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## A Book of Abstract Algebra, Second Edition (1982)

CHAPTER 2

A3This is not an operation on , since a * b is not uniquely defined for any a, b, a ≠ 0, and b ≠ 0. If a ≠ 0 and b ≠ 0, then the equation x2a2b2 = 0 has two roots—namely, x = a * b = ± ab.

B7

(ii)Associative law:

(iii)Identity element: To find an identity element (if one exists), we try to solve the equation x * e = x for e:

Cross multiplying gives xe = x2 + xe + x, and thus 0 = x(x + 1). Consequently, this equation can be solved only for x = 0 and x = – 1. For positive values of x there is no solution.

C3We shall examine only one of the operations—namely, the one whose table is

To say that the operation is associative is to say that the equation

x * (y * z) = (x * y) * z

is true for all possible choices of x, of y, and of z. Each of the variables may be equal to either a or b, yielding eight cases to be checked:

1.a * (a * a) = a * b = a, (a * a) * a = b * a = a

2.a * (a * b) = a * a = b, (a * a) * b = b * b = a

3.a * (b * a) = a * a = b, (a * b) * a = a * a = b

4.a * (b * b) = a * a = b, (a * b) * b = a * b = a

5.b * (a * a) = b * b = a, (b * a) * a = a * a = b

6.b * (a * b) = b * a = a, (b * a) * b = a * b = a

7.b * (b * a) = b * a = a, (b * b) * a = a * a = b

8.b * (b * b) = b * a = a, (b * b) * b = a * b = a

Since a * (a * b) ≠ (a * a) * b, * is not associative.

By the way, this operation is commutative, since a * b = a = b * a. Using commutativity, we need not check all eight cases above; in fact, equality in cases 1, 3, 6, and 8 follows using commutativity. For example, in case 3, it follows from commutativity that a * (b * a) = (b * a) * a = (a * b) * a. Thus, for commutative operations, only four of the eight cases need to be checked. If you are able to show that it is sufficient to check only cases 2 and 4 for commutative operations, your work will be further reduced.

We have checked only one of the 16 operations. Check the remaining 15.

CHAPTER 3

A4(ii) Associative law:

(iii) Identity element: Solve x * e = x for e.

If

then e(1 − x2) = 0, so e = 0. Now check that x * 0 = 0 * x = x.

(iv) Inverses: Solve x * x′ = 0 for x′. (You should get x′ = − x.) Then check that x * (− x) = 0 = (− x) * x.

B2(ii) Associative law:

(a, b) * [(c, d) * (e, f)] = (a, b) * (ce, de + f) = (ace, bce + de + f)

[(a, b) * (c, d)] * f) = (ac, bc + d) * (e, f) = (ace, bce + de + f)

(iii) Identity element: Solve (a, b) * (e1, e2) = (a, b) for (e1, e2). Suppose

(a, b) * (e1, e2) = (ae, be1 + e2) = (a, b)

This implies that ae1 = a and be1 + e2 = b, so e1 = 1 and e2 = 0. Thus, (e1, e2) = (l, 0). Now check that (a, b) * (1, 0) = (1, 0) * (a, b) = (a, b).

(iv) Inverses: Solve (a, b) * (a′, b′) = (1, 0) for (x′, b′). You should get a′ = 1/a and b′ = − b/a. Then check.

G1 The equation a4 = a1 + a3 means that the fourth digit of every codeword is equal to the sum of the first and third digits. We check this fact for the eight codewords of C1 in turn: 0 = 0 + 0, 1 = 0 + 1, 0 = 0 + 0, 1 = 0 + 1, 1 = 1 + 0, 0=1 + 1, 1 = 1 + 0, and 0 = 1 + 1.

G2(a) As stated, the first three positions are the information positions. Therefore there are eight codewords; omitting the numbers in the last three positions (the redundancy positions), the codewords are 000, 001, 010, 011, 100, 101, 110, and 111. The numbers in the redundancy positions are specified by the parity-check equations given in the exercise. Thus, the complete codewords are 000000, 001001, …. (Complete the list.)

G6Let ak, bk, xk denote the digits in the kth position of a, b, and x, respectively. Note that if xkak and xkbk, then ak = bk. But if xkak and xk = bk, then akbk. And if xk = ak and xkbk, then akbk. Finally, if xk = ak and xk = bk, then ak = bk. Thus, a differs from b in only those positions where a differs from x or b differs from x. Since a differs from x in t or fewer positions, and b differs from x in t or fewer positions, a cannot differ from b in more than 2t positions.

CHAPTER 4

A3Take the first equation, x2a = bxc1, and multiply on the right by c:

x2ac = (bxc 1)c = bx(c 1c) = bxe = bx

From the second equation, x2ae = x(xac) = xacx. Thus,

xaca = bx

By the cancellation law (cancel x on the right),

xac = b

Now multiply on the right by (ac) 1 to complete the problem.

C1From Theorem 3, a 1 b 1 = (ba) 1 and b 1a 1 = (ab) 1.

D2From Theorem 2, if (ab)c = e, then c is the inverse of ab.

G3Let G and H be abelian groups. To prove that G × H is abelian, let (a, b) and (c, d) be any two elements of G × H (so that aG, cG, bH, and dH) and show that (a, b) · (c, d) = (c, d) · (a, b).

PROOF:

CHAPTER 5

B5Suppose f, gH. Then df/dx = a and dg/dx = b for constants a and b. From calculus, d(f + g)dx = df/dx + dg/dx = a + b, which is constant. Thus, f + gH.

C5PROOF: (i) K is closed under products. Let x, yK. Then there are integers n, m > 0 such that xnH and ymH. Since H is a subgroup of G, it is closed under products—and hence under exponentiation (which is repeated multiplication of an element by itself). Thus (xn)mH and (xm)nH. Set q = mn. Since G is abelian, (xy)q = xqyq = xmnymn = (xn)m(ym)nH since both (xn)m and (ym)n are in H. Complete the problem.

D5S = {a1 …, an} has n elements. The n products a1a1, a1a2, …, a1an are elements of S (why?) and no two of them can be equal (why?). Hence every element of S is equal to one of these products. In particular, a1 = a1ak for some k. Thus, a1e = a1ak, and hence e = ak. This shows that eS. Now complete the problem.

D7(a) Suppose xK. Then

(i) if aH, then xax 1H, and

(ii) if xbx 1H, then bH.

We shall prove that x 1K: we must first show that if aH, then x 1a(x 1) 1 = x 1 axH. Well, a = x(x 1ax)x 1 and if x(x 1 ax)x 1H, then x 1axH by (ii) above. [Use (ii) with x1ax replacing b.] Conversely, we must show that if x 1axH, then aH. Well, if x 1axH, then by (i) above, x(x 1ax)x 1 = aH.

E7We begin by listing all the elements of 2 × 4 obtained by adding (1, 1) to itself repeatedly: (1, 1); (1, 1) + (1, 1) = (0, 2); (1, 1) + (1, 1) + (1, 1) = (1, 3); (1, 1) + (1, 1) + (1, 1) + (1, 1) = (0, 0). If we continue adding (1, 1) to multiples of itself, we simply obtain the above four pairs over and over again. Thus, (1, 1) is not a generator of all of 2 × 4.

This process is repeated for every element of 2 × 4. None is a generator of 2 × 4; hence 2 × 4 is not cyclic.

F1The table of G is as follows:

Using the defining equations a2 = e, b3 = e, and ba = ab2, we compute the product of ab and ab2 in this way:

(ab)(ab2) = a(ba)b2 = a(ab2)b2 = a2b3b = eeb = b

Complete the problem by exhibiting the computation of all the table entries.

H3Recall the definition of the operation + in Chapter 3, Exercise F: x + y has s in those positions where x and y differ, and 0s elsewhere.

CHAPTER 6

A4From calculus, the function f(x) = x3 – 3x is continuous and unbounded. Its graph is shown below. [f is unbounded because f(x) = x(x2 − 3) is an arbitrarily large positive number for sufficiently large positive values of x, and an arbitrarily large negative number (large in absolute value) for sufficiently large negative values of x.] Because f is continuous and unbounded, the range of f is . Thus, f is surjective. Now determine whether f is injective, and prove your answer.

Graph of f (x) = x3 − 3x

A6 f is injective: To prove this, note first that if x is an integer then f(x) is an integer, and if x is not an integer, then f(x) is not an integer. Thus, if f(x) = f(y), then x and y are either both integers or both nonintegers. Case 1, both integers: then f(x) = 2x, f(y) = 2y, and 2x = 2y; so x = y. Case 2, both nonintegers: ⋯ (Complete the problem. Determine whether f is surjective.)

F5Let A = {a1, a2, …, an). If f is any function from A to A, there are n possible values for f(ax), namely, a1, a2, …, an. Similarly there are n possible values for f(a2). Thus there are n2 pairs consisting of a value of f(a1) together with a value of f(a2). Similarly there are n3 triples consisting of a value of f(a1), a value of f(a2), and a value of f(a3). We may continue in this fashion and conclude as follows: Since a function f is specified by giving a value for f(a1), a value for f(a2), and so on up to a value for f(an), there are nn functions from A to A. Now, by reasoning in a similar fashion, how many bijective functions are there?

H1The following is one example (though not the only possible one) of a machine capable of carrying out the prescribed task:

A = {a, b, c, d} S = {s0, s1, s2, s3, s4}

The next-state function is described by the following table:

To explain why the machine carries out the prescribed function, note first that the letters b, c, and d never cause the machine to change its state—only a does. If the machine begins reading a sequence in state s0, it will be in state s3 after exactly three a’s have been read. Any subsequent a’s will leave the machine in state s4. Thus, if the machine ends in s3, the sequence has read exactly three a’s. The machine’s state diagram is illustrated below:

I4M1 has only two distinct transition functions, which we shall denote by To and Te (where o may be any sequence with on odd number of 1s, and e any sequence with an even number of 1s). To and Te may be described as follows:

 To(s0), = s1, To(s1) = s0 Te(s0), = s0, Te(s1) = s1

Now, TeTo = Toe by part 3. Since e is a sequence with an even number of 1s and o is a sequence with an odd number of 1s, oe is a sequence with an odd number of 1s; hence Toe = To. Thus, TeTo = To. Similarly

TeTe = Te, ToTe = To, and ToTo = Te

In brief, the table of (M1) is as follows:

The table shows that (M1) is a two-element group. The identity element is Te, and To is its own inverse.

CHAPTER 7

D2fn + m is the function defined by the formula

fn + m(x) = x + n + m

Use this fact to show that fnfm = fn + m.

In order to show that f n is the inverse of fn, we must verify that fnf n = ε and f nfn = ε. We verify the first equation as follows: f n(x) = x + (− n); hence

[fnf n](x) = fn(f n(x)) = fn(xn) = xn + n = x = ε(x)

Since [fnf n](x) = ε(x) for every x in , it follows that fnf n = ε.

E3To prove that f1/a, b/a is the inverse of fa, b, we must verify that

fa, bf1/a, b/a = ε and f1/a, b/afa, b = ε

To verify the first equation, we begin with the fact that f1/a, b/a(x) = x/ab/a. Now complete the problem.

H2If f, gG, then f moves a finite number of elements of A, say a1, …, an, and g moves a finite number of elements of A, say b1, …, bm. Now if x is any element of A which is not one of the elements a1, …, an, b1, …, bm, then

[fg](x) = f(g(x)) = f(x) = x

Thus, fg moves at most the finite set of elements {a1, …, an, b1, …, bm}.

CHAPTER 8

A1(e)

A4(f)γ3 = γγγ = (1 2 3 4 5); α 1 = (4 1 7 3); thus; γ3α 1 = γ3α 1 = 1 (1 2 3 4 5) ∘ (4 1 7 3) = (1 7 4 2 3 5)

B2

and so on. Note that α2(a1) = a3, α2(a2) = a4, …, α2(as 2) = as. Finally, α2(as 1) = al and α2(as) = a2. Thus, α2(ai) = a? Complete the problem using addition modulo s, page 27.

B4Let α = (a1 a2as) where s is odd. Then α2 = (a1 a3 a5as a2 a4as 1). If s is even, then α2 = ? Complete the problem.

E2If α and β are cycles of the same length, α = (a1as) and β = (b1bs, let π be the following permutation: π(k) = bi for i = 1, …, s and π(k) = k for ka1, …, as, b1 …, bs. Finally, let π map distinct elements of {b1 …, bs) − {a1, …, as) to distinct elements of {a1, …, as} – {b1, …, bs}. Now complete the problem, supplying details.

F1 when k is a positive integer, k < s

For what values of k can you have αk = ε?

H2Use Exercise H1 and the fact that (ij) = (1i)(1j)(1i).

CHAPTER 9

C1The group tables for G and H are as follows:

G and H are not isomorphic because in G every element is its own inverse (VV = I, HH = I, and DD = I), whereas in H there are elements not equal to their inverse; for example, (-i)(-i) = −1 ≠ 1. Find at least one other difference which shows that G H.

C4 The group tables for G and H are as follows:

G and H are isomorphic. Indeed, let the function f : GH be defined by

By inspection, f transforms the table of G into the table of H, Thus, f is an isomorphism from G to H.

ElShow that the function f: E given by f(n) = 2n is bijective and that f(n + m) = f(n) + f(m).

F1Check that (2 4)2 = ε, (1 2 3 4)4 = ε, and (1 2 3 4)(2 4) = (2 4)(1 2 3 4)3. Now explain why GG′.

H2Let fG: G1G2 and fH : H1H2 be isomorphisms, and find an isomorphism f : G1 × H1G2 × H2.

CHAPTER 10

A1(c)If m < 0 and n < 0, let m = −k and n = l, where k, l > 0. Then m + n = − (k + l). Now,

am = a k = (a 1)k

and

an = a 1 = (a 1)l

Hence

aman = (a 1)k (a 1)l = (a 1)k + l = a (k + l) = am + n

B3The order of f is 4. Explain why.

C4For any positive integer k, if ak = e, then

Conversely, if (bab 1k = bakb 1 = e, then ak = e. (Why?) Thus, for any positive integer k, ak = e iff (bab 1)k = e. Now complete the problem.

D2 Let the order of a be equal to n. Then (ak)n = ank = (an)k = ek = e. Now use Theorem 5.

F2The order of a8 is 3. Explain why.

H2The order is 24. Explain why.

CHAPTER 11

A6If k is a generator of , this means that consists of all the multiples of k; that is, k, 2k, 3k, etc., as well as 0 and −k, −2k, −3k, etc.:

B1Let G be a group of order n, and suppose G is cyclic, say G = ⟨a⟩. Then a, the generator of G, is an element of order n. (This follows from the discussion on the first two pages of this chapter.) Conversely, let G be a group of order n (that is, a group with n elements), and suppose G has an element a of order n. Prove that G is cyclic.

C4By Exercise B4, there is an element b of order m in ⟨a⟩, and bCm. Since Cm is a subgroup of ⟨a⟩, which is cyclic, we know from Theorem 2 that Cm is cyclic. Since every element x in Cm satisfies xm = e, no element in Cm can have order greater than m. Now complete the argument.

C7First, assume that ord (ar) = m. Then (ar)m = arm = e. Use Theorem 5 of Chapter 10 to show that r = kl for some integer l. To show that l and m are relatively prime, assume on the contrary that l and m have a common factor q; that is,

l = jq m = hq h < m

Now raise ar to the power h and derive a contradiction. Conversely, suppose r = lk where l and m are relatively prime. Complete the problem.

D6Let (a) be a cyclic group of order mn. If ⟨a⟩ has a subgroup of order n, this subgroup must be cyclic, and generated by one of the elements in ⟨a⟩. Say ⟨ak⟩ is a subgroup of order n. Then ak has order n, so (ak)n = akn = e. By Theorem 5 of Chapter 10, kn = mnq for some integer q (explain why); hence k = mq and therefore ak = (am)q ∈ ⟨am⟩.

Use this information to show that (i) ⟨a⟩ has a subgroup of order n and (ii) ⟨a⟩ has only one subgroup of order n.

F3If there is some j such that am = (aj)k = ajk, then e = am(ajk) 1 = am jk; so by Theorem 5 of Chapter 10, mjk = nq for some integer q. (Explain why.) Thus, m = jk + nq. Now complete the problem, supplying all details.

CHAPTER 12

A2Let x be any rational number (any element of ), and suppose xAi. and xAj, where i and j are integers. Then ix < i + 1 and jx < j + 1. Now if i < j, then i + 1 ≥ j; so x < i + 1 implies that x < j, which is a contradiction. Thus we cannot have i < j, nor can we have j < i; hence i = j. We have shown that if Ai. and Aj have a common element x, then Ai = Ai: this is the first condition in the definition of partition. For the second condition, note that every rational number is an integer or lies between two integers: if ix < i + 1, then x is in Ai

C1Ar consists of all the points (x, y) satisfying y = 2x + r; that is, Ar is the line with slope 2 and y intercept equal to r. Thus, the classes of the partition are all the lines with slope 2.

For the corresponding equivalence relation, we say that two points (x, y) and (u, υ) are “equivalent,” that is,

(x, y) ~ (u, υ)

if the two points lie on the same line with slope 2:

Complete the solution, supplying details.

D5If ab 1 commutes with every x in G, then we can show that ba 1 commutes with every x in G:

ba 1 x = (x 1ab 1)’ 1 = (ab 1x 1) 1 (why?)

CHAPTER 13

B1Note first that the operation in the case of the group is addition. The subgroup (3) consists of all the multiples of 3, that is,

⟨3⟩ = {. . ., −9, −6, −3, 0, 3, 6, 9, …}

The cosets of (3) are (3) + 0 = (3), as well as

⟨3⟩ + 1 = {. . ., −8, −5, −2, 1, 4, 7, 10, …}

⟨3⟩ + 2 = {. . ., −7, −4, −1, 2, 5, 8, 11, …}

Note that ⟨3⟩ + 3 = ⟨3⟩, ⟨3⟩ + 4 = ⟨3⟩ + 1, and so on; hence there are only three cosets of ⟨3⟩, namely,

⟨3⟩ = ⟨3⟩ + 0 ⟨3⟩ + 1 ⟨3⟩ + 2

C6 Every element a of order p belongs in a subgroup ⟨a⟩. The subgroup ⟨a⟩ has p – 1 elements (why?), and each of these elements has order p (why?). Complete the solution.

D6For one part of the problem, use Lagrange’s theorem. For another part, use the result of Exercise F4, Chapter 11.

E4To say that aH = Ha is to say that every element in aH is in Ha and conversely. That is, for any hH, there is some kH such that ah = ka and there is some lH such that ha = al. (Explain why this is equivalent to aH = Ha.) Now, an arbitrary element of a 1H is of the form a 1h = (h 1a) 1. Complete the solution.

J3O(1) = O(2) = {1, 2, 3, 4}; G1 = {ε, β}; G2 = {ε, αβα}. Complete the problem, supplying details.

CHAPTER 14

A6We use the following properties of sets: For any three sets X, Y and Z,

(i) (XY) ∩ Z = (XZ) ∪ (YZ)

(ii) (XY) ∩ Z = (XZ) − (YZ)

Now here is the proof that h is a homomorphism: Let C and D be any subsets of A; then

Now complete, using (i) and (ii).

C2Let f be injective. To show that K = {e}, take any arbitrary element xK and show that necessarily x = e. Well, since xK, f(x) = e = f(e). Now complete, and prove the converse: Assume K = {e} ….

D6Consider the following family of subsets of G: {Hi : iI}, where each Hi is a normal subgroup of G. Show that H = Hi is a normal subgroup of G. First, show that H is closed under the group operation: well, if a, bH, then aHi and bHi for every iI. Since each Hi is a subgroup of G, abHi for every iI; hence ab Hi. Now complete.

E1If H has index 2, then g is partitioned into exactly two right cosets of H; also G is partitioned into exactly two left cosets of H. One of the cosets in each case is H.

E6First, show that if x ∈ S and y ∈ S, then xy ∈ S. Well, if x ∈ S, then x ∈ Ha = aH for some a ∈ G. And if y ∈ S, then y ∈ Hb = bH for some b ∈ G. Show that xyH(ab) and that H(ab) = (ab)H and then complete the problem.

I4It is easy to show that aHa 1H. Show it. What does Exercise I2 tell you about the number of elements in aHa 1

I8Let X = {aHa 1: aG} be the set of all the conjugates of H, and let Y = {aN : aG} be the set of all the cosets of N. Find a function f : XY and show that f is bijective.

CHAPTER 15

C4Every element of G/H is a coset Hx. Assume every element of G/H has a square root: this means that for every xG,

Hx = (Hy)2

for some yG. Avail yourself of Theorem 5 in this chapter.

D4Let H be generated by {h1, …, hn and let G/H be generated by {Ha1, …, Ham). Show that G is generated by

{a1, …, am, h1, …, hn}

that is, every x in G is a product of the above elements and their inverses.

E6Every element of / is a coset + (m/n).

G6If G is cyclic, then necessarily Gp2. (Why?)

If G is not cyclic, then every element xe in G has order p. (Why?) Take any two elements ae and be in G where b is not a power of a. Complete the problem.

CHAPTER 16

D1Let f Aut(G); that is, let f be an isomorphism from G onto G. We shall prove that f 1 ∈ Aut(G); that is, f 1 is an isomorphism from G onto G. To begin with, it follows from the last paragraph of Chapter 6 that f 1 is a bijective function from G onto G. It remains to show that f 1 is a homomorphism. Let f (c) = a and f 1(d) = b, so that c = f(a) and d = f(b). Then cd = f(a)f(b) = f(ab), whence f 1(cd = ab. Thus,

f 1(cd) = ab = f 1(c)f 1(d)

which shows that f 1 is a homomorphism.

F2If a, bHK, then a = h1k1 and b = h2k2, where h1, h2H and k1, k2K. Then ab = h1k1h2k2 = .

G3Note that the range of h is a group of functions. What is its identity element?

H1From calculus, cos(x + y) = cos x cos y − sin x sin y, and sin (x + y) = sin x cos y + cos x sin y.

L4 The natural homomorphism (Theorem 4, Chapter 15) is a homomorphism f : GG/⟨a⟩ with kernel ⟨a⟩. Let S be the normal subgroup of G/⟨a⟩ whose order is pm 1. (The existence of S is assured by part 3 of this exercise set.) Referring to Exercise J, show that S* is a normal subgroup of G, and that the order of S* is pm.

CHAPTER 17

A3We prove that is associative:

Thus, (a, b)[(c, d)(p, q)] = [(a, b)(c, d)](p, q).

B2A nonzero function f is a divisor of zero if there exists some nonzero function g such that fg = 0, where 0 is the zero function (page 46). The equation fg = 0 means that f(x)g(x) = 0(x) for every x. Very precisely, what functions f have this property?

D1For the distributive law, refer to the diagram on page 30, and show that A ∩ (B + C) = (AB) + (AC):

B + C consists of the regions 2, 3, 4, and 7; A ∩ (B + C) consists of the regions 2 and 4. Now complete the problem.

E3

G4(a, b) is an invertible element of A × B iff there is an ordered pair (c, d) in A × B satisfying (a, b) · (c, d) = (1, 1). Now complete.

H6If A is a ring, then, as we have seen, A with addition as the only operation is an abelian group: this group is called the additive group of the ring A. Now, suppose the additive group of A is a cyclic group, and its generator is c. If a and b are any two elements of A, then

a = c + c + ⋯ + c (m terms)

and

b = c + c + ⋯ + c (n terms)

for some positive integers m and n.

J2If ab is a divisor of 0, this means that ab ≠ 0 and there is some x ≠ 0 such that abx = 0. Moreover, a ≠ 0 and b ≠ 0, for otherwise ab = 0.

M3Suppose am = 0 and bn = 0. Show that (a + b)m + n = 0. Explain why, in every term of the binomial expansion of (a + b)m + n, either a is raised to a power ≥ m, or b is raised to a power ≥ n.

CHAPTER 18

A4From calculus, the sum and product of continuous functions are continuous.

B3The proof hinges on the fact that if k and a are any two elements of n, then

ka = a + a + · + a (k terms)

C4If the cancellation law holds in A, it must hold in B. (Why?) Why is it necessary to include the condition that B contains 1?

C5Let B be a subring of a field F. If b is any nonzero element of B, then b 1 is in F, though not necessarily in B. (Why?) Complete the argument.

E5f(x, y)f(u, υ) = = f[x, y)(u, υ)] =

Complete the problem.

H3If anJ and bmJ, show that (a + b)n + mJ. (See the solution of Exercise M3, Chapter 17.) Complete the solution.

CHAPTER 19

E1To say that the coset J + x has a square root is to say that for some element y in A, J + x = (J + y)(J + y) = J + y2.

E6A unity element of A/J is a coset J + a such that for any xA,

(J + a)(J + x) = J + x and (J + x)(J + a) = J + a

G1To say that aJ is equivalent to saying that J + aJ; that is, J + a is not the zero element of A/J. Explain and complete.

CHAPTER 20

E5Restrict your attention to A with addition alone, and see Chapter 13, Theorem 4.

E6For n = 2 you have

Prove the required formula by reasoning similarly and using induction: assume the formula is true for n = k, and prove for n = k + 1.

CHAPTER 21

B5Use the product (a − 1)(b − 1).

C8In the induction step, you assume the formula is true for n = k and prove it is true for n = k + 1. That is, you assume

Fk + 1 Fk + 2FkFk + 3 = (–1)k

and prove

Fk + 2 Fk + 3Fk + 1 Fk + 4 = (− 1)k + 1

Recall that by the definition of the Fibonacci sequence, Fn + 2 = Fn + 1 + Fn for every n > 2. Thus, Fk + 2 = Fk + 1 + Fk and Fk + 4 = Fk + 3 + Fk + 2. Substitute these in the second of the equations above.

E5An elegant way to prove this inequality is by use of part 4, with a + b in the place of a, and |a| + |b| in the place of b.

E8This can be proved easily using part 5.

F2You are given that m = nq + r and q = kq1 + r1 where 0 ≤ r < n and 0 ≤ r1 < k. (Explain.) Thus,

m = n(kq1 + r1 + r = (nk)q1 + (nr1 + r)

You must show that nr1 + r < nk, (Why?) Begin by noting that k − r1 > 0; hence k – r1 ≥ 1, so n(k – r1) ≥ n.

G5For the induction step, assume k · a = (k · 1)a, and prove (k + 1) · a = [(k + 1) · 1]a. From (ii) in this exercise, (k + 1) · 1 = k · 1 + 1.

CHAPTER 22

B1Assume a > 0 and a|b. To solve the problem, show that a is the greatest common divisor of a and b. First, a is a common divisor of a and b: a|a and a|b. Next, suppose t is any common divisor of a and b: t|a and t|b. Explain why you are now done.

D3From the proof of Theorem 3, d is the generator of the ideal consisting of all the linear combinations of a and b.

E1Suppose a is odd and b is even. Then a + b is odd and ab is odd. Moreover, if t is any common divisor of ab and a + b, then t is odd. (Why?) Note also that if ris a common divisor of ab and a + b, then t divides the sum and difference of a + b and ab.

F3 If l = lcm(ab, ac), then l = abx = acy for some integers x and y. From these equations you can see that a is a factor of l, say l = am. Thus, the equations become am = abx = acy. Cancel a, then show that m = lcm(b, c).

G8Look at the proof of Theorem 3.

CHAPTER 23

A4(f)3x2 − 6x + 6 = 3(x2 − 2x + 1) + 3 = 3(x − 1)2 + 3. Thus, we are to solve the congruence 3(x − 1)2 ≡ −3 (mod 15). We begin by solving 3y ≡ −3 (mod 15), then we will set y = (x − 1)2.

We note first that by Condition (6), in a congruence ax = b (mod n), if the three numbers a, b, and n have a common factor d, then

(That is, all three numbers a, b, and n may be divided by the common factor d.) Applying this observation to our congruence 3y ≡ −3 (mod 15), we get

3y ≡ −3 (mod 15) is equivalent to y ≡ −1 (mod 5)

This is the same as y ≡ 4 (mod 5), because in 5 the negative of 1 is 4.

Thus, our quadratic congruence is equivalent to

(x − 1)2 ≡ 4 (mod 5)

In 5, 22 = 4 and 32 = 4; hence the solutions are x − 1 = 2 (mod 5) and x − 1 ≡ 3 (mod 5), or finally,

x ≡ 3 (mod 5) and x ≡ 4 (mod 5)

A6(d)We begin by finding all the solutions of 30z + 24y = 18, then set z = x2. Now,

30z + 24y = 18 iff 24y = 18 − 30z iff 30z ≡ 18 (mod 24)

By comments in the previous solution, this is equivalent to 5z ≡ 3 (mod4). But in 4, = ; hence 5z = 3 (mod 4) is the same as z ≡ 3 (mod 4).

Now set z = x2. Then the solution is x2 ≡ 3 (mod 4). But this last congruence has no solution, because in 4, is not the square of any number. Thus, the Diophantine equation has no solutions.

B3Here is the idea: By Theorem 3, the first two equations have a simultaneous solution; by Theorem 4, it is of the form xc (mod t), where t = lcm(m1, m2). To solve the latter simultaneously with xc3 (mod m3), you will need to know that c3c [mod gcd(t, m3)]. But gcd(t, m3) = lcm(d13, d23). (Explain this carefully, using the result of Exercise H4 in Chapter 22.) From Theorem 4 (especially the last paragraph of its proof), it is clear that since c3c1(mod d13) and c3c2 (mod d23), therefore c3c [mod lcm(d13, d23)].

Thus, c3c [mod gcd(t, m3)], and therefore by Theorem 3 there is a simultaneous solution of xc (mod t) and xc3 (mod m3). This is a simultaneous solution of xc1 (mod m1), xc2 (mod m2), and xc3 (mod m3). Repeat this process k times.

An elegant (and mathematically correct) form of this proof is by induction on k, where k is the number of congruences.

B5(a)Solving the pair of Diophantine equations is equivalent to solving the pair of linear congruences 4x ≡ 2 (mod 6) and 9x = 3 (mod 12). Now see the example at the beginning of Exercise B.

D6Use the definitions, and form the product (ab − 1)(ac − 1)(bc − 1).

E4Use the product (pq 1 − 1)(qp 1 − 1).

E6Use Exercise D5.

E8(a)Note the following:

(i) 133 = 7 × 19.

(ii) 18 is a multiple of both 7-1 and 19-1.

Now use part 6(b).

F8Consider (nϕ(m) − 1)(mϕ(n) − 1).

H2In any commutative ring, if a2 = b2, then (a + b)(ab) = 0; hence a = ±b. Thus, for any a ≠ 0, the two elements a and − a have the same square; and no x ≠ ± a can have the same square as ± a.

H4() = − 1 because 17 is not a quadratic residue mod 23.

CHAPTER 24

A1We compute a(x)b(x) in 6[x]:

a(x)b(x) = (2x2 + 3x + 1)(x3 + 5x2 + x) = 2x5 + 13x4 + 18x3 + 8x2 + x

But the coefficients are in 6, and in 6

13 = 1 18 = 0 and 8 = 2

Thus, a(x)b(x) = 2x5 + x4 + 2x2 + x in 6[x].

Note that while the coefficients are in 6, the exponents are positive integers and are not in 6. Thus, for example, in 6[x] the terms x8 and x2 are not the same.

B3In 5[x], there are 5 × 5 × 5 = 125 different polynomials of degree 2 or less. Explain why. There are 5 × 5 = 25 polynomials of degree 1 or 0; hence there are 125 − 25 = 100 different polynomials of degree 2. Explain, then complete the problem.

C7In 9[x], x2 = (x + 3)(x + 6) = (2x + 3)(5x + 6) = etc. Systematically list all the ways of factoring x2 into factors of degree 1, and complete the problem.

D6 After proving the first part of the problem, you can prove the second part by induction on the number n of terms in the polynomial. For n = 2, it follows from the first part of the problem. Next, assume the formula for k terms,

and prove for k + 1 terms:

[(a0 + a1x + ⋯ akxk) + ak + 1xk + 1]p = (Complete.)

E5If a(x) = a0 + a1x + ⋯ anxnJ, then a0 + a1 + ⋯ + an = 0. If b(x) is any element in A[x], say b(x) = b0 + b1x + ⋯ + bmxm, let B = b0 + b1 + ⋯ + bm. Explain why the sum of all the coefficients in a(x)b(x) is equal to (a0 + a1 ⋯ + an)B. Supply all remaining details.

G3If h is surjective, then every element of B is of the form h(a) for some a in A. Thus, any polynomial with coefficients in B is of the form h(a0) + h(a1)x + ⋯ + h(an)xn.

CHAPTER 25

A4Assume there are a, b5 such that (x + a)(x + b) = x2 + 2. Note that in 5, only 1 and 4 are squares.

D4Let ⟨p(x)⟩ ⊂ J where J is an ideal of F[x], and assume there is a polynomial a(x) ∈ J where a(x) ∉ ⟨p(x)⟩. Then a(x) and p(x) are relatively prime. (Why?) Complete the solution.

F2The gcd of the given polynomials is x + 1.

CHAPTER 26

A2To find the roots of x100 −1 in 7[x], reason as follows: From Fermat’s theorem, if a7, then in 7, a69 = 1 for every integer q. In particular, a96 = 1; hence a100 = a96a4 = a4. Thus any root (in 7) of x100 − 1 is a root of x4 − 1.

B1Any rational root of 9x3 + 18x2 − 4x − 8 is a fraction s/t where

s = ± 1, ± 8, ± 2, or ± 4

and

t = ± 1, ± 9, or ± 3

Thus, the possible roots are ± 1, ± 2, ± 4, ± 8, ± 1/9, ± 1/3, ± 8/9, ± 8/3, ± 2/9, ± 2/3, ± 4/9, and ± 4/3. Once a single root has been found by substitution into the polynomial, the problem may be simplified. For example, −2 is one root of the above. You may now divide the given polynomial by x + 2. The remaining roots are roots of the quotient, which is a quadratic.

C5 Prove by induction: For n = 1, we need to show that if a(x) is a monic polynomial of degree 1, say a(x) = xc, and a(x) has one root c1, then a(x) = xc1. Well, if c1 is a root of a(x) = xc, then a(c1) = c1c = 0, so c = c1; thus, a(x) = xc1.

For the induction step, assume the theorem is true for degree k and k roots, and prove it is true for degree k + 1 and k + 1 roots.

C8If x2x = x(x − 1) = 0, then x = 0 or x − 1 = 0. Does this rule hold in both 10 and 11? Why?

D3Note that if p is a prime number, and 0 < k < p, then the binomial coefficient is a multiple of p. (See page 202.)

F1The fact that a(x) is monic is essential. Explain why the degree of (a(x)) is the same as the degree of a(x).

H3Assume there are two polynomials p(x) and q(x), each of degree ≤ n, satisfying the given condition. Show that p(x) = q(x).

I4If a(x) and b(x) determine the same function, then a(x) − b(x) is the zero function; that is, a(c) – b(c) = 0 for every cF.

CHAPTER 27

A1(e)Let ; then a2 = i and a4 = −1 − i + 2 = 1− 2i. Thus, a4 − 1 = −2i, and (a4 − 1)2 = −8. Therefore a is a root of the polynomial (x4 – 1)2 + 8, that is,

x8 −2x4 + 9

B1(d)Of the six parts, (a) − (f), of this exercise, only (d) involves a polynomial of degree higher than 4; hence it is the most painstaking. Let a = ; then a2 = 2 + 31/3 and a2 − 2 = 31/3. Thus, (a2 − 2)3 = 3, so a is a root of the polynomial

p(x) = (x2 − 2)3 − 3 = x6x6 − 12x2 − 11

The bulk of the work is to ensure that this polynomial is irreducible. We will use the methods of Chapter 26, Exercises F and E. By the first of these methods, we transform p(x) into a polynomial in Z3[x]:

x6 − 2 = x6 + 1

Since none of the three elements 0, 1, 2 in 3 is a root of the polynomial, the polynomial has no factor of degree 1 in 3[x]. So the only possible factorings into non constant polynomials are

x6 + 1 = (x3 + ax2 + bx + c)(x3 + dx2 + ex + f)

or

x6 + 1 = (x4 + ax3 + bx2 + cx + d)(x2 + ex + f)

From the first equation, since corresponding coefficients are equal, we have:

From (1), c = f = ±1, and from (5), a + d = 0. Consequently, af + cd = c(a + d) = 0, and by (3), eb = 0. But from (2) (since c = f), b + e = 0, and therefore b = e = 0. It follows from (4) that c + f = 0, which is impossible since c= f = ±1. We have just shown that x6 + 1 cannot be factored into two polynomials each of degree 3. Complete the solution.

C4From part 3, the elements of 2(c) are 0, 1, c, and c + 1. (Explain.) When adding elements, note that 0 and 1 are added as in 2, since 2(c) is an extension of 2. Moreover, c + c = c(1 + 1) = c0 = 0. When multiplying elements, note the following example: (c + 1)2 = c2 + c + c + 1 = c (because c2 + c + 1 = 0).

D1See Exercise E3, this chapter.

D7In (), 1 + 1 is a square; so if () ≅ (), then 1 + 1 is a square in (), that is, (). But all the elements of () are of the form a + b for a, b ∈ ∈. (Explain why.) So we would have = a + b. Squaring both sides and solving for (supply details) shows that is a rational number. But it is known that is irrational.

G2Let Q denote the field of quotients of {a(c) : a(x) ∈ F[x]}. Since F(c) is a field which contains F and c, it follows that F(c) contains every a(c) = a0 + a1c + ⋯ + ancn where a0, …, anF. Thus, QF(c). Conversely, by definition F(c) is contained in any field containing F and c; hence F(c) ⊆ Q. Complete the solution.

CHAPTER 28

B1Let U = {(a, b, c): 2a − 3b + c = 0}. To show that U is closed under addition, let u, vU. Then u = (a1, b1, c1) where 2a1 −3b1 + c1 = 0, and v = (a2, b2, c2) where 2a2 − 3b2 + c2 = 0. Adding,

u + v = (a1 + a2, b1 + b2, c1 + c2)

and

2(a1 + a2) − 3(b1 + b2) + (c1 + c2) = 0

C5(a)S1 = {(x, y, z): z = 2y − 3x}. A suggested basis for S1 would consist of the two vectors (1, 0, ?) and (0, 1, ?) where the third component satisfies z = 2y − 3x, that is,

(1, 0, −3) and (0, 1, 2)

Now show that the set of these two vectors is linearly independent and spans S1.

D6Suppose there are scalars k, l, and m such that

k(a + b) + l(b + c) + m(a + c) = 0

Use the fact that {a, b, c} is independent to show that k = l = m = 0.

E5Let kr + 1, kr + 2, …, kn be scalars such that

kr + 1 h(ar + 1) + ⋯ + kn h(an) = b

Hence

h(kr + 1ar + 1 + ⋯ + knan) = 0

Then kr + 1ar + 1 + ⋯ + knan. Recall that {a1, …, ar} is a basis for , and complete the proof.

F2Use the result of Exercise E7.

G2Assume that every cV can be written, in a unique manner, as c = a + b where aT and bU. Thus, every vector in V is an element of T + U, and conversely, since T and U are subspaces of V, every vector in T + U is an element of V. Thus,

V = T + U

In order to show that V = TU, it remains to show that T U = {0}; that is, if cT U then c = 0. Complete the solution.

CHAPTER 29

A3Note first that a2 − = ; hence (a) and therefore (a) = (, a). (Explain.) Now x32 (which is irreducible over by Eisenstein’s criterion) is the minimum polynomial of ; hence [() : ] = 3. Next, in (), a satisfies a2 − 1 − = 0, so a is a root of the polynomial x2 − (1 + ). This quadratic is irreducible over ()[x] (explain why); hence x2 − (1 + ) is the minimum polynomial of a over ()[x]. Thus, [(, a) : (] = 2. Use Theorem 2 to complete.

A4This is similar to part 3; adjoin first , then a.

C2Note that x2 + x + 1 is irreducible in 2[x].

D2Suppose there is a field L properly between F and K, that is, F L K. As vector spaces over the field F, L is a subspace of K. (Why?) Use Chapter 28, Exercise D1.

F4The relationship between the fields may be sketched as follows:

[K(b) : F] = [K(b) : F(b)] · [F(b) : F] = [K(b) : K] · [K : F]

F5Reasoning as in part 4 (and using the same sketch), show that [K(b) : K] = [F(b) : F], Here, b is a root of p(x).

CHAPTER 30

B3If (a, b) ∈ × , then a and are rational numbers. By Exercise A5 and the definition of , (a, 0) and (b, 0) are constructible from {O, I}. With the aid of a compass (mark off the distance from the origin to b along the y axis), we can construct the point (0, b). From elementary geometry there exists a ruler-and-compass construction of a perpendicular to a given line through a specified point on the line. Construct a perpendicular to the x axis through (a, 0) and a perpendicular to the y axis through (0, b). These lines intersect at (a, b); hence (a, b) is constructible from {O, I}.

C6Describe a ruler-and-compass construction of the 30-60-90° triangle.

D1From geometry, the angle at each vertex of a regular n-gon is equal to the supplement of 2π/n, that is, π − (2π/n).

G2A number is constructible iff it is a coordinate of a constructible point. (Explain, using Exercise A.) If P is a constructible point, this means there are some points, say n points P1, P2, …, Pn = P, such that each Pi is constructible in one step from × {P1, …, Pi 1. In the proof of the lemma of this chapter it is shown that the coordinates of Pi can be expressed in terms of the coordinates of Px, …, Pi 1 using addition, subtraction, multiplication, division, and taking of square roots. Thus, the coordinates of P are obtained by starting with rational numbers and sucessively applying these operations.

Using these ideas, write a proof by induction of the statement of this exercise.

CHAPTER 31

A6Let be a real cube root of 2. () is not a root field over because it does not contain the complex cube roots of 2.

B5First, p(x) = x3 + x2 + x + 2 is irreducible over 3 because, by successively substituting x = 0, 1, 2, one verifies that p(x) has no roots in 3. [Explain why this fact proves that p(x) is irreducible over 3.] If u is a root of p(x) in some extension of 3, then 3(u) is an extension of 3 of degree 3. (Why?) Thus, 3(u) consists of all elements au2 + bu + c, where a, b, c3. Hence 3(u) has 27 elements. Dividing p(x) = x3 + x2 + x + 2 by xu gives

p(x) = (xu)[x2 + (u + 1)x + (u2 + u + 1)]

where q(x) = x2 + (u + l)x + (u2 + u + 1) is in Z3(u)[x]. (Why?)

In 3(u)[x], q(x) is irreducible: this can be shown by direct substitution of each of the 27 elements of 3(u), successively, into q(x). So if w denotes a root of q(x) in some extension of 3(u), then Z3(u, w) includes u and both roots of q(x). (Explain.) Thus, 3(u, w) is the root field of p(x) over 3. It is of degree 6 over 3. (Why?)

C5We may identify F[x]/⟨p(x)⟩ with F(c), where c is a root of p(x). Then F(c) is an extension of degree 4 over F. (Why?) Now complete the square:

Form the iterated extension F(d1, d2), where d1 is a root of x2 − [(a2/4) − b] and d2 is a root of x2 − [(a/2) + d1]. Explain why (a) F(d1, d2) = F(c) and (b) F(d1, d2) contains all the roots of p(x).

D3From page 313, (, ) = ( + ); hence (, , ) = ( + , ). As in the illustration for Theorem 2, taking t = 1 gives c = + + . (Provide details.)

D5Use the result of part 3. The degree of (, , ) over is 8 (explain why). Now find a polynomial p(x) of degree 8 having + + as a root: if p(x) is monic and of degree 8, p(x) must be the minimum polynomial of + + over . (Why?) Hence p(x) must be irreducible. Explain why the roots of p(x) are ± ± ± , all of which are in (, , ).

E4For n = 6, we have x6 − 1 = (x − 1)(x + 1)(x2x + 1)(x2 + x + 1). From the quadratic formula, the roots of the last two factors are in ().

H2Note that if c(x) = c0 + c1x + ⋯ + cnxn then

and

h(c(a)) = h(c0) + h(c1)b + ⋯ + h(cn)bn

For d(x) = d0 + d1x + ⋯ + dnxn, we have similar formulas. Show that h(c(a)) = h(d(a)) iff b is a root of hc(x) − hd(x).

I4The proof is by induction on the degree of a(x). If deg a(x) = 1, then K1 = F1 and K2 = F2 (explain why), and therefore K1K2. Now let deg a(x) = n, and suppose the result is true for any polynomial of degree n − 1.

Let p(x) be an irreducible factor of a(x); let u be a root of p(x) and υ a root of hp(x). If F1(u) = K1, then by parts 1 and 2 we are done. If not, let = F1(u) and = F2(υ); h can be extended to an isomorphism h: , with h(u) = υ. In [x], a(x) = (xu)a1(x), and in [x], ha(x) = a(x) = (xυ)ha1(x), where deg a1(x) = deg ha1(x) = n − 1. Complete the solution.

J2Explain why any monomorphism : F(c) → , which is an extension of h, is fully determined by the value of (c), which is a root of hp(x).

J4Since h(1) = 1, begin by proving by induction that for any positive integer n, h(n) = n. Then complete the solution.

CHAPTER 32

A2In the first place, () is of degree 2 over . Next, x2 + 1 is irreducible over (). (Explain why.) Complete the solution.

D1The complex fourth roots of 1 are ±1 and ±i. Thus, the complex fourth roots of 2 are (1), (− 1), (i), and (− i), that is:

Explain why any field containing the fourth roots of 2 contains α and i, and conversely.

E1See Chapter 31, Exercise E.

E6Let a α be a real sixth root of 2. Then [(α) : ] = 6. Explain why x2 + 3 is irreducible over (α) and why [(a, i) : (α)] = 2. If ω = (1/2) + (/2)i (which is a primitive sixth root of 1), then the complex sixth roots of 2 are α, αω, αω2, αω3, αω4, and αω5. Any automorphism of (α, i) fixing maps sixth roots of 2 to sixth roots of 2, at the same time mapping i to ±i (and hence mapping ω to ω5). Provide details and complete the solution.

H4Suppose ah(a), say a < h(a). Let r be a rational number such that a < r < h(a).

I4Every subgroup of an abelian group is abelian; every homomorphic image of an abelian group is abelian.

CHAPTER 33

A2(a)The polynomial is irreducible over by Eisenstein’s criterion, and it has three real roots and two complex roots. (Explain why.) Argue as in the example of page 340 and show that the Galois group is S5.

B3It must be shown that for each i, i = 0,1, …, n – 1, the quotient group Ji + 1/Ji is abelian. It must be shown that Ji contains xyx 1y 1 for all x and y in Ji + 1. Provide details and complete.

C3F has an extension K which contains all the roots d1, d2, …, dp of xpa. In K, xpa factors into linear factors:

xpa = (xd1)(xd2) ⋯ (xdp)

By uniqueness of factorization, p(x) is equal to the product of m of these factors, while f(x) is the product of the remaining factors.

D3If f: GG/K is the natural homomorphism, then = f 1 () = {xG: f(x) ∈ }.

D7Prove this statement by induction on the order of G/H. Let |G/H| = n, and assume the statement is true for all groups H G′, where |G′/H′| < n. If G has no normal subgroup J such that HJG (HJG), then H is a maximal normal subgroup of G; so we are done by parts 4 and 5. Otherwise, |G/J| < n and |J/H| < n. Complete the solution.

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