<<

. 13
( 14)



>>

100
S = °0 3 0»
006
20

2. The matrix P ’1 is the product of the elementary column matrices in the order in which
they appeared. Thus
® ® ® 
1 ’3 0 1 ’3 0
100
P ’1 = °1 1 0» °0 1 0» = °1 ’2 0»
001 001 001
® 
’2 3 0
P = °’1 1 0»
0 01
The matrix Q is the product of the elementary row matrices in opposite order (i.e., if
R1 appears ¬rst, followed by R2 and R3 , then Q = R3 R2 R1 ). In this case there is only
one matrix, so
® 
100
Q = °0 1 0»
0 ’4 1
A direct computation shows that QAP ’1 = S.
The new basis is given by Y = P X, i.e., y1 = ’2x1 + 3x2 , y2 = ’x1 + x2 , y3 = x3 .
3.
The new set of generators is given by V = SY , i.e., v1 = y1 , v2 = 3y2 , v3 = 6y3 .
Let di = a1 · · · ai . Then di is the gcd of the i — i minors of S, and hence of A. The
4.
ai are recoverable from the di via a1 = d1 and ai = di /di’1 , i > 1. Thus the ai are
determined by the matrix A and do not depend on any particular sequence leading to
a Smith normal form.
5. If A and B have the same Smith normal form S, then A and B are each equivalent to
S and therefore equivalent to each other. If A and B are equivalent, then by the result
stated before Problem 4, they have the same gcd of i—i minors for all i. By Problem 4,
they have the same invariant factors and hence the same Smith normal form.
6. Here are the results, in sequence:
second row is now (3 2 ’13 2)
1. The
¬rst row is (3 2 ’13 2) and the second row is (6 4 13 5)
2. The
3. The second row is (0 0 39 1) and the third row is (0 0 51 4)
4. The third row becomes (0 0 12 3)
5. The second row is (0 0 12 3) and the third row is (0 0 39 1)
third row is (0 0 3 ’8)
6. The
second row is (0 0 3 ’8) and the third row is (0 0 12 3)
7. The
8. The third row is (0 0 0 35)
¬rst row is now (3 2 2 ’38)
9. The
10. The ¬nal matrix is
® 
3 2 2 32
°0 0 3 27» .
0 0 0 35
21

7. We see from the Hermite normal form that we can take x = 0, y = 7, z = 9, provided
0 and 35 are congruent mod m. Thus m must be 5, 7 or 35.


Section 4.6
1. 441 = 32 — 72 , and since there are two partitions of 2, there are 2 — 2 = 4 mutually
nonisomorphic abelian groups of order 441, with the following invariant factors:
(1) a1 = 32 72 , G ∼ Z441
=
(2) a1 = 30 71 , a2 = 32 71 , G ∼ Z7 • Z63
=
(3) a1 = 3 7 , a2 = 3 7 , G ∼ Z3 • Z147
10 12
=
(4) a1 = 31 71 , a2 = 31 71 , G ∼ Z21 • Z21
=
2. 40 = 23 — 51 , and since there are three partitions of 3 and one partition of 1, there
are 3 — 1 = 3 mutually nonisomorphic abelian groups of order 40, with the following
invariant factors:
(1) a1 = 23 51 , G ∼ Z40
=
(2) a1 = 21 50 , a2 = 22 51 , G ∼ Z2 • Z20
=
(3) a1 = 21 50 , a2 = 21 50 , a3 = 21 51 , G ∼ Z2 • Z2 • Z10
=
3. The steps in the computation of the Smith normal form are
®  ®  ®  ® 
153 1 5 3 1 0 0 10 0
°2 ’1 7» ’ °0 ’11 1 » ’ °0 ’11 1 » ’ °0 1 ’11»
0 ’11 ’7 0 ’11 ’7 0 ’7 ’11
342
®  ® 
10 0 1 0 0
’ °0 1 ’11» ’ °0 0»
1
0 0 ’88 0 0 88

Thus G ∼ Z1 • Z1 • Z88 ∼ Z88 .
= =
4. Cancelling a factor of 2 is not appropriate. After the relations are imposed, the group
is no longer free, so that 2y = 0 does not imply that y = 0. Another di¬culty is
that the submodule generated by 2x1 + 2x2 + 8x3 is not the same as the submodule
generated by x1 + x2 + 4x3 .
5. Take M = •∞ Mn , where each Mn is a copy of Z. Take N = Z and P = 0. Since the
n=1
union of a countably in¬nite set and a ¬nite set is still countably in¬nite, we have the
desired result.
6. If N and P are not isomorphic, then the decompositions of N and P will involve
di¬erent sequences of invariant factors. But then the same will be true for M • N and
M • P , so M • N and M • P cannot be isomorphic.


Section 4.7
1. If u is another solution, then f u = f u (= vf ), and since f is injective, u = u .
22

2. By commutativity, wgf a = g vf a, and by exactness, gf = 0. Thus vf a ∈ ker g =
im f by exactness.
3. For commutativity, we must have f ua = vf a = f a , so ua = a . Note that a is
unique because f is injective. Checking that u is a homomorphism is routine, e.g., if
vf ai = f ai , i = 1, 2, then vf (a1 + a2 ) = f (a1 + a2 ), so u(a1 + a2 ) = ua1 + ua2 , etc.
4. uc = ugb = g vb.
5. Suppose c = gb1 = gb2 . Then b1 ’ b2 ∈ ker g = im f by exactness, so b1 ’ b2 = f a.
Then f wa = vf a = v(b1 ’ b2 ). By exactness, 0 = g f wa = g v(b1 ’ b2 ), and the
result follows.
6. Add a vertical identity map at the left side of the diagram and apply (ii) of the four
lemma.
7. Add a vertical identity map at the right side of the diagram and apply (i) of the four
lemma.
8. Add a vertical identity map w at the right side of the diagram and apply (ii) of the
four lemma, shifting the notation [s ’ t, t ’ u, u ’ v, v ’ w].
9. Since u and g are surjective, v must be also, by commutativity.
10. Since f and u are injective, f t, hence t, must be also, by commutativity.
11. Add a vertical identity map s at the left side of the diagram, and apply (i) of the four
lemma, shifting notation [t ’ s, u ’ t, v ’ u, w ’ v].
12. If vb = 0, then b = gm, hence 0 = vgm = g um. Thus um ∈ ker g = im f , say
um = f a . Since t is surjective, a = ta, so uf a = f ta = f a . Therefore um and
uf a are both equal to f a . Since u is injective, m = f a, so b = gm = gf a = 0,
proving that v is injective.
13. Let a ∈ A . Since u is surjective, f a = um, so vgm = g um = g f a = 0. Since v
is injective, gm = 0, hence m ∈ ker g = im f , so m = f a. Thus um = uf a = f ta.
Therefore f a and f ta are both equal to um. Since f is injective, a = ta, proving
that t is surjective.


Section 5.1
1. The kernel of any homomorphism is a (normal) subgroup. If g ∈ ker ¦ then
g(xH) = xH for every x ∈ G, so by (1.3.1), x’1 gx ∈ H. Take x = g to get g ∈ H.
2. By Problem 1, ker ¦ is a normal subgroup of G, necessarily proper since it is contained
in H. Since G is simple, ker ¦ = {1}, and hence ¦ is injective. Since there are n left
cosets of H, ¦ maps into Sn .
3. If [G : H] = n < ∞, then by Problem 2, G can be embedded in Sn , so G is ¬nite, a
contradiction.
4. g(xH) = xH i¬ x’1 gx ∈ H i¬ g ∈ xHx’1 .
5. If x ∈ G, then K = xKx’1 ⊆ xHx’1 , and since x is arbitrary, K ⊆ N .
’1
6. g1 (H © K) = g2 (H © K) i¬ g2 g1 ∈ H © K i¬ g1 H = g2 H and g1 K = g2 K, proving
both assertions.
23

7. Since [G : H] and [G : K] are relatively prime and divide [G : H © K] by (1.3.5), their
product divides, and hence cannot exceed, [G : H © K].
8. By the ¬rst isomorphism theorem, G/N is isomorphic to a group of permutations of L,
the set of left cosets of H. But |L| = [G : H] = n, so by Lagrange™s theorem, |G/N |
divides |Sn | = n!.
9. Since n > 1, H is a proper subgroup of G, and since N is a subgroup of H, N is a
proper subgroup of G as well. If N = {1}, then |G| = [G : N ], so by Problem 8,
G divides n!, contradicting the hypothesis. Thus {1} < N < G, and G is not simple.


Section 5.2
1. For arbitrary σ and π, we have πσπ ’1 (π(i)) = πσ(i). In the cycle decomposition of σ,
i is followed by σ(i), and in the cycle decomposition of πσπ ’1 , π(i) is followed by
πσ(i), exactly as in the given numerical example.
2. If g ∈ CG (S) and x ∈ S then gxg ’1 = x, so gSg ’1 = S, hence CG (S) ¤ NG (S).
If g ∈ NG (S) and x ∈ S, then gxg ’1 ∈ S, and the action is legal. As in (5.1.3),
Example 3, the kernel of the action consists of all elements of NG (S) that commute
with everything in S, that is, NG (S) © CG (S) = CG (S).
3. We have z ∈ G(gx) i¬ zgx = gx i¬ g ’1 zgx = x i¬ g ’1 zg ∈ G(x) i¬ z ∈ gG(x)g ’1 .
’1 ’1
4. We have g1 G(x) = g2 G(x) i¬ g2 g1 ∈ G(x) i¬ g2 g1 x = x i¬ g1 x = g2 x, proving that
Ψ is well-de¬ned and injective. If the action is transitive and y ∈ X, then for some x,
y = gx = Ψ(gG(x)) and Ψ is surjective.
5. If g, h ∈ G, then h takes gx to hgx. In the coset action, the corresponding statement is
that h takes gG(x) to hgG(x). The formal statement is that Ψ is a“G-set isomorphism”.
In other words, Ψ is a bijection of the space of left cosets of G(x) and X, with Ψ(hy) =
hΨ(y) for all h ∈ G and y in the coset space. Equivalently, the following diagram is
commutative.
G hgx
gx
y y
Ψ Ψ

G hgG(x)
gG(x)

6. The two conjugacy classes are {1} and G \ {1}. Thus if |G| = n > 1, the orbit sizes
under conjugacy on elements are 1 and n ’ 1. But each orbit size divides the order of
the group, so n ’ 1 divides n. Therefore n = k(n ’ 1), where k is a positive integer.
Since k = 1 is not possible, we must have k ≥ 2, so n ≥ 2(n ’ 1), so n ¤ 2.
7. If gi is an element in the ith conjugacy class, 1 ¤ i ¤ k, then by the orbit-stabilizer
theorem, the size of this class is |G|/|CG (gi )|. Since the orbits partition G, the sum of
the class sizes is |G|, and
k
1
=1
xi
i=1
24

where xi = |CG (gi )|. If, say, g1 = 1, so that x1 = |G|, the result follows from the
observation that each xi , in particular x1 , is bounded by N (k).


Section 5.3
1. The group elements are I, R = (1, 2, 3, 4), R2 = (1, 3)(2, 4), R3 = (1, 4, 3, 2),
F = (1)(3)(2, 4), RF = (1, 2)(3, 4), R2 F = (1, 3)(2)(4), R3 F = (1, 4)(2, 3). Thus
the number of distinct colorings is

14 14
n + n + n2 + n + n3 + n 2 + n3 + n2 = n + 2n3 + 3n2 + 2n .
8 8

2. Yes. If the vertices of the square are 1, 2, 3, 4 in counterclockwise order, we can identify
vertex 1 with side 12, vertex 2 with side 23, vertex 3 with side 34, and vertex 4 with
side 41. This gives a one-to-one correspondence between colorings in one problem and
colorings in the other.
3. If the vertices of the square are 1, 2, 3, 4 in counterclockwise order, then W GGW
will mean that vertices 1 and 4 are colored white, and vertices 2 and 3 green. The
equivalence classes are

{W W W W }, {GGGG}, {W GGG, GW GG, GGW G, GGGW },
{GW W W, W GW W, W W GW, W W W G},
{W W GG, GW W G, GGW W, W GGW }, {W GW G, GW GW }.

4. Label (’1, 0) as vertex 1, (0, 0) as vertex 2, and (1, 0) as vertex 3. Then I = (1)(2)(3)
and σ = (1, 3)(2). Thus the number of distinct colorings is 1 (n3 + n2 ).
2
5. We have free choice of color in two cycles of I and one cycle of σ. The number of
distinct colorings is 1 (n2 + n).
2
6. We can generate a rotation by choosing a face of the tetrahedron to be placed on a
table or other ¬‚at surface, and then choosing a rotation of 0,120 or 240 degrees. Thus
there are 12 rotations, and we have enumerated all of them. By examining what each
rotation does to the vertices, we can verify that all permutations are even. Since A4
has 4!/2 = 12 members, G must coincide with A4 , up to isomorphism.
7. The members of A4 are (1, 2, 3), (1, 3, 2), (1, 2, 4), (1, 4, 2), (1, 3, 4), (1, 4, 3), (2, 3, 4),
(2, 4, 3), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3), and the identity. Counting cycles of length
1, we have 11 permutations with 2 cycles and one permutation with 4 cycles. The
1
number of distinct colorings is 12 (n4 + 11n2 ).
8. In the above list of permutations, the ¬rst 8 have no ¬xed colorings. In the next 3, we
can pick a cycle to be colored B, and pick a di¬erent color for the other cycle. This
gives 2 — 3 = 6 ¬xed colorings. For the identity, we can pick two vertices to be colored
B, and then choose a di¬erent color for each of the other two vertices. The number
of ¬xed colorings is 4 32 = 54. The number of distinct colorings of the vertices is
2
[(6x3) + 54)]/12 = 6.
25

9. As in Problem 6, a rotation can be generated by choosing a face of the cube to be
placed on a table, and then choosing a rotation of 0,±90 or 180 degrees. Thus there
are 24 rotations, and we have enumerated all of them. Alternatively, there is a one-
to-one correspondence between rotations and permutations of the 4 diagonals of the
cube. Since there are 4! = 24 permutations of a set with 4 elements, there can be no
additional rotations. The correspondence between rotations and permutations of the
diagonals yields an isomorphism of G and S4 .
10. Any permutation of the faces except the identity has a cycle of length 2 or more, and
each of the faces within that cycle must receive the same color, which is a contradic-
tion. Thus f (π) = 0 for π = I. Now I ¬xes all legal colorings, and since there are 6
colors and 6 faces, the number of legal colorings is 6! = 720. The number of distinct
colorings is therefore 720/24 = 30.
Remark This problem can be solved directly without using the heavy machinery
of this section. Without loss of generality, choose any particular color for a particular
face, and move the cube so that this face is at the bottom. Choose one of the remaining
5 colors for the top face. The number of allowable colorings of the 4 remaining sides
of the cube is the number of circular permutations of 4 objects, which is 3! = 6. The
number of distinct colorings is 5 — 6 = 30.
11. The group G = {1, R, R2 , . . . , Rp’1 } is cyclic of order p. Since p is prime, each Ri ,
i = 1, . . . , p ’ 1, has order p, and therefore as a permutation of the vertices consists
of a single cycle. Thus the number of distinct colorings is
1p
[n + (p ’ 1)n] .
p
12. Since the result of Problem 11 is an integer, np + (p ’ 1)n = np ’ n + np is a multiple
of p, hence so is np ’ n. Thus for any positive integer n, np ≡ n mod p. It follows
that if n is not a multiple of p, then np’1 ≡ 1 mod p.


Section 5.4
1. Let G act on subgroups by conjugation. If P is a Sylow p-subgroup, then the stabilizer
of P is NG (P ) (see (5.2.2), Example 4). By (5.2.3), the index of NG (P ) is np .
2. Since P is normal in NG (P ) (see (5.2.2), Example 4), P Q = QP ¤ G by (1.4.3). By
(5.2.4), P Q is a p-subgroup.
3. The Sylow p-subgroup P is contained in P Q, which is a p-subgroup by Problem 2.
Since a Sylow p-subgroup is a p-subgroup of maximum possible size, we have P = P Q,
and therefore Q ⊆ P .
4. (a) By de¬nition of normalizer, we have gP g ’1 ¤ gNG (P )g ’1 ¤ gHg ’1 = H. Thus
P and gP g ’1 are subgroups of H, and since they are p-subgroups of maximum
possible size, they are Sylow p-subgroups of H.
(b) Since H is always a subgroup of its normalizer, let g ∈ NG (H). By (a), P and
gP g ’1 are conjugate in H, so for some h ∈ H we have gP g ’1 = hP h’1 . Thus
(h’1 g)P (h’1 g)’1 = P , so h’1 g ∈ NG (P ) ¤ H. But then g ∈ H, and the result
follows.
26

5. By the second isomorphism theorem, [N : P © N ] = [P N : P ] = |P N |/|P |. Since |P | is
the largest possible power of p for p-subgroups of G, [P N : P ] and p must be relatively
prime. Therefore [N : P © N ] and p are relatively prime, so P © N is a p-subgroup
of N of maximum possible size, i.e., a Sylow p-subgroup of N .
6. By the third isomorphism theorem, [G/N : P N/N ] = [G : P N ] = |G|/|P N |. Since
|G|/|P | and p are relatively prime and P ¤ P N , it follows that |G|/|P N | and p are
relatively prime. The result follows as in Problem 5.
7. Since f is an automorphism, f (P ) is a subgroup of G and has the same number of
elements as P , in other words, f (P ) is a Sylow p-subgroup. By hypothesis, f (P ) = P .
8. By (1.3.5), [G : N ] = [G : H][H : N ] = p[H : N ], and since [G : N ] divides p! =
p(p ’ 1)!, the result follows.
9. If q is a prime factor of [H : N ], then by Problem 8, q is a divisor of some integer
between 2 and p ’ 1, in particular, q ¤ p ’ 1. But by Lagrange™s theorem, q divides
|H|, hence q divides |G|. This contradicts the fact that p is the smallest prime divisor
of |G|. We conclude that there are no prime factors of [H : N ], which means that
[H : N ] = 1, Thus H = N , proving that H is normal in G.


Section 5.5
1. This follows from (5.5.6), part (iii), with p = 3 and q = 5.
2. Let Z(G)a be a generator of G/Z(G). If g1 , g2 ∈ G, then Z(G)g1 = Z(G)ai for some i,
so g1 a’i = z1 ∈ Z(G), and similarly g2 a’j = z2 ∈ Z(G). Thus g1 g2 = ai z1 aj z2 =
z1 z2 ai+j = z2 z1 aj+i = z2 aj z1 ai = g2 g1 .
3. By (5.5.3), the center Z(G) is nontrivial, so has order p or p2 . In the latter case,
G = Z(G), so G is abelian. If |Z(G)| = p, then |G/Z(G)| = p, and G/Z(G) has order p
and is therefore cyclic. By Problem 2, G is abelian (and |Z(G)| must be p2 , not p).
4. Each Sylow p-subgroup is of order p and therefore has p ’ 1 elements of order p, with
a similar statement for q and r. If we include the identity, we have 1 + np (p ’ 1) +
nq (q ’ 1) + nr (r ’ 1) distinct elements of G, and the result follows.
5. G cannot be abelian, for if so it would be cyclic of prime order. By (5.5.5), np , nq
and nr are greater than 1. We know that np divides qr and np > 1. But np can™t be
q since q ≡ 1 mod p (because p > q). Similarly, np can™t be r, so np = qr. Now nq
divides pr and is greater than 1, so as above, nq must be either p or pr (it can™t be r
because q > r, so r ≡ 1 mod q). Thus nq ≥ p. Finally, nr divides pq and is greater
than 1, so nr is p, q, or pq. Since p > q, we have nr ≥ q.
6. Assume that G is simple. Substituting the inequalities of Problem 5 into the identity
of Problem 4, we have
pqr ≥ 1 + qr(p ’ 1) + p(q ’ 1) + q(r ’ 1).
Thus
0 ≥ pq ’ p ’ q + 1 = (p ’ 1)(q ’ 1),
a contradiction.
27

7. Since |P | = pr with r ≥ 1 and m > 1, we have 1 < |P | < |G|. Since G is simple, P is
not normal in G. By (5.5.4), n > 1. By Problem 9 of Section 5.1, |G| divides n!.
8. Assume G simple, and let n = np with p = 5. By Sylow (2), n divides 24 = 16 and
n ≡ 1 mod 5. The only divisors of 16 that are congruent to 1 mod 5 are 1 and 16, and
1 is excluded by Problem 7. Thus the only possibility is n = 16, and by Problem 7,
24 56 divides 16!, hence 56 divides 16!. But in the prime factorization of 16!, 5 appears
with exponent 3 (not 6), due to the contribution of 5,10 and 15. We have reached a
contradiction, so G cannot be simple.


Section 5.6
1. Apply the Jordan-H¨lder theorem to the series 1
o N G.
2. Z has no composition series. By Section 1.1, Problem 6, each nontrivial subgroup of
Z consists of multiples of some positive integer, so the subgroup is isomorphic to Z
itself. Thus Z has no simple subgroups, so if we begin with {0} and attempt to build
a composition series, we cannot even get started.
3. We have the composition series 1 Z2 Z2 • Z3 ∼ Z6 (or 1 Z3 Z2 • Z3 ∼ Z6 ) and
= =
1 A3 S3 .
4. An consists of products of an even number of transpositions, and the result follows
from the observation that (a, c)(a, b) = (a, b, c) and (c, d)(a, b) = (a, d, c)(a, b, c).
5. If (a, b, c) ∈ N and (d, e, f ) is any 3-cycle, then for some permutation π we have
π(a, b, c)π ’1 = (d, e, f ). Explicitly, we can take π(a) = d, π(b) = e, π(c) = f ; see
Section 5.2, Problem 1. We can assume without loss of generality that π is even, for
if it is odd, we can replace it by (g, h)π, where g and h are not in {d, e, f }. (We use
n ≥ 5 here.) Since N is normal, (d, e, f ) ∈ N .
6. If N contains (1, 2, 3, 4), then it contains (1, 2, 3)(1, 2, 3, 4)(1, 3, 2) = (1, 4, 2, 3), and
hence contains (1, 4, 2, 3)(1, 4, 3, 2) = (1, 2, 4), contradicting Problem 5. If N con-
tains (1, 2, 3, 4, 5), then it contains (1, 2, 3)(1, 2, 3, 4, 5)(1, 3, 2) = (1, 4, 5, 2, 3), and so
contains (1, 4, 5, 2, 3)(1, 5, 4, 3, 2) = (1, 2, 4), a contradiction. The analysis for longer
cycles is similar. [Actually, we should have assumed that N contains a permuta-
tion π whose disjoint cycle decomposition is · · · (1, 2, 3, 4) · · · . But multiplication by
π ’1 = · · · (1, 4, 3, 2) · · · cancels the other cycles.]
7. If N contains (1, 2, 3)(4, 5, 6), then it must also contain (3, 4, 5)(1, 2, 3)(4, 5, 6)(3, 5, 4) =
(1, 2, 4)(3, 6, 5). Thus N also contains (1, 2, 4)(3, 6, 5)(1, 2, 3)(4, 5, 6) = (1, 4, 3, 2, 6),
which contradicts Problem 6. If the decomposition of a permutation σ in N contains
a single 3-cycle, then σ 2 is a 3-cycle in N , because a transposition is its own inverse.
This contradicts Problem 5.
8. If, (1, 2)(3, 4) ∈ N , then (1, 5, 2)(1, 2)(3, 4)(1, 2, 5) = (1, 5)(3, 4) belongs to N , and so
does (1, 5)(3, 4)(1, 2)(3, 4) = (1, 2, 5), contradicting Problem 5.
9. If N contains (1, 2)(3, 4)(5, 6)(7, 8), then it contains

(2, 3)(4, 5)(1, 2)(3, 4)(5, 6)(7, 8)(2, 3)(4, 5) = (1, 3)(2, 5)(4, 6)(7, 8).
28

Therefore N contains

(1, 3)(2, 5)(4, 6)(7, 8)(1, 2)(3, 4)(5, 6)(7, 8) = (1, 5, 4)(2, 3, 6),

contradicting Problem 7.
10. We can reproduce the analysis leading to the Jordan-H¨lder theorem, with appropriate
o
notational changes. For example, we replace the “subnormal” condition Gi Gi+1
by the “normal” condition Gi G.
11. We say that N is a minimal normal subgroup of H if {1} < N H and there
is no normal subgroup of H strictly between {1} and N . In a chief series, there
can be no normal subgroup of G strictly between Gi and Gi+1 . Equivalently, by
the correspondence theorem, there is no normal subgroup of G/Gi strictly between
Gi /Gi = {1} and Gi+1 /Gi . Thus Gi+1 /Gi is a minimal normal subgroup of G/Gi .


Section 5.7
1. S3 is nonabelian and solvable (1 A3 S3 ).
2. Let 1 = G0 G1 · · · Gr = G be a composition series, with all Gi /Gi’1 cyclic of
prime order (see (5.7.5)). Since |Gi | = |Gi /Gi’1 ||Gi’1 | and G0 is ¬nite, an induction
argument shows that G is ¬nite.
3. The factors of a composition series are simple p-groups P , which must be cyclic of
prime order. For 1 Z(P ) P , so Z(P ) = P and P is abelian, hence cyclic of prime
order by (5.5.1). [The trivial group is solvable with a derived series of length 0.]
4. S3 is solvable by Problem 1, but is not nilpotent. Since S3 is nonabelian, a central
series must be of the form 1 H S3 with H ⊆ Z(S3 ) = 1, a contradiction.
5. If Sn is solvable, then so is An by (5.7.4), and this contradicts (5.7.2).
6. By (5.5.3), P has a nontrivial center. Since Z(P ) is normal in P and P is simple,
Z(P ) = P and P is abelian. By (5.5.1), P is cyclic of prime order, and since P is a
p-group, the only possibility is |P | = p.
7. Let N be a maximal proper normal subgroup of P . (N exists because P is ¬nite
and nontrivial, and 1 P .). Then the p-group P/N is simple (by the correspondence
theorem). By Problem 6, |P/N | = p.
8. If P is a Sylow p-subgroup of G, then |P | = pr and by Problem 7, P has a subgroup
Q1 of index p, hence of order pr’1 . If Q1 is nontrivial, the same argument shows that
Q1 has a subgroup Q2 of order pr’2 . An induction argument completes the proof.
9. Let G = D6 , the group of symmetries of the equilateral triangle. Take N = {I, R, R2 },
where R is rotation by 120 degrees. Then N has index 2 in G and is therefore normal.
(See Section 1.3, Problem 6, or Section 5.4, Problem 9.) Also, N has order 3 and
G/N has order 2, so both N and G/N are cyclic, hence abelian. But G is not abelian,
since rotations and re¬‚ections do not commute.
10. It follows from the splicing technique given in the proof of (5.7.4) that dl(G) ¤
dl(N ) + dl(G/N ).
29

Section 5.8
1. Let H = a | an , and let Cn be a cyclic group of order n generated by a. Then
an = 1 in Cn , and since an+j = aj , we have |H| ¤ n = |Cn |. The result follows as
in (5.8.6).
2. The discussion in Example 4 of (2.1.3), with i = a and j = b, shows that the quater-
nion group Q satis¬es all the relations. Since ab = ba’1 , it follows as in (1.2.4) that
every element of the given presentation H is of the form br as , r, s ∈ Z. Since b2 = a2 ,
we can restrict r to 0 or 1, and since a4 = 1, we can restrict s to 0, 1, 2 or 3. Thus
|H| ¤ 8, and the result follows as in (5.8.6).
3. Take a = (1, 2, 3) and b = (1, 2) to show that S3 satis¬es all the relations. Since
ba = a’1 b, each element of H is of the form ar bs , r, s ∈ Z. Since a3 = b2 = 1,
|H| ¤ 3 — 2 = 6, and the result follows as in (5.8.6).
4. No. There are many counterexamples; an easy one is Cn = a | an = 1, a2n = 1 , the
cyclic group of order n.
5. n’1 n1 = h2 h’1 ∈ N © H = 1.
2 1
6. Take ψ to be the inclusion map. Then πψ(h) = π(h) = π(1h) = h. To show that π is
a homomorphism, note that n1 h1 n2 h2 = n1 (h1 n2 h’1 )h1 h2 and h1 n2 h’1 ∈ N .
1 1
7. If g ∈ G then g = (gπ(g)’1 )π(g) with π(g) ∈ H and π(gπ(g)’1 ) = π(g)π(g)’1 = 1, so
gπ(g)’1 ∈ N . [Remember that since we are taking ψ to be inclusion, π is the identity
on H.] Thus G = N H. If g ∈ N © H, then g ∈ ker π and g ∈ H, so g = π(g) = 1,
proving that H © N = 1.
8. If we de¬ne π(n, h) = (1, h), i(n, 1) = (n, 1), and ψ(1, h) = (1, h), then the sequence
of Problem 6 is exact and splits on the right.
9. We have (n1 h1 )(n2 h2 ) = n1 (h1 n2 h’1 )h1 h2 , so we may take f (h) to be the inner
1
automorphism of N given by conjugation by h ∈ H.
10. Consider the sequence

G C3 G S3 G C2 G1
i π
1

where C3 consists of the identity 1 and the 3-cycles (1, 2, 3) and (1, 3, 2), and C2 con-
sists of the identity and the 2-cycle (1, 2). The map i is inclusion, and π takes each
2-cycle to (1, 2) and each 3-cycle to the identity. The identity map from C2 to S3
gives a right-splitting, but there is no left splitting. If g were a left-splitting map
from S3 to C3 , then g(1, 2) = (1, 2, 3) is not possible because g(1) = g(1, 2)g(1, 2) =
(1, 2, 3)(1, 2, 3) = (1, 3, 2), a contradiction. Similarly, g(1, 2) = (1, 3, 2) is impossible,
so g(1, 2) = 1, so g —¦ i cannot be the identity. Explicitly, g(2, 3) = g((1, 2)(1, 2, 3)) =
g(1, 2, 3) = (1, 2, 3), and g(1, 3) = g((1, 2)(1, 3, 2)) = g(1, 3, 2) = (1, 3, 2). Conse-
quently, g(1, 3, 2) = g((1, 3)(2, 3)) = 1, a contradiction.
11. In the exact sequence of Problem 6, take G = Zp2 , N = Zp , H = G/N ∼ Zp , i the =
inclusion map, and π the canonical epimorphism. If f is a right-splitting map, its
image must be a subgroup with p elements (since f is injective), and there is only one
such subgroup, namely Zp . But then π —¦ f = 0, a contradiction.
30

12. If g ∈ G, then gP g ’1 ⊆ gN g ’1 = N , so P and gP g ’1 are both Sylow p-subgroups
of N . By Sylow (3), they are conjugate in N (the key point). Thus for some n ∈ N we
have P = n(gP g ’1 )n’1 . But then by de¬nition of normalizer we have ng ∈ NG (P ),
hence g ∈ N NG (P ).
13. The multiplication table of the group is completely determined by the relations an = 1,
b2 = 1, and bab’1 = a’1 . The relations coincide with those of D2n , with a = R
and b = F .
14. The relation an = 1 disappears, and we have a, b | b2 = 1, bab’1 = a’1 .
Solutions Chapters 6“10

Section 6.1
1. We have r1 = 2, r2 = 1, r3 = 1 so t1 = 1, t2 = 0, t3 = 1. The algorithm terminates in
one step after after subtraction of (X1 + X2 + X3 )(X1 X2 X3 ). The given polynomial
can be expressed as e1 e3 .
2. We have r1 = 2, r2 = 1, r3 = 0 so t1 = 1, t2 = 1, t3 = 0. At step 1, subtract
(X1 + X2 + X3 )(X1 X2 + X1 X3 + X2 X3 ). The result is ’3X1 X2 X3 + 4X1 X2 X3 =
X1 X2 X3 . By inspection (or by a second step of the algorithm), the given polynomial
can be expressed as e1 e2 + e3 .
3. Equation (1) follows upon taking σ1 (h) outside the summation and using the linear
dependence. Equation (2) is also a consequence of the linear dependence, because
σi (h)σi (g) = σi (hg).
4. By hypothesis, the characters are distinct, so for some h ∈ G we have σ1 (h) = σ2 (h).
Thus in (3), each ai is nonzero and

=0 if i = 1;
σ1 (h) ’ σi (h)
=0 if i = 2.

This contradicts the minimality of r. (Note that the i = 2 case is important, since
there is no contradiction if σ1 (h) ’ σi (h) = 0 for all i.)
5. By (3.5.10), the Galois group consists of the identity alone. Since the identity ¬xes all

elements, the ¬xed ¬eld of G is Q( 3 2).
6. Since C = R[i], an R-automorphism σ of C is determined by its action on i. Since σ
must permute the roots of X 2 + 1 by (3.5.1), we have σ(i) = i or ’i. Thus the Galois
group has two elements, the identity automorphism and complex conjugation.
7. The complex number z is ¬xed by complex conjugation if and only if z is real, so the
¬xed ¬eld is R.


Section 6.2
p
1. The right side is a subset of the left since both Ei and Ei+1 are contained in Ei+1 . Since
p
Ei is contained in the set on the right, it is enough to show that ±i+1 ∈ Ei (Ei+1 ). By

1
2

p
hypothesis, ±i+1 is separable over F , hence over Ei (±i+1 ). By Section 3.4, Problem 3,
p p
±i+1 ∈ Ei (±i+1 ) ⊆ Ei (Ei+1 ).
p
2. Apply Section 3.4, Problem 7, with E = F (E p ) replaced by Ei+1 = Ei (Ei+1 ), to
conclude that Ei+1 is separable over Ei . By the induction hypothesis, Ei is separable
over F . By transitivity of separable extensions (Section 3.4, Problem 8), Ei+1 is
separable over F . By induction, E/F is separable.
3. Let fi be the minimal polynomial of ±i over F . Then E is a splitting ¬eld for f =
f1 · · · fn over F , and the result follows.
4. This is a corollary of part 2 of the fundamental theorem, with F replaced by Ki’1 and
G replaced by Gal(E/Ki’1 ) = Hi’1 .
5. E(A) is a ¬eld containing E ≥ F and A, hence E(A) contains E and K, so that by
de¬nition of composite, EK ¤ E(A). But any ¬eld (in particular EK) that contains
E and K contains E and A, hence contains E(A). Thus E(A) ¤ EK.
6. If σ ∈ G, de¬ne Ψ(σ)(„ (x)) = „ σ(x), x ∈ E. Then ψ(σ) ∈ G . [If y = „ (x) ∈ F with
x ∈ F , then Ψ(σ)y = Ψ(σ)„ x = „ σ(x) = „ (x) = y.] Now Ψ(σ1 σ2 )„ (x) = „ σ1 σ2 (x) and
Ψ(σ1 )Ψ(σ2 )„ (x) = Ψ(σ1 )„ σ2 (x) = „ σ1 σ2 (x), so Ψ is a group homomorphism. The
inverse of Ψ is given by Ψ (σ )„ ’1 y = „ ’1 σ (y), σ ∈ G , y ∈ E . To see this, we
compute

Ψ (Ψ(σ))„ ’1 y = „ ’1 Ψ(σ)y = „ ’1 Ψ(σ)„ x = „ ’1 „ σ(x) = σ(x) = σ(„ ’1 y).

Thus Ψ Ψ is the identity on G.
7. Since H is a normal subgroup of G, its ¬xed ¬eld L = F (H ) is normal over F , so
by minimality of the normal closure, we have N ⊆ L. But all ¬xed ¬elds are sub¬elds
of N , so L ⊆ N , and consequently L = N .
8. If σ ∈ H , then σ ¬xes everything in the ¬xed ¬eld N , so σ is the identity. Thus the
largest normal subgroup of G that is contained in H is trivial. But this largest normal
subgroup is the core of H in G, and the resulting formula follows from Problems 4
and 5 of Section 5.1.


Section 6.3
1. G = {σ1 , . . . , σn } where σi is the unique F -automorphism of E that takes ± to ±i .
2. We must ¬nd an ± such that 1, ±, . . . , ±n’1 is a basis for E/Q. If ± = b1 x1 + · · · +
bn xn , we can compute the various powers of ± and write ±i = ci1 x1 + · · · + cin xn ,
i = 0, 1, . . . , n ’ 1, where each cij is a rational number. The powers of ± will form
a basis i¬ det[cij ] = 0. This will happen “with probability 1”; if a particular choice
of the bi yields det[cij ] = 0, a slight perturbation of the bi will produce a nonzero
determinant.
3. By (6.3.1), we may regard G as a group of permutations of the roots ±1 , . . . , ±n of f ,
and therefore G is isomorphic to a subgroup H of Sn . Since G acts transitively on
the ±i (see (6.3.1)), the natural action of H on {1, 2, . . . , n} is transitive. [For an
earlier appearance of the natural action, see the discussion before (5.3.1).]
3

4. The Galois group G must be isomorphic to a transitive subgroup of S2 , which is cyclic
of order 2. There is only one transitive subgroup of S2 , namely S2 itself, so G is a
cyclic group of order 2.
√ √√ √
5. Since [Q( √ : Q] =√ and [Q( 2, 3) : Q( 2)] = 2, the Galois group G has order 4.
2) 2 √ √
[Note that 3 ∈ Q( 2) because a + b 2 √ never be 3 for a, b ∈ Q.] An automor-
/ can
√ √ √
phism σ in G must take 2 to ± 2 and 3 to ± 3. Thus σ is either the identity or
has order 2. Now a group in which every element has order 1 or 2 must be abelian,
regardless of the size of the group [(ab)(ab) = 1, so ab = b’1 a’1 = ba]. Since G is not
cyclic, it must be isomorphic to the four group Z2 • Z2 . (See the analysis in (4.6.4).)
6. Let H be the subgroup generated by H1 and H2 , that is, by H1 ∪ H2 . If σ ∈ H1 ∪ H2 ,
then σ ¬xes K1 © K2 = K. Since H consists of all ¬nite products (= compositions)
of elements in H1 or H2 , everything in H ¬xes K, so that K ⊆ F(H). On the other
hand, if x ∈ F(H) but x ∈ K, say x ∈ K1 . Then some „ ∈ H1 ⊆ H fails to ¬x x, so
/ /
x ∈ F(H), a contradiction. Therefore K = F(H).
/
7. The ¬xed ¬eld is K1 K2 , the composite of K1 and K2 . For if σ ¬xes K1 K2 , then
it ¬xes both K1 and K2 , so σ belongs to H1 © H2 . Conversely, if σ ∈ H1 © H2 ,
then σ is the identity on both K1 and K2 . But by the explicit form of K1 K2 (see
Section 3.1, Problem 1 and Section 6.2, Problem 5), σ is the identity on K1 K2 . Thus
F(H1 © H2 ) = K1 K2 .
8. We have E = F (±1 , . . . , ±n ), where the ±i are the roots of f . Since min(±i , F ) divides
the separable polynomial f , each ±i is separable over F . By Section 6.2, Problem 1,
E is separable over F .
9. Since [Q(θ, i) : Q] = [Q(θ) : Q][Q(θ, i) : Q(θ)] = 4 — 2 = 8, we have |G| = 8. Any
σ ∈ G must map θ to a root of f (4 choices), and i to a root of X 2 + 1 (2 choices, i
or ’i). Since σ is determined by its action on θ and i, we have found all 8 members
of G.
10. Let σ(θ) = iθ, σ(i) = i, and let „ (θ) = θ, „ (i) = ’i. Then σ 4 = 1, „ 2 = 1, and the
automorphisms 1, σ, σ 2 , σ 3 , „, σ„, σ 2 „, σ 3 „ are distinct (by direct veri¬cation). Also,
we have σ„ = „ σ ’1 = „ σ 3 . The result follows from the analysis of the dihedral group
in Section 5.8.

11. By direct veri¬cation, every member of N ¬xes iθ2 = i 2. Since N has index 2 in G, √
the ¬xed ¬eld of N has degree 2 over Q. But the minimal polynomial of i 2 over Q

is X 2 + 2, and it follows that F(N ) = Q(i 2}. F(N ) is the splitting ¬eld of X 2 + 2
over Q and is therefore normal over Q, as predicted by Galois theory.


Section 6.4
1. We have ±4 = 1 + ± + ±2 + ±3 and ±5 = 1. Thus the powers of ± do not exhaust the
nonzero elements of GF (16).
2. We may assume that E = GF (pn ) and that E contains F = GF (pm ), where n = md.
Then [E : F ] = [E : Fp ]/[F : Fp ] = n/m = d. Since E/F is separable, we have
E = F (±) by the theorem of the primitive element. The minimal polynomial of ±
over F is an irreducible polynomial of degree d.
4

3. Exactly as in (6.4.5), carry out a long division of X n ’ 1 by X m ’ 1. The division
will be successful i¬ m divides n.
4. Since the bi belong to L, we have K ⊆ L, and since h ∈ L[X], it follows that g|h. But
g ∈ K[X] by de¬nition of K, so h|g. Since g and h are monic, they must be equal.
In particular, they have the same degree, so [E : L] = [E : K]. Since K ⊆ L, we have
L = K.
5. Since L = K, L is completely determined by g. But if f = min(±, F ), then g divides f .
Since f has only ¬nitely many irreducible divisors, there can only be ¬nitely many
intermediate ¬elds L.
6. Since there are ¬nitely many intermediate ¬elds between E and F , the same is true
between L and F . By induction hypothesis, L = F (β) for some β ∈ L. Thus
E = L(±n ) = F (β, ±n ).
7. By hypothesis, there are only ¬nitely many ¬elds of the form F (cβ + ±n ), c ∈ F . But
there are in¬nitely many choices of c, and the result follows.
8. Since E = F (β, ±n ), it su¬ces to show that β ∈ F (cβ + ±n ). This holds because
(cβ + ±n ) ’ (dβ + ±n )
β= .
c’d
9. Let σ : F ’ F be the Frobenius automorphism, given by σ(x) = xp . Let f =
min(±, Fp ) and g = min(±p , Fp ). Then f (±p ) = f (σ(±)) = σ(f (±)) since σ is a
monomorphism, and σ(f (±)) = σ(0) = 0. Thus g divides the monic irreducible
polynomial f , so g = f .
10. By Problem 9, the subsets are {0}, {1, 3, 9}, {2, 6, 5}, {4, 12, 10}, and {7, 8, 11}. [For
example, starting with 2, we have 2 — 3 = 6, 6 — 3 = 18 ≡ 5 mod 13, 5 — 3 = 15 ≡
2 mod 13.] In the second case, we get

{0}, {1, 2, 4, 8}, {3, 6, 9, 12}, {5, 10}, {7, 14, 13, 11}.


Section 6.5
1. Ψn (X p ) = i (X p ’ωi ) where the ωi are the primitive nth roots of unity. But the roots
of X p ’ ωi are the pth roots of ωi , which must be primitive npth roots of unity because
p is prime and p divides n. The result follows. (The map θ ’ θp is a bijection between
primitive npth roots of unity and primitive nth roots of unity, because •(np) = p•(n).)
2. By (6.5.1) and (6.5.6), the Galois group of the nth cyclotomic extension of Q can be
identi¬ed with the group of automorphisms of the cyclic group of nth roots of unity.
By (6.5.6), the Galois group is isomorphic to Un , and the result follows.
3. The powers of 3 mod 7 are 3, 9 ≡ 2, 6, 18 ≡ 4, 12 ≡ 5, 1.
4. This follows from Problem 3 and (1.1.4).
5. σ6 (ω + ω 6 ) = ω 6 + ω 36 = ω + ω 6 , so ω + ω 6 ∈ K. Now ω + ω 6 = ω + ω ’1 = 2 cos 2π/7,
so ω satis¬es a quadratic equation over Q(cos 2π/7). By (3.1.9),

[Q7 : Q] = [Q7 : K][K : Q(cos 2π/7)][Q(cos 2π/7) : Q]
5

where the term on the left is 6, the ¬rst term on the right is | σ6 | = 2, and the second
term on the right is (by the above remarks) 1 or 2. But [K : Q(cos 2π/7)] cannot be 2
(since 6 is not a multiple of 4), so we must have K = Q(cos 2π/7).
6. σ2 (ω + ω 2 + ω 4 ) = ω 2 + ω 4 + ω 8 = ω + ω 2 + ω 4 , so ω + ω 2 + ω 4 ∈ L;
σ3 (ω + ω 2 + ω 4 ) = ω 3 + ω 6 + ω 12 = ω 3 + ω 5 + ω 6 = ω + ω 2 + ω 4 , so ω + ω 2 + ω 4 ∈ Q.
/
3 5 6 2 4
[If ω + ω + ω = ω + ω + ω , then we have two distinct monic polynomials of degree
6 satis¬ed by ω (the other is Ψ7 (X)), which is impossible.]
7. By the fundamental theorem, [L : Q] = [G : σ2 ] = 2, so we must have L = Q(ω +
ω 2 + ω 4 ).
8. The roots of Ψq are the pr th roots of unity that are not pr’1 th roots of unity. Thus
r
Xp ’ 1 tp ’ 1
Ψq (X) = pr’1 =
t’1
’1
X
and the result follows.
9. By Problem 1,
Ψ18 (X) = Ψ(3)(6) (X) = Ψ6 (X 3 ) = X 6 ’ X 3 + 1.


Section 6.6
1. f is irreducible by Eisenstein, and the Galois group is S3 . This follows from (6.6.7) or
via the discriminant criterion of (6.6.3); we have D(f ) = ’27(4) = ’108, which is not
a square in Q.
2. f is irreducible by the rational root test, and D(f ) = ’4(’3)3 ’ 27 = 108 ’ 27 = 81,
a square in Q. Thus the Galois group is A3 .
3. f is irreducible by Eisenstein. The derivative is f (X) = 5X 4 ’ 40X 3 = 5X 3 (X ’ 8).
We have f (x) positive for x < 0 and for x > 8, and f (x) negative for 0 < x < 8. Since
f (0) > 0 and f (8) < 0, graphing techniques from calculus show that f has exactly 3
real roots. By (6.6.7), G = S5 .
4. f is irreducible by the rational root test. By the formula for the discriminant of a
general cubic with a = 3, b = ’2, c = 1, we have D = 9(’8) ’ 4(’8) ’ 27 ’ 18(6) =
’175. Alternatively, if we replace X by X ’ a = X ’ 1, the resulting polynomial is
3
g(X) = X 3 ’ 5X + 5, whose discriminant is ’4(’5)3 ’ 27(25) = ’175. In any event,
D is not a square in Q, so G = S3 . (Notice also that g is irreducible by Eisenstein, so
we could have avoided the rational root test at the beginning.)
5. If f is reducible, then it is the product of a linear factor and a quadratic polynomial g.
If g is irreducible, then G is cyclic of order 2 (Section 6.3, Problem 4). If g is reducible,
then all roots of f are in the base ¬eld, and G is trivial.
6. Let the roots be a, b + ic and b ’ ic. Then
∆ = (a ’ b ’ ic)(a ’ b + ic)2ic = ((a ’ b)2 + c2 )2ic
and since i2 = ’1, we have D < 0. Since D cannot be a square in Q, the Galois group
is S3 . [This also follows from (6.6.7).]
6

7. If the roots are a, b and c, then D = (a ’ b)2 (a ’ c)2 (b ’ c)2 > 0. The result follows
from (6.6.3).


Section 6.7

Q( m)/Q is Z2 for m = 2, 3, 5, 7. It follows that the
1. By (6.7.2), the Galois group of√
√√√
Galois group of Q( 2, 3, 5, 7)/Q is Z2 — Z2 — Z2 — Z2 . See (6.7.5), and note that
Q contains a primitive square root of unity, namely ’1. (It is not so easy to prove
that the Galois group has order 16. One approach is via the texhnique of Section 7.3,
Problems 9 and 10.)
2. Yes. Let E be the pth cyclotomic extension of Q, where p is prime. If p > 2, then
Q does not contain a primitive pth root of unity. By (6.5.6), the Galois group is
isomorphic to the group of units modp, which is cyclic.
3. Since the derivative of X n ’ a is nX n’1 = 0, it follows as in (6.5.1) that f has n
’n
distinct roots β1 , . . . , βn in E. Since βi = a and βi = a’1 , there are n distinct
n
’1 ’1 ’1
nth roots of unity in E, namely 1 = β1 β1 , β2 β1 , . . . , βn β1 . Since the group of nth
roots of unity is cyclic, there must be a primitive nth root of unity in E.
4. Each root of g is of the form ω i θ, so g0 = ω k θd for some k. Since ω p = 1, we have
p
g0 = θdp . But c = θp since θ is also a root of f , and the result follows.
5. By Problem 4 we have
ap
c = c1 = cad cbp = g0 cbp = (g0 cb )p
a


with g0 cb ∈ F . Thus g0 cb is a root of f in F .
a a

6. [E : F (ω)] divides p and is less than p by (6.7.2); note that E is also a splitting ¬eld
for f over F (ω). Thus [E : F (ω)] must be 1, so E = F (ω).
7. F contains a primitive pth root of unity ω i¬ E(= F (ω)) = F i¬ X p ’ c splits over F .
8. By induction, σ j (θ) = θ +j, 0 ¤ j ¤ p’1. Thus the subgroup of G that ¬xes θ, hence
¬xes F (θ), consists only of the identity. By the fundamental theorem, E = F (θ).
9. We have σ(θp ’ θ) = σ(θ)p ’ σ(θ) = (θ + 1)p ’ (θ + 1) = θp ’ θ in characteristic p.
Thus θp ’ θ belongs to the ¬xed ¬eld of G, which is F . Let a = θp ’ θ, and the result
follows.
10. Since f (θ) = 0, min(θ, F ) divides f . But the degree of the minimal polynomial is
[F (θ) : F ] = [E : F ] = p = deg f . Thus f = min(θ, F ), which is irreducible.
11. Since θp ’θ = a, we have (θ+1)p ’(θ+1) = θp ’θ = a. Inductively, θ, θ+1, . . . , θ+p’1
are distinct roots of f in E, and since f has degree p, we have found all the roots and
f is separable. Since E is a splitting ¬eld for f over F , we have E = F (θ).
12. By Problem 11, every root of f generates the same extension of F , namely E. But
any monic irreducible factor of f is the minimal polynomial of at least one of the
roots of f , and the result follows.
13. [E : F ] = [F (θ) : F ] = deg(min(θ, F )) = deg f = p. Thus the Galois group has prime
order p and is therefore cyclic.
7

Section 6.8
1. Take the real part of each term of the identity to get

cos 3θ = cos3 θ + 3 cos θ(i sin θ)2 = cos3 θ ’ 3 cos θ(1 ’ cos2 θ);

thus cos 3θ = 4 cos3 θ ’ 3 cos θ. If 3θ = π/3, we have

cos π/3 = 1/2 = 4±3 ’ 3±

so 8±3 ’ 6± ’ 1 = 0. But 8X 3 ’ 6X ’ 1 is irreducible over Q (rational root test), so
± is algebraic over Q and [Q(±) : Q] = 3 (not a power of 2), a contradiction.
√ √
2. X 3 ’ 2 is irreducible by Eisenstein, so [Q( 3 2) : Q] = 3 and 3 2 is not constructible.
√ √
3. The side of such a square would be π, so π, hence π, would be algebraic over Q,
a contradiction.
4. ω is a root of X 2 ’ 2(cos 2π/n)X + 1 since cos 2π/n = 1 (ω + ω ’1 ) and
2
ω 2 ’ (ω + ω ’1 )ω + 1 = 0. The discriminant of the quadratic polynomial is nega-
tive, proving irreducibility over R ⊇ Q(cos 2π/n).
5. By (6.5.2), (6.5.5) and (3.1.9),

•(n) = [Q(ω) : Q] = [Q(ω) : Q(cos 2π/n)][Q(cos 2π/n) : Q].

By Problem 4, [Q(ω) : Q(cos 2π/n)] = 2, and if the regular n-gon is constructible,
then [Q(cos 2π/n) : Q] is a power of 2. The result follows.
6. By hypothesis, G = Gal(Q(ω)/Q) is a 2-group since its order is •(n). Therefore
every quotient group of G, in particular Gal(Q(cos 2π/n)/Q), is a 2-group. [Note
that by (6.5.1), G is abelian, hence every subgroup of G is normal, and therefore
every intermediate ¬eld is a Galois extension of Q. Thus part 2c of the fundamental
theorem (6.2.1) applies.]
7. By the fundamental theorem (speci¬cally, by Section 6.2, Problem 4), there are ¬elds
Q = K0 ¤ K1 ¤ · · · ¤ Kr = Q(cos 2π/n) with [Ki : Ki’1 ] = 2 for all i = 1, . . . , r.
Thus cos 2π/n is constructible.
8. If n = pe1 · · · per , then (see Section 1.1, Problem 13)
r
1

•(n) = pe1 ’1 (p1 ’ 1) · · · per ’1 (pr ’ 1).
r
1

If pi = 2, we must have ei = 1, and in addition, pi ’ 1 must be a power of 2. The
result follows.
9. If m is not a power of 2, then m can be factored as ab where a is odd and 1 < b < m.
In the quotient (X a + 1)/(X + 1), set X = 2b . It follows that (2m + 1)/(2b + 1) is an
integer. Since 1 < 2b + 1 < 2m + 1, 2m + 1 cannot be prime.
10. The ei belong to E and are algebraically independent over K, so the transcendence
detree of E over K is at least n. It follows that the ±i are algebraically independent
over K, and the transcendence degree is exactly n. Therefore any permutation of
the ±i induces a K-automorphism of E = K(±1 , . . . , ±n ) which ¬xes each ei , hence
¬xes F . Thus the Galois group of f consists of all permutations of n letters.
8

11. Since Sn is not solvable, the general equation of degree n is not solvable by radicals
if n ≥ 5. In other words, if n ≥ 5, there is no sequence of operations on e1 , . . . , en
involving addition, subtraction, multiplication, division and extraction of mth roots,
that will yield the roots of f .


Section 6.9
1. If S is not maximal, keep adding elements to S until a maximal algebraically indepen-
dent set is obtained. If we go all the way to T , then T is algebraically independent and
spans E algebraically, hence is a transcendence basis. (Trans¬nite induction supplies
the formal details.)
2. For the ¬rst statement, take T = E in Problem 1. For the second statement, take
S = ….
3. (i) implies (ii): Suppose that ti satis¬es f (ti ) = b0 + b1 ti + · · · + bm tm = 0, with
i
bj ∈ F (T \ {ti }). By forming a common denominator for the bj , we
may assume that the bj are polynomials in F [T \ {ti }] ⊆ F [T ]. By (i),
bj = 0 for all j, so f = 0.
(ii) implies (iii): Note that F (t1 , . . . , ti’1 ) ⊆ F (T \ {ti }).
(iii) implies (i): Suppose that f is a nonzero polynomial in F [X1 , . . . , Xm ] such that
f (t1 , . . . , tm ) = 0, where m is as small as possible. Then f = h0 +
h1 Xm + · · · + hr Xm where the hj belong to F [X1 , . . . , Xm’1 ]. Now
r

f (t1 , . . . , tm ) = b0 + b1 tm + · · · + br tr where bj = hj (t1 , . . . , tm’1 ).
m
If the bj are not all zero, then tm is algebraic over F (t1 , . . . , tm’1 ),
contradicting (iii). Thus bj ≡ 0, so by minimality of m, hj ≡ 0,
so f = 0.
4. If S ∪ {t} is algebraically dependent over F , then there is a positive integer n and
a nonzero polynomial f in F [X1 , . . . , Xn , Z] such that f (t1 , . . . , tn , t) = 0 for some
t1 , . . . , tn ∈ S. Since S is algebraically independent over F , f must involve Z. We
may write f = b0 + b1 Z + · · · + bm Z m where bm = 0 and the bj are polynomials in
F [X1 , . . . , Xn ]. But then t is algebraic over F (S).
Conversely, if t is algebraic over F (S), then for some positive integer n, there are
elements t1 , . . . , tn ∈ S such that t is algebraic over F (t1 , . . . , tn ). By Problem 3,
{t1 , . . . , tn , t} is algebraically dependent over F , hence so is S ∪ {t}.
5. Let A = {s1 , . . . , sm , t1 , . . . , tn } be an arbitrary ¬nite subset of S ∪ T , with si ∈ S and
tj ∈ T . By Problem 3, si is transcendental over F (s1 , . . . , si’1 ) and tj is transcendental
over K(t1 , . . . , tj’1 ), hence over F (s1 , . . . , sm , t1 , . . . , tj’1 ) since S ⊆ K. Again by
Problem 3, A is algebraically independent over F . Since A is arbitrary, S ∪ T is
algebraically independent over F . Now if t ∈ K then {t} is algebraically dependent
over K (t is a root of X ’ t). But if t also belongs to T , then T is algebraically
dependent over K, contradicting the hypothesis. Thus K and T , hence S and T , are
disjoint.
6. By Problem 5, S ∪T is algebraically independent over F . By hypothesis, E is algebraic
over K(T ) and K is algebraic over F (S). Since each t ∈ T is algebraic over F (S)(T ) =
9

F (S ∪ T ), it follows that K(T ) is algebraic over F (S ∪ T ). By (3.3.5), E is algebraic
over F (S ∪ T ). Therefore S ∪ T is a transcendence basis for E/F .
7. If T is algebraically independent over F , the map f (X1 , . . . , Xn ) ’ f (t1 , . . . , tn ) ex-
tends to an F -isomorphism of F (X1 , . . . , Xn ) and F (t1 , . . . , tn ). Conversely, assume
that F (T ) is F -isomorphic to the rational function ¬eld. By Problem 2, there is a tran-
scendence basis B for F (T )/F such that B ⊆ T . By (6.9.7), the transcendence degree
of F (T )/F is |T | = n. By (6.9.5) or (6.9.6), B = T , so T is algebraically independent
over F .
8. The “if” part is clear since [K(z) : K] can™t be ¬nite; if so, [F : K] < ∞. For the “only
if” part, z is algebraic over K(x), so let

z n + •n’1 (x)z n’1 + · · · + •0 (x) = 0, •i ∈ K(x).

Clear denominators to get a polynomial f (z, x) = 0, with coe¬cients of f in K. Now
x must appear in f , otherwise z is not transcendental. Thus x is algebraic over K(z),
so [K(z, x) : K(z)] < ∞. Therefore

[F : K(z)] = [F : K(z, x)][K(z, x) : K(z)].

The ¬rst term on the right is ¬nite since K(x) ⊆ K(z, x), and the second term is ¬nite,
as we have just seen. Thus [F : K(z)] < ∞, and the result follows. ™
9. We have tr deg(C/Q) = c, the cardinality of C (or R). For if C has a countable
transcendence basis z1 , z2 , . . . over Q, then C is algebraic over Q(z1 , z2 , . . . ). Since a
polynomial over Q can be identi¬ed with a ¬nite sequence of rationals, it follows that
|C| = |Q|, a contradiction.


Section 7.1
1. Replace (iii) by (iv) and the proof goes through as before. If R is a ¬eld, then in (iii)
implies (i), x is an eigenvalue of C, so det(xI ’ C) = 0.
2. Replace (iii) by (v) and the proof goes through as before. [Since B is an A[x]-module,
in (iii) implies (i) we have xβi ∈ B; when we obtain [det(xI ’ C)]b = 0 for every b ∈ B,
the hypothesis that B is faithful yields det(xI ’ C) = 0.]
3. Multiply the equation by an’1 to get

a’1 = ’(cn’1 + · · · + c1 an’2 + c0 an’1 ) ∈ A.

4. Since A[b] is a subring of B, it is an integral domain. Thus if bz = 0 and b = 0,
then z = 0.
5. Any linear transformation on a ¬nite-dimensional vector space is injective i¬ it is
surjective. Thus if b ∈ B and b = 0, there is an element c ∈ A[b] ⊆ B such that bc = 1.
Therefore B is a ¬eld.
6. P is the preimage of Q under the inclusion map of A into B, so P is a prime ideal.
The map a + P ’ a + Q is a well-de¬ned injection of A/P into B/Q, since P = Q © A.
Thus A/P can be viewed as a subring of B/Q.
10

7. If b + Q ∈ B/Q, then b satis¬es an equation of the form

xn + an’1 xn’1 + · · · + a1 x + a0 = 0, ai ∈ A.

By Problem 6, b + Q satis¬es the same equation with ai replaced by ai + P for all i.
Thus B/Q is integral over A/P .
8. By Problems 3“5, A/P is a ¬eld if and only if B/Q is a ¬eld, and the result follows.
(Note that since Q is a prime ideal, B/Q is an integral domain, as required in the
hypothesis of the result just quoted.)


Section 7.2

1. By the quadratic formula, L = Q( b2 ’ 4c). Since b2 ’ 4c ∈ Q, we may write
b2 ’ 4c = s/t = st/t2 for relatively prime integers s and t. We also have s = uy 2
and t = vz 2 where√ v, y, z ∈ Z, with u and v relatively prime and square-free. Thus
u,

L = Q( uv) = Q( d).
√ √
√ √
2. If Q( d) = Q( e), then d = a + b e for rational numbers a and b. Thus d =
√ √
a2 + b2 e + 2ab e, so e is rational, a contradiction (unless a = 0 and b = 1).
√ √
√ √
3. Any isomorphism of Q( d) and Q( e) must carry d into a+b e for rational numbers

a and b. Thus d is mapped to a2 + b2 e + 2ab e. But a Q-isomorphism maps d to d,
and we reach a contradiction as in Problem 2.
4. Since ωn = ω2n we have ωn ∈ Q(ω2n ), so Q(ωn ) ⊆ Q(ω2n ). If n is odd then n + 1 = 2r,
2

so

ω2n = ’ω2n = ’(ω2n )r = ’ωn .
2r 2 r


Therefore Q(ω2n ) ⊆ Q(ωn ).
5. Let f be a monic polynomial over Z with f (x) = 0. If f is factorable over Q, then it is
factorable over Z by (2.9.2). Thus min(x, Q) is the monic polynomial in Z[X] of least
degree such that f (x) = 0.
√ √
6. Q( ’3) = Q(ω) where ω = ’ 1 + 1 ’3 is a primitive cube root of unity.
2 2
7. If n = [L : Q], then an integral basis consists of n elements of L that are linearly inde-
pendent over Z, hence over Q. (A linear dependence relation over Q can be converted
to one over Z by multiplying by a common denominator.)


Section 7.3

1. The Galois group of E/Q consists of the identity and the automorphism σ(a+b d) =

a ’ b d. By (7.3.6), T (x) = x + σ(x) = 2a and N (x) = xσ(x) = a2 + db2 .
2. A basis for E/Q is 1, θ, θ2 , and

θ2 θ = θ3 = 3θ ’ 1, θ2 θ2 = θ4 = θθ3 = 3θ2 ’ θ.
θ2 1 = θ2 ,
11

Thus
® 
0 ’1 0
m(θ2 ) = °0 3 ’1»
10 3

and we have T (θ2 ) = 6, N (θ2 ) = 1. Note that if we had already computed the norm
of θ (the matrix of θ is
® 
0 0 ’1
m(θ) = °1 0 3 »
01 0

and T (θ) = 0, N (θ) = ’1), it would be easier to calculate N (θ2 ) as [N (θ)]2 =
(’1)2 = 1.
3. The cyclotomic polynomial Ψ6 has only two roots, ω and its complex conjugate ω.
By (7.3.5),

T (ω) = ω + ω = eiπ/3 + e’iπ/3 = 2 cos(π/3) = 1.

4. By (7.3.6), N (x) = xσ(x) · · · σ n’1 (x) and T (x) = x + σ(x) + · · · + σ n’1 (x). If
x = y/σ(y), then σ(x) = σ(y)/σ 2 (y), . . . , σ n’1 (x) = σ n’1 (y)/σ n (y) = σ n’1 (y)/y,
and the telescoping e¬ect gives N (x) = 1. If x = z ’ σ(z), then σ(x) = σ(z) ’ σ 2 (z),
. . . , σ n’1 (x) = σ n’1 (z) ’ z, and a similar telescoping e¬ect gives T (x) = 0.
5. Choose v ∈ E such that

y = v + xσ(v) + xσ(x)σ 2 (v) + · · · + xσ(x) · · · σ n’2 (x)σ n’1 (v) = 0

and hence

σ(y) = σ(v) + σ(x)σ 2 (v) + σ(x)σ 2 (x)σ 3 (v) + · · · + σ(x)σ 2 (x) · · · σ n’1 (x)σ n (v).

We are assuming that N (x) = xσ(x) · · · σ n’1 (x) = 1, and it follows that the last
summand in σ(y) is x’1 σ n (v) = x’1 v. Comparing the expressions for y and σ(y), we
have xσ(y) = y, as desired.
6. Since T (x) = 0, we have ’x = σ(x) + · · · + σ n’1 (x), so the last summand of σ(w)
is ’xu. Thus

w ’ σ(w) = x(u + σ(u) + σ 2 (u) + · · · + σ n’1 (u)) = xT (u).

7. We have
w ’ σ(w)
w σ(w)
z ’ σ(z) = ’ =
T (u) σ(T (u)) T (u)

since T (u) belongs to F and is therefore ¬xed by σ (see (7.3.3)). By Problem 6,
z ’ σ(z) = x.
12

8. No. We have y/σ(y) = y /σ(y ) i¬ y/y = σ(y/y ) i¬ y/y is ¬xed by all automor-
phisms in the Galois group G i¬ y/y belongs to the ¬xed ¬eld of G, which is F .
Similarly, z ’ σ(z) = z ’ σ(z ) i¬ z ’ z ∈ F .
9. We have min(θ, Q) = X 4 ’ 2, min(θ2 , Q) = X 2 ’ 2, min(θ3 , Q) = X 4 ’ 8,

min( 3θ, Q) = X 4 ’ 18. (To compute the last two minimal polynomials, note that

(θ3 )4 = (θ4 )3 = 23 = 8, and ( 3θ)4 = 18.) Therefore, all 4 traces are 0.

10. Suppose that 3 = a + bθ + cθ2 + dθ3 with a, b, c ∈ Q. Take the trace of both sides to

conclude that a = 0. (The trace of 3 is 0 because its minimal polynomial is X 2 ’ 3.)
√ √
Thus 3 = bθ + cθ2 + dθ3 , so 3θ = bθ2 + cθ3 + 2d. Again take the trace of both sides
to get d = 0. The same technique yields b = c = 0, and we reach a contradiction.


Section 7.4
1. If l(y) = 0, then (x, y) = 0 for all x. Since the bilinear form is nondegenerate, we
must have y = 0.
2. Since V and V — have the same dimension n, the map y ’ l(y) is surjective.
3. We have (xi , yj ) = l(yj )(xi ) = fj (xi ) = δij . Since the fj = l(yj ) form a basis, so do
the yj .
n
4. Write xi = k=1 aik yk , and take the inner product of both sides with xj to conclude
that aij = (xi , xj ).
5. The “if” part was done in the proof of (7.4.10). If det C = ±1, then C ’1 has coe¬-
cients in Z by Cramer™s rule.

6. If d ≡ 1 mod 4, then by (7.2.3), 1 and d form an integral basis. Since the trace of

a + b d is 2a (Section 7.3, Problem 1), the ¬eld discriminant is

2 0
D = det = 4d.
0 2d

If d ≡ 1 mod 4, then 1 and 1 (1 + d) form an integral basis, and
2

√ 1 d 1√
2
1
(1 + d) = ++ d.
2 442

Thus
2 1
D = det = d.
d+1
1 2

7. The ¬rst statement follows because multiplication of each element of a group G by a
particular element g ∈ G permutes the elements of G. The plus and minus signs are
balanced in P + N and P N , before and after permutation. We can work in a Galois
extension of Q containing L, and each automorphism in the Galois group restricts to
one of the σi on L. Thus P + N and P N belong to the ¬xed ¬eld of the Galois group,
which is Q.
13

8. Since the xj are algebraic integers, so are the σi (xj ), as in the proof of (7.3.10). By
(7.1.5), P and N , hence P + N and P N , are algebraic integers. By (7.1.7), Z is
integrally closed, so by Problem 7, P + N and P N belong to Z.
9. D = (P ’ N )2 = (P + N )2 ’ 4P N ≡ (P + N )2 mod 4. But any square is congruent
to 0 or 1 mod 4, and the result follows.
n
10. We have yi = j=1 aij xj with aij ∈ Z. By (7.4.3), D(y) = (det A)2 D(x). Since D(y)
is square-free, det A = ±1, so A has an inverse with entries in Z. Thus x = A’1 y, as
claimed.
11. Every algebraic integer is a Z-linear combination of the xi , hence of the yi by Prob-
lem 10. Since the yi form a basis for L over Q, they are linearly independent and the
result follows.

12. No. For example, let L = Q( d), where d is a square-free integer with d ≡ 1 mod 4.
(See Problem 6). The ¬eld discriminant is 4d, which is not square-free.
13. This follows from the proof of (7.4.7).


Section 7.5
1. A0 ‚ A1 ‚ A2 ‚ · · ·
2. Let a/pn ∈ B, where p does not divide a. There are integers r and s such that
ra + spn = 1. Thus ra/pn = 1/pn in Q/Z, and An ⊆ B. If there is no upper bound
on n, then 1/pn ∈ B for all n (note 1/pn = p/pn+1 = p2 /pn+2 , etc.), hence B = A. If
there is a largest n, then for every m > n, B © Am ⊆ An by maximality of n. Therefore
B = An .
3. Let x1 , x2 , . . . be a basis for V . Let Mr be the subspace spanned by x1 , . . . , xr , and
Lr the subspace spanned by the xj , j > r. If V is n-dimensional, then V = L0 > L1 >
· · · > Ln’1 > Ln = 0 is a composition series since a one-dimensional subspace is a
simple module. [V = Mn > Mn’1 > · · · > M1 > 0 is another composition series.]
Thus V is Noetherian and Artinian. If V is in¬nite-dimensional, then M1 < M2 < · · ·
violates the acc, and L0 > L1 > L2 > · · · violates the dcc. Thus V is neither
Noetherian nor Artinian. [Note that if V has an uncountable basis, there is no problem;
just take a countably in¬nite subset of it.]
4. l(M ) is ¬nite i¬ M has a composition series i¬ M is Noetherian and Artinian i¬ N
and M/N are Noetherian and Artinian i¬ l(N ) and l(M/N ) are ¬nite.
5. By Problem 4, the result holds when l(M ) = ∞, so assume l(M ), hence l(N ) and
l(M/N ), ¬nite. Let 0 < N1 < · · · < Nr = N be a composition series for N , and let
N/N < (M1 + N )/N < · · · < (Ms + N )/N = M/N be a composition series for M/N .
Then

0 < N1 < · · · < Nr < M1 + N < · · · < Ms + N = M

is a composition series for M . (The factors in the second part of the series are simple
by the third isomorphism theorem.) It follows that l(M ) = r + s = l(N ) + l(M/N ).
14

6. By (7.5.9), R is a Noetherian S-module, hence a Noetherian R-module. (Any R-
submodule T of R is, in particular, an S-submodule of R. Therefore T is ¬nitely
generated.)
7. Yes. Map a polynomial to its constant term and apply the ¬rst isomorphism theorem
to show that R ∼ R[X]/(X). Thus R is a quotient of a Noetherian R-module, so is
=
Noetherian by (7.5.7).
8. If there is an in¬nite descending chain of submodules Mi of M , then the intersection
N = ©i Mi cannot be expressed as the intersection of ¬nitely many Mi . By the corre-
spondence theorem, ©i (Mi /N ) = 0, but no ¬nite intersection of the submodules Mi /N
of M/N is 0. Thus M/N is not ¬nitely cogenerated. Conversely, suppose that M/N is
not ¬nitely cogenerated. By the correspondence theorem, we have ©± M± = N , but no
¬nite intersection of the M± is N . Pick any M± and call it M1 . If M1 ⊆ M± for all ±,
then M1 = N , a contradiction. Thus we can ¬nd M± = M2 such that M1 ⊃ M1 © M2 .
Continue inductively to produce an in¬nite descending chain.


Section 7.6
1. The “only if” part follows from (7.6.2). If the given condition is satis¬ed and ab ∈ P ,
then (a)(b) ⊆ P , hence (a) ⊆ P or (b) ⊆ P , and the result follows.
2. If xi ∈ Pi for some i, then xi ∈ I \ ∪n Pj and we are ¬nished.
/ j=1
3. Since I is an ideal, x ∈ I. Say x ∈ P1 . All terms in the sum that involve x1 belong
to P1 by Problem 2. The remaining term x2 · · · xn is the di¬erence of two elements in
P1 , hence x2 · · · xn ∈ P1 . Since P1 is prime, xj ∈ P1 for some j = 1, contradicting the
choice of xj .
4. The product of ideals is always contained in the intersection. If I and J are relatively
prime, then 1 = x + y with x ∈ I and y ∈ J. If z ∈ I © J, then z = z1 = zx + zy ∈ IJ.
The general result follows by induction, along with the computation
R = (I1 + I3 )(I2 + I3 ) ⊆ I1 I2 + I3 .
Thus I1 I2 and I3 are relatively prime.
5. See (2.6.9).
6. Assume that R is not a ¬eld, equivalently, {0} is not a maximal ideal. Thus by (7.6.9),
every maximal idea is invertible.
7. Let r be a nonzero element of R such that rK ⊆ R, hence K ⊆ r’1 R ⊆ K. Thus
K = r’1 R. Since r’2 ∈ K we have r’2 = r’1 s for some s ∈ R. But then r’1 = s ∈ R,
so K ⊆ R and consequently K = R.
8. R = Rr = (P1 + P2 )r ⊆ P1 + P2 . Thus P1 and P2 are relatively prime for all r ≥ 1.
r r
r s
Assuming inductively that P1 and P2 are relatively prime, we have
P2 = P2 R = P2 (P1 + P2 ) ⊆ P1 + P2
s+1
s s s r r


so
R = P1 + P2 ⊆ P1 + (P1 + P2 ) = P1 + P2
s+1 s+1
r s r r r
15

completing the induction.


Section 7.7

1. By Section 7.3, Problem 1, the norms are 6, 6, 4 and 9. Now if x = a + b ’5 and
x = yz, then N (x) = a2 + 5b2 = N (y)N (z). The only algebraic integers of norm 1
are ±1, and there are no algebraic integers of norm 2 or 3. Thus there cannot be a

nontrivial factorization of 1 ± ’5, 2 or 3.
√ √
2. If (a + b ’5)(c + d ’5) = 1, take norms to get (a2 + 5b2 )(c2 + 5d2 ) = 1, so b = d = 0,
a = ±1, c = ±1.
3. By Problem 2, if two factors are associates, then the quotient of the factors is ±1,
which is impossible.
4. This is a nice application of the principle that divides means contains. The greatest
common divisor is the smallest ideal containing both I and J, that is, I + J. The least
common multiple is the largest ideal contained in both I and J, which is I © J.
5. If I is a fractional ideal, then by (7.7.1) there is a fractional ideal I such that II = R.
By de¬nition of fractional ideal, there is a nonzero element r ∈ R such that rI is an
integral ideal. If J = rI , then IJ = Rr, a principal ideal of R.
This is done just as in Problems 1“3, using the factorization 18 = (2)(32 ) = (1 +
6. √ √
’17)(1 ’ ’17).
√ √
7. By (7.2.2), the algebraic integers are of the form a + b ’3, a, b ∈ Z, or u + v ’3
2 2
where u and v are odd integers. If we require that the norm be 1, we only get ±1 in
the ¬rst case. But in the second case, we have u2 + 3v 2 = 4, so u = ±1, v = ±1. Thus
if ω = ei2π/3 , the algebraic integers of norm 1 are ±1, ±ω, and ±ω 2 .


Section 7.8
1. If Rx and Ry belong to P (R), then (Rx)(Ry)’1 = (Rx)(Ry ’1 ) = Rxy ’1 ∈ P (R), and
the result follows from (1.1.2).
2. If C(R) is trivial, then every integral ideal I of R is a principal fractional ideal Rx, x ∈
K. But I ⊆ R, so x = 1x must belong to R, proving that R is a PID. The converse
holds because every principal ideal is a principal fractional ideal.
√ √ √ √
3. 1 ’ ’5 = 2 ’ (1 + ’5) ∈ P2 , so (1 + ’5)(1 ’ ’5) = 6 ∈ P2 . 2

4. Since 2 ∈ P2 , it follows that 4 ∈ P2 , so by Problem 3, 2 = 6 ’ 4 ∈ P2 .
2 2
√ √ √ √ √ √
5. (2, 1+ ’5)(2, 1+ ’5) = (4, 2(1+ ’5), (1+ ’5)2 ), and (1+ ’5)2 = ’4+2 ’5.
2
Thus each of the generators of the ideal P2 is divisible by 2, hence belongs to (2).
Therefore P2 ⊆ (2).
2

6. x2 +5 ≡ (x+1)(x’1) mod 3, which suggests that (3) = P3 P3 , where P3 = (3, 1+ ’5)

and P3 = (3, 1 ’ ’5).
√ √
7. P3 P3 = (3, 3(1’ ’5), 3(1+ ’5), 6) ⊆ (3) since each generator of P3 P3 is divisible by
3. But 3 ∈ P3 © P3 , hence 9 ∈ P3 P3 , and therefore 9 ’ 6 = 3 ∈ P3 P3 . Thus (3) ⊆ P3 P3 ,
and the result follows.
16

Section 7.9
1. Using (1), the product is z = 4 + 2p + 4p2 + p3 + p4 . But 4 = 3 + 1 = 1 + p and
4p2 = p2 + 3p2 = p2 + p3 . Thus z = 1 + 3p + p2 + 2p3 + p4 = 1 + 2p2 + 2p3 + p4 .
Using (2), we are multiplying x = {2, 5, 14, 14, . . . } by y = {2, 2, 11, 11, . . . }. Thus
z0 = 4, z1 = 10, z2 = 154, z3 = 154, z4 = 154, and so on. But 4 ≡ 1 mod 3, 10 ≡
1 mod 9, 154 ≡ 19 mod 27, 154 ≡ 73 mod 81, 154 ≡ 154 mod 243. The standard form
is {1, 1, 19, 73, 154, 154, . . . }. As a check, the product is (2 + 3 + 9)(2 + 9) = 154, whose
base 3 expansion is 1 + 0(3) + 2(9) + 2(27) + 1(81) as found above.
2. We have a0 = ’1 and an = 0 for n ≥ 1; equivalently, xn = ’1 for all n. In standard
form, x0 = p’1, x1 = p2 ’1, x2 = p3 ’1, . . . .Since (pr ’1)’(pr’1 ’1) = (p’1)(pr’1 ),
the series representation is
(p ’ 1) + (p ’ 1)p + (p ’ 1)p2 + · · · + (p ’ 1)pn + · · · .

The result can also be obtained by multiplying by -1 on each side of the equation
1 = (1 ’ p)(1 + p + p2 + · · · ).

3. Let x be a nonzero element of GF (q). By (6.4.1), xq’1 = 1, so |x|q’1 = 1. Thus |x| is
a root of unity, and since absolute values are nonnegative real, we must have |x| = 1,
and the result follows.
4. If the absolute value is nonarchimedian, then S is bounded by (7.9.6). If the absolute
value is archimedian, then by (7.9.6), |n| > 1 for some n. But then |nk | = |n|k ’ ∞
as k ’ ∞. Therefore S is unbounded.
5. A ¬eld of prime characteristic p has only ¬nitely many integers 0, 1, . . . , p ’ 1. Thus
the set S of Problem 4 must be bounded, so the absolute value is nonarchimedian.
6. The “only if” part is handled just as in calculus. For the “if” part, note that by (iv)
of (7.9.5), we have |zm + zm+1 + · · · + zn | ¤ max{|zi | : m ¤ i ¤ n} ’ 0 as m, n ’ ∞.
Thus the nth partial sums form a Cauchy sequence, which must converge to an element
in Qp .
7. Since n! = 1 · 2 · · · p · · · 2p · · · 3p · · · , it follows from (7.9.2) and (7.9.3) that if rp ¤ n <
(r + 1)p, then |n!| = 1/pr . Thus |n!| ’ 0 as n ’ ∞.
8. No. Although |pr | = 1/pr ’ 0 as r ’ ∞, all integers n such that rp < n < (r + 1)p
have p-adic absolute value 1, by (7.9.2). Thus the sequence of absolute values |n|
cannot converge, hence the sequence itself cannot converge.


Section 8.1
1. If x ∈ V and f1 (x) = 0, then f2 (x) must be 0 since f1 f2 ∈ I(V ); the result follows.
2. By Problem 1, V ⊆ V (f1 ) ∪ V (f2 ). Thus
V = (V © V (f1 )) ∪ (V © V (f2 )) = V1 ∪ V2 .
Since f1 ∈ I(V ), there exists x ∈ V such that f1 (x) = 0. Thus x ∈ V1 , so V1 ‚ V ;
/ /
similarly, V2 ‚ V .
17

3. I(V ) ⊇ I(W ) by (4). If I(V ) = I(W ), let V = V (S), W = V (T ). Then IV (S) =
IV (T ), and by applying V to both sides, we have V = W by (6).
4. Let x ∈ V ; if f1 (x) = 0, then since f1 ∈ I(V1 ), we have x ∈ V1 . But then x ∈ V2 , and
/
therefore f2 (x) = 0 (since f2 ∈ I(V2 )). Thus f1 f2 = 0 on V , so f1 f2 ∈ I(V ).
5. If V is reducible, then V is the union of proper subvarieties V1 and V2 . If V1 is
reducible, then it too is the union of proper subvarieties. This decomposition process
must terminate in a ¬nite number of steps, for otherwise by Problems 1“4, there
would be a strictly increasing in¬nite sequence of ideals, contradicting the fact that
k[X1 , . . . , Xn ] is Noetherian.
6. If V = i Vi = j Wj , then Vi = j (Vi Wj ), so by irreducibility, Vi = Vi Wj for
some j. Thus Vi ⊆ Wj , and similarly Wj ⊆ Vk for some k. But then Vi ⊆ Vk , hence
i = k (otherwise we would have discarded Vi ). Thus each Vi can be paired with a
corresponding Wj , and vice versa.
7. By hypothesis, An = ∪(An \ V (Ii )). Taking complements, we have ©V (Ii ) = …. But by
(8.1.2), ©V (Ii ) = V (∪Ii ) = V (I), so by the weak Nullstellensatz, I = k[X1 , . . . , Xn ].
Thus the constant polynomial 1 belongs to I.
8. Suppose that the open sets An \ V (Ii ) cover An . By Problem 7, 1 ∈ I, hence 1
belongs to a ¬nite sum i∈F Ii . Since 1 never vanishes, V ( i∈F Ii ) = …. By (8.1.2),
©i∈F Vi = …, where Vi = V (Ii ). Taking complements, we have ∪i∈F (An \ Vi ) = An .
Thus the original open covering of An has a ¬nite subcovering, proving compactness.


Section 8.2
1. If a ∈ (a1 , . . . , ak ), then g or some other element of I would extend the inductive
/
process to step k + 1.
2. In going from di to di+1 we are taking the minimum of a smaller set.
3. By minimality of m, a ∈ (a1 , . . . , am’1 ), hence fm and g satisfy conditions 1 and 2.
/
By choice of fm we have dm ¤ d. (If m = 1, then d1 ¤ d by choice of f1 .)
4. Let f be the unique ring homomorphism from R[X1 , . . . , Xn ] to S such that f is the
identity on R and f (Xi ) = xi , i = 1, . . . , n. (For example, if a ∈ R, then aX1 X4 ’27

ax2 x7 .) Since the image of f contains R and {x1 , . . . , xn }, f is surjective and the result
14
follows.
5. By the Hilbert basis theorem, R[X1 , . . . , Xn ] is a Noetherian ring, hence a Noetherian
R[X1 , . . . , Xn ]-module. By (7.5.7), S is a Noetherian R[X1 , . . . , Xn ]-module. But the
submodules of S considered as an R[X1 , . . . , Xn ]-module coincide with the submodules
of S as an S-module. (See Section 4.2, Problems 6 and 7; note that the kernel of the
homomorphism f of Problem 4 annihilates S.) Thus S is a Noetherian S-module, that
is, a Noetherian ring.


Section 8.3
1. Suppose that xy ∈ J with x ∈ J and y ∈ J. By maximality of J, the ideal J + (x)
/ /
contains an element s ∈ S. Similarly, J + (y) contains an element t ∈ S. But then
18

st ∈ (J + (x))(J + (y)) ⊆ J + (xy) ⊆ J, so S © J = …, a contradiction.

2. Let S = {1, f, f 2 , . . . , f r , . . . }. Then I © S = … since f ∈ I. By Problem 1, I is
/
contained in a prime ideal P disjoint from S. But f ∈ S, so f cannot belong to P , and
the result follows.
3. The “if” part follows because f and f r have the same zero-set. Conversely, if
V (f ) = V (g), then by the Nullstellensatz, (f ) = (g), and the result follows.
4. W ⊆ V since (t4 )2 = (t3 )(t5 ) and (t5 )2 = (t3 )2 (t4 ); L ⊆ V by direct veri¬cation.
Conversely, if y 2 = xz and z 2 = x2 y, let t = y/x. (If x = 0, then y = z = 0 and
we can take t = 0.) Then z = y 2 /x = (y/x)2 x = t2 x, and z 2 = x2 y. Therefore
z 2 = t4 x2 = x2 y, hence y = t4 . If t = 0 then y = 0, hence z = 0 and (x, y, z) ∈ L.
Thus assume t = 0. But then x = y/t = t3 and z = t2 x = t5 .
5. We will show that I(V ) is a prime ideal (see the exercises in Section 8.1). If f g ∈ I(V ),
then f g vanishes on V . Using the parametric form, we have f (t, t2 , t3 )g(t, t2 , t3 ) = 0
for all complex numbers t. Since we are now dealing with polynomials in only one
variable, either f (t, t2 , t3 ) = 0 for all t or g(t, t2 , t3 ) = 0 for all t. Thus f ∈ I(V )
or g ∈ I(V ).
6. (a) x = 2t/(t2 + 1), y = (t2 ’ 1)/(t2 + 1)
(b) x = t2 , y = t3
(c) x = t2 ’ 1, y = t(t2 ’ 1)
7. (Following Shafarevich, Basic Algebraic Geometry, Vol.1, page 2.) We can assume that
x appears in f with positive degree. Viewing f and g as polynomials in k(y)[x], a PID,
f is still irreducible because irreducibility over an integral domain implies irreducibility
over the quotient ¬eld. If g = f h where h is a polynomial in x with coe¬cients in k(y),
then by clearing denominators we see that f must divide g in k[x, y], a contradiction.
(Since f is irreducible, it must either divide g or a polynomial in y alone, and the
latter is impossible because x appears in f .) Thus f does not divide g in k(y)[x].
Since f and g are relatively prime, there exist s, t ∈ k(y)[x] such that f s + gt = 1.
Clearing denominators, we get u, v ∈ k[x, y] such that f u + gv = a, where a is a
nonzero polynomial in y alone. Now if ±, β ∈ k and f (±, β) = g(±, β) = 0, then
a(β) = 0, and this can only happen for ¬nitely many β. For any ¬xed β, consider
f (x, β) = 0. If this polynomial in x is not identically 0, then there are only ¬nitely
many ± such that f (±, β) = 0, and we are ¬nished. Thus assume f (x, β) ≡ 0. Then
f (x, y) = f (x, y) ’ f (x, β) = (y ’ β)h in k(x)[y], contradicting the irreducibility of f .


Section 8.4
1. Since f = 0 i¬ some fi = 0, V (f ) is the union of the V (fi ). Since each fi is irreducible,
the ideal Ii = (fi ) is prime by (2.6.1), hence V (Ii ) = V (fi ) is an irreducible subvariety

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