, β) ∈ {e} ×B assumed→ (a, b) ∈ A×B ∧ (e, β) ∈ {e} ×B ∧ (a−1, b−1) ∈ A×B inverse property of A×B→ (a, b)(e, β)(a−1, b−1) ∈ A×B closure property of A×B→ (a, b)(e, β)(a−1, b−1) = (ae, bβ)(a−1, b−1) algebra→ (a, b)(e, β)(a−1, b−1) = (aea−1, bβb−1) algebra→ (a, b)(e, β)(a−1, b−1) = (e, bβb−1) algebra→ (a, b)(e, β)(a−1, b−1) = (e, bβb−1) ∧ (e, bβb−1) ∈ {e} ×B definition of {e} ×B→ (a, b)(e, β)(a−1, b−1) ∈ {e} ×B algebraic replacement

21.25 Theorem 15.1 tells us that this intersection is a subgroup. We need only verify that it is normal.

c ∈⋃C ∧ g ∈ G assumed

→ (∀Ci ∈⋃C)(c ∈ Ci ∧ g ∈ G) definition of membership in

⋃C

→ (∀Ci ∈⋃C)(c ∈ Ci ∧ g ∈ G ∧ Ci / G) definition of

⋃C

→ (∀Ci ∈⋃C)(gcg−1 ∈ Ci) definition of normalcy

→ gcg−1 ∈⋃C definition of membership in

⋃C

21.26 Recall the definitions of cosets: Ng = {ng : n ∈ N}, gN = {gn : n ∈ N}. If N is normal, then these twocosets are subsets of each other:

N / G assumed↔ (∀g ∈ G,n ∈ N)(∃m ∈ N)(m = gng−1) definition of normalcy↔ (∀g ∈ G,n ∈ N)(∃m ∈ N)(mg = gn) algebra↔ (∀g ∈ G) [(∀n ∈ N)(∃m ∈ n)(gn = mg)] rearrangement of quantifiers↔ (∀g ∈ G)(gN ⊆ Ng)

N / G assumed↔ (∀g ∈ G,n ∈ N)(∃m ∈ N)(m = g−1ng) consequence of exercise 12.20↔ (∀g ∈ G,n ∈ N)(∃m ∈ N)(gm = ng) algebra↔ (∀g ∈ G) [(∀n ∈ N)(∃m ∈ n)(ng = gm)] rearrangement of quantifiers↔ (∀g ∈ G)(Ng ⊆ gN)

21.27 i) Because [G : N ] = 2, we know that |N | (the number of elements in N) is exactly one-half that of |G|.ii) Let a be any element of G that is not in N . From the properties of cosets, we know that Na is disjoint

from Ne and also that aN is disjoint from eN . But since Ne = eN = N , this means that both Naand aN are disjoint from N . Another property of cosets tells us that |Na| = |N | = |aN |.

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iii) Because Na is disjoint from N and |Na| + |N | = |G|, we know that Na is the set of all elements ofG that aren’t in N :i.e., Na = G−N . For the same reasons, we know that aN = G−N . ThereforeNa = aN .

iv) From problem 21.26, we seen that Na = aN implies N / G.

21.29 n ∈ H ∩N ∧ g ∈ H assumed→ n ∈ H ∧ n ∈ N ∧ g ∈ H definition of intersection→ n ∈ H ∧ n ∈ N ∧ g ∈ H ∧ g ∈ G property of subgroups→ (n ∈ N ∧ g ∈ G) ∧ (n ∈ H ∧ g ∈ H) rearrangement of terms→ (gng−1 ∈ N) ∧ (n ∈ H ∧ g ∈ H) from N / G→ (gng−1 ∈ N) ∧ (n ∈ H ∧ g ∈ H ∧ g−1 ∈ H) inverse property of H→ (gng−1 ∈ N) ∧ (gng−1 ∈ H) closure property of H→ gng−1 ∈ H ∩N definition of intersection

21.31 If every element of G can be expressed as some power of an, then every element of θ(G) can be expressedas θ(an). By theorem 18.2(c) this is equivalent to saying that every element of θ(G) can be expressed asθ(a)n, which means that the image of θ is equal to 〈θ(a)〉.

21.32 From property 18.2(a), we know that θ must map the identity element of Z2 onto the identity element ofZ3: so θ([0]2) = [0]3. In order to preserve homomorphism, it must be the case that θ([0]2) = θ([1]2⊕[1]2) =θ([1]2)θ([1]2) = [0]3. This means that θ([1]2) must be its own inverse in Z3, which necessitates thatθ([1]2) = [0]3. (see exercise 11.17(b) for a full proof of this). So there is only one homomorphism, and thatis defined by θ([0]2) = θ([1]2) = [0]3.

21.33θ(a+ b) = eH = eHeH = θ(a)θ(b)

21.34 Note that the operation on Z × Z is the direct product operation, and the operation on the group Z isaddition instead of multiplication. So with these two groups, (a, b)(c, d) = (a+ c, b+ d).

θ((a, b)(α, β)) = θ((a+ α, b+ β)) = a+ α+ b+ β = (a+ b) + (α+ β) = θ((a, b)) + θ((α, β))

21.35 If θ is a homomorphic function Z→ Z, then it must be the case that θ(x+ ∆x) = θ(x) + θ(∆x) But thisindicates that θ must be a linear function (of the form θ(x) = ax+ b), because:

θ(x+ ∆x)− θ(x)

∆x=θ(∆x)

∆x

which means that θ(a+ ∆x) = θ(b+ ∆x) for any possible values of a, b. But we also know that θ must bean odd function (as opposed to an even one), because:

θ(0) = θ(x+ (−x)) = θ(x) + θ(−x) = 0

which means that θ(−x) = −θ(x). So not only must θ be a linear function, it must also be an odd linearfunction. So the set of homomorphisms can be completely described by:

{θ : θ(x) = ax, a ∈ Z}

22 Quotient Groups

22.5 Let Na and Nb be arbitrary elements of G/N , with a, b ∈ G.

(Na)(Nb)= N(ab) definition of quotient group multiplication= N(ba) abelianism of G= (Nb)(Na) definition of quotient group multiplication

36

22.6 G is cyclic→ (∃a ∈ G)(G = 〈a〉) definition of cyclic→ (∃a ∈ G)(∀g ∈ G)(∃n ∈ N)(g = an) definition of 〈a〉→ (∃a ∈ G)(∀g ∈ G)(∃n ∈ N)(Ng = N(an)) N is well-defined→ (∃a ∈ G)(∀g ∈ G)(∃n ∈ N)(Ng = (Na)n) definition of quotient group multiplication→ (∃Na ∈ G/N)(∀Ng ∈ G/N)(∃n ∈ N)(Ng = (Na)n) change of variable→ (∃Na ∈ G/N)(∀Ng ∈ G/N)(Ng ∈ 〈Na〉) definition of 〈a〉→ (∃Na ∈ G/N)(G/N = 〈Na〉) definition of set equality→ G/N is cyclic definition of cyclic

22.7 G consists of n cosets of size |〈n〉|, or of m cosets of size |〈m〉|. So n|〈n〉| = m|〈m〉|, which is equivalent tonm = |〈m〉|

|〈n〉| = |〈m〉/〈n〉|.

22.8 Na(NbNc)= Na(N(bc)) definition of the quotient product operation= N(a(bc)) definition of the quotient product operation= N(ab)c) associativity of the group G= N(ab)Nc definition of the quotient product operation= (NaNb)Nc definition of the quotient product operation

22.9a [G : N ] represents the order of the group G/N . If it is prime, then G/N is isomorphic to Z[G:N ] bytheorem 19.3. And each group Zn is cyclic.

22.9b Let G = Z8 and let N = 〈[4]〉. In this case, [G : N ] is not prime even though G/N is cyclic (G/N isgenerated by 〈N [1]〉).

22.10 [G : N ] = 12. In example 22.3, 〈[3]〉× 〈[2]〉 represented the direct product of two subgroups: one of order4 and one of order 2. But 〈([3], [2])〉 is a single subgroup of order 4.

22.11 This is a direct consequence of problem 21.17. By theorem 22.2, we know that there is a homomorphismη : G → G/N defined as η(a) = Na. We are told that a ∈ G. Therefore, by exercise 21.17, o(η(a))|o(a),which means o(Na)|o(a).

22.12a Note that the normal group in this problem is Z, not N . So the coset containing a is indicated by Zainstead of Na.

Za ∈ Q/Z assumed→ (∃q ∈ Q)(Za = Zq) definition of membership in Q/Z→ (∃r, s ∈ Z)(Za = Z r

s ) definition of membership in Q→ (∃r, s ∈ Z)((Za)s = (Z r

s )s) coset operation is well-defined→ (∃r, s ∈ Z)((Za)s = Z( rs )s) definition of coset operation→ (∃r, s ∈ Z)((Za)s = {z + ( rs )s : z ∈ Z}) definition of Z r

s

Remember that we are dealing with an additive group, so exponents are actually repeated addition:in an additive group, 23 = 2 + 2 + 2 = 6. Technically, I should haved used the notation r−s torepresent division instead of a fraction, but it made the flow of the proof less clear. But, despite thesloppy notation, in an additive group it’s true that ( rs )s = r∗s

s = r. So this last step implies:

→ (∃r, s ∈ Z)((Za)s = {z + r : z ∈ Z}) algebra→ (∃s ∈ Z)((Za)s = Z) closure of addition in Z→ o(Za)|s Theorem 14.3 (Z is the identity element of Q/Z)→ o(Za) ≤ s→ o(Za) is finite

The difficult part of this problem is understanding exactly what “the order of an element of Q/Z” is evenreferring to. Q/Z is a group consisting of an infinite number of elements. Each of these elements is of theform Za = {z + a : z ∈ Z, a ∈ Q}, so each Za ∈ Q/Z also has an infinite number of elements. But Zais not generally a group, so it’s order is not the number of elements it contains: the order of Za is the

37

number of distinct cosets in generated group 〈Za〉. And the number of distinct cosets depends on what a is.

If a = 0, for example, then Z0 = {z + 0 : z ∈ Z} = Z and the order of Z0 is shown to be 1. Ifa = 1/3, then Za = {z + 1

3 : z ∈ Z, a ∈ Q}, (Za)2 = {z1 + z2 + 23 : z1, z2 ∈ Z, a ∈ Q}, and

(Za)3 = {z1 + z2 + z3 + 1 : z1, z2, z3 ∈ Z} = Z: so o(Z 13 ) = 3. But no matter what a is chosen, the

order of Za is still finite.

22.12b Let n be some arbitrary integer, and let q = 1n . For the reasons given above, o(Nq) = n.

22.13 Let Za be an arbitrary element of R/Z. Remember that the only operation defined on this group isaddition, so a3 is equivalent to 3a.

o(Za) is finite assumed→ (∃p ∈ Z)(o(Za) = p) definition of finite→ (∃p ∈ Z)((Za)p = e) theorem 14.3→ (∃p ∈ Z)(Z(ap) = e) theorem 22.2 and 18.2(c)→ (∃p ∈ Z)(Z(ap) = Z) Z is the identity of R/Z→ (∃p ∈ Z)({z + ap : z ∈ Z, a ∈ R} = Z) definition of Za→ (∃p ∈ Z)(ap ∈ Z) closure property of Z→ (∃p ∈ Z, q ∈ Z)(ap = q) definition of ∈ Z

Just as in the previous problem, note that exponents in additive groups are actually repeatedaddition: in an additive group, 23 = 2 + 2 + 2 = 6. So the statement ap = q is equivalent to thestatement a ∗ p = q. So this last step implies:

→ (∃p ∈ Z, q ∈ Z)(a ∗ p = q)→ (∃p ∈ Z, q ∈ Z)(a = q

p ) algebra

→ a ∈ Q definition of Q→ Za ∈ Q/Z definition of Q/Z

So we have shown that all the finite members of R/Z are members of Q/Z. Q is a subset of R. Andproblem 22.12 showed that all members of Q/Z are finite. Therefore, the set of all finite members of R/Zis Q/Z.

22.14 G/N is abelian↔ (∀a, b ∈ G)(N(a)N(b) = N(b)N(a)) definition of abelian↔ (∀a, b ∈ G)(N(ab) = N(ba)) definition of quotient group operation↔ (∀a, b ∈ G)(N(ab)N(ba)−1 = Ne) Ne is the identity element of G/N↔ (∀a, b ∈ G)(N(ab)N((ba)−1) = Ne) Theorems 22.2 and 18.2(c)↔ (∀a, b ∈ G)(N(ab)N(a−1b−1) = Ne) algebra↔ (∀a, b ∈ G)(N(aba−1b−1) = Ne) definition of quotient group operation↔ (∀a, b ∈ G)(N(aba−1b−1) = N) definition of Ne↔ (∀a, b ∈ G)(aba−1b−1 ∈ N) closure property of N

22.15 Let n be an arbitrary element of N and let g be an arbitrary element of G.

N(e) = N(n) n ∈ N iff N(n) = N(e)= N(gg−1) identity property of G= N(g)N(g−1) def. of quotient group operation, N is well-defined= N(ge)N(g−1) identity property of G= N(g)N(e)N(g−1) def. of quotient group operation, N is well-defined= N(g)N(n)N(g−1) n ∈ N iff N(n) = N(e)= N(gng−1) def. of quotient group operation, N is well-defined

→ N(e) = N(gng−1) summary of the above equalities→ gng−1 ∈ N n ∈ N iff N(n) = N(e)

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23 Fundamental Homomorphism Theorem

23.7 (φ ◦ η)(a) = φ(η(a)) = φ(Na) = θ(a)

23.8 Let θ : G → G be defined as the identity function, θ(g) = g. The kernel of this function is just K = {e},the identity element of G. So, by the fundamental homomorphism theorem, G/K ≈ G.

23.11 i) Let G be a simple abelian group. By the definition of “simple”, the only normal subgroups of G are{e} and G itself. But all subgroups of an abelian group are normal (n = ngg−1 = gng−1), so the onlysubgroups of G, normal or otherwise, are {e} and G.

ii) But this means that G cannot be isomorphic to any direct product Zm × Zn,m, n > 1 other than{e} ×G. If it were, then G would have subgroups of size m and n.

iii) And we care about this because the fundamental theorem of finite abelian groups (section 19), tells usthat every abelian group G is isomorphic to the direct product of cyclic groups of prime order: i.e.,that G = Zp1 × . . .× Zpn.

iv) So we see that G is isomorphic to a direct product of cyclic groups of prime order, and that the onlydirect product isomorphic to G is G×{e}. This implies that both G and e are cyclic groups of primeorder. And since G is obviously isomorphic to G, this means that G is isomorphic to a cyclic groupof prime order, which is what we wanted to prove.

23.12 Let θ : Z18 → Z3 be defined as θ([a]18) = [a]3.

θ is well defined[a]18 = [b]18 assumed→ (∃n ∈ Z)(a = 18n+ b) definition of modular equivalence→ (∃n ∈ Z)(a = 3(6n) + b) algebra[a]3 = [b]3 definition of modular equivalenceθ([a]18) = θ([b]18) definition of θ

θ is a homomorphismθ([a]18 ⊕ [b]18)= θ([a+ b]18) definition of ⊕= [a+ b]3 definition of θ= [a]3 ⊕ [b]3 definition of ⊕= θ([a]18)⊕ θ([b]18) definition of θ

θ is onto H(trivial)

kernel of θ = 〈[3]〉θ([a]18) = [0]3 assumed→ [a]3 = [0]3 definition of θ→ (∃n ∈ Z)(a = 3n+ 0) definition of modular equivalence→ Ker(θ) = {[3n] : n ∈ Z} definition of kernel→ Ker(θ) = 〈[3]〉 definition of 〈x〉

This shows that there is a homomorphic function θ : Z18 → Z3 with a kernel of 〈[3]〉. Therefore, bythe fundamental homomorphism theorem, Z18/〈[3]〉 ≈ Z3.

23.13 Let θ : Zn → Zk be defined as θ([a]n) = [a]k.

θ is well defined[a]n = [b]n assumed→ (∃r ∈ Z)(a = nr + b) definition of modular equivalence→ (∃r, s ∈ Z)(a = (ks)r + b) from the fact that k|n→ (∃r, s ∈ Z)(a = k(sr) + b) associative property[a]k = [b]k definition of modular equivalenceθ([a]n) = θ([b]n) definition of θ

39

θ is a homomorphismθ([a]n ⊕ [b]n)= θ([a+ b]n) definition of ⊕= [a+ b]k definition of θ= [a]k ⊕ [b]k definition of ⊕= θ([a]n)⊕ θ([b]n) definition of θ

θ is onto H(trivial)

kernel of θ = 〈[k]〉θ([a]n) = [0]k assumed→ [a]k = [0]k definition of θ→ (∃r ∈ Z)(a = rk + 0) definition of modular equivalence→ Ker(θ) = {[rk] : r ∈ Z} definition of kernel→ Ker(θ) = 〈[k]〉 definition of 〈x〉

This shows that there is a homomorphic function θ : Zn → Zk with a kernel of 〈[k]〉. Therefore, bythe fundamental homomorphism theorem, Zn/〈[k]〉 ≈ Zk.

23.16 We know that A is a subgroup from the proof in exercise 21.10. So we need only prove normalcy of A/G:

n ∈ A ∧ g ∈ G assumed→ θ(n) ∈ B ∧ g ∈ G definition of membership in A→ θ(n) ∈ B ∧ θ(g) ∈ H range of θ is H→ θ(n) ∈ B ∧ θ(g) ∈ H ∧ θ(g)−1 ∈ H closure property of H→ θ(n) ∈ B ∧ θ(g) ∈ H ∧ θ(g−1) ∈ H 18.2(c)→ θ(g)θ(n)θ(g−1) ∈ B normalcy of B / H→ θ(gng−1) ∈ B property of homomorphism of θ→ gng−1 ∈ A definition of A

23.17 Proof by contrapositive. Assume that G/N is not simple. This means that there exists some normalsubgroup M ⊂ G/N such that M 6= G/N and M 6= Ne (the identity element of G/N). Because M is asubgroup of G/N , it must contain the identity element of G/N : so N ∈M .

Because η : G → G/N is a homomorphism and M / G/N , exercise 23.16 tells us that we can construct anormal set A = {g ∈ G : η(g) ∈ M}. This tells us that A is a normal subgroup of G that contains everyelement from every coset in M . But we know that N is one of the cosets in M , so N ⊆ A. And A is anormal subgroup of G, so we know that A ⊆ G. So we know that N ⊆ A ⊆ G.

In order to prove the contrapositive and show that A is a subgroup strictly between N and G, we mustnow rule out the possibility that A = N or that A = G. Because M 6= G/N , we know that there are someelements of G that aren’t in any of the cosets in M . And since A is just the set of all elements of all cosetsof M , this means that A 6= G. For similar reasons, because M 6= Ne, we know that A 6= N . ThereforeN ⊂ A ⊂ G.

So by assuming that G/N is not simple, we’ve shown that there is a normal subset strictly between N andG. This is the contrapositive of what we wanted to prove.

23.18 If G is simple, the only normal subgroups of G are G and eG. The kernel of θ(G) is a normal subgroupof G by theorem 21.1, so the kernel is either G or eG. If the kernel is eG, then θ is one-to-one and ontoθ(G) which means that G ≈ θ(G). If the kernel is G then θ(G) = {0} which means that o(θ(G)) = 1.

23.19 b) This is a direct consequence of theorems 21.2 and 23.1.

c) A is an extension of R by R# because A contains a normal subgroup B such that R ≈ B and A/B ≈ R(It can be shown that R ≈ B with the function θ : B → R defined as θ(α1,b) = b).

d) C contains all the same subgroups that R does, but C 6≈ R (section 32).

23.20 a) Both H and K must contain the identity element of G, so eGeG ∈ HK. This is the identity elementof HK, since (hk)(eGeG) = hk = (eGeG)(hk) for all hk ∈ HK. And each hk ∈ HK has an inverse:

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hk ∈ HK ∧ h ∈ H ∧ k ∈ K assumed→ h−1 ∈ H ∧ k ∈ K existence of inverses in group H→ h−1 ∈ H ∧ hkh−1 ∈ K from normalcy of K→ h−1 ∈ H ∧ (hkh−1)−1 ∈ K existence of inverses in group K→ h−1 ∈ H ∧ hk−1h−1 ∈ K theorem 18.2(e)→ h−1hk−1h−1 ∈ HK definition of HK→ k−1h−1 ∈ HK cancellation of inverses→ (hk)−1 ∈ HK theorem 18.2(e)

Note that ab ∈ HK does not necessarily imply that a ∈ H or b ∈ K. The most we can assume isthat there exist h ∈ H, k ∈ K such that hk = ab. For this reason both the preceeding and proceedingproofs explicitly state not only that hk ∈ HK, but also that h ∈ H and k ∈ K. Next, proof that HK isclosed:

→ (h1k1 ∈ HK) ∧ (h2k2 ∈ HK) ∧ (h1, h2 ∈ H) ∧ (k1, k2 ∈ K) assumed→ (h1 ∈ H) ∧ (h2 ∈ H) ∧ (k1 ∈ K) ∧ (k2 ∈ K)→ (h1h2 ∈ H) ∧ (k1 ∈ K) ∧ (k2 ∈ K) closure property of H→ (h1h2 ∈ H) ∧ (h−12 k1h2 ∈ K) ∧ (k2 ∈ K) normalcy of K, exercise 21.20→ (h1h2 ∈ H) ∧ (h−12 k1h2k2 ∈ K) closure property of K→ h1h2h

−12 k1h2k2 ∈ K definition of HK

→ h1k1h2k2 ∈ K cancellation of inverses→

Finally, we show that K /HK.

(hk1 ∈ HK) ∧ (h ∈ H) ∧ (k ∈ K) assumed→ (∀j ∈ K)[(h ∈ H) ∧ (k ∈ K) ∧ (j ∈ K)]→ (∀j ∈ K)[(h−1 ∈ H) ∧ (k ∈ K) ∧ (j−1 ∈ K)] existence of inverses in groups→ (∀j ∈ K)[(h−1 ∈ H) ∧ (j−1kj−1 ∈ K)] closure property of K→ (∀j ∈ K)(h−1j−1kj−1 ∈ HK) definition of HK→ (∀j ∈ K)((h−1j−1)kj−1 ∈ HK) associativity of operation on G→ (∀j ∈ K)((jh)kj−1 ∈ HK) theorem 18.2(e)→ (∀j ∈ K)(j(hk)j−1 ∈ HK) associativity of operation on G→ K /HK defintion of normalcy

23.20 b)θ(ab) = K(ab) = K(a)K(b) = θ(a)θ(b)

23.20 c) Proof that ker θ ⊆ H ∩K:

x ∈ ker θ assumed→ x ∈ H ∧ θ(x) = K domain of θ is H, identity of codomain is K→ x ∈ H ∧Kx = K definition of θ→ x ∈ H ∧Kx ⊆ K partial definition of set equality→ x ∈ H ∧ (∀k ∈ K)(∃j ∈ K)(kx = j) quantification of ⊆→ x ∈ H ∧ (∀k ∈ K)(∃j ∈ K)(k−1kx = k−1j) algebra→ x ∈ H ∧ (∀k ∈ K)(∃j ∈ K)(x = k−1j) cancellation of inverses→ x ∈ H ∧ x ∈ K closure property of K→ x ∈ H ∩K definition of intersection

Proof that H ∩K ⊆ ker θ:

x ∈ H ∩K assumed→ x ∈ H ∧ x ∈ K definition of intersection→ x ∈ H ∧ x ∈ K ∧ θ(x) = Kx definition of θ→ x ∈ H ∧ θ(x) = K lemma 16.1, from fact that x ∈ K→ x ∈ ker θ domain of θ is H, identity of the range is K.

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Given the properties that were proven in steps (a-c), the isomorphism of the last two groups is a di-rect consequence of the fundamental homomorphism theorem.

25 Integral Domains and Subrings

*Double-check 25.18. The first assumption doesn’t seem to be valid.

25.10 Let p be prime, and let [a] and [b] be arbitrary elements of Zp such that [a]⊗ [b] = 0. We can prove thatZp is an integral domain by showing that one of these two elements must be the zero of Zp:

(∃a, b ∈ Zp)([a]p ⊗ [b]p = [0]p) assumed→ (∃a, b ∈ Zp)([ab]p = [0]p) definition of the operation ⊗→ (∃a, b, n ∈ Z)(ab = np+ 0) definition of modular equivalence→ (∃a, b ∈ Zp)(p|ab) definition of divisibility→ (∃a, b ∈ Zp)(p|a ∨ p|b) property of p being prime→ (∃a, b,m, n ∈ Z)(pn = a ∨ pm = b) definition of divisibility→ (∃a, b ∈ Zp)([0]p = [a]p ∨ [0]p = [b]p) definition of modular equivalence

25.11 Z[√

2] is a subset of R. If there were zero divisors in Z[√

2], these would also be zero divisors of R. Andsince we are given that R is an integral domain, there must not be any such zero divisors. Therefore Z[

√2]

is an integral domain.

25.12 Let f and g be piecewise defined functions such that f(1) = 1, f(x) = 0 otherwise and g(2) = 1, g(x) = 0otherwise. Both of these functions are in M(R), neither of them are the zero function 0(x), but theirproduct is equivalent to the zero function.

25.13 Let R be a ring with zero element 0R, and let S be a ring with zero element 0S .

case i) If S is not an integral domain, then there are nonzero elements s1, s2 such that s1s2 = 0S . Thismeans that the nonzero elements (0R, s1)(0R, s2) = (0R, 0S) and therefore R × S is not an integraldomain. Similar results occur if R is not an integral domain.

case ii) If R is an integral domain and S contains some other element than 0S , then the nonzero elements(0R, s1)(r1, 0S) = (0R, 0S) and therefore R × S is not an integral domain. Similar results occur if Sis an integral domain and R contains a nonzero element.

case iii) If R is an integral domain and S contains only the zero element 0S , then (r1, 0S)(r2, 0S) =(r1r2, 0S). (r1r2, 0S) is the zero element of R × S when and only when one of the terms on theleft-hand side of the equation is the zero element of R×S: this means that R×S has no zero divisors.R × S also has the nonzero unity element (1R, 0S) where 1R is the unity element of R. So in thiscase, R × S is an integral domain. Similar results occur if S is an integral domain and R containsonly the zero element 0R.

25.14 Let D be a commutative ring with a, b, c ∈ D, a 6= 0, and the property of left cancellation (ab = ac →b = c). Proof by contradiction:

D has at least one zero divisor→ (∃a, b ∈ D)(ab = 0 ∧ a 6= 0 ∧ b 6= 0) definition of zero divisor→ (∃a, b ∈ D)(ab = a0 ∧ a 6= 0 ∧ b 6= 0) property of zero→ (∃a, b ∈ D)(b = 0 ∧ a 6= 0 ∧ b 6= 0) left cancellation

And this last statement is contradictory. So if all of our assumptions are true, it cannot be the casethat D has zero divisors. Note that this does not necessarily prove that D is an integral domain, since wehave not shown that it has a nonzero unity.

25.16 C(R) is nonempty because it contains 0x. For each continuous function f(x), its negative function −f(x)is also continuous and therefore −f(x) ∈ C(R). The sum or product of two continuous functions is alsocontinuous, but I haven’t taken an analysis course and therefore have no idea how to prove this.

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25.17 From theorem 15.1 we know that the intersection⋂Ci is a subgroup of R, so we need only prove ring

properties.

nonemptinesseR ∈ R necessary property of groups→ (∀Ci ∈

⋂Ci)(eR ∈ Ci) property of subgroups, lemma 7.1

→ eR ∈⋂Ci definition of intersection

closed under its operationsa ∈

⋂Ci ∧ b ∈

⋂Ci assumed

→ (∀Ci ∈⋂Ci)(a ∈ Ci ∧ b ∈ Ci) definition of intersection

→ (∀Ci ∈⋂Ci)(ab ∈ Ci ∧ (a+ b) ∈ Ci) closure property of rings

→ ab ∈⋂Ci ∧ (a+ b) ∈

⋂Ci closure property of rings

existence of negativesa ∈ R necessary property of groups→ (∀Ci ∈

⋂Ci)(a ∈ Ci) property of subgroups, lemma 7.1

→ (∀Ci ∈⋂Ci)(−a ∈ Ci) property of subgroups

→ −a ∈⋂Ci definition of intersection

25.18 Let S be any subset of a ring R, and let 〈S〉 denote the intersection of all of the subrings of R thatcontain S. Then 〈S〉 is the unique smallest subring of R that contains S in the sense that

(a) 〈S〉 contains S

(b) 〈S〉 is a subring

(c) if T is any subring of R that contains S, then T contains S.

The proof of this follows directly from Theorem 15.2 and exercise 25.17. Because rings are just additivegroups with an extra operation defined on them, theorem 15.2 tells us that 〈S〉 is the smallest uniquesubgroup of R that contains S. Exercise 25.17 tells us that this subgroup is also a subring. Together, theytell us that 〈S〉 is the unique smallest subring.

25.19 Let⋂Ci be an intersection of integral domains of ring R. We proved that this intersection is a subring

of R in exercise 25.17, so we need only prove that it is also an integral domain.

lemma 1:⋂Ci is commutative

a ∈⋂Ci ∧ b ∈

⋂Ci assumed

→ a ∈⋂Ci ∧ b ∈

⋂Ci ∧ ab ∈

⋂Ci closure of subrings

→ (∀Ci ∈⋂Ci)(a ∈ Ci ∧ b ∈ Ci ∧ ab ∈

⋂Ci) definition of intersection

→ (∀Ci ∈⋂Ci)(ab ∈ Ci ∧ ab ∈

⋂Ci) closure of subrings

→ (∀Ci ∈⋂Ci)(ba ∈ Ci ∧ ab = ba ∧ ab ∈

⋂Ci) commutativity of integral domains

→ ba ∈⋂Ci ∧ ab = ba ∧ ab ∈

⋂Ci definition of intersection

→⋂Ci is commutative definition of commutative

lemma 2:⋂Ci has no zero divisors (by contradiction)⋂

Ci has at least one zero divisor assumed→ (∃a, b ∈

⋂Ci)(ab = 0 ∧ a 6= 0 ∧ b 6= 0) definition of zero divisor

→ (∀Ci ∈⋂Ci)(∃a, b ∈ Ci)(ab = 0 ∧ a 6= 0 ∧ b 6= 0) definition of intersection

→ (∀Ci ∈⋂Ci)(Ci has at least one zero divisor) definition of zero divisor

The second lemma shows, by contradiction, that if none of the elements Ci ∈⋂Ci contain a zero di-

visor, then⋂Ci itself has no zero divisors. So the fact of whether or not

⋂Ci is an integral domain rests

entirely on whether or not it contains a unity element. It turns out that it does: each element Ci, beingan integral domain, has a unity element and the following proof shows that these unity elements must allbe identical:

Let C1 and C2 be two elements of⋂Ci. Let e1 and e2 be their respective unity elements (which they must

43

have as a consequence of being integral domains):

(∀a ∈ C1, b ∈ C2)(a = ae1 ∧ b = be2) definition of unity→ (∀a ∈ C1, b ∈ C2)(ab = ab ∧ a = ae1 ∧ b = be2)→ (∀a ∈ C1, b ∈ C2)(ae1b = abe2) algebraic replacement on ab = ab→ (∀a ∈ C1, b ∈ C2)(abe1 = abe2) commutativity from lemma 2

Note that we can’t just use the left-cancellation property of groups here, since left-cancellationdepends on the existence of inverses and integral domains don’t necessarily have multiplicativeinverses. But it turns out that left cancellation also works in integral domains (albeit for a differentreason):

→ (∀a ∈ C1, b ∈ C2)((ab(e1 − e2) = 0) algebra→ (∀a ∈ C1, b ∈ C2)(ab = 0 ∨ e1 − e2 = 0) lemma 2→ (∀a ∈ C1, b ∈ C2)(a = 0 ∨ b = 0 ∨ e1 − e2 = 0) lemma 2 again→ a = 0 ∨ b = 0 ∨ e1 = e2 algebra

And it can’t be the case that (∀a ∈ C1)(a = 0): if every element were zero, then it wouldn’t havea unity element and therefore wouldn’t be an integral domain. For the same reason, it can’t be thecase that (∀b ∈ C2)(b = 0). This means that this last statement implies:

→ e1 = e2 algebra

Therefore each element of⋂Ci contains the same unity element e, which means that

⋂Ci contains e.

This shows that⋂Ci meets all criteria for an integral domain.

25.20 From exercise 25.18, we see that the smallest subring containing n is 〈n〉. The only such subrings thatcontains the multiplicative unity are 〈1〉 and 〈−1〉, so these are the only subrings that are integral domains.

25.21 This subring does not contain zero divisors in Z. I had to use Mathematica to determine this. I’m surethere’s some elegant proof that treats the roots as isomorphic to Z3, but I can’t find it.

25.22 Let C represent the center of R.

existence of an identity elementeR ∈ R identity element of group R→ (∀a ∈ R)(eRr = reR) property of the identity element→ eR ∈ C definition of the center

closure of multiplicationa ∈ C ∧ b ∈ C assumed→ (∀c ∈ R)(ac = ca ∧ bc = cb) definition of the center→ (∀c ∈ R)(acb = cab ∧ abc = acb) left and right multiplication→ (∀c ∈ R)(abc = cab) algebraic replacement→ ab ∈ C definition of the center

closure of additiona ∈ C ∧ b ∈ C assumed→ (∀c ∈ R)(ac = ca ∧ bc = cb) definition of the center→ (∀c ∈ R)(ac+ bc = ca+ cb) algebra→ (∀c ∈ R)((a+ b)c = c(a+ b)) left and right distributive property→ (a+ b) ∈ C definition of the center

44

existence of negativesa ∈ C assumed→ (∀c ∈ R)(ac = ca) definition of the center→ (∀c ∈ R)(ac = ca ∧ c = c)→ (∀c ∈ R)(ac = ca ∧ 0c = c0) property of the zero element→ (∀c ∈ R)(ac = ca ∧ (a− a)c = c(a− a)) property of negatives→ (∀c ∈ R)(ac = ca ∧ ac− ac = ca− ca) distributive property of rings→ (∀c ∈ R)(ca− ac = ca− ca) algebraic replacement→ (∀c ∈ R)(−(ac) = −(ca)) left cancellation→ (∀c ∈ R)((−a)c = c(−a)) 24.2(b)→ −a ∈ C definition of the center

The center is a commutative ring R is R itself.

25.23 The center is the subring consisting of all matrices of the form

[x 00 y

].

25.24 M(R) is its own center because its multiplicative operation is commutative.

25.25 A subset S of integral domain R is itself an integral domain, by definition, if it is a commutative subringof R with no zero divisors and a unity element. But as long as S is a ring, it inherits commutativity andlack of zero divisors from R. So it is sufficient that

(a) S is a subring of R

(b) S contains a unity element

26 Fields

26.11 By definition, a field is a commutative ring whose nonzero elements form a group with respect to mul-tiplication. If R is an integral domain, then it is by definition a commutative ring with a nonzero unityelement. Because R is a ring, multiplication is closed and associative. The only way R could fail to be afield is if the multiplicative operation is not closed on the set of nonzero elements. But if we assume thateach nonzero element has an inverse, then the operation is closed on this set. Proof by contradiction:

R− {0} is not closed under multiplication hypothesis to be contradicted→ (∃a, b ∈ R− {0})(ab = 0) definition of closure→ (∃a, b, b−1 ∈ R− {0})(ab = 0 ∧ bb−1 = e) given:each nonzero element has an inverse→ (∃a, b, b−1 ∈ R− {0})(ab = 0 ∧ abb−1 = a) algebra→ (∃a, b, b−1 ∈ R− {0})(ab = 0 ∧ 0b−1 = a) algebraic replacement→ (∃a, b, b−1 ∈ R− {0})(ab = 0 ∧ 0 = a) property of zero→ (∃a ∈ R− {0})(a = 0)

This last statement is a contradiction, which means that our assumption is false: if R is an integraldomain with an inverse for each nonzero element, then the nonzero elements in R are closed with respectto multiplication. And this was the final property needed for R to be a field.

26.12 As in the last problem, we need only prove that the multiplicative operation is closed on the set of nonzeroelements. And if there is a unique solution for ax = b, a 6= 0 then the operation is closed on this set. Proofby contradiction:

R− {0} is not closed under multiplication hypothesis to be contradicted→ (∃a, b ∈ R− {0})(ab = 0) definition of closure→ (∃a, b ∈ R− {0})(ab = 0 ∧ a0 = 0) property of zero→ (∃b ∈ R− {0})(b = 0) unique solution for ax = b

This last statement is a contradiction, which means that our assumption is false: if R is an integraldomain with a unique solution for ax = b, then the nonzero elements in R are closed with respect tomultiplication. And this was the final property needed for R to be a field.

26.13 Z#n is a group with respect to � when n is prime.

45

26.15 The commutativity of the field along with the properties of zero and unity determine every productinvolving 0 and e. The only other products are aa, bb, and ab. It cannot be the case that aa = 0, ba = 0, ab =0, or bb = 0 (nonzero elements are closed w/r/t multiplication) or that aa = a, ba = a, ab = a, ba = b, ab = b,or bb = b (uniqueness of the unity element). By elimination it must be the case that ab = ba = e. Thismeans that it cannot be the case that aa = e or bb = e (uniqueness of inverses), so by elimination it mustbe the case that aa = b, bb = a.

26.16 ⊕ (0,0) (0,1) (1,0) (1,1)(0,0) (0,0) (0,1) (1,0) (1,1)(0,1) (0,1) (0,0) (1,1) (1,0)(1,0) (1,0) (1,1) (0,0) (0,1)(1,1) (1,1) (1,0) (0,1) (0,0)

+ 0 e a b0 0 e a be e 0 b aa a b 0 eb b a e 0

These two functions are isomorphic under the function θ(0) = (0, 0), θ(e) = (0, 1), θ(a) = (1, 0), θ(b) =(1, 1).

26.17 ([1], [0]) and ([0], [1]) are two nonzero elements of R × R, but their product is zero. This means thatthe ring R × R contains a zero divisor. This isn’t in conflict with 26.16 because we’ve merely created anisomorphism between the additive group of the ring R × R and the additive group of the field. But thefunction θ is not an isomorphism between their multiplicative groups, so there is no reason to expect thatR× R should be a field.

26.18 Let F = F1 × F2 × · · · × Fn be a direct product of fields. Choose a ∈ F such that a = (0, a2, a3, . . . , an)where each element ai is an arbitrary element of Fi. Choose b ∈ F such that b = (b1, 0, 0, . . . , 0). Theseare both nonzero elements, but their product is zero.

26.19 If a set K meets all of the properties listed in theorem 26.2, then it already meets the definition ofa subring of F . This means that K inherits multiplicative associativity and commutativity from themultiplicative group of F and additive associativity from the additive group of F . It also inherits thedistributive property and a lack of zero divisors from F . This meets all the requirements for K being afield.

26.20 Because each the field F is also a ring, we know from problem 25.17 that⋂Ci is a subring. As explained

in problem 26.19, this means that it inherits most of the necessary field properties. In order to show that⋂Ci is a subfield, we need only to show that it contains the zero and unity of F , is closed under both

operations, and that it contains negatives and inverses of each of its elements. Since every subfield in⋂Ci

contains the unity and zero of F , so must⋂Ci itself. Proofs of the remaining properties:

additive and multiplicative closurea ∈

⋂Ci ∧ b ∈

⋂Ci assumed

→ (∀Cn ∈⋂Ci)(a ∈ Cn ∧ b ∈ Cn) definition of intersection

→ (∀Cn ∈⋂Ci)(ab ∈ Cn ∧ a+ b ∈ Cn) closure of both groups in the field Cn

→ (ab ∈⋂Ci ∧ a+ b ∈

⋂Ci) definition of intersection

negatives and inversesa ∈

⋂Ci assumed

→ (∀Cn ∈⋂Ci)(a ∈ Cn) definition of intersection

→ (∀Cn ∈⋂Ci)(−a ∈ Cn ∧ a−1 ∈ Cn) completeness of inverses and negatives in the field Cn

→ −a ∈⋂Ci ∧ a−1 ∈

⋂Ci definition of intersection

26.21 Let S be a subset of a field F , and let 〈S〉 denote the intersection of all of the subfields of F that containS. Then 〈S〉 is the unique smallest subfield of F that contains S, in that

(a) 〈S〉 contains S

(b) 〈S〉 is a subfield

(c) If T is any subfield of F that contains S, then T contains 〈S〉

Property (a) is true, since each element of 〈S〉 contains S by definition. Property (b) was proven in exercise26.20. Property (c) is true by the property of intersections that states (∀S ∈

⋂Si)(

⋂Si ⊆ S).

46

26.22 Let F be a field containing Z. It must contain multiplicative inverses, so (∀n ∈ Z)( 1n ∈ F ). And it must

be closed under multiplication, so (∀m,n ∈ Z)(mn ∈ F ). This field describes Q.

26.23a We’re asked to show that the set of invertible elements is a subgroup of the multiplicative group ofR. By theorem 7.1, we need to show that the set of invertible elements is nonempty, closed under mul-tiplication, and contains inverses of each of its elements. Let H represent the set of invertible elements:H = {a ∈ R : (∃b ∈ R)(ab = e)}.

nonemptye ∈ R given property of R→ ee = e property of unity→ e ∈ H definition of H

closed under multiplicationa ∈ H ∧ b ∈ H assumed→ (∃r, s ∈ R)(ar = e ∧ bs = e) definition of H→ (∃r, s ∈ R)(ar = e ∧ bs = e ∧ rs ∈ R) multiplicative closure of R→ (∃r, s ∈ R)(ar(bs) = e ∧ rs ∈ R) multiply both equations→ (∃r, s ∈ R)(ab(rs) = e ∧ rs ∈ R) commutativity of R→ (∃rs ∈ R)(ab(rs) = e) change of variable→ ab ∈ H definition of H

invertibility of all elementsa ∈ H assumed→ (∃r ∈ R)(ar = e ∧ a ∈ H) definition of H→ (∃r ∈ R)(ar = e ∧ a ∈ R) H is a subset of R→ (∃r ∈ R)(ar = e ∧ ar ∈ R ∧ a ∈ R) closure property of R→ (∃r ∈ R)(ar = e ∧ (ar)−1 ∈ R ∧ a ∈ R) existence of inverses in R→ (∃r ∈ R)(ar = e ∧ (ar)−1 ∈ R ∧ arr−1 ∈ R) existence of inverses in R→ (∃r ∈ R)(ar(ar)−1 = e ∧ arr−1 ∈ R) definition of inverses→ (∃r ∈ R)(arr−1a−1 = e ∧ arr−1 ∈ R) theorem 14.1→ (∃r ∈ R)(a−1(arr−1) = e ∧ arr−1 ∈ R) commutativity of R→ a−1 ∈ H definition of H

26.24 Let a be an arbitrary element of commutative ring R Proof by contradiction:

a is invertible and a is a zero divisor assumed→ (∃b, c ∈ R− {0})(ac = e ∧ ab = 0) definition of invertible, zero divisor→ (∃b, c ∈ R− {0})(bac = b ∧ ab = 0) algebra→ (∃b, c ∈ R− {0})(abc = b ∧ ab = 0) commutativity of R→ (∃b, c ∈ R− {0})(0c = b) algebraic replacment→ (∃b ∈ R− {0})(b = 0) property of zero

This last statement is a contradiction: it cannot be the case that b ∈ R− {0} and b = 0.

26.25a Ker(ρr) 6= {0} iff (∃a ∈ R − {0})(ar = 0) iff (r = 0 or r is a zero divisor). The “iff”s here are true byconsequence of the definitions of ρr, 0, and zero divisors.

26.25b The field R, by definition, has no zero divisors. If r 6= 0, then Ker(ρr) = {0} by the result of 26.25a.From theorem 21.1, this means that the function ρr is a one-to-one function. In order to show that it isan isomorphism, then, we need only show that the function is homomorphic:

ρ(a+ b) = (a+ b)r = ar + br = ρ(a) + ρ(b)

27 Isomorphism and Characteristic

27.1 If θ is an isomorphism between two rings R and S, then by definition it is also an isomorphism betweentheir multiplicative groups. So, by theorem 18.2, θ(eR) = eS . Note that θ is also an isomorphism betweenthe two additive groups, so the same theorem tells us that θ(0R) = 0S .

47

27.2 θ : R→ S is an isomorphism and R is commutative assumed→ (∀s, t ∈ S)(∃a, b ∈ R)(θ(a) = s ∧ θ(b) = t ∧ ab = ba) definition of onto, commutativity→ (∀s, t ∈ S)(∃a, b ∈ R)(θ(a) = s ∧ θ(b) = t ∧ θ(ab) = θ(ba)) θ is well-defined→ (∀s, t ∈ S)(∃a, b ∈ R)(θ(a) = s ∧ θ(b) = t ∧ θ(a)θ(b) = θ(b)θ(a)) homomorphism of θ→ (∀s, t ∈ S)(∃a, b ∈ R)(θ(a) = s ∧ θ(b) = t ∧ st = ts) algebraic replacement→ (∀s, t ∈ S)(st = ts) removal of unnecessary quantifiers→ S is commutative definition of commutative

27.3 We know that S must contain a unity from exercise 27.1 We need only show that S contains no zerodivisors. Proof by contrapositive:

The ring S contains a zero divisor assumed→ (∃s, t ∈ S − {0S})(st = 0S) definition of a zero divisor→ (∃s, t ∈ S − {0S}, a, b ∈ R− {0}R)(st = 0S ∧ θ(a) = s ∧ θ(b) = t) onto-ness of θ→ (∃s, t ∈ S − {0S}, a, b ∈ R− {0}R)(st = 0S ∧ θ(a)θ(b) = 0S) algebraic replacement→ (∃s, t ∈ S − {0S}, a, b ∈ R− {0}R)(st = 0S ∧ θ(ab) = 0S) homomorphism of θ→ (∃s, t ∈ S − {0S}, a, b ∈ R− {0}R)(st = 0S ∧ ab = 0R) exercise 27.2→ (∃a, b ∈ R− {0}R)(ab = 0R) removal of unnecessary quantifiers→ the ring R contains a zero divisor definition of a zero divisor

By contrapositive we have shown that if R contains no zero divisors, then S contains no zero divisors.

27.4 S is a field if it is an integral domain with multiplicative inverses for each nonzero element. From exercise27.3, θ guarantees that S is an integral domain. We need only show that it contains multiplicative inverses.

s ∈ S assumed→ (∃a ∈ R)(θ(a) = s) onto-ness of θ→ (∃a ∈ R)(θ(a) = s ∧ a−1 ∈ R) existence of inverses in R→ (∃a ∈ R)(θ(a) = s ∧ θ(a−1) ∈ S) θ : R→ S→ (∃a ∈ R)(θ(a) = s ∧ θ(a)−1 ∈ S) 18.2(b)→ (∃a ∈ R)(θ(a) = s ∧ s−1 ∈ S) algebraic replacement→ s−1 ∈ S removal of unnecessary quantifiers

27.9 a ∈ E assumed→ aa−1 = eE existence of inverses in E→ θ(aa−1) = θ(eE) θ is well-defined, a 6= 0→ θ(a)θ(a−1) = θ(eE) θ homomorphic→ θ(a)θ(a−1) = eF 18.2(a)→ θ(a)−1θ(a)θ(a−1) = θ(a)−1eF algebra→ θ(a−1) = θ(a)−1 properties of inverses and unity

27.10 From theorem 25.2, we need only show that θ(R) is closed under both operations and that it containsthe negative of each of its elements.

multiplicative closurea ∈ θ(R) ∧ b ∈ θ(R) assumed→ (∃s, t ∈ R)(θ(s) = a ∧ θ(t) = b) definition of θ(R)→ (∃s, t ∈ R)(θ(s) = a ∧ θ(t) = b ∧ st ∈ R) closure of R→ (∃s, t ∈ R)(θ(s)θ(t) = ab ∧ st ∈ R) algebra→ (∃s, t ∈ R)(θ(st) = ab ∧ st ∈ R) homomorphism of θ→ (∃st ∈ R)(θ(st) = ab) change of variable→ ab ∈ θ(R) definition of θ(R)

48

additive closurea ∈ θ(R) ∧ b ∈ θ(R) assumed→ (∃s, t ∈ R)(θ(s) = a ∧ θ(t) = b) definition of θ(R)→ (∃s, t ∈ R)(θ(s) = a ∧ θ(t) = b ∧ s+ t ∈ R) closure of R→ (∃s, t ∈ R)(θ(s) + θ(t) = a+ b ∧ s+ t ∈ R) algebra→ (∃s, t ∈ R)(θ(s+ t) = a+ b ∧ s+ t ∈ R) homomorphism of θ→ (∃(s+ t) ∈ R)(θ(s+ t) = a+ b) change of variable→ a+ b ∈ θ(R) definition of θ(R)

existence of negativesa ∈ θ(R) assumed→ (∃s ∈ R)(θ(s) = a) definition of θ(R)→ (∃s ∈ R)(θ(s) = a ∧ ss−1 = eR) closure of R→ (∃s ∈ R)(θ(s) = a ∧ θ(ss−1) = θ(eR)) θ is well-defined→ (∃s ∈ R)(θ(s) = a ∧ θ(s)θ(s−1) = θ(eR)) homomorphism of θ→ (∃s ∈ R)(θ(s) = a ∧ θ(s)θ(s−1) = eS) 18.2(a)→ (∃s ∈ R)(aθ(s−1) = eS) algebraic replacement→ (∃s ∈ R)(θ(s−1) = a−1) uniqueness of inverses→ a−1 ∈ θ(R) definition of θ(R)

27.11 Let the function θ : Zp → D be defined as θ([a]) = ae. We wish to show that the image of θ is a subringof D isomorphic to Zp. Exercise 27.10 showed that the image of a ring homomorphism is itself a subring,so we need only the isomorphism Zp ≈ θ(Zp). In order to do this, we must show that θ is well-defined,one-to-one and onto the image of θ, and preserves the operations of D.

θ is well-defined and one-to-one[a]p = [b]p assumed↔ (∃r ∈ Z)(a = pr + b) definition of modular equivalence↔ (∃r ∈ Z)(a− b = pr) algebra↔ (∃r ∈ Z)((a− b)e = pre) property of unity↔ (a− b)e = 0 D has characteristic p↔ ae− be = 0 distributive property of D↔ ae = be algebra↔ θ([a]) = θ([b]) definition of θ

Note that the right arrow of each step (the “if” direction) shows that θ is well-defined, and the leftarrow (the “only if” direction) shows that θ is onto.

θ preserves additionθ([a]⊕ [b])= θ([a+ b]) definition of ⊕= (a+ b)e definition of θ= ae+ be distributive property of D= θ([a]) + θ([b])

θ preserves multiplicationθ([a]� [b])= θ([ab]) definition of �= (ab)e definition of θ= (ae)(be) exercise 27.6θ([a])θ([b]) definition of θ

And since θ is onto its image by definition, this is all that is necessary to prove that θ(Zp) is a subring ofD isomorphic to Z.

27.12 Theorem 19.2 can be applied to both the additive and multiplicative groups of rings.

27.13 Commutativity, lack of zero divisors, existence of a unity element, being an integral domain, havingcharacteristic 0

49

27.14 It must have either characteristic 1 (if it is isomorphic to Z1) or 2, since 2x = (x+ x) = (x− x) = 0.

27.15 Let θ : R→ S be an isomorphism between two rings.

R has characteristic m assumed→ (∀r ∈ R)(mr = 0) definition of characteristic→ (∀r ∈ R)(mr = 0) ∧ (∀s ∈ S)(∃r ∈ R)(θ(r) = s) θ is onto→ (∀r ∈ R)(mr = 0) ∧ (∀s ∈ S)(∃r ∈ R)(mθ(r) = ms) algebra→ (∀r ∈ R)(mr = 0) ∧ (∀s ∈ S)(∃r ∈ R)(θ(mr) = ms) 18.2(c) for additive groups→ (∀s ∈ S)(θ(0) = ms) algebraic replacment of mr = 0→ (∀s ∈ S)(ms = 0) 18.2(a)

Because isomorphism is reflexive, the same steps can be used to show that if S has characteristic n,then (∀r ∈ R)(nr = 0). This shows that {m : (∀r ∈ R)(mr = 0)} = {n : (∀s ∈ S)(ns = 0)}. The leastsuch shared element is the characteristic of both S and R.

27.16 Z2 × Z2 has characteristic 2, while Z4 does not.

27.17θ([a]� [b]) = θ([ab]) = ([ab]2, [ab]3) = ([a]2, [a]3)⊗ ([b]2, [b]3) = θ([a])⊗ θ([b])

27.18 In Z4: (2[2] = [2]⊕ [2] = [2 + 2] = [0]) but (2[1] = [1]⊕ [1] = [1 + 1] = [2]). Or, more trivially, let a = 0.

27.19 The subring {[0], [2], [4]} of Z6 is a ring of characteristic 3, but it is not an integral domain and its nonzeroelements are not closed under the multiplicative operation.

27.20a Let m,n be integers and let z represent their least common multiple. Define the function θ : Zz →Zm × Zn as θ([a]z) = ([a]m, [a]n).

([a]m, [a]n) = ([b]m, [b]n) assumed↔ [a]m = [b]m ∧ [a]n = [b]n definition of ordered pair equivalence↔ m|(a− b) ∧ n|(a− b) definition of modular equivalence↔ z|(a− b) z is the lcm: theorem 12.3↔ [a]z = [b]z definition of modular equivalence

The “if” portion of this proof shows that the function is well-defined, and the “only if” portion showsthat it is one-to-one. We showed that θ preserves both the additive and multiplicative operations in ex-ercise 27.17. Because θ is one-to-one, it is onto if and only if |Zz| = |Zm × Zn|. Since mn = gcd(m,n)×lcm(m,n) (exercise 13.22), this means that θ is onto if gcd(m,n) = 1. This proves that θ is an isomorphismif m and n are relatively prime.

27.20b The proof from 27.20a shows that Zm × Zn ≈ Zz, where z is the least common multiple of m and n.If the gcd(m,n) 6= 1, then the lcm(m,n) 6= mn (because mn = gcd(m,n)lcm(m,n)). But if z 6= mn,obviously Zz 6≈ Zmn. And since Zm × Zn ≈ Zz and isomorphism is an equivalence relation, we know thatZm × Zn 6≈ Zmn

27.21 lemma: x = (-x)x = x2 definition of boolean ring= (−x)2 theorem 24.2(c)= −x definition of boolean ring

commutativitya+ b = a+ b→ (a+ b)2 = a+ b definition of boolean ring→ (a2 + ab+ ba+ b2) = a+ b left and right distributive properties→ (a+ ab+ ba+ b) = a+ b definition of boolean ring→ (ab+ ba) = 0 left and right additive cancellation→ (ab− ba) = 0 lemma→ ab = ba

50

2x = 02x= x+ x= x+ (−x) lemma= 0 additive inverses

27.22 Let R be a finite ring of order p and let m be its characteristic.

(∃a, b ∈ Z)(pa+mb = gcd(p,m)) theorem 12.2(∀r ∈ R)(∃a, b ∈ Z)((pa+mb)r = (gcd(p,m))r) algebra(∀r ∈ R)(∃a, b ∈ Z)(p(ar) +m(br) = (gcd(p,m))r) algebra(∀r ∈ R)(∃a ∈ Z)(p(ar) + 0 = (gcd(p,m))r) R is characteristic m(∀r ∈ R)(0 = (gcd(p,m))r) p(ar) = 0 by 14.2 and Lagrange’s theorem: o(a)|p

So gcd(p,m) is a zero divisor for all elements of R. Because it is a divisor of m, we know that 1 ≤gcd(p,m) ≤ m. And since m, by the definition of characteristic, is the least integer that is a zero divisorfor all elements of R, it must be the case that m = gcd(p,m). This of course means that m divides p = |R|,which is what we wanted to prove.

27.23θ(a+ b

√2)θ(c+ d

√2) = (a+ b

√3)(c+ d

√3) = (ac+ 3bd+ (bc+ ad)

√3)

θ((a+ b√

2)(c+ d√

2)) = θ(ac+ 2bd+ (bc+ ad)√

2) = (ac+ 2bd+ (bc+ ad)√

3)

27.24a x2 = eR + eR assumed→ θ(x2) = θ(eR + eR) θ is well-defined→ θ(x)θ(x) = θ(eR) + θ(eR) additive and multiplicative homomorphisms→ [θ(x)]2 = θ(eR) + θ(eR) algebra→ [θ(x)]2 = eS + eS theorem 18.2a or exercise 27.1→ (∃s ∈ S)(s2 = eS + eS) from the onto-ness of θ

27.24b (0 + 1√

2)2 = 2

27.24c Proof by contradiction: Assume that there exists an isomorphism θ : R→ S:

(∃x, y ∈ Q)(θ(1 + 1√

2) = (x+ y√

3)) θ is well-defined

(∃x, y ∈ Q)(θ(1) + θ(1√

2) = (x+ y√

3)) homomorphism of θ

(∃x, y ∈ Q)(θ(1) + θ(1)√

2 = (x+ y√

3)) 18.2(c) for additive groups

(∃x, y ∈ Q)(1 +√

2 = (x+ y√

3)) 18.2(a) or exercise 27.1

(∃x, y ∈ Q)(1− x = y√

3−√

2) algebra

The left-hand side of this last equality is rational and the right-hand side contains the irrational squareroot of 2, so there is no way for this equality to ever be valid. By contradiction, our assumption that anisomorphism exists must be false.

27.26 Modify the proof in exercise 27.25, but use Zp instead of Z.

28 Ordered Integral Domains

28.1 x ∈ D assumed→ x2 ∈ Dp ∨ x2 = 0 theorem 28.2(f)→ x2 + e ∈ Dp ∨ x2 + e = e closure of Dp

→ x2 + e 6= 0 trichotomy

28.2 a > b ∧ c < 0 assumed→ (a− b) ∈ Dp ∧ (0− c) ∈ Dp definition of <→ (a− b)(0− c) ∈ Dp multiplicative closure of Dp

→ (−ac+ bc) ∈ Dp distributivity of rings→ (bc− ac) ∈ Dp additive commutativity of rings→ ac < bc definition of <

51

28.3 lemma: a ∈ Dp ∧ ab ∈ Dp → b ∈ Dp

case i)a ∈ Dp ∧ (−b) ∈ Dp assumed→ a(−b) ∈ Dp closure of Dp

→ −(ab) ∈ Dp 14.1→ ab 6∈ Dp trichotomy

case ii)a ∈ Dp ∧ b = 0 assumed→ ab = 0 property of zero→ ab 6∈ Dp trichotomy

case iii)a ∈ Dp ∧ b ∈ Dp assumed→ ab ∈ Dp closure of Dp

proof: ac > bc ∧ c > 0 assumed→ (ac− bc) ∈ Dp ∧ c ∈ Dp definition of >→ (a− b)c ∈ Dp ∧ c ∈ Dp distributivity of rings→ (a− b) ∈ Dp lemma

28.4 a < 0 ∧ b < 0 assumed→ (0− a) ∈ Dp ∧ (0− b) ∈ Dp definition of <→ (0− a)(0− b) ∈ Dp multiplicative closure of Dp

→ (−a)(−b) ∈ Dp property of zero→ ab ∈ Dp theorem 14.1→ ab > 0 definition of >

28.5 a > b assumed→ (a− b) ∈ Dp definition of >→ (a+ (−b)) ∈ Dp

→ (−(−a) + (−b)) ∈ Dp

→ (−b− (−a)) ∈ Dp

→ −b > −a definition of >

28.6 a > b ∧ a ∈ Dp ∧ b ∈ Dp assumed→ (a− b) ∈ Dp ∧ a ∈ Dp ∧ b ∈ Dp definition of >→ (a− b) ∈ Dp ∧ a+ b ∈ Dp additive closure of Dp

→ (a− b)(a+ b) ∈ Dp multiplicative closure of Dp

→ a2 + ab− ba− b2 ∈ Dp left and right distributivity of rings→ a2 + ab− ab− b2 ∈ Dp commutativity of integral domains→ a2 − b2 ∈ Dp

→ a2 > b2 definition of >

28.8 Let Ep = E ∩Dp.

52

additive and multiplicative closure of Ep

a ∈ Ep ∧ b ∈ Ep assumed→ a ∈ E ∧ b ∈ E ∧ a ∈ Dp ∧ b ∈ Dp definition of Ep

→ ab ∈ E ∧ ab ∈ Dp ∧ (a+ b) ∈ E ∧ (a+ b) ∈ Dp closures of E and Dp

→ ab ∈ E ∩Dp ∧ (a+ b) ∈ E ∩Dp definition of intersection→ ab ∈ Ep ∧ (a+ b) ∈ Ep definition of Ep

trichotomy of Ep

a ∈ E assumed→ a ∈ E ∧ a ∈ D property of subrings→ a ∈ E ∧ (a ∈ Dp ∨ a = 0 ∨ (−a) ∈ Dp) trichotomy of D→ (a ∈ E ∧ a ∈ Dp) ∨ (a ∈ E ∧ a = 0) ∨ (a ∈ E ∧ (−a) ∈ Dp)→ (a ∈ Ep) ∨ (a = 0) ∨ (a ∈ E ∧ (−a) ∈ Dp) definition of Ep

→ (a ∈ Ep) ∨ (a = 0) ∨ ((−a) ∈ E ∧ (−a) ∈ Dp) existence of negatives in ring E→ (a ∈ Ep) ∨ (a = 0) ∨ ((−a) ∈ Ep) definition of Ep

28.9 The only subsets of D that are integral domains are Zp where p is prime (example 25.2, exercise 25.10)or D itself. But for each nonzero a ∈ Zp, pa = 0 and p(−a) = 0 (theorem 14.2 and Lagrange’s theorem).So Dp is not closed under multiplication for these finite groups. Therefore the only subset of D that is anordered integral domain is the one with characteristic 0: that is, the set of integers. And if D = Z, thenDp = N.

28.10b R is not ordered because R is not an integral domain: it is not an integral domain because it has zerodivisors. For each f ∈ M(R), define a function f0 for which f0(x) = 0 iff f(x) 6= 0. If f(x) = 0 for anyvalue of x, then f0 is a zero divisor of f .

28.12 Disproof: if a ∈ Dp, then a > −a but a2 6> (−a)2.

28.13 If b 6= −a, then (b+ a) 6= 0 and the following proof holds:

a > b assumed→ (a− b) ∈ Dp definition of >→ (a− b) ∈ Dp ∧ (a+ b)2 ∈ Dp ∧ (a− b)2 ∈ Dp 28.2(f)→ (a− b) ∈ Dp ∧ (a2 + b2 + 2ab) ∈ Dp ∧ (a2 − 2ab+ b2) ∈ Dp algebra→ (a− b) ∈ Dp ∧ (a2 + b2) > −2ab ∧ (a2 + b2) > 2ab definition of >→ (a− b) ∈ Dp ∧ (a2 + b2) > |2ab| definition of absolute value→ (a− b) ∈ Dp ∧ (a2 + b2) > |2ab| > |ab|→ (a− b) ∈ Dp ∧ (a2 + b2) > |ab| transitivity of >→ (a− b) ∈ Dp ∧ (a2 + b2) > −ab partial definition of absolute value→ (a− b) ∈ Dp ∧ (a2 + b2 + ab) ∈ Dp definition of >→ (a− b)(a2 + b2 + ab) ∈ Dp closure of Dp

→ (a3 − b3) ∈ Dp algebra, commutativity of integral domains→ a3 > b3 definition of >

If b = −a, then a > b implies a > −a, which means that a ∈ Dp. It must then be the case thata3 > (−a)a2, from which algebraic replacement shows that a3 > b3.

29 The Integers

29.1 Assume that there is some least positive element q ∈ Q. Then q ∈ Dp and 12 ∈ D

p since they are bothpositive elements of Q. But then multiplicative closure of Dp requires q

2 ∈ Dp, which contradicts ourassumption that q is the least element of Dp.

29.2 Theorem 29.1 tells us that if D were an ordered integral domain, then it would be isomorphic to Z. Proofby contradiction that there can exist no isomorphism θ : Z[

√2]→ Z:

53

There exists an isomorphism θ : Z[√

2]→ Z assumed

→ (∀a, b ∈ Z)(∃c ∈ Z)(θ(a+ b√

2) = c) onto-ness of θChoose a, b such that a = 0 and b = 1.

→ (∃c ∈ Z)(θ(1√

2) = c) onto-ness of θ

→ (∃c ∈ Z)(θ(1)√

2 = c) 18.2(d) for additive groups

→ (∃c ∈ Z)(1√

2 = c) 18.2(a)

But√

2 is irrational and c is rational, so this last statement must be false. By contradiction, then, therecan exist no isomorphism between the two groups.

29.3 a ∈ D assumeda ∈ D ∧ e ∈ Dp lemma 28.1→ a ∈ D ∧ 0 + e ∈ Dp existence of additive identity in rings→ a+ (−a) + e ∈ Dp existence of additive inverses in rings→ a+ (−a) + (−(−e)) ∈ Dp 14.1→ a− (a+ (−e)) ∈ Dp 14.1→ a > a− e definition of Dp

29.4 Proof by contradiction:

a > a2 assumed→ a− a2 ∈ Dp definition of >→ a(e− a) ∈ Dp left distributivity of rings→ (a ∈ Dp ∧ e− a ∈ Dp) ∨ (−a ∈ Dp ∧ −(e− a) ∈ Dp) lemma from exercise 28.3→ (a ∈ Dp ∧ e− a ∈ Dp) ∨ (−a ∈ Dp ∧ −(e− a) + e+ e ∈ Dp) closure of Dp

→ (a ∈ Dp ∧ e− a ∈ Dp) ∨ (−a ∈ Dp ∧ e− (−a) ∈ Dp) algebra→ (a ∈ Dp ∧ a < e) ∨ (−a ∈ Dp ∧ (−a) < e) definition of >

This last statement is false: neither of the conditions can be true, since they both imply the existence ofa member of Dp smaller than e, which is the smallest member of Dp by definition. So, by contradiction,it cannot be the case that a > a2. Either a = a2 (when a = 0 or a = 1) or a < a2 (all other cases).

This proof is only valid for well-ordered integral domains. For example, Q is an ordered integral domainand it is not the case that ( 1

2 )2 > 12 .

29.5 For every nonempty subset S of Dn, define a complementary set T = {−s : s ∈ S}. Because T is anonempty subset of Dp, it must have a least element. Let −a represent this least element of T . Proof bycontradiction that a is the greatest element of S:

(∃a, b ∈ S)(b > a) hypothesis of contradiction→ (∃a, b ∈ S)(−b < −a) exercise 28.5→ (∃ − a,−b ∈ T )(−b < −a) definition of T

But this last statement cannot be true, since we defined −a as the least element of T . So our initialassumption that there was some element in Dn greater than a is false: a is the greatest element of Dn.

29.6 Proof by contradiction that there is no integer between n and n+ 1:

(∃m ∈ Z)(n+ 1 > m > n) hypothesis of contradiction→ (∃m ∈ Z)(n+ 1−m ∈ Dp ∧m− n ∈ Dp definition of >→ (∃m ∈ Z)(1− (m− n) ∈ Dp ∧m− n ∈ Dp commutativity of addition→ (∃m ∈ Z)(1 > m− n ∧m− n ∈ Dp definition of >

This last statement cannot be true, or 1 would not be the smallest element of Dp. So by contradictionthere cannot be any integer between n and n+ 1.

54

29.8 θ must be the identity mappingθ(x)= θ(xe) property of the identity element= xθ(e) 18.2(d) for additive groups= xe 18.2(a)= x

θ(x) = 2x is an additive isomorphismθ(x+ y) = 2(x+ y) = 2x+ 2y = θ(x) + θ(y)

30 Field of Rational Numbers

30.1 ([2], [1]) ∼ ([0], [2]) and ([0], [2]) ∼ ([0], [1]). If ∼ were an equivalence relation on Z4 ×Z#4 then transitivity

should guarantee that ([2], [1]) ∼ ([0], [1]) but this is not the case.

30.2 Let F = R× R′. (5, 1) ∼ (15, 3), but (5, 1) 6= (15, 3).

30.5 reflexive(a, b) ∈ D ×D′ assumed→ a ∈ D ∧ b ∈ D definition of cartesian product and D→ ab = ba commutativity of integral domain D→ (a, b) ∼ (a, b) definition of ∼

symmetric(a, b) ∼ (c, d) assumed→ ad = bc definition of ∼→ cb = da commutativity of integral domain D→ (c, d) ∼ (a, b) definition of ∼

transitive(a, b) ∼ (m,n) ∧ (m,n) ∼ (x, y) assumed→ an = bm ∧my = nx definition of ∼→ an(my) = bm(nx) algebra→ ay(nm) = bx(nm) commutativity of integral domain D→ ay = bx theorem 25.1

note: this is not right-cancellation, since right-cancellation requires theexistence of inverse elements and multiplication is not necessarily a group.This step would not be legal for rings that were not integral domains.)

→ (a, b) ∼ (x, y) definition of ∼

30.6 [a1, b1] = [a2, b2] ∧ [c1, d1] = [c2, d2] assumed→ (a1, b1) ∼ (a2, b2) ∧ (c1, d1) ∼ (c2, d2) definition of = equivalence→ (a1b2 = b1a2) ∧ (c1d2 = d1c2) definition of ∼ equivalence→ (a1b2)(c1d2) = (b1a2)(d1c2) algebra→ (a1c1)(b2d2) = (b1d1)(a2c2) associativity→ (a1c1, b1d1) ∼ (a2c2, b2d2) definition of ∼ equivalence→ [a1c1, b1d1] ∼ [a2c2, b2d2] definition of = equivalence

55

30.7 zero of FD is [0, eD][0, e] + [a, b]= [0b+ ea, eb] definition of addition in FD= [ea, eb] property of zero in multiplicative groups= [a, b] property of the unity element of D= [ae, be] property of the unity element of D= [ae+ b0, be] property of zero in multiplicative groups= [a, b] + [0, e] definition of addition in FD

negative of [a, b] is [−a, b][a, b] + [−a, b]= [ab+ (−a)b, bb] definition of addition in FD= [ab− ab, bb] 14.1= [0, bb]= [0, e] from (0, bb) ∼ (0, e)

addition is associative[a, b] + ([m,n] + [[x, y])= [a, b] + [my + nx, ny] definition of addition in FD= [any + b(my + nx), bny] definition of addition in FD= [any + bmy + bnx, bny] left distributivity of the ring D= [(an+ bm)y + bnx, bny] right distributivity of the ring D= [an+ bm, bn] + [x, y] definition of addition in FD= ([a, b] + [m,n]) + [x, y] definition of addition in FD

addition is abelian[a, b] + [x, y]= [ax+ by, by] definition of addition in FD= [xa+ yb, yb] commutativity of integral domain D= [x, y] + [a, b] definition of addition in FD

The previous four properties show that the additive operation on FD is an abelian group.

multiplication is associative[a, b]([m,n][x, y])= [a, b]([mx, ny]) definition of multiplication in FD= [a(mx), b(ny)] definition of multiplication in FD= [(am)x, (bn)y] associativity of multiplication in ring D= [am, bn][x, y] definition of multiplication in FD= ([a, b][m,n])[x, y] definition of multiplication in FD

multiplication is commutative[a, b][x, y]= [ax, by] definition of multiplication in FD= [xa, yb] commutativity of multiplication in integral domain D= [x, y][a, b] definition of multiplication in FD

left distributive property[a, b]([m,n] + [x, y])= [a, b][my + nx, ny] definition of addition in FD= [a(my + nx), bny] definition of multiplication in FD= [amy + anx, bny] distributive property of ring D= [b(amy + anx), b(bny)] from (x, y) ∼ (nx, ny)= [bamy + banx, bbny] distributive property of ring D= [am(by) + bn(ax), bn(by)] commutative property of integral domain D= [am, bn] + [ax, by] definition of addition in FD= [a, b][m,n] + [a, b][x, y] definition of multiplication in FD

56

right distributive property([m,n] + [x, y])[a, b]= [my + nx, ny][a, b] definition of addition in FD= [(my + nx)a, nyb] definition of multiplication in FD= [mya+ nxa, nyb] distributive property of ring D= [b(mya+ nxa), b(nyb)] from (x, y) ∼ (nx, ny)= [bmya+ bnxa, bnyb] distributive property of ring D= [ma(yb) + nb(xa), nb(yb)] commutative property of integral domain D= [ma, nb] + [xa, yb] definition of addition in FD= [m,n][a, b] + [x, y][a, b] definition of multiplication in FD

The previous four properties show that the FD is a commutative ring

unity of [a, b] is [eD, eD][a, b][e, e]= [ae, be] definition of multiplication in FD= [a, b] property of the unity element of D= [ea, eb] property of the unity element of D= [e, e][a, b] definition of multiplication in FD

FD contains no zero divisors[a, b][c, d] = [0, e] assumed→ [ac, bd] = [0, e] definition of multiplication in FD→ (ac, bd) ∼ (0, e) definition of = equivalence in FD→ ace = bd0 definition of ∼ equivalence in FD→ ac = 0 properties of unity and zero→ a = 0 ∨ c = 0 integral domain D has no zero divisors→ a = b0 ∨ c = d0 property of zero→ ae = b0 ∨ ce = d0 property of unity→ (a, b) ∼ (0, e) ∨ (c, d) ∼ (0, e) definition of ∼ equivalence in FD→ [a, b] = [0, e] ∨ [c, d] = [0, e] definition of = equivalence in FD

The previous two properties show that FD is an integral domain

inverse of [a, b] is [b, a][a, b][b, a]= [ab, ba] definition of multiplication in FD= [ab, ab] commutativity of integral domain D= [e, e] from (ab, ab) ∼ (e, e)= [ba, ba] from (e, e) ∼ (ba, ba)= [ba, ab] commutativity of integral domain D= [b, a][a, b] definition of multiplication in FD

This shows that FD is a field.

30.8 A subring of a field cannot have any zero divisors: if it did, they would also be zero divisors of the field. Z6

has zero divisors, and the existence of zero divisors is something that is preserved by isomorphism (section27).

30.9 Theorem 30.1 tell us that the field of quotients FD is the smallest field into which the integral domain Dcan be embedded. If D is already a field, then the smallest field into which it can be embedded is D itself,in which case FD ≈ D.

30.10 (∀m ∈ D,n ∈ D′)([n, e][m,n] = [m, e]

30.11 If K is any field of prime characteristic p, then K contains a subfield isomorphic to Qp (the quotientfield of Zp). Proof: Theorem 27.3 tells us that if K is a field of prime characteristic p, then K contains aring isomorphic to Zp. So by theorem 30.1, K must also contain a subfield isomorphic to the equivalenceclasses of Zp × Z′p. And by definition, this set of equivalence classes is Qp.

57

30.12 [1, 2] = [2, 4] and [0, 1] = [0, 1], but ([1, 2] + [0, 1]) = [1, 3] 6= [2, 5] = ([2, 4] + [0, 1]).

34 Polynomials: Definition and Elementary Properties

34.8 In order for R[x] to be a field, it would have to have multiplicative inverses for each of its nonzero elements.Let f(x) = 0 + r1x. In order for some other polynomial to be its multiplicative inverse, there would haveto be some (a0 +a1x+ . . . anx

n) such that ((a00) +(a1r1)x+ . . . (an0)xn) = e. But there can be no a0 ∈ Rsuch that a00 = e, so there can be no multiplicative inverse for f(x).

34.9a False. Consider the sum of two nth degree polynomials anxn +−anxn.

34.9b False. Consider two polynomials in Z4[x]. ([2]x5)([2]x5) is equal to [0], which certainly does not havedegree 10.

34.10 Let R be a commutative ring and let R[x] be an integral domain. If R is not an integral domain, thenthere is some nonzero a, b ∈ R such that ab = 0. But ab is also represents the product of two nonzeropolynomials of degree zero in R[x]: any zero divisor of R is also a zero divisor of R[x].

This means that (R is not an integral domain → R[x] is not an integral domain). By contrapositive, IfR[x] is an integral domain, then R is an integral domain.

34.11 If p is a zero divisor for every element of R, then it is a zero divisor for every element of R[x]. Theconverse is also true. Proof:

p is a zero divisor of R assumed↔ (∀r ∈ R)(pr = 0) definition of zero divisor↔ (∀f ∈ R[x])(∀ri ∈ f)(pr0 + pr1x

1 + . . .+ prnxn = 0) definition of a polynomial in R[x]

↔ (∀f ∈ R[x])(∀ri ∈ f)(p(r0 + r1x1 + . . .+ rnx

n) = 0) distributive property of rings↔ (∀f ∈ R[x])(pf = 0) definition of a polynomial in R[x]↔ p is a zero divisor of R[x]

Since R and R[x] share all zero divisors, they also share the least such divisor. This least zero divisor isthe characteristic by definition.

34.12 Let θ : R → S represent the isomorphism between R and S. Defined a second function φ : R[x] → S[x]as φ(a0 + a1x

1 + . . . + anxn) = θ(a0) + θ(a1)x1 + . . . + θ(an)xn. The fact that φ is a bijection follows

directly from the bijection of θ. Proof of homomorphism:

additive homomorphismφ((a0 + a1x

1 + . . .+ anxn) + (b0 + b1x

1 + . . .+ bnxn))

= φ((a0 + b0) + (a1 + b1)x1 + . . .+ (an + bn)xn))

The previous step required the associativity and commutativity of addition on R and the distributiveproperty of R.

= θ(a0 + b0) + θ(a1 + b1)x1 + . . .+ θ(an + bn)xn) definition of φ= θ(a0) + θ(b0) + θ(a1)x1 + θ(b1)x1 + . . .+ θ(an)xn + θ(bn)xn additive homomorphism of θ= θ(a0) + θ(a1)x1 + . . .+ θ(an)xn + θ(b0) + θ(b1)x1 + . . .+ θ(bn)xn additive associativity of S→ φ(a0 + a1x

1 + . . .+ anxn) + φ(b0 + b1x

1 + . . .+ bnxn)) definition of φ

multiplicative homomorphismφ((a0 + . . . amx

m)(b0 + . . . bnxn))

= φ(a0b0 + . . . (horrifying binomial)x? . . .+ ambnxm+n)

= θ(a0b0) + . . . θ(horrifying binomial) . . .+ θ(ambn)xm+n = φ(a0 + . . . amxm)φ(b0 + . . . bnx

n)

34.14a D[x]p is closed under the operations of addition and multiplication: If p, q are two elements of D[x]p

then they are of finite order with highest terms pm, qn ∈ Dp (true by definition). So their sum is of degreemax(m,n) with a highest term of either pm, qn, or pm + qn: all of which are ∈ Dp (true by the additiveclosure property of integral domain D). Their product is of degree m+n with highest term pmqn which is

58

in Dp (true by the multiplicative closure property of integral domain D). Because a polynomial is memberof D[x]p iff its highest coefficient is a member of Dp, the trichotomy of D[x]p follows directly from thetrichotomy of D.

34.14b Proof by contradiction:

there exists a positive polynomial less than 1 hypothesis of contradiction→ (∃f ∈ Z[x]p)(f < 1) formalization of hypothesis→ (∃f ∈ Z[x]p)(1− f ∈ Z[x]p) definition of >→ (∃a0, . . . , an−1 ∈ Z, an ∈ Zp)(1− (a0 + a1x

1 . . .+ anxn) ∈ Z[x]p) definition of f ∈ Z[x]p

→ (∃a0, . . . , an−1 ∈ Z, an ∈ Zp)(1− a0 − a1x1 . . .− anxn) ∈ Z[x]p) distribute property

If the degree n of f is greater than 0, then the highest term of (1− f) is −an for some an ∈ Zp: this meansthat (1− f) 6∈ Z[x]p. If the degree n of f is 0, then the highest term is (1− a0). This can only be positiveif a0 < 1, which is not possible since a0 ∈ Zp (it has to be to make f a member of Z[x]p) and 1 is theleast member of Zp by definition. So no matter what polynomial we choose for f , (1 − f) 6∈ Z[x]p, whichcontradicts our initial assumption that 1 > f .

34.14c If Z[x] were a well-ordered domain, then every nonempty subset of Z[x]p would have a least element.Let S be the set of all polynomials of degree 2 in Z[x]. There can be no least element of S: If a0 + a1xwere the least positive element, then there is an immediate contradiction since a0 + a1x > (a0 − 1) + a1xand (a0 − 1) + a1x ∈ Z[x]p. In other words, S has no least element because a0 ∈ Z and Z has no leastelement.

34.16 Define a function θ : R[x] → R[X] as θ(a0 + a1x1 + . . . + anx

n) = (a0, a1, . . . , an). The proof thatθ is a bijection follows directly from the definitions of equality for polynomials and sequences. Proof ofhomomorphism:

additive homomorphismθ[(a0 + a1x

1 + . . .+ anxn) + (b0 + b1x

1 + . . .+ bnxn)]

= θ((a0 + b0) + (a1 + b1)x1 + . . .+ (an + bn)xn)

The previous step required the associativity and commutativity of addition on R and the distributiveproperty of R.

= (a0 + b0, a1 + b1, . . . , an + bn) definition of θ= (a0, a1, . . . , an) + (b0, b1, . . . , bn) definition of sequence additionθ(a0 + a1x

1 + . . .+ anxn) + θ(b0 + b1x

1 + . . .+ bnxn) definition of θ

multiplicative homomorphismθ((a0 + . . . amx

m)(b0 + . . . bnxn))

= θ(a0b0 + . . . (horrifying binomial)x? . . .+ ambnxm+n) distributivity of R

= (a0b0, . . . ,horrifying binomial coefficients, . . . , ambn) definition of θ= (a0, . . . , am)(b0, . . . , bn) definition of sequence multiplication= θ(a0 + . . . amx

m)θ(b0 + . . . bnxn) definition of θ

35 Division Algorithm

35.11 Because the highest term of g(x) is bnxn, the highest term of the product g(x)amb

−1n xm−n is amx

m.Since this is also the highest term of f(x), we know that the degree of f(x)− g(x)amb

−1n xm−n is less than

or equal to the degree of f(x).

35.13 f([2]) = [30], which is equal to [0] for p ∈ {3, 5}.

59

35.14 Let FD represent the set of equivalence classes of Zp[x] × Zp[x]′. By lemma 30.3, FD is a field. Proofthat FD and Zp share all zero divisors:

m is a zero divisor of every element in FD assumed↔ (∀(f(x), g(x)) ∈ FD)(m(f(x), g(x)) ≈ (0, e)) lemma 30.3: (0,e) is the zero of FD↔ (∀f(x) ∈ FD)(m(f(x)) = 0)) lemma 30.1: definition of ∼↔ (∀f(x) ∈ FD)(∀[ai] ∈ f(x))(m[a0] +m[a1]x+ . . .+m[an]xn = 0) definition of zero polynomial

This last statement tells us that ma = 0 every coefficient a for every polynomial f(x) in FD. Butthe set of every possible coefficient is just Zp, so we know that:

↔ (∀a ∈ Zp)(ma = 0)↔ m is a zero divisor of every element in Zp definition of zero divisor

Because FD and Zp share all elements with the “characteristic property”, they also share a least suchelement: by definition, this least zero divisor is their common characteristic which we know to be p (theo-rem 27.3).

35.15 Let f(x) = 3, g(x) = 2. In order for the division algorithm to be true, there would have to be some a ∈ Zsuch that 3 = 2a+ r, where the deg(r) < deg(2). The only integer with degree less than deg(2) is 0, andthere is no a ∈ Z such that 3 = 2a+ 0.

35.16 Let f = a0 and g = b0 be two arbitrary nonzero polynomials of degree 0 in D[x]. In order for the divisionalgorithm to be true, it must be the case that:

(∀a0, b0 ∈ D)(∃r, s ∈ D)(a0 = b0r + s),deg(s) < deg(a0)

But since the degree of a0 is zero, this means that s = 0. If we then choose a 1 as a specific value for a0,the previous statement reduces to:

(∀b0 ∈ D)(∃r ∈ D)(1 = b0r)

This tells us that every nonzero element b0 in the integral domain D has an inverse, which makes D a fieldby definition.

35.17 From Lagrange’s theorem, we know that the multiplicative group of Z5 is cyclic with order 5 and that,for any [a] ∈ Z5, 〈[a]5〉 = [a]. So the equation f([a]) = [a]6 − [a] will be zero for any [a], which means thatevery [a] is a root of f([a]) = [a]6 − [a].

35.18 See exercise 35.17.

37 Unique Factorization Domains

37.7a N(a+ bi) = |a+ bi|2 = a2 + b2, and the last term must be ≥ 0 from the lemma of theorem 28.1.

37.7b z = 0 + 0i assumed↔ N(z) = |0 + 0i|2 definition of N↔ N(z) = 02 + 02 definition of complex norm↔ N(z) = 0

37.7c N((a+ bi)(c+ di))= N((ac− bd) + (ad+ bc)i) distributivity and commutativity of C= (ac− bd)2 + (ad+ bc)2 definition of N = (ac)2 − 2(abcd) + (bd)2 + (ad)2 + 2(abcd) + (bc)2 algebra= (ac)2 + (ad)2 + (bc)2 + (bd)2 algebra= (a2 + b2)(c2 + d2) algebra= N(a+ bi)N(c+ di) definition of N

37.8 Proof by contradiction:

60

2±√−5 is reducible in Z[

√−5] hypothesis of contradiction

→ (∃z, w ∈ Z[√−5])(zw = 2±

√−5) definition of reducibility

→ (∃z, w ∈ Z[√−5])(N(zw) = N(2±

√−5)) N is well-defined

→ (∃z, w ∈ Z[√−5])(N(zw) = 22 + 5 = 9) definition of N

→ (∃z, w ∈ Z[√−5])(N(z)N(w) = 9) N is a multiplicative function

→ (∃N(z), N(w) ∈ Z)(N(z)N(w) = 9) The range of N is Z

The only factors of 9 in Z are 1,3,and 9. In order for N(zw) =9 to be reducible, it must have a divisor that is neither a unitof Z nor an associate of N(zw) = 9. 1 is a unit and 9 isan associate, so in order to be reducible, N(zw) must have afactor of 3.

→ (∃N(z), N(w) ∈ Z)(N(z) = 3)→ (∃a, b ∈ Z)(a2 + b2 = 3) definition of N

And this last statement is false: there are no such integers. So, by contradiction, 2±√−5 is irreducible.

37.10 No matter how the distance function d is defined, d(a) = d(b) for all a, b in the field F . Proof:

d(a) ≤ d(aa−1b)→ d(a) ≤ d(b)d(b) ≤ d(bb−1a)→ d(b) ≤ d(a)

This is odd, but it doesn’t prove that F isn’t a Euclidean domain: it just shows that d(x) is a constantfunction. However, the second property of Euclidean domains is that all nonzero elements a, b ∈ F can beexpressed as a = bq + r where d(r) < d(b). But we know that d(r) = d(b) for any choice of b, r. This stilldoesn’t prove that F isn’t a Euclidean domain: it might be the case that we can always find a = bq withr = 0. And in a field, we can always do this: just choose q = b−1a.

37.11a The unity of Z[x] is 1, which is divisible by ±1.

37.11b Requirement (b) is not met because of the lack of multiplicative inverses in Z. Consider the polynomialsa(x) = 3x and b(x) = 2x. There are no polynomials in Z[x] such that 2x = 3xg(x) + h(x),deg(h) < 1.

37.12 a is a unit assumed→ a|e definition of a unit→ (∃b ∈ D)(ab = e) definition of divisibility→ (∃b ∈ D)(ab = e ∧ d(a) ≤ d(ab) ∧ d(ab) ≤ d(aab)) property of Euclidean domains→ (∃b ∈ D)(d(a) ≤ d(e) ∧ d(e) ≤ d(ae)) algebraic replacement→ (∃b ∈ D)(d(a) ≤ d(e) ∧ d(e) ≤ d(a)) property of unity→ d(a) = d(e)

37.13 ae = a, so a|a. a0 = 0, so a|0. And for any x ∈ D, if x|a ∧ x|0, then obviously x|a. And so bothconditions are met for a to be the gcd(a, 0).

61

37.14 a and b are associates assumed→ a|b ∧ b|a definition of associate→ (∃r, s ∈ D)(ar = b ∧ bs = a) definition of divisiblity→ (∃r, s ∈ D)(bsr = b ∧ ars = a) algebraic replacement from last step→ (∃r, s ∈ D)(b(sr − e) = 0 ∧ a(sr − e) = 0) distributivity, negatives. commutativity

We can’t do any type of cancellation to simplify this equation because we can’t assume the existence ofmultiplicative inverses. Instead, note that D is an integral domain: by definition, there are no zero divisors.In order for b(sr − e) = a(sr − e) = 0, it must be the case that:

→ (∃r, s ∈ D)([a = b = 0] ∨ [sr − e = 0])→ (∃r, s ∈ D)([a = be] ∨ [sr = e]) algebra→ (∃r, s ∈ D)([a = be]∨(s and r are units)) definition of unit

So either a = be or a = rs. But both e and r are units of D. So in either case, the lemma is true.

37.15 The GCD exists in Euclidean domains by virtue of the fact that the division algorithm applies to Euclideandomains. See page 66 for the proof.

37.16 d1 = gcd(a, b)→ d1|a ∧ d1|b definition of gcd→ d1| gcd(a, b) second part of the definition of gcd→ d1|d2 d2 = gcd(a, b)

The same proof with the variables exchanged shows d2|d1. Because d1 and d2 divide each other, theyare associates by definition.

37.17 b is a unit↔ (∃c ∈ D)(bc = e) definition of unit↔ (∃c ∈ D)(bc = e ∧ d(ab) ≤ d(abc) ∧ d(a) ≤ d(ab)) property of Euclidean domains

The “only if” (→) direction of this step is justified by algebraic replacement. To justify the “if” (←) step,note that d(ab) ≤ d(a) is only true if d(a) = d(abc) for some c.

↔ d(ab) ≤ d(a) ∧ d(a) ≤ d(ab) algebraic replacement↔ d(ab) = d(a)

37.18 gcd(a, b) = e ∧ a|bc assumed→ (∃r, s ∈ D)(e = ar + bs ∧ a|bc) lemma 37.2→ (∃r, s, t ∈ D)(e = ar + bs ∧ at = cb) definition of divisibility,commutativity→ (∃r, s, t ∈ D)(ce = car + cbs) ∧ at = cb)→ (∃r, s, t ∈ D)(c = car + ats) algebraic replacement→ (∃r, s, t ∈ D)(c = a(cr + ts)) commutativity, distributivity→ a|c definition of divisibility

37.19 Because p is irreducible, all of its divisors are either units or associates of p.

x|p→ [(∃r ∈ D)(xr = e) ∨ p|x]

Let x = gcd(p, a). Because the gcd is a divisor of p, it also must either be a unit or an associate of p.

case i) If x is an associate of p, then p|x and x|a (the former from x being an associate of p, the latterfrom x being the gcd) which means p|a.

case ii) If x is a unit:

62

x is a unit assumed→ (∃r ∈ D)(rx = e) definition of unit→ (∃r ∈ D)(rx = e) ∧ (∃m,n ∈ D)(x = pm+ an) property of gcd→ (∃r ∈ D)(rx = e) ∧ (∃m,n ∈ D)(rx = rpm+ ran) distributivity→ (∃r,m, n ∈ D)(e = rpm+ ran) algebraic replacement→ (∃r,m, n ∈ D)(be = brpm+ bran) distributivity→ (∃r,m, n ∈ D)(b = p(brm) + ab(rn)) commutativity,property of unity

We see that p divides both terms on the right-hand side of this last equation: p|p(brm), for obvious reasons,and we are told that p|ab in the description of the problem. This means that p|b. We’ve shown that eitherp|a or p|b, which is what we set out to prove.

37.21 u is a unit assumed→ (∃r ∈ D)(ur = e) definition of unit→ (∃r ∈ D)(ura = a) algebra→ u|a definition of divides

37.22 Let U represent the set of all units of D. Each element of U has an inverse: (∃r ∈ D)(ru = e) is boththe necessary condition for being a unit and the necessary condition for having an inverse. The unity ofD is the unity of U , since ee = e. And U is closed under multiplication:

closurea ∈ U ∧ b ∈ U assumed→ a and b are units definition of membership in U→ (∃r, s ∈ D)(ar = e ∧ bs = e) definition of unit→ (∃r, s ∈ D)(ar(e) = e ∧ bs = e) property of unity→ (∃r, s ∈ D)(ar(bs) = e) algebraic replacement→ (∃r, s ∈ D)(ab(rs) = e) commutativity of D→ ab is a unit definition of unit→ ab ∈ U definition of membership in U

Therefore U meets all the requirements of being a multiplicative group.

38 Homomorphisms of Rings

38.8b π1(a, b) = 0 iff a = 0, b ∈ S. So the kernel of π1 is {0} × S. Define a function θ : S → {0} × S to beθ(s) = (0, s). Proof that this function is isomorphic:

θ is well-defined and one-to-ones1 = s2↔ s1 = s2 ∧ 0 = 0↔ (0, s1) = (0, s2) definition of ordered pair equality↔ θ(0, s1) = θ(0, s2) definition of θ

θ is onto(0, s1) ∈ {0} × S assumed→ s1 ∈ S definition of cartesian product→ θ(s1) = (0, s1) definition of θ→ (∃s ∈ S)(θ(s) = (0, s1)

θ is homomorphicθ(s1 + s2) = (0, s1 + s2) = (0, s1) + (0, s2) = θ(s1) + θ(s2)θ(s1s2) = (0, s1s2) = (0, s1)(0, s2) = θ(s1)θ(s2)

38.9 Let R be a commutative ring and let θ : R→ S be a homomorphic function.

63

a ∈ θ(R) ∧ b ∈ θ(R) assumed→ (∃r, s ∈ R)(θ(r) = a ∧ θ(s) = b) definition of image→ (∃r, s ∈ R)(θ(r)θ(s) = ab)→ (∃r, s ∈ R)(θ(rs) = ab) homomorphism of θ→ (∃r, s ∈ R)(θ(sr) = ab) commutativity of R→ (∃r, s ∈ R)(θ(s)θ(r) = ab) homomorphism of θ→ (ba = ab) algebraic replacement from θ(s) = b, θ(r) = a

38.10 Let θ : R→ S be a homomorphic function.

(∀s ∈ θ(R))(∃r ∈ R)(θ(r) = s) definition of image→ (∀s ∈ θ(R))(∃r ∈ R)(θ(r) = s ∧ re = r = er) unity of R→ (∀s ∈ θ(R))(∃r ∈ R)(θ(r) = s ∧ θ(re) = θ(r) = θ(er) θ is well-defined→ (∀s ∈ θ(R))(∃r ∈ R)(θ(r) = s ∧ θ(r)θ(e) = θ(r) = θ(e)θ(r) homomorphism of θ→ (∀s ∈ θ(R))(sθ(e) = s = θ(e)s algebraic replacement→ θ(e) is a unity of θ(R) definition of unity

The existence of a unity for θ(R) is also an immediate consequence of theorem 18.2

38.11 a) {n2 : n ∈ Z}b) {2n : n ∈ Z}c) {k2n : n, k ∈ Z}d) Q (see example 38.4)

e) Q. Because of additive closure, the smallest subfield must contain every integer. And because a fieldcontains multiplicative inverses for each element, it must contain a−1 for every integer a. And becauseof multiplicative closure, it must contain ab−1 for every integer a, b ∈ Z. And this is the definition ofQ from section 30.

38.12 Let S be a subring of Z.

s ∈ S assumed

→ (∀n ∈ Z)(

n∑1

s ∈ S) additive closure

→ (∀n ∈ Z)(ns ∈ S) definition of integer multiplication→ S is an ideal of Z definition of ideal

38.13 The fact that the constant polynomials of Z[x] form a ring follows directly from the fact that Z forms aring. That the constant polynomials aren’t an ideal of Z[x] can be shown by the fact that 3 is a constantpolynomial, x ∈ Z[x], but 3x is not a constant polynomial.

38.14 We can show that (a) is the smallest ideal containing a by showing that (a) must be a subset of anyother ideal containing a. Let I be any ideal of R containing a. Proof that (a) is a subset of I:

b ∈ (a) assumed→ (∃r ∈ R)(b = ar) definition of (a)→ (∃r ∈ R)(b = ar ∧ ar ∈ I) definition of ideal I→ b ∈ I algebraic replacement

38.15 For each nonzero element a ∈ R, (a) is an ideal containing a. Since a ∈ (a), we know that (a) 6= {0}, soit must be the case that (a) = R. We will prove that R is a field by proving that (a) is a field. First, proofthat each element in (a) contains a multiplicative inverse:

64

(∀a ∈ R− {0})((a) = R) true for reasons outlined above→ (∀a ∈ R− {0})(R ⊆ (a)) partial definition of set equality→ (∀a ∈ R− {0})(x ∈ R→ x ∈ (a)) definition of subset→ (∀a, x ∈ R− {0})(x ∈ (a)) quantificationBecause this last statement is true for all x ∈ R, it is still true when we choose any particular value of x.Choose x to be the unity element of R (which is guaranteed to exist from the text of the exercise):→ (∀a ∈ R− {0})(e ∈ (a)) quantification→ (∀a ∈ R− {0})(∃r ∈ R)(e = ra) definition of membership in (a)

And this last statement is just the quantification of “every nonzero a ∈ R has an inverse in R”. Sowe know that every element of R has an inverse, that R has a unity, and that multiplication on R isassociative (guaranteed from the properties of rings): this means that multiplication is a group on R. Inorder for R to be a field, though, we would have to show that the set of nonzero elements forms a group:i.e., we need to show that R has no zero divisors. But this is true by virtue of the fact that each nonzeroelement a has an inverse:

ab = 0 ∧ a 6= 0 assumed→ a−1ab = a−10 each nonzero a has an inverse→ b = 0 properties of unity and zero

We’ve shown that R is a commutative ring in which the set of nonzero elements form a multiplicativegroup: this is the definition of a field.

38.17 If I 6= (0), then it must contain some nonzero element a. Since the ideal is a ring, it must also contain−a. One of these two must be positive (from the trichotomy property of ordered integral domains). So theset of positive ideals is nonempty, which means that it must contain a least element (from the well-orderedproperty of Z, section 29). Let n represent this least positive integer in I.

Since ideals are subrings by definition, we know that I is closed under addition. This tells us that allmultiples of n are in I: (n) ⊆ I. But it’s also true that I ⊆ (n):

x ∈ I assumed→ (∃r, s ∈ Z)(x = rn+ s, 0 ≤ s < n) division algorithm→ (∃r, s ∈ Z)(x = rn+ s, s = 0) n is the least positive element→ (∃r ∈ Z)(x = rn) property of zero→ x ∈ (n) definition of (n)

And since (n) ⊆ I ∧ I ⊆ (n), we know that (n) = I.

38.18a f ∈ I→ (∃ai ∈ Z)(f = 2a0 + a1x+ a2x

2 + . . .) definition of membership in I→ (∀g ∈ Z[x])(∃ai, bi ∈ Z)(fg = 2a0b0 + horrifying binomial expansion)→ (∀g ∈ Z[x])(fg ∈ I) definition of membership in I

We’ve shown that (∀f ∈ I, g ∈ Z[x])(fgıI) which is the definition of I being an ideal.

38.18b (x+ 2) and (x+ 4) are both members of I and are irreducible in Z[x]. Assume that I is equal to someprincipal ideal (a). By the definition of principal ideal, both (x+2) and (x+4) would have to be multiplesof (a). But since (x + 2) and (x + 4) are irreducible, their only common divisor is 1, so 1 ∈ (a). But 1clearly doesn’t have an even number as the constant term, so 1 6∈ I. So we have shown that no principalideal can contain all the members of I without also containing some members no in I: I cannot be aprincipal ideal.

38.19 We will disprove this by constructing a homomorphism that is not an ideal. Define θ : Z[x] → Z[x] asθ(a0 + a1x + . . . + anx

n) = (a0). The domain of this function is all polynomials, and its codomain is theset of constant polynomials. That θ is a homomorphism is trivially verified, and exercise 38.13 shows thatthe codomain is a subring that is not an ideal.

38.20 From theorem 38.1(b), we know that the kernel of θ is an ideal of F . From example 38.4, we know thatthe only ideals of F are {0} and F itself. If the kernel is {0}, then θ is one-to-one (theorem 28.1(c)). Andif the kernel is F , then by definition of kernel θ(a) = 0 for all a ∈ F .

65

38.21 From the definitions of Q given in section 30, we know that each element in Q can be expressed asab−1, a ∈ Z, b ∈ Z− {0}. Let ab−1 be an arbitrary element of Q:

α(ab−1)= α(a)α(b−1) homomorphism of α= α(a)α(b)−1 theorem 18.2(b)= β(a)β(b)−1 algebraic replacement with α(x) = β(x)= β(a)β(b−1) theorem 18.2(b)= β(ab−1) homomorphism of β

38.22 Let I represent the set of all nilpotent elements in R. We need to prove that I is a subring and also thatI is an ideal.

I is nonempty02 = 0→ 0 ∈ I

I is closed under both operationsa ∈ I ∧ b ∈ I assumed→ (∃m,n ∈ Z)(am = 0 ∧ bn = 0)→ (a+ b)mn = 0 ∧ (ab)mn = 0

This previous step is justified because every term in the binomial expansion of (a+ b)mn contains either am

or bn as a factor, which means that every term is zero. And the sole term of the multiplicative expansion of(ab)nm also obviously contains a factor of am. Note that both of these facts rely on the commutativity of R.

→ (a+ b) ∈ I ∧ (ab) ∈ I definition of I

I contains negatives for each nonzero elementa ∈ I assumed→ (∃m ∈ N)(am = 0) definition of membership in I→ (∃m ∈ N)(a2m = 0)→ (∃m ∈ N)((a2)m = 0)→ (∃m ∈ N)(((−a)2)m = 0) theorem 28.1→ (∃m ∈ N)((−a)2m = 0)→ −a ∈ I definition of membership in I

39 Quotient Rings

39.6 (I + a)(I + b) = (I + b)(I + a) definition of commutativity↔ (I + ab) = (I + ba) definition of addition in R/I↔ (I + ab)− (I + ba) = (I + i ∈ I) zero element of R/I↔ (I + (ab− ba)) = (I + i ∈ I) definition of subtraction in R/I↔ ab− ba ∈ I

So the set of elements a, b for which (I + a) and (I + b) commute is identical to the set of elementsa, b for which ab− ba ∈ I. So R/I is commutative iff ab− ba ∈ I for every pair of elements in I.

39.7 Lemma: x ∈ (n) iff n|x:By the definition of principal ideal, (n) is the set of all multiples of (n). So x ∈ (n) iff rn = x for some r.And this is also the definition of divisibility, so x ∈ (n) iff n|x.

i) Proof by contrapositive that n is prime if (n) is a prime ideal:n is not prime→ (∃a, b ∈ Z)(n = ab ∧ n 6 |a ∧ n 6 |b) definition of prime→ (∃a, b ∈ Z)(n|ab ∧ n 6 |a ∧ n 6 |b) x|x for all x→ (∃a, b ∈ Z)(ab ∈ (n) ∧ a 6∈ (n) ∧ b 6∈ (n)) lemma→ (n) is not a prime ideal definition of prime ideal

66

ii) Proof by contrapositive that (n) is a prime ideal if n is prime:(n) is not a prime ideal→ (∃a, b ∈ Z)(ab ∈ (n) ∧ a 6∈ (n) ∧ b 6∈ (n)) definition of prime ideal→ (∃a, b ∈ Z)(n|ab ∧ n 6 |a ∧ n 6 |b) lemma→ (∃a ∈ Z)(gcd(n, a) 6= 1 ∧ n 6 |a) contrapositive of theorem 13.1→ n is not prime definition of prime

39.8 The proof is an immediate consequence of the defintions of prime ideal and integral domain. Rememberingthat the zeroes ofR/P are all elements of the from (P+p ∈ P ): R/P is an integral domain

↔ (P + a)(P + b) = (P + p ∈ P )→ [(P + a) = (P + p ∈ P ) ∨ (P + b) = (P + p ∈ P )] definition of integraldomain

↔ (P + ab) = (P + p ∈ P )→ [(P + a) = (P + p ∈ P ) ∨ (P + b) = (P + p ∈ P )] definition of multi-plication in R/P

↔ (ab ∈ P )→ [a ∈ P ∧ b ∈ P ] definition of equiva-lence in R/P

↔ P is a prime ideal

39.9 To show that II∩J ≈

I+JJ , define a function θ : I → (I + J)/J as θ(i) = Ji. The domain of this function is

I and the function is equal to 0 whenever i ∈ J . So the kernel of this function is I∩J . By the fundamentalhomomorphism theorem for rings, then, I

I∩J ≈I+JJ .

40 Quotient Rings of F[X]

40.2 p(x) is irreducible in F [x] assumed→ divisors of p(x) must be units or associates definition of irreducible→ (∀g(x) ∈ F [x])(g(x)|p(x)→ g(x)|e ∨ p(x)|g(x)) quantification of previous step

Choose g(x) such that g(x) = gcd(f, p). Since the gcd obviously divides p(x), the right-hand side of theprevious conditional gives us:

→ (gcd(f, p)|e ∨ p(x)| gcd(f, g))→ (gcd(f, p)|e ∨ p(x)| gcd(f, g)) ∧ gcd(f, p)|f(x) definition of the gcd→ (gcd(f, p)|e ∧ gcd(f, p)|f(x)) ∨ (p(x)| gcd(f, g) ∧ gcd(f, p)|f(x)) logical distributivity→ (gcd(f, p)|e ∧ gcd(f, p)|f(x)) ∨ (p(x)|f(x)) transitivity of divisibility→ (gcd(f, p)|e) ∨ (p(x)|f(x)) p ∧ q → p→ gcd(f, p)|e) we are told that p(x) 6 |f(x)→ gcd(f, p) = e) definition of gcd from theorem 36.1

The last step (moving from gcd(f, p)|e to gcd(f, p) = e) is justified by the definition of gcd given intheorem 36.1. By definition of gcd for polynomials, the greatest common divisor must be monic: and bydefinition of monic, the highest coefficient must be e.

40.11 (f) = (g) assumed→ (f) ⊆ (g) ∧ (g) ⊆ (f) definition of set equality→ (∀a ∈ (f))(a ∈ (g)) ∧ (∀b ∈ (g))(b ∈ (f)) definition of subset

f ∈ (f) and g ∈ (g), so we can choose a to be f and choose b to be g.

→ (f ∈ (g)) ∧ (g ∈ (f))→ (∃q ∈ F [x])(f = gq) ∧ (∃r ∈ F [x])(g = fr) definition of (f), (g)→ g|f ∧ f |g definition of divisibility→ g and f are associates definition of associate

This proves that (f) = (g) only if g and f are associates. For the “if” portion of the proof, the same stepscan be used in reverse order. The same justifications can be used for all of the steps except for the step

67

from f ∈ (g) to (f) ⊆ (g). This step is justified because if f ∈ (g), then all multiples of f are in (g), whichmeans that (f) ⊆ (g).

40.12 Theorem 40.3 tells us that there is some m ∈ F [x] such that (m) = I, but the theorem does not guaranteethis this polynomial is monic or that it is unique.

To show that there is at least one monic polynomial a such that (a) = I, note that theorem 36.1 tellsus that gcd(m,m) exists and is monic. From the properties of the gcd, we know that (gcd(m,m) ⊆ (m)(because gcd(m,m)|m) and (m) ⊆ (gcd(m,m) (because m| gcd(m,m)). So (gcd(m,m)) = (m) = I andgcd(m,m) is monic.

To show that this monic polynomial is unique, assume that there a and b are two monic polynomials and(a) = I = (b). Exercise 40.11 showed that a and b must be associates: (∃r, s ∈ F [x])(ar = b ∧ a = bs).Because a and b are of the same degree, both r and s must be constant polynomials of degree 0 (note: thisfact depends on F being an integral domain). And because the highest term of a and b are both 1, thismeans that r = s = e. Therefore a = b.

40.13a If we aren’t supposed to assume that these are polynomials in a field F [x], then this is false. Let R[x]be a subring of Z4[x] consisting of {[0], [2]x, [2]x2, [2]x2 + [2]x}. [2]x is not the same degree as ([2]x2), but([2]x) = ([2]x2):

([2]x) = {[4]x2, [4]x3, [4]x3 + [4]x2} = {[0]}

([2]x2) = {[4]x3, [4]x4, [4]x4 + [4]x3} = {[0]}

This would not work in a field, since there would be no zero divisors.

40.13b False. In the commutative ring of integers, 1 and 2 both have degree 0, but 1 6∈ (2) so (1) 6= (2).

40.14 True for the same reason as above: 2 ∈ (1), deg(2) = deg(1), but (2) 6= (1).

41 Factorization and Ideals

41.1 Let I represent any ideal in Euclidean domain D. Because the division algorithm holds for Euclideandomains, for any two elements a, b ∈ I we find that gcd(a, b) ∈ I. Let x = gcd({i ∈ I}), the greatestelement that divides every element of I. We want to show that (x) = I. First, assume that a ∈ (x).Then a = rx for some r ∈ D (from the definition of (x)). But since x ∈ I, rx ∈ I (from the definitionof an ideal). And rx = a, so a ∈ I. Thus (x) ⊆ I. Next, Assume that a ∈ I. By the division algorithm,a = xq + r, d(r) < d(x) or, alternatively, a − xq = r, d(r) < d(x). But x|a (from being the gcd of everyelement of I) and x|xq (from the definition of division) so x|a− xq = r. But it can’t be the case that x|rand d(r) < d(x) unless r = 0 (otherwise it would be the case that d(r = xm) < d(x), which would makeD not a Euclidean domain). So the fact that we could write a as a = xq+ r means that r = 0 and a = xq.And this means that a ∈ (x). Thus I ⊆ (x). We’ve shown that I ⊆ (x) ⊆ I, so I = (x).

41.2a (b) ⊆ (a) assumed→ b ∈ (a) (b) is an ideal containing b→ (∃r ∈ D)(b = ar) definition of (a)→ a|b definition of divisibility

a|b assumed→ (∃r ∈ D)(b = ar) definition of divisibility→ (∀s ∈ D)(∃r ∈ D)(bs = ars) algebra→ (∀s ∈ D)(∃r ∈ D)(bs = a(rs)) associativity of rings→ (∀s ∈ D)(bs ∈ (a)) definition of (a)→ (b) ⊆ (a) definition of (b)

41.2b a and b are associates assumed→ a|b ∧ b|a definition of associate→ (b) ⊆ (a) ∧ (a) ⊆ (b) exercise 41.2a→ (a) = (b) definition of set equality

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41.3a Principal ideal domains are commutative. So, for all r ∈ D, i ∈ I :

ri = r(ax+ by) = rax+ rby = a(rx) + b(ry)

So ri ∈ I as long as rx and ry are members of D. And because x, y, and r are members of D, closure ofD tells us that rx and ry are in D.

41.3b) This is true by definition, because D is a principal ideal domain.

41.3c) Proof that d is a divisor of both a and b:

I ⊆ (d) true from part (b) of this exercise→ (∀i ∈ I)(∃r ∈ D)(i = rd) definition of subset and of (d)→ (∀x, y ∈ D)(∃r ∈ D)(ax+ by = rd) definition of membership in I

We could choose x = 0, y = e to give us b = rd. Or we could choose x = e, y = 0 to give us x = rd.And because we could make either choice, this implies:

→ (∃r ∈ D)(b = rd) ∧ (∃s ∈ D)(a = sd)→ d|b ∧ d|a definition of divisibility

Proof that d is the greatest such divisor:

c|a ∧ c|b assumed→ c|a ∧ c|b ∧ (d) ⊆ I true from part (b) of this exercise→ c|a ∧ c|b ∧ (∀r ∈ D)(∃i ∈ I)(dr = i) definition of (d)→ c|a ∧ c|b ∧ (∀r ∈ D)(∃x, y ∈ D)(dr = ax+ by) definition of membership in I→ (∃m,n ∈ D)(cm = a ∧ cn = b) ∧ (∀r ∈ D)(∃x, y ∈ D)(dr = ax+ by) definition of divisibility→ (∀r ∈ D)(∃m,n, x, y ∈ D)(dr = cmx+ cny) algebraic replacement

Choose r to be e:→ (∃m,n, x, y ∈ D)(d = cmx+ cny)→ (∃m,n, x, y ∈ D)(d = c(mx+ ny)) distributivity→ c|d definition of divisibility

41.4 From problem 41.3, we know that the set I = {ar + bs : r, s ∈ D} = (gcd(a, b)). So we can prove thate = ar + bs if we can show that e ∈ gcd(a, b). Although we don’t know for certain that e = gcd(a, b),we know that e|a and e|b, so e| gcd(a, b). But this means that er = gcd(a, b) for some r ∈ D which, bydefinition, means that e ∈ (gcd(a, b)).

41.5 If p is irreducible, then its only divisors are units and associates. Let g = gcd(p, a). Since g is a divisorof p, then g must either be a unit (g|e) or an associate of p (g|p ∧ p|g). Assume g is an associate. Thenp|g (from being an associate) and g|a (from being the gcd(p, a)). So, because divisibility is transitive, p|a.Assume that g is a unit. Then g|e, which implies:

g|e assumed→ g|e ∧ (∃x, y ∈ D)(g = px+ ay) lemma 2 of theorem 37.1→ (∃r ∈ D)(gr = e) ∧ (∃x, y ∈ D)(g = px+ ay) definition of divisibility→ (∃r ∈ D)(gr = e) ∧ (∃x, y ∈ D)(gr = (px+ ay)r) algebra→ (∃r, x, y ∈ D)(e = pxr + ayr) algebraic replacement, distributivity→ (∃r, x, y ∈ D)(be = bpxr + bayr) algebra→ (∃r, x, y ∈ D)(b = pbxr + abyr) commutativity

We are told p|ab, so (∃S ∈ D)(ps = ab). Using this replacement:→ (∃r, s, x, y ∈ D)(b = pbxr + psyr) algebraic replacement→ (∃r, s, x, y ∈ D)(b = p(bxr + syr)) distributivity of rings→ p|b definition of divisibility

So the fact that g is either a unit or an associate implies that either p|a or p|b.

41.6 Proof by induction. Let p be irreducible and let S = {n : p|a1a2 . . . an → (∃i)(p|ai)}. 1 ∈ S for trivialreasons and exercise 41.5 showed that 2 ∈ S. Now, assume that n ∈ S. Then p|a1a2 . . . (anan+1) impliesthat p either divides ai for some i < n, or p|anan+1. And if the latter is true, then 2 ∈ S tells us that p|ano p|an+1. So n+ 1 ∈ S. Therefore, by induction, S = N.

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41.8 Exercise 41.7 shows that all nonzero, non-unit elements of a principal ideal domain can be written as aproduct of irreducible elements. Exercise 41.6 can be used to prove uniqueness via the same steps as theproof of the fundamental theorem of arithmetic on pp70-71.

41.9 a,b,c) Every field is a Euclidean domain (example 38.4)

d,g,h) F [x] is a Euclidean domain for any field F . (p175, below definition)

e) This is a Euclidean domain (example 37.2)

f) This is an integral domain but not a unique factorization domain (example 37.1)

41.10 {1,3,4,5,9}

42 Simple Extensions

42.a This is an expanded proof of theorem 42.1. Define a function θ : F [x] → F [a] by θ(p(x)) = p(a). Thisfunction can easily be shown to be a homomorphism, so by the fundamental homomorphism theory forquotient rings, we know that:

F [x]/Ker(θ) ≈ θ(F [x])

θ(F [x]), the range of θ, is the set of all linear combinations of elements of F and the element a. This rangeis, by definition, the field extension F (a). So we see that:

F [x]/Ker(θ) ≈ F (a)

The kernel of θ is the set of all polynomials in F [x] for which p(a) = 0. And we know that kernels arealways normal subgroups and ideal subrings. And since F [x] is a principal ideal domain if F is a field, weknow that the kernel is an ideal domain: it is equal to (p(x)) for some p(x) ∈ F [x]. So:

F [x]/(p(x)) ≈ F (a)

And since F (a) is a field, that means that F [x]/(p(x)) is a field and therefore, by theorem 40.1, p(x) isirreducible over F .

42.2 R(a+ bi) ⊆ Cx ∈ R(a+ bi) assumed→ (∃r ∈ R)(x = r(a+ bi)) unique representation from theorem 42.3 corollary 2→ (∃r ∈ R)(x = ra+ rbi) distributive property of R→ (∃ra, rb ∈ R)(x = (ra) + (rb)i) multiplicative closure of R→ x ∈ C definition of C

C ⊆ R(a+ bi) : x ∈ C assumed→ (∃m,n ∈ R)(x = m+ ni) definition of C→ (∃m,n ∈ R)(x = n

b (a+ bi) + mb−nab(a2−b2) (a+ bi)2) tons of algebra

→ x ∈ R(a+ bi) unique representation from theorem 42.3 corollary 2Note that the requirement that a+bi is imaginary forces b to be nonzero, so the fractions in the penultimatestep are all defined.

42.3 If Q(√

2) = Q(√

3), then every element of Q(√

2) (including√

2 itself) would have to be expressible as a2nd-degree linear combination of p

r

√3 (corollary 2 of theorem 42.3). Proof by contradiction that this is

not possible:

Q(√

2) = Q(√

3) assumed

→ Q(√

2) ⊆ Q(√

3) partial definition of set equality

→√

2 ∈ Q(√

3)√

2 is an element of Q(√

2)

→ (∃r, s ∈ Q)(√

2 = r√

3 + s) theorem 42.3 corollary 2

→ (∃r, s ∈ Q)(2 = 3r2 + 2rs√

3 + s2) algebra: square both sides

→ (∃r, s ∈ Q)(2− 3r2 − s2 = 2rs√

3) algebra

→ (∃r, s ∈ Q)( 2−3r2−s22rs =

√3) algebra

The left-hand side of this equation is also a rational number, so:

→ (∃t ∈ Q)(t =√

3)

→√

3 is rational

70

And we know, from proving it a thousand times in every class since discrete math, that√

3 is not rational.

42.4 a) 0 + 7a

b) −5− ac) 8 + 2a

d) − 16 + 1

6a

42.5 a) x2 − 2

b) x4 − 4

c) x4 − 2x2 − 9

d) x2 − 2√

2x+ 3

42.6 f(x) ∈ Q 3√

5

→ (∃ai ∈ Q)(f(x) = a0 + a1( 3√

5) + . . .+ an( 3√

5)n)

→ (∃ai ∈ Q)(f(x) = (∑a3k( 3√

5)3k) + (∑a3k+1( 3

√5)3k+1) + (

∑a3k+2( 3

√5)3k+2))

→ (∃ai ∈ Q)(f(x) = (∑a3k(5)k) + (

∑a3k+1(5)k) 3

√5 + (

∑a3k+2(5)k) 3

√25)

→ (∃a, b, c ∈ Q)(f(x) = a+ b 3√

5 + c 3√

25)

42.7 If a2 is algebraic over F , then there exists some function f(x) such that f(a2) = 0. Define a new functiong(x) = f(x2). Then g(a) = f(a2) = 0, so a is algebraic over F .

42.8 The field of quotients of F [x] is defined in section 30 as the unique smallest field that contains a solutionin F [x] for the equation f(x)λ(x) = g(x) for all nonzero elements f(x), g(x) in F [x]. This field is clearlythe field containing all elements of the form f(x)/g(x).

42.12 By theorem 42.5, we can prove that F (α) ≈ F (β) just by showing that there exists an isomorphismF ≈ F . And this is trivially done by the function θ(a) = a.

42.12 (alternate proof) From theorem 42.1, we know that Q(α) ≈ Q[x]/(p(x) = x2−2) and Q(β) ≈ Q[x]/(q(x) =x2−4x+ 2). But note that x2−4x+ 2 is equivalent to (x−2)2−2. So (p(x)) ≈ (q(x)) under the mappingθ(f(x)) = f(x− 2). And so, by the definition of extension at the start of the chapter, Q(α) ≈ Q(β).

42.13 x3 − 5 = (x− 3√

5)(x+ x 3√

5 + 3√

25), which has two complex roots.

44 Splitting Fields

44.1 1,5,7, and 11 are roots. Z12 is not a field (it has zero divisors), so theorem 44.1 does not apply.

44.2 (x− [3])2(x− [1]) = ([1]x3 + [−7]x2 + [15]x+ [−9]) = ([1]x3 + [3]x+ [1])

44.3 (x− i)2(x+ 1) = (x3 − ix2 + x− i) = (x3 + x)− (x2 + 1)i

44.5 Use theorem 44.7 to generate the set of all possible rational roots, and then prove by exhaustion that noneof these possible rational roots are roots of 12x3 − 3x+ 2.

44.6 a) f(x) = (x− 32 )(x+ 1

2 )(x+ 1), roots are 32 ,−

12 ,−1.

b) roots are 2, 12 ,13 (found using theorem 44.7).

c) f(x) = 2(x− 12 )(x2 + 4), 1

2 is the only root in Q.

44.7 a) f(x) = (x− 1)(x2 − 5), 1 is the root in Qb) f(x) = 3(x− 2

3 )(x2 + 1), 23 is the root in Q

c) f(x) = x(x2 − 2x+ 2), 0 is the root in Q

44.8 a) f(x) = (x− 1)(x2 − 5), 1,√

5,−√

5 are the roots in Rb) f(x) = 3(x− 2

3 )(x2 + 1), 23 is the root in R

c) f(x) = x(x2 − 2x+ 2), 0 is the root in R

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44.9 a) f(x) = (x− 1)(x2 − 5), 1,√

5,−√

5 are the roots in Cb) f(x) = 3(x− 2

3 )(x2 + 1), 23 are the roots in C

c) f(x) = x(x2 − 2x+ 2), 0 is the only root in C

44.11 If f(x) has an imaginary root of degree two, then (a+bi)2 is a root for some a, b ∈ R. From exercise 42.7,it is also the case that (a+ bi) is a root. And from the corollary to theorem 44.5, the complex conjugatesof (a+ bi)2 and (a+ bi) are also roots. So the degree of f(x) is at least 4.

44.12 x2 + (2 + 2i)x+ 2i = (x+ (1 + i))2, so −1− i is a root of multiplicity 2.

44.13 [1]x3 + [1]x2 + [1]x+ [1]

44.14 F (c1, . . . , cn) is by definition the smallest extension of F that contains all of the elements (c1, . . . , cn).Since F (c1, . . . , cn) is the smallest field containing those roots, we know that F (c1, . . . , cn) ⊆ H. So if it isalso true that H ⊆ F (c1, . . . , cn), then H = F (c1, . . . , cn).

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