Skip to main content
Logo image

Appendix B Hints and Answers to Selected Exercises

1 Preliminaries
1.4 Exercises

1.4.1.

Hint.
(a) AB={2}; (b) BC={5}.

1.4.2.

Hint.
(a) A×B={(a,1),(a,2),(a,3),(b,1),(b,2),(b,3),(c,1),(c,2),(c,3)}; (d) A×D=.

1.4.6.

Hint.
Observe that xAB if and only if xA or xB. Equivalently, xB or xA, which is the same as xBA. Therefore, AB=BA.

1.4.10.

Hint.
(AB)(AB)(BA)=(AB)(AB)(BA)=[A(BB)](BA)=A(BA)=(AB)(AA)=AB.

1.4.14.

Hint.
A(BC)=A(BC)=(AA)(BC)=(AB)(AC)=(AB)(AC).

1.4.17.

Hint.
(a) Not a map since f(2/3) is undefined; (b) this is a map; (c) not a map, since f(1/2)=3/4 but f(2/4)=3/8; (d) this is a map.

1.4.18.

Hint.
(a) f is one-to-one but not onto. f(R)={xR:x>0}. (c) f is neither one-to-one nor onto. f(R)={x:1x1}.

1.4.20.

Hint.
(a) f(n)=n+1.

1.4.22.

Hint.
(a) Let x,yA. Then g(f(x))=(gf)(x)=(gf)(y)=g(f(y)). Thus, f(x)=f(y) and x=y, so gf is one-to-one. (b) Let cC, then c=(gf)(x)=g(f(x)) for some xA. Since f(x)B, g is onto.

1.4.23.

Hint.
f1(x)=(x+1)/(x1).

1.4.24.

Hint.
(a) Let yf(A1A2). Then there exists an xA1A2 such that f(x)=y. Hence, yf(A1) or f(A2). Therefore, yf(A1)f(A2). Consequently, f(A1A2)f(A1)f(A2). Conversely, if yf(A1)f(A2), then yf(A1) or f(A2). Hence, there exists an x in A1 or A2 such that f(x)=y. Thus, there exists an xA1A2 such that f(x)=y. Therefore, f(A1)f(A2)f(A1A2), and f(A1A2)=f(A1)f(A2).

1.4.25.

Hint.
(a) The relation fails to be symmetric. (b) The relation is not reflexive, since 0 is not equivalent to itself. (c) The relation is not transitive.

1.4.28.

Hint.
Let X=N{2} and define xy if x+yN.

2 The Integers
2.4 Exercises

2.4.1.

Hint.
The base case, S(1):[1(1+1)(2(1)+1)]/6=1=12 is true. Assume that S(k):12+22++k2=[k(k+1)(2k+1)]/6 is true. Then
12+22++k2+(k+1)2=[k(k+1)(2k+1)]/6+(k+1)2=[(k+1)((k+1)+1)(2(k+1)+1)]/6,
and so S(k+1) is true. Thus, S(n) is true for all positive integers n.

2.4.3.

Hint.
The base case, S(4):4!=24>16=24 is true. Assume S(k):k!>2k is true. Then (k+1)!=k!(k+1)>2k2=2k+1, so S(k+1) is true. Thus, S(n) is true for all positive integers n.

2.4.11.

Hint.
The base case, S(0):(1+x)01=00=0x is true. Assume S(k):(1+x)k1kx is true. Then
(1+x)k+11=(1+x)(1+x)k1=(1+x)k+x(1+x)k1kx+x(1+x)kkx+x=(k+1)x,
so S(k+1) is true. Therefore, S(n) is true for all positive integers n.

2.4.17. Fibonacci Numbers.

Hint.
For (a) and (b) use mathematical induction. (c) Show that f1=1, f2=1, and fn+2=fn+1+fn. (e) Use part (b) and Exercise 2.4.16.

2.4.19.

Hint.
Use the Fundamental Theorem of Arithmetic.

2.4.23.

Hint.
Use the Principle of Well-Ordering and the division algorithm.

2.4.27.

Hint.
Since gcd(a,b)=1, there exist integers r and s such that ar+bs=1. Thus, acr+bcs=c.

2.4.29.

Hint.
Every prime must be of the form 2, 3, 6n+1, or 6n+5. Suppose there are only finitely many primes of the form 6k+5.

3 Groups
3.5 Exercises

3.5.1.

Hint.
(a) 3+7Z={,4,3,10,}; (c) 18+26Z; (e) 5+6Z.

3.5.2.

Hint.
(a) Not a group; (c) a group.

3.5.6.

Hint.
157111157115511177711151111751

3.5.8.

Hint.
Pick two matrices. Almost any pair will work.

3.5.15.

Hint.
There is a nonabelian group containing six elements.

3.5.16.

Hint.
Look at the symmetry group of an equilateral triangle or a square.

3.5.17.

Hint.
The are five different groups of order 8.

3.5.18.

Hint.
Let
σ=(12na1a2an)
be in Sn. All of the ais must be distinct. There are n ways to choose a1, n1 ways to choose a2,, 2 ways to choose an1, and only one way to choose an. Therefore, we can form σ in n(n1)21=n! ways.

3.5.25.

Hint.
(aba1)n=(aba1)(aba1)(aba1)=ab(aa1)b(aa1)bb(aa1)ba1=abna1.

3.5.31.

Hint.
Since abab=(ab)2=e=a2b2=aabb, we know that ba=ab.

3.5.35.

Hint.
H1={id}, H2={id,ρ1,ρ2}, H3={id,μ1}, H4={id,μ2}, H5={id,μ3}, S3.

3.5.41.

Hint.
The identity of G is 1=1+02. Since (a+b2)(c+d2)=(ac+2bd)+(ad+bc)2, G is closed under multiplication. Finally, (a+b2)1=a/(a22b2)b2/(a22b2).

3.5.46.

Hint.
Look at S3.

3.5.49.

Hint.
ba=a4b=a3ab=ab

4 Cyclic Groups
4.5 Exercises

4.5.1.

Hint.
(a) False; (c) false; (e) true.

4.5.2.

Hint.
(a) 12; (c) infinite; (e) 10.

4.5.3.

Hint.
(a) 7Z={,7,0,7,14,}; (b) {0,3,6,9,12,15,18,21}; (c) {0}, {0,6}, {0,4,8}, {0,3,6,9}, {0,2,4,6,8,10}; (g) {1,3,7,9}; (j) {1,1,i,i}.

4.5.4.

Hint.
(a)
(1001),(1001),(0110),(0110).
(c)
(1001),(1110),(1110),(0111),(0111),(1001).

4.5.10.

Hint.
(a) 0; (b) 1,1.

4.5.11.

Hint.
1,2,3,4,6,8,12,24.

4.5.15.

Hint.
(a) 3+3i; (c) 4318i; (e) i

4.5.16.

Hint.
(a) 3+i; (c) 3.

4.5.17.

Hint.
(a) 2cis(7π/4); (c) 22cis(π/4); (e) 3cis(3π/2).

4.5.18.

Hint.
(a) (1i)/2; (c) 16(i3); (e) 1/4.

4.5.22.

Hint.
(a) 292; (c) 1523.

4.5.27.

Hint.
|gh|=1.

4.5.31.

Hint.
The identity element in any group has finite order. Let g,hG have orders m and n, respectively. Since (g1)m=e and (gh)mn=e, the elements of finite order in G form a subgroup of G.

4.5.37.

Hint.
If g is an element distinct from the identity in G, g must generate G; otherwise, g is a nontrivial proper subgroup of G.

5 Permutation Groups
5.4 Exercises

5.4.1.

Hint.
(a) (12453); (c) (13)(25).

5.4.2.

Hint.
(a) (135)(24); (c) (14)(23); (e) (1324); (g) (134)(25); (n) (17352).

5.4.3.

Hint.
(a) (16)(15)(13)(14); (c) (16)(14)(12).

5.4.4.

Hint.
(a1,a2,,an)1=(a1,an,an1,,a2)

5.4.5.

Hint.
(a) {(13),(13)(24),(132),(134),(1324),(1342)} is not a subgroup.

5.4.8.

Hint.
(12345)(678).

5.4.11.

Hint.
Permutations of the form
(1),(a1,a2)(a3,a4),(a1,a2,a3),(a1,a2,a3,a4,a5)
are possible for A5.

5.4.17.

Hint.
Calculate (123)(12) and (12)(123).

5.4.25.

Hint.
Consider the cases (a,b)(b,c) and (a,b)(c,d).

5.4.29.

Hint.
Show that the center of Dn consists of the identity if n is odd and consists of the identity and a 180 rotation if n is even.

5.4.30.

Hint.
For (a), show that στσ1(σ(ai))=σ(ai+1).

6 Cosets and Lagrange’s Theorem
6.5 Exercises

6.5.1.

Hint.
The order of g and the order h must both divide the order of G.

6.5.2.

Hint.
The possible orders must divide 60.

6.5.3.

Hint.
This is true for every proper nontrivial subgroup.

6.5.4.

Hint.
False.

6.5.5.

Hint.
(a) 8, 1+8, 2+8, 3+8, 4+8, 5+8, 6+8, and 7+8; (c) 3Z, 1+3Z, and 2+3Z.

6.5.7.

Hint.
4ϕ(15)481(mod15).

6.5.12.

Hint.
Let g1gH. Show that g1Hg and thus gHHg.

6.5.19.

Hint.
Show that g(HK)=gHgK.

7 Introduction to Cryptography
7.4 Exercises

7.4.1.

Hint.
LAORYHAPDWK

7.4.3.

Hint.
Hint: V = E, E = X (also used for spaces and punctuation), K = R.

7.4.4.

Hint.
26!1

7.4.7.

Hint.
(a) 2791; (c) 11213525032442.

7.4.9.

Hint.
(a) 31 (c) 14.

7.4.10.

Hint.
(a) n=1141; (c) n=87794327.

8 Algebraic Coding Theory
8.6 Exercises

8.6.2.

Hint.
This cannot be a group code since (0000)C.

8.6.3.

Hint.
(a) 2; (c) 2.

8.6.4.

Hint.
(a) 3; (c) 4.

8.6.6.

Hint.
(a) dmin=2; (c) dmin=1.

8.6.7.

Hint.
  1. (00000),(00101),(10011),(10110)
    G=(0100100111)
  2. (000000),(010111),(101101),(111010)
    G=(100110110111)

8.6.9.

Hint.
Multiple errors occur in one of the received words.

8.6.11.

Hint.
(a) A canonical parity-check matrix with standard generator matrix
G=(11001).
(c) A canonical parity-check matrix with standard generator matrix
G=(10011110).

8.6.12.

Hint.
(a) All possible syndromes occur.

8.6.15.

Hint.
(a) C, (10000)+C, (01000)+C, (00100)+C, (00010)+C, (11000)+C, (01100)+C, (01010)+C. A decoding table does not exist for C since this is only a single error-detecting code.

8.6.19.

Hint.
Let xC have odd weight and define a map from the set of odd codewords to the set of even codewords by yx+y. Show that this map is a bijection.

8.6.23.

Hint.
For 20 information positions, at least 6 check bits are needed to ensure an error-correcting code.

9 Isomorphisms
9.4 Exercises

9.4.1.

Hint.
Every infinite cyclic group is isomorphic to Z by Theorem 9.7.

9.4.2.

Hint.
Define ϕ:CGL2(R) by
ϕ(a+bi)=(abba).

9.4.3.

Hint.
False.

9.4.6.

Hint.
Define a map from Zn into the nth roots of unity by kcis(2kπ/n).

9.4.8.

Hint.
Assume that Q is cyclic and try to find a generator.

9.4.11.

Hint.
There are two nonabelian and three abelian groups that are not isomorphic.

9.4.16.

Hint.
(a) 12; (c) 5.

9.4.19.

Hint.
Draw the picture.

9.4.20.

Hint.
True.

9.4.25.

Hint.
True.

9.4.27.

Hint.
Let a be a generator for G. If ϕ:GH is an isomorphism, show that ϕ(a) is a generator for H.

9.4.38.

Hint.
Any automorphism of Z6 must send 1 to another generator of Z6.

9.4.45.

Hint.
To show that ϕ is one-to-one, let g1=h1k1 and g2=h2k2 and consider ϕ(g1)=ϕ(g2).

10 Normal Subgroups and Factor Groups
10.4 Exercises

10.4.1.

Hint.
(a)
A4(12)A4A4A4(12)A4(12)A4(12)A4A4
(c) D4 is not normal in S4.

10.4.8.

Hint.
If aG is a generator for G, then aH is a generator for G/H.

10.4.11.

Hint.
For any gG, show that the map ig:GG defined by ig:xgxg1 is an isomorphism of G with itself. Then consider ig(H).

10.4.12.

Hint.
Suppose that g is normal in G and let y be an arbitrary element of G. If xC(g), we must show that yxy1 is also in C(g). Show that (yxy1)g=g(yxy1).

10.4.14.

Hint.
(a) Let gG and hG. If h=aba1b1, then
ghg1=gaba1b1g1=(gag1)(gbg1)(ga1g1)(gb1g1)=(gag1)(gbg1)(gag1)1(gbg1)1.
We also need to show that if h=h1hn with hi=aibiai1bi1, then ghg1 is a product of elements of the same type. However, ghg1=gh1hng1=(gh1g1)(gh2g1)(ghng1).

11 Homomorphisms
11.4 Exercises

11.4.2.

Hint.
(a) is a homomorphism with kernel {1}; (c) is not a homomorphism.

11.4.4.

Hint.
Since ϕ(m+n)=7(m+n)=7m+7n=ϕ(m)+ϕ(n), ϕ is a homomorphism.

11.4.5.

Hint.
For any homomorphism ϕ:Z24Z18, the kernel of ϕ must be a subgroup of Z24 and the image of ϕ must be a subgroup of Z18. Now use the fact that a generator must map to a generator.

11.4.9.

Hint.
Let a,bG. Then ϕ(a)ϕ(b)=ϕ(ab)=ϕ(ba)=ϕ(b)ϕ(a).

11.4.17.

Hint.
Find a counterexample.

12 Matrix Groups and Symmetry
12.4 Exercises

12.4.1.

Hint.
12[x+y2+x2y2]=12[x+y,x+yx2y2]=12[x2+2x,y+y2x2y2]=x,y.

12.4.3.

Hint.
(a) is in SO(2); (c) is not in O(3).

12.4.5.

Hint.
(a) x,y=y,x.

12.4.7.

Hint.
Use the unimodular matrix
(5221).

12.4.10.

Hint.
Show that the kernel of the map det:O(n)R is SO(n).

12.4.13.

Hint.
True.

12.4.17.

Hint.
p6m

13 The Structure of Groups
13.4 Exercises

13.4.1.

Hint.
There are three possible groups.

13.4.4.

Hint.
(a) {0}63Z12; (e) {(1)}×{0}{(1),(123),(132)}×{0}S3×{0}S3×2S3×Z4.

13.4.7.

Hint.
Use the Fundamental Theorem of Finitely Generated Abelian Groups.

13.4.12.

Hint.
If N and G/N are solvable, then they have solvable series
N=NnNn1N1N0={e}G/N=Gn/NGn1/NG1/NG0/N={N}.

13.4.16.

Hint.
Use the fact that Dn has a cyclic subgroup of index 2.

13.4.21.

Hint.
G/G is abelian.

14 Group Actions
14.5 Exercises

14.5.2.

Hint.
(a) X(1)={1,2,3}, X(12)={3}, X(13)={2}, X(23)={1}, X(123)=X(132)=. G1={(1),(23)}, G2={(1),(13)}, G3={(1),(12)}.

14.5.3.

Hint.
(a) O1=O2=O3={1,2,3}.

14.5.6.

Hint.
The conjugacy classes for S4 are
O(1)={(1)},O(12)={(12),(13),(14),(23),(24),(34)},O(12)(34)={(12)(34),(13)(24),(14)(23)},O(123)={(123),(132),(124),(142),(134),(143),(234),(243)},O(1234)={(1234),(1243),(1324),(1342),(1423),(1432)}.
The class equation is 1+3+6+6+8=24.

14.5.8.

Hint.
(34+31+32+31+32+32+33+33)/8=21.

14.5.11.

Hint.
The group of rigid motions of the cube can be described by the allowable permutations of the six faces and is isomorphic to S4. There are the identity cycle, 6 permutations with the structure (abcd) that correspond to the quarter turns, 3 permutations with the structure (ab)(cd) that correspond to the half turns, 6 permutations with the structure (ab)(cd)(ef) that correspond to rotating the cube about the centers of opposite edges, and 8 permutations with the structure (abc)(def) that correspond to rotating the cube about opposite vertices.

14.5.15.

Hint.
(126+324+423+222+221)/12=13.

14.5.17.

Hint.
(128+326+224)/6=80.

14.5.22.

Hint.
Use the fact that xgC(a)g1 if and only if g1xgC(a).

15 The Sylow Theorems
15.4 Exercises

15.4.1.

Hint.
If |G|=18=232, then the order of a Sylow 2-subgroup is 2, and the order of a Sylow 3-subgroup is 9.

15.4.2.

Hint.
The four Sylow 3-subgroups of S4 are P1={(1),(123),(132)}, P2={(1),(124),(142)}, P3={(1),(134),(143)}, P4={(1),(234),(243)}.

15.4.5.

Hint.
Since |G|=96=253, G has either one or three Sylow 2-subgroups by the Third Sylow Theorem. If there is only one subgroup, we are done. If there are three Sylow 2-subgroups, let H and K be two of them. Therefore, |HK|16; otherwise, HK would have (3232)/8=128 elements, which is impossible. Thus, HK is normal in both H and K since it has index 2 in both groups.

15.4.8.

Hint.
Show that G has a normal Sylow p-subgroup of order p2 and a normal Sylow q-subgroup of order q2.

15.4.10.

Hint.
False.

15.4.17.

Hint.
If G is abelian, then G is cyclic, since |G|=3517. Now look at Example 15.14.

15.4.23.

Hint.
Define a mapping between the right cosets of N(H) in G and the conjugates of H in G by N(H)gg1Hg. Prove that this map is a bijection.

15.4.26.

Hint.
Let aG,bGG/G. Then (aG)(bG)=abG=ab(b1a1ba)G=(abb1a1)baG=baG.

16 Rings
16.7 Exercises

16.7.1.

Hint.
(a) 7Z is a ring but not a field; (c) Q(2) is a field; (f) R is not a ring.

16.7.3.

Hint.
(a) {1,3,7,9}; (c) {1,2,3,4,5,6}; (e)
{(1001),(1101),(1011),(0110),(1110),(0111),}.

16.7.4.

Hint.
(a) {0}, {0,9}, {0,6,12}, {0,3,6,9,12,15}, {0,2,4,6,8,10,12,14,16}; (c) there are no nontrivial ideals.

16.7.7.

Hint.
Assume there is an isomorphism ϕ:CR with ϕ(i)=a.

16.7.8.

Hint.
False. Assume there is an isomorphism ϕ:Q(2)Q(3) such that ϕ(2)=a.

16.7.13.

Hint.
(a) x17(mod55); (c) x214(mod2772).

16.7.16.

Hint.
If I{0}, show that 1I.

16.7.18.

Hint.
(a) ϕ(a)ϕ(b)=ϕ(ab)=ϕ(ba)=ϕ(b)ϕ(a).

16.7.26.

Hint.
Let aR with a0. Then the principal ideal generated by a is R. Thus, there exists a bR such that ab=1.

16.7.28.

Hint.
Compute (a+b)2 and (ab)2.

16.7.33.

Hint.
Let a/b,c/dZ(p). Then a/b+c/d=(ad+bc)/bd and (a/b)(c/d)=(ac)/(bd) are both in Z(p), since gcd(bd,p)=1.

16.7.37.

Hint.
Suppose that x2=x and x0. Since R is an integral domain, x=1. To find a nontrivial idempotent, look in M2(R).

17 Polynomials
17.5 Exercises

17.5.2.

Hint.
(a) 9x2+2x+5; (b) 8x4+7x3+2x2+7x.

17.5.3.

Hint.
(a) 5x3+6x23x+4=(5x2+2x+1)(x2)+6; (c) 4x5x3+x2+4=(4x2+4)(x3+3)+4x2+2.

17.5.5.

Hint.
(a) No zeros in Z12; (c) 3, 4.

17.5.7.

Hint.
Look at (2x+1).

17.5.8.

Hint.
(a) Reducible; (c) irreducible.

17.5.10.

Hint.
One factorization is x2+x+8=(x+2)(x+9).

17.5.13.

Hint.
The integers Z do not form a field.

17.5.14.

Hint.
False.

17.5.16.

Hint.
Let ϕ:RS be an isomorphism. Define ϕ:R[x]S[x] by ϕ(a0+a1x++anxn)=ϕ(a0)+ϕ(a1)x++ϕ(an)xn.

17.5.20. Cyclotomic Polynomials.

Hint.
The polynomial
Φn(x)=xn1x1=xn1+xn2++x+1
is called the cyclotomic polynomial. Show that Φp(x) is irreducible over Q for any prime p.

17.5.26.

Hint.
Find a nontrivial proper ideal in F[x].

18 Integral Domains
18.4 Exercises

18.4.1.

Hint.
Note that z1=1/(a+b3i)=(ab3i)/(a2+3b2) is in Z[3i] if and only if a2+3b2=1. The only integer solutions to the equation are a=±1,b=0.

18.4.2.

Hint.
(a) 5=i(1+2i)(2+i); (c) 6+8i=i(1+i)2(2+i)2.

18.4.4.

Hint.
True.

18.4.9.

Hint.
Let z=a+bi and w=c+di0 be in Z[i]. Prove that z/wQ(i).

18.4.15.

Hint.
Let a=ub with u a unit. Then ν(b)ν(ub)ν(a). Similarly, ν(a)ν(b).

18.4.16.

Hint.
Show that 21 can be factored in two different ways.

19 Lattices and Boolean Algebras
19.5 Exercises

19.5.2.

Hint.
A graph with 30 at the top level which is connected to 10 and 15 at the second level.  The third level has 2, which is connected to 30 and 10, and 5, which is connected to 10 and 15, and 3 which is connected to 15.  The bottom level is 1 which is connected to 2, 3, and 5.

19.5.4.

Hint.
What are the atoms of B?

19.5.5.

Hint.
False.

19.5.6.

Hint.
(a) (aba)a
A graph from left to right which splits into three paths, a b, and b’ and then rejoins into a single path and goes through a.
(c) a(ab)
A graph from left to right which splits into two paths and then rejoins.  The top path is a then b.  The bottom path is a.

19.5.8.

Hint.
Not equivalent.

19.5.10.

Hint.
(a) a[(ab)b]=a(ab).

19.5.14.

Hint.
Let I,J be ideals in R. We need to show that I+J={r+s:rI and sJ} is the smallest ideal in R containing both I and J. If r1,r2I and s1,s2J, then (r1+s1)+(r2+s2)=(r1+r2)+(s1+s2) is in I+J. For aR, a(r1+s1)=ar1+as1I+J; hence, I+J is an ideal in R.

19.5.18.

Hint.
(a) No.

19.5.20.

Hint.
(). a=b(ab)(ab)=(aa)(aa)=OO=O. (). (ab)(ab)=Oab=(aa)b=a(ab)=a[I(ab)]=a[(aa)(ab)]=[a(ab)][a(ab)]=a[(ab)(ab)]=a0=a. A symmetric argument shows that ab=b.

20 Vector Spaces
20.5 Exercises

20.5.3.

Hint.
Q(2,3) has basis {1,2,3,6} over Q.

20.5.5.

Hint.
The set {1,x,x2,,xn1} is a basis for Pn.

20.5.7.

Hint.
(a) Subspace of dimension 2 with basis {(1,0,3),(0,1,2)}; (d) not a subspace

20.5.10.

Hint.
Since 0=α0=α(v+v)=α(v)+αv, it follows that αv=α(v).

20.5.12.

Hint.
Let v0=0,v1,,vnV and α00,α1,,αnF. Then α0v0++αnvn=0.

20.5.15. Linear Transformations.

Hint.
(a) Let u,vker(T) and αF. Then
T(u+v)=T(u)+T(v)=0T(αv)=αT(v)=α0=0.
Hence, u+v,αvker(T), and ker(T) is a subspace of V.
(c) The statement that T(u)=T(v) is equivalent to T(uv)=T(u)T(v)=0, which is true if and only if uv=0 or u=v.

20.5.17. Direct Sums.

Hint.
(a) Let u,uU and v,vV. Then
(u+v)+(u+v)=(u+u)+(v+v)U+Vα(u+v)=αu+αvU+V.

21 Fields
21.5 Exercises

21.5.1.

Hint.
(a) x4(2/3)x262/9; (c) x42x2+25.

21.5.2.

Hint.
(a) {1,2,3,6}; (c) {1,i,2,2i}; (e) {1,21/6,21/3,21/2,22/3,25/6}.

21.5.3.

Hint.
(a) Q(3,7).

21.5.5.

Hint.
Use the fact that the elements of Z2[x]/x3+x+1 are 0, 1, α, 1+α, α2, 1+α2, α+α2, 1+α+α2 and the fact that α3+α+1=0.

21.5.8.

Hint.
False.

21.5.14.

Hint.
Suppose that E is algebraic over F and K is algebraic over E. Let αK. It suffices to show that α is algebraic over some finite extension of F. Since α is algebraic over E, it must be the zero of some polynomial p(x)=β0+β1x++βnxn in E[x]. Hence α is algebraic over F(β0,,βn).

21.5.22.

Hint.
Since {1,3,7,21} is a basis for Q(3,7) over Q, Q(3,7)Q(3+7). Since [Q(3,7):Q]=4, [Q(3+7):Q]=2 or 4. Since the degree of the minimal polynomial of 3+7 is 4, Q(3,7)=Q(3+7).

21.5.27.

Hint.
Let βF(α) not in F. Then β=p(α)/q(α), where p and q are polynomials in α with q(α)0 and coefficients in F. If β is algebraic over F, then there exists a polynomial f(x)F[x] such that f(β)=0. Let f(x)=a0+a1x++anxn. Then
0=f(β)=f(p(α)q(α))=a0+a1(p(α)q(α))++an(p(α)q(α))n.
Now multiply both sides by q(α)n to show that there is a polynomial in F[x] that has α as a zero.

22 Finite Fields
22.4 Exercises

22.4.1.

Hint.
Make sure that you have a field extension.

22.4.4.

Hint.
There are eight elements in Z2(α). Exhibit two more zeros of x3+x2+1 other than α in these eight elements.

22.4.5.

Hint.
Find an irreducible polynomial p(x) in Z3[x] of degree 3 and show that Z3[x]/p(x) has 27 elements.

22.4.7.

Hint.
(a) x51=(x+1)(x4+x3+x2+x+1); (c) x91=(x+1)(x2+x+1)(x6+x3+1).

22.4.8.

Hint.
True.

22.4.11.

Hint.
(a) Use the fact that x71=(x+1)(x3+x+1)(x3+x2+1).

22.4.12.

Hint.
False.

22.4.17.

Hint.
If p(x)F[x], then p(x)E[x].

22.4.18.

Hint.
Since α is algebraic over F of degree n, we can write any element βF(α) uniquely as β=a0+a1α++an1αn1 with aiF. There are qn possible n-tuples (a0,a1,,an1).

22.4.24. Wilson’s Theorem.

Hint.
Factor xp11 over Zp.

23 Galois Theory
23.5 Exercises

23.5.1.

Hint.
(a) Z2; (c) Z2×Z2×Z2.

23.5.2.

Hint.
(a) Separable over Q since x3+2x2x2=(x1)(x+1)(x+2); (c) not separable over Z3 since x4+x2+1=(x+1)2(x+2)2.

23.5.3.

Hint.
If
[GF(729):GF(9)]=[GF(729):GF(3)]/[GF(9):GF(3)]=6/2=3,
then G(GF(729)/GF(9))Z3. A generator for G(GF(729)/GF(9)) is σ, where σ36(α)=α36=α729 for αGF(729).

23.5.5.

Hint.
(a) Q(i)

23.5.7.

Hint.
Let E be the splitting field of a cubic polynomial in F[x]. Show that [E:F] is less than or equal to 6 and is divisible by 3. Since G(E/F) is a subgroup of S3 whose order is divisible by 3, conclude that this group must be isomorphic to Z3 or S3.

23.5.9.

Hint.
G is a subgroup of Sn.

23.5.16.

Hint.
True.

23.5.20.

Hint.
  1. Clearly ω,ω2,,ωp1 are distinct since ω1 or 0. To show that ωi is a zero of Φp, calculate Φp(ωi).
  2. The conjugates of ω are ω,ω2,,ωp1. Define a map ϕi:Q(ω)Q(ωi) by
    ϕi(a0+a1ω++ap2ωp2)=a0+a1ωi++cp2(ωi)p2,
    where aiQ. Prove that ϕi is an isomorphism of fields. Show that ϕ2 generates G(Q(ω)/Q).
  3. Show that {ω,ω2,,ωp1} is a basis for Q(ω) over Q, and consider which linear combinations of ω,ω2,,ωp1 are left fixed by all elements of G(Q(ω)/Q).