Selected Solutions to Artin’s Algebra, Second Ed.
Takumi Murayama
July 22, 2014
These solutions are the result of taking MAT323 Algebra in the Spring of 2012,
and also TA-ing for MAT346 Algebra II in the Spring of 2014, both at Princeton
University. This is not a complete set of solutions; see the List of Solved Exercises
at the end. Please e-mail takumim@umich.edu with any corrections.
Contents
2 Groups
2.1 Laws of Composition . . . . .
2.2 Groups and Subgroups . . . .
2.4 Cyclic Groups . . . . . . . . .
2.6 Isomorphisms . . . . . . . . . .
2.8 Cosets . . . . . . . . . . . . . .
2.10 The Correspondence Theorem
2.11 Product Groups . . . . . . . .
2.12 Quotient Groups . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
6 Symmetry
6.3 Isometries of the Plane . . . . . . . . . . . . . . . .
6.4 Finite Groups of Orthogonal Operators on the Plane
6.5 Discrete Groups of Isometries . . . . . . . . . . . . .
6.6 Plane Crystallographic Groups . . . . . . . . . . . .
6.7 Abstract Symmetry: Group Operations . . . . . . .
6.8 The Operation on Cosets . . . . . . . . . . . . . . .
6.9 The Counting Formula . . . . . . . . . . . . . . . .
6.10 Operations on Subsets . . . . . . . . . . . . . . . . .
6.11 Permutation Representations . . . . . . . . . . . . .
1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
4
4
4
5
7
8
9
9
10
.
.
.
.
.
.
.
.
.
13
13
14
16
17
17
18
19
19
20
6.12 Finite Subgroups of the Rotation Group . . . . . . . . . . . . . . . .
6.M Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . .
11 Rings
11.1 Definition of a Ring . . . .
11.2 Polynomial Rings . . . . .
11.3 Homomorphisms and Ideals
11.4 Quotient Rings . . . . . . .
11.5 Adjoining Elements . . . .
11.6 Product Rings . . . . . . .
11.7 Fractions . . . . . . . . . .
11.8 Maximal Ideals . . . . . . .
11.9 Algebraic Geometry . . . .
11.M Miscellaneous Problems . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
12 Factoring
12.1 Factoring Integers . . . . . . .
12.2 Unique Factorization Domains
12.3 Gauss’s Lemma . . . . . . . .
12.4 Factoring Integer Polynomials
12.5 Gauss Primes . . . . . . . . . .
12.M Miscellaneous Problems . . . .
13 Quadratic Number Fields
13.1 Algebraic Integers . . . . . .
13.2 Factoring Algebraic
Integers
√
13.3 Ideals in Z[ −5] . . . . . . .
13.4 Ideal Multiplication . . . . .
13.5 Factoring Ideals . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
21
23
.
.
.
.
.
.
.
.
.
.
24
24
25
27
32
33
35
37
38
40
42
.
.
.
.
.
.
45
45
46
50
51
56
59
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
61
61
62
62
63
63
14 Linear Algebra in a Ring
14.1 Modules . . . . . . . . . . . . .
14.2 Free Modules . . . . . . . . . . .
14.4 Diagonalizing Integer Matrices .
14.5 Generators and Relations . . . .
14.7 Structure of Abelian Groups . .
14.8 Applications to Linear Operators
14.M Miscellaneous Problems . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
64
64
65
66
66
67
67
69
.
.
.
.
.
2
15 Fields
15.2 Algebraic and Transcendental Elements
15.3 The Degree of a Field Extension . . . .
15.4 Finding the Irreducible Polynomial . . .
15.6 Adjoining Roots . . . . . . . . . . . . .
15.7 Finite Fields . . . . . . . . . . . . . . .
15.8 Primitive Elements . . . . . . . . . . .
15.9 Function Fields . . . . . . . . . . . . . .
15.10 The Fundamental Theorem of Algebra .
15.M Miscellaneous Problems . . . . . . . . .
16 Galois Theory
16.1 Symmetric Functions . . . . . . .
16.2 The Discriminant . . . . . . . . .
16.3 Splitting Fields . . . . . . . . . . .
16.4 Isomorphisms of Field Extensions
16.5 Fixed Fields . . . . . . . . . . . .
16.6 Galois Extensions . . . . . . . . .
16.7 The Main Theorem . . . . . . . .
16.8 Cubic Equations . . . . . . . . . .
16.9 Quartic Equations . . . . . . . . .
16.10 Roots of Unity . . . . . . . . . . .
16.11 Kummer Extensions . . . . . . . .
16.12 Quintic Equations . . . . . . . . .
16.M Miscellaneous Problems . . . . . .
3
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
70
70
71
72
73
74
76
77
77
78
.
.
.
.
.
.
.
.
.
.
.
.
.
81
81
83
86
86
87
88
88
91
92
94
95
95
96
2
2.1
Groups
Laws of Composition
Exercise 2.1.2. Prove the properties of inverses that are listed near the end of the
section.
Remark. The properties are listed on p. 40 as the following:
(a) If an element a has both a left inverse l and a right inverse r, i.e., if la = 1
and ar = 1, then l = r, a is invertible, r is its inverse.
(b) If a is invertible, its inverse is unique.
(c) Inverses multiply in the opposite order: If a and b are invertible, so is the
product ab, and (ab)−1 = b−1 a−1 .
(d) An element a may have a left inverse or a right inverse, though it is not
invertible.
Proof of (a). We see l = lar = r.
Proof of (b). Let b, b′ be inverses of a. Then, b = bab′ = b′ , by (a).
Proof of (c). Consider ab. We see that b−1 a−1 is the inverse of ab since (b−1 a−1 )(ab) =
b−1 a−1 ab = b−1 b = 1 by associativity. Uniqueness follows by (b).
Proof of (d). Consider Exercise 2.1.3 below. s is not invertible since it does not have
a two-sided inverse, but it does have a left inverse.
Exercise 2.1.3. Let N denote the set {1, 2, 3, . . .} of natural numbers, and let s : N →
N be the shift map, defined by s(n) = n + 1. Prove that s has no right inverse, but
that it has infinitely many left inverses.
Proof. s does not have a right inverse since s does not map any element of N back
to 1; however, we can define a left inverse rk (n) = n − 1 for n > 1, and rk (1) = k for
some k ∈ N; we see that this is a left inverse of s, i.e., that rk ◦ s = idN . Since k is
arbitrary this implies that there is an infinite number of rk ’s.
2.2
Groups and Subgroups
Exercise 2.2.3. Let x, y, z, and w be elements of a group G.
(a) Solve for y, given that xyz −1 w = 1.
(b) Suppose that xyz = 1. Does it follow that yzx = 1? Does it follow that
yxz = 1?
4
Solution for (a). We claim that y = x−1 w−1 z. This follows since
x(x−1 w−1 z)z −1 w = xx−1 w−1 zz −1 w = 1.
Solution for (b). Suppose xyz = 1. This implies x−1 = yz, and by Exercise 2.1.2(a),
this left inverse is a right inverse, and so 1 = xyz = x(yz) = (yz)x = yzx.
Now consider yxz; the example
1 −1
1 0
0 1
0 1
1 1
, yxz =
, xyz =
, z=
, y=
x=
1 0
0 1
1 −1
1 0
0 1
in GL2 (R) shows that xyz = 1 does not imply yxz = 1.
Exercise 2.2.4. In which of the following cases is H a subgroup of G?
(a) G = GLn (C) and H = GLn (R).
(b) G = R× and H = {1, −1}.
(c) G = Z+ and H is the set of positive integers.
(d) G = R× and H is the set of positive reals. a 0
, with a 6= 0.
(e) G = GL2 (R) and H is the set of matrices
0 0
Solution for (a). H is a subset since R ⊂ C implies H ⊂ GLn (R). H is a subgroup
since GLn (R) is a group, hence contains an identity and is closed under multiplication
and inversion.
Solution for (b). H is a subgroup since it is clearly a subset, contains the identity 1,
and −1 × −1 = 1 implies H is closed under multiplication and inversion.
Solution for (c). H is not a subgroup since −1 ∈
/ H, though it is the inverse of 1.
Solution for (d). H is a subgroup since it is clearly a subset, contains 1, is closed
under multiplication since the product of two positive real numbers is a positive real
number, and since x ∈ H has inverse 1/x ∈ H, which is still positive and real.
Solution for (e). H is not a subgroup since it is not even a subset of G.
2.4
Cyclic Groups
Exercise 2.4.1. Let a and b be elements of a group G. Assume that a has order 7
and that a3 b = ba3 . Prove that ab = ba.
Proof. ab = aba7 = a(ba3 )a4 = a(a3 b)a4 = a4 (ba3 )a = a4 (a3 b)a = ba.
5
Exercise 2.4.3. Let a and b be elements of a group G. Prove that ab and ba have
the same order.
Proof. Suppose (ab)n = 1. We note that b = b(ab)n = (ba)n b, but this implies
(ba)n = 1, and so both have order n.
Exercise 2.4.6.
(a) Let G be a cyclic group of order 6. How many of its elements generate G?
Answer the same question for cyclic groups of order 5 and 8.
(b) Describe the number of elements that generate a cyclic group of arbitrary orders
n.
Solution for (b). By Prop. 2.4.3, if x generates G a cyclic group of order n, another
element xi ∈ G generates G if and only if gcd(i, n) = 1 for 1 ≤ i ≤ n, since then
|xi | = n. Thus, the number of elements that generate G is equal to the number of
numbers less than n that are coprime to n.
Solution for (a). By (b), it suffices to count the number of numbers less than n that
are coprime to n. For 6, {1, 5} are coprime to 6, hence two elements generate the
cyclic group of order 6. For 5, {1, 2, 3, 4} are coprime to 5, hence four elements
generate the cyclic group of order 5. For 8, {1, 3, 5, 7} are coprime to 8, hence four
elements generate the cyclic group of order 8.
Exercise 2.4.9. How many elements of order 2 does the symmetric group S4 contain?
Solution. The order 2 elements of S4 consist of 42 = 6 two-cycles, and 42 × 12 = 3
products of disjoint two-cycles, and so there are 9 elements of order 2.
Exercise 2.4.10. Show by example that the product of elements of finite order in a
group need not have finite order. What if the group is abelian?
Solution. Consider GL2 (R), and the following matrices in GL2 (R):
−1 0
−1 1
.
,
B=
A=
0 1
0 1
We see that A2 = B 2 = 1, and so
1
AB =
0
they are of order 2, whereas
1 n
1
n
,
=⇒ (AB) =
0 1
1
and so AB has infinite order.
Now suppose the group is abelian. Suppose a, b are our elements of finite order,
of order n, m respectively. Then, (ab)nm = anm bnm = (an )m (bm )n = 1, and so ab is
necessarily of finite order.
6
2.6
Isomorphisms
Exercise 2.6.2. Describe all homomorphisms ϕ : Z+ → Z+ . Determine which are
injective, which are surjective, and which are isomorphisms.
Solution. By the definition of homomorphism, for all positive n ∈ Z+ , we have
ϕ(n) = ϕ(1) + · · · + ϕ(1),
|
{z
}
ϕ(−n) = −ϕ(n),
ϕ(0) = ϕ(n) + ϕ(−n) = 0.
n times
Thus, ϕ is fully determined by what 1 maps to. By the above, we then have that ϕn :
z
nz for n ∈ Z+ are all the homomorphisms of Z+ . The injective homomorphisms
consist of those ϕn for n 6= 0. The surjective homomorphisms consist of those ϕn for
n = ±1; these are also the isomorphisms of Z+ since they are injective.
Exercise 2.6.3. Show that the functions f = 1/x, g = (x − 1)/x generate a group of
functions, the law of composition being composition of functions, that is isomorphic
to the symmetric group S3 .
Solution. We define
1
1
, f3 = 1 − x, f4 =
,
x
1−x
Then, we can construct the multiplication table:
f1 = x,
f2 =
f1
f2
f3
f4
f5
f6
f1
f1
f2
f3
f4
f5
f6
f2
f2
f1
f6
f5
f4
f3
f3
f3
f4
f1
f2
f6
f5
f4
f4
f3
f5
f6
f2
f1
f5
f5
f6
f4
f3
f1
f2
f5 =
x
,
x−1
f6 =
x−1
.
x
f6
f6
f5
f2
f1
f3
f4
This proves closure since every combination of factors is accounted for, identity since
every row/column contains e = f1 , and associativity since associativity holds for
composition of rational functions. We claim that this is isomorphic to S3 . This
follows since if we let f1
e, f2
(12), f6
(123), we get the following table:
e
(12) (13) (132) (23) (123)
e
e
(12) (13) (132) (23) (123)
(12) (12)
e
(132) (13) (123) (23)
e
(23) (132) (12)
(13) (13) (123)
e
(132) (132) (23) (12) (123) (13)
e
(13)
(23) (23) (132) (123) (12)
(123) (123) (13) (23)
e
(12) (132)
7
This proves it is a homomorphism since all of the multiplications are accurate, and
is an isomorphism since every element in S3 is mapped to, with inverse defined by
matching entries. This shows f2 , f6 generate the group of functions since (12), (123)
generate S3 as on p. 42.
1
1 1
conjugate elements of the group
,
Exercise 2.6.6. Are the matrices
1 1
1
GL2 (R)? Are they conjugate elements of SL2 (R)?
Solution. We explicitly calculate the conjugation for the conjugation matrix A:
a11 a12 1 1
1 0 a11 a12
=
a21 a22 0 1
1 1 a21 a22
a11 a11 + a12
a11
a12
=
a21 a21 + a22
a11 + a21 a12 + a22
This equality requires a11 = 0, a12 = a21 ; however, we see that then det A < 0 in this
case, so the matrices are not conjugate in SL2 (R).
We see that they are, however, conjugate elements of the group GL2 (R), since
0 1 1 1
1 0 0 1
=
.
1 0 0 1
1 1 1 0
2.8
Cosets
Exercise 2.8.4. Does a group of order 35 contain an element of order 5? of order
7?
Solution. Any element in G has order in {1, 5, 7, 35} by Cor. 2.8.10. Suppose G
had no elements of order 5; then, all non-identity elements must have order 7, for if
|x| = 35, then |x7 | = 5. Let h have order 7, and H = hhi; since |H| = 7, pick g ∈
/ H.
Then, g 6= e and has order 7. The left cosets H, gH, g 2 H, . . . , g 6 H must be disjoint,
for, if g a hi = g b hj , then g a−b = hi−j , and so picking r such that r(a − b) ≡ 1 mod 7,
we have that g = g r(a−b) = hr(i−j) ∈ H, a contradiction. But this contradicts the
counting formula (2.8.8), since |G| = 35 6= 49 = 7 · 7 = |H|[G : H], and so G contains
an element of order 5.
Now suppose G had no elements of order 7; then all non-identity elements have
order 5 as before. Letting h have order 5 and H, g as before, the same argument gives
that H, gH, g 2 H, . . . , g 4 H are disjoint left cosets in G. This contradicts the counting
formula (2.8.8) again, since |G| = 35 6= 25 = 5 · 5 = |H|[G : H], hence G contains an
element of order 7.
8
Exercise 2.8.8. Let G be a group of order 25. Prove that G has at least one subgroup
of order 5, and that if it contains only one subgroup of order 5, then it is a cyclic
group.
Proof. Any element in G has order in {1, 5, 25} by Cor. 2.8.10. If G had no elements
of order 5, it must have an element x of order 25, but then |x5 | = 5, hence hx5 i is a
subgroup of order 5.
Now suppose H is the only subgroup of order 5 in G, and pick x ∈
/ H. x has
order 5 or 25 by Cor. 2.8.10 again, and since e ∈ H. But |x| = 5 implies H = hxi,
hence x ∈ H, and so we know |x| = 25, i.e., G = hxi is cyclic.
Exercise 2.8.10. Prove that every subgroup of index 2 is a normal subgroup, and
show by example that a subgroup of index 3 need not be normal.
Proof. If H 6 G with [G : H] = 2, we see that, picking a ∈
/ H, G = H ∐ aH =
H ∐ Ha, since this is the only we can form cosets of a in G. Thus, aH = Ha, hence
H ⊳ G by Prop. 2.8.17.
Now we consider S3 ; the 2-cycle y = (12) generates a subgroup H of order 2, and
therefore of index 3 by the counting formula (2.8.8). However, by the multiplication
table constructed last week, we see that, from (2.8.4) and (2.8.16),
xH = {x, xy} 6= {x, x2 y} = Hx.
2.10
The Correspondence Theorem
Exercise 2.10.3. Let G and G′ be cyclic groups of orders 12 and 6, generated by
elements x and y, respectively, and let ϕ : G → G′ be the map defined by ϕ(xi ) = y i .
Exhibit the correspondence referred to in the Correspondence Theorem explicitly.
Solution. We note that K = ker ϕ = {e, x6 }. The subgroups of G that contain these
are hxi, hx2 i, hx3 i, hx6 i, and each correspond to hyi, hy 2 i, hy 3 i, e respectively, which
are all the subgroups of G′ .
2.11
Product Groups
Exercise 2.11.1. Let x be an element of order r of a group G, and let y be an
element of G′ of order s. What is the order of (x, y) in the product group G × G′ ?
Solution. The order is lcm(r, s), since (x, y)n = (xn , y n ) = (e, e) implies r, s | n.
9
Exercise 2.11.3. Prove that the product of two infinite cyclic groups is not infinite
cyclic.
Proof. Recall that a cyclic group must be generated by a single element; since all
infinite cyclic groups are isomorphic to Z, we can consider Z × Z. Suppose that (a, b)
is this single element. But then, we see that (2a, b) cannot be obtained from adding
(a, b) to itself, which implies that Z × Z is not infinite cyclic.
2.12
Quotient Groups
Exercise 2.12.2. In the general linear group GL3 (R), consider the subsets




1 0 ∗
1 ∗ ∗
H = 0 1 ∗ , and K = 0 1 0 ,
0 0 1
0 0 1
where ∗ represents an arbitrary real number. Show that H is a subgroup of GL3 , that
K is a normal subgroup of H, and identify the quotient group H/K. Determine the
center of H.
Proof. H 6 GL3 since clearly


1 b11
1 a11 a21
0 1 a22  0 1
0 0
0 0
1
and since
I ∈ H,

 
1 a11 + b11 a21 + b21 + a11 b22
b21
 ∈ H,
1
a22 + b22
b22  = 0
0
0
1
1


1 −a11 −a21 + a11 a22
1 a11 a21
 = I.
0 1 a22  0
1
−a22
0
0
1
0 0
1

We now show that K is a normal subgroup of H. Define ka as the matrix in K
that has the parameter a; we see k0 = I ∈ K, ka kb = ka+b , ka k−a = I, and so it
only remains to show normality by showing hk = kh for h ∈ H, k ∈ K, which is
equivalent to hkh−1 = k:


 
 


1 0 a
1 b11 b21
1 b11 a + b21
1 b11 b21
1 0 a
0 1 0 0 1 b22  = 0 1
b22  = 0 1 b22  0 1 0 .
0 0 1
0 0
1
0 0
1
0 0
1
0 0 1
By the additive nature of the ka relations, we see that K ≈ R+ .
10
The quotient group H/K is then represented by matrices of the form


1 b 0
0 1 c  ,
0 0 1
i.e., the cosets are these times K, since multiplying by ka only alters the top-right
entry, and so multiplying by ka will keep b, c constant, and therefore remain in the
same coset. Moreover, this generates the entire space since we can multiply arbitrary
ka to a coset with arbitrary parameters b, c.
The center of H contains K, since K commutes with elements of H as shown
above; we claim the center is solely K. If A, B ∈ H, then by the above AB = BA is


 
1 a11 + b11 a21 + b21 + a22 b11
1 a11 + b11 a21 + b21 + a11 b22
.
0
 = 0
1
a22 + b22
1
a22 + b22
0
0
1
0
0
1
This implies a11 b22 = a22 b11 . But if A is fixed and B is an arbitrary matrix in H, then
a11 b22 = a22 b11 must hold for all matrices B, and so a11 = a22 = 0, hence A ∈ K.
Thus, the center of H is K ≈ R+ .
Exercise 2.12.4. Let H = {±1, ±i} be the subgroup of G = C× of fourth roots of
unity. Describe the cosets of H in G explicitly. Is G/H isomorphic to G?
Solution. By (2.8.5), cosets aH, bH are equal if and only if b = ah for some h ∈ H,
and so if a = re2πiθ , b = se2πiη for r, s ∈ R>0 , θ, η ∈ [0, 1), then aH = bH if and
only if r = s and θ − η ∈ {0, 1/4, 1/2, 3/4}. Hence the cosets of H are {re2πθ H | r ∈
R>0 , θ ∈ [0, 1/4)}.
Now consider the map ϕ : G → G, x
x4 . Then, this is trivially a homomorphism of G; it is moreover surjective since any nonzero complex number has a fourth
root. We see that ker ϕ = H, and so G/H ≈ G by the isomorphism theorem.
a b
, with
Exercise 2.12.5. Let G be the group of upper triangular real matrices
0 d
a and d different from zero. For each of the following subsets, determine whether or
not S is a subgroup, and whether or not S is a normal subgroup. If S is a normal
subgroup, identify the quotient group G/S.
(i) S is the subset defined by b = 0.
(ii) S is the subset defined by d = 1.
(iii) S is the subset defined by a = d.
11
Solution for (i). S 6 G since the diagonal entries would multiply with each other
and not affect the other entries of the matrices, I ∈ S, and the inverse can be found
by letting the inverse have entries a−1 , d−1 . S is not a normal subgroup since
a 0 1 1
a a
a d
1 1 a 0
.
=
6=
=
0 d 0 1
0 d
0 d
0 1 0 d
Solution for (ii). This forms a subgroup since
−1 a b
1/a −b/a
,
=
0 1
0
1
implies it is closed under inversion, and
′
aa ab′ + b
a b a′ b′
,
=
0
1
0 1 0 1
implies it is closed under multiplication; the identity is trivially in S.
We now see that it is a normal subgroup:
′
a (b + ab′ − a′ b)/d
a b a′ b′ 1/a −b/ad
∈ S.
=
0
1
0
1/d
0 d 0 1
We see that the quotient group G/S would be represented by matrices of the form
1 0
,
0 d
for d ∈ R× since this would cover all of G:
a b
1 0 a b
,
=
0 d
0 d 0 1
and since each matrix of the form above gives a different coset because multiplying
by elements in S keep d constant.
Solution for (iii). This is a
a
0
inverse:
subgroup since it satisfies closure:
ac bc + ad
b c d
∈ S,
=
0
ac
a 0 c
−1 a b
1/a −b/a2
=
∈ S,
0 a
0
1/a
12
and clearly the identity is in S.
We now check this subgroup is normal:
−1 ′
a b a′ b′ a b
a ab′ /d
=
∈ S.
0 d 0 a′ 0 d
0
a′
The quotient group G/S would be represented by matrices of the form
1 0
,
0 d
for d ∈ R× since this would cover all of G:
1 0 a b
a b
,
=
0 ad
0 d 0 a
and since each matrix of the form above gives a different coset since multiplying by
elements in S cannot give another matrix of the same form unless d is the same.
6
Symmetry
6.3
Isometries of the Plane
Exercise 6.3.2. Let m be an orientation-reversing isometry. Prove algebraically
that m2 is a translation.
Proof. By Thm. 6.3.2 and (6.3.3), m2 = tv ρθ rtv ρθ r = tv ρθ rtv rρ−θ = tv ρθ tv′ ρ−θ =
tv+v′′ , where v ′′ = ρθ (r(v)).
Exercise 6.3.6.
(a) Let s be the rotation of the plane with angle π/2 about about the point (1, 1)t .
Write the formula for s as a product ta ρθ .
(b) Let s denote reflection of the plane about the vertical axis x = 1. Find an
isometry g such that grg −1 = s, and write s in the form ta ρθ r.
Solution for (a). By (6.3.3),
s = t(1,1) ρπ/2 t(−1,−1) = t(1,1) tρπ/2 (−1,−1) ρπ/2 = t(2,0) ρπ/2 .
Solution for (b). By (6.3.3), letting g = t1,0 ρπ/2 ,
s = grg −1 = t(1,0) ρπ/2 rρ−π/2 t(−1,0) = t(1,0) ρπ rt(−1,0) = t(1,0) ρπ t(−1,0) r = t(2,0) ρπ r.
13
6.4
Finite Groups of Orthogonal Operators on the Plane
Exercise 6.4.2.
(a) List all subgroups of the dihedral group D4 , and decide which ones are normal.
(b) List the proper normal subgroups N of the dihedral group D15 , and identify the
quotient groups D15 /N .
(c) List the subgroups of D6 that do not contain x3 .
Solution for (a). We claim the subgroups of D4 form the lattice diagram
{e}
hyi
hx2 yi
hx2 i
hx3 yi
hx2 , yi
hxi
hx2 , xyi
hxyi
D4
By Lagrange’s theorem (Thm. 2.8.9), any subgroup H 6 D4 must have |H| ∈
{1, 2, 4, 8}. {e} is the unique order 1 subgroup. The order 2 subgroups are generated by one order 2 element, hence are given by the second row above. The order
4 subgroups are either generated by an order 4 element or by two order 2 elements;
these give the third row. Finally, D4 is the unique order 8 subgroup.
We claim the normal subgroups of D4 are
{e}, hxi, hx2 i, hx2 , yi, hx2 , xyi, D4 .
hx2 i is normal since x2 commutes with all other elements of D4 ; the other proper
subgroups are normal because they are of index 2 by the counting formula (2.8.8),
and by Exercise 2.8.10. Lastly, the other four subgroups in the second row of the
diagram above are not normal since they are not closed under conjugation by x.
Solution for (b). Any proper subgroup H 6 D15 must have |H| ∈ {2, 3, 5, 15} by
Lagrange’s theorem (Thm. 2.8.9). The order 2 subgroups are of the form hxi yi; these
are not normal since they are not closed under conjugation by x. Since 2 ∤ 3, 5, 15,
we know no other subgroup contains an element of the form xi y. Thus, every normal
subgroup of D15 must be of the form hxi i for i = 1, 3, 5, since any other i gives a
subgroup equal to one of these three. These are all in fact normal subgroups since
they are kernels of homomorphisms D15 → Di for i = 1, 3, 5 mapping x
x, y
y.
14
By the first isomorphism theorem (Thm. 2.12.10), this implies the quotient groups
for each nontrivial N are isomorphic to
D15 /hxi ≈ D1 ,
D15 /hx3 i ≈ D3 ,
D15 /hx5 i ≈ D5 .
Solution for (c). By Lagrange’s theorem (Thm. 2.8.9), any subgroup H 6 D6 must
have |H| ∈ {1, 2, 3, 4, 6, 12}. {e} is the unique order 1 subgroup, and does not contain
x3 . The order 2 subgroups are hx3 i and hxi yi for 0 ≤ i ≤ 5; only the first contains
x3 . The unique order 3 subgroup is hx2 i, which does not contain x3 . The order 4
subgroups not containing x3 are of the form hxi y, xj yi for i 6= j since D6 contains no
elements of order 4, but this subgroup is of order 4 if and only if xj−i = x3 = xi−j ,
hence there are no order 4 subgroups not containing x3 . The order 6 subgroups must
be generated by an order 2 and an order 3 element, hence those not containing x3
are of the form hx2 , xi yi. But different choices i, j for i give the same subgroup if
and only if i ≡ j mod 2, hence the only subgroups of order 6 not containing x3 are
hx2 , yi and hx2 , xyi. The unique order 12 subgroup D16 contains x3 . In summary,
the subgroups that do not contain x3 are
{e}, hyi, hxyi, hx2 yi, hx3 yi, hx4 yi, hx5 yi, hx2 i, hx2 , yi, hx2 , xyi.
Exercise 6.4.3.
(a) Compute the left cosets of the subgroup H = {1, x5 } in the dihedral group D10 .
(b) Prove that H is normal and that D10 /H is isomorphic to D5 .
(c) Is D10 isomorphic to D5 × H?
Solution for (a). The cosets are represented by e, x, x2 , x3 , x4 , y, xy, x2 y, x3 y, x4 y.
Solution for (b). To show that H is normal, it suffices to show x5 commutes with
every element of D10 : it trivially commutes with x, and with y since yx5 y = yyx5 =
x5 . Thus H ⊳ D10 .
Now the surjective homomorphism D10 → D5 sending x
x, y
y has kernel
H, hence D10 /H ≈ D5 by the first isomorphism theorem (Thm. 2.12.10).
Solution for (c). Note D5 ֒→ D10 by having
x2 , hence to show D10 ≈ D5 × H,
T x
it suffices by Prop. 2.11.4(d) to show H D5 = {e}, HD5 = G, and H, D5 ⊳ G with
this embedding of D5 . The normality of both groups follows by (b) and since D5 is
of index 2 in G (Exercise 2.8.10). Their intersection is e since x5 ∈
/ D5 . HD5 = G
5
since x commutes with every element of D5 ⊂ D10 by (b), and so we can get any
element of the form xi , xi y for 0 ≤ i ≤ 9.
15
6.5
Discrete Groups of Isometries
Exercise 6.5.5. Prove that the group of symmetries of the frieze pattern ⊳ ⊳ ⊳ ⊳
⊳ ⊳ ⊳ is isomorphic to the direct product C2 × C∞ of a cyclic group of order 2 and
an infinite cyclic group.
Proof. Let G 6 M be the group of symmetries. By Thm. 6.3.2, any symmetry arises
as tv ρθ or tv ρθ r where tv are translations, ρθ are rotations, and r are reflections across
the x-axis. In both cases, θ must be zero to be a symmetry, and v must be an integer
multiple of the length of ⊳. Hence the group C2 of reflections across the x-axis and
the group C∞ of translations by integer multiples of lengths of ⊳ generate G.
We claim C2 × C∞ ≈ G. Since clearly C2 ∩ C∞ = {e}, by Prop. 2.11.4(d) it only
remains to show C2 , C∞ ⊳ G. C∞ ⊳ G since it has index 2 in G by the classification
of symmetries above, and then by Exercise 2.8.10. C2 ⊳ G since our translations all
lie on the x-axis, and then by using (6.3.3).
Exercise 6.5.9. Let G be a discrete subgroup of M whose translation group is not
trivial. Prove that there is a point p0 in the plane that is not fixed by any element of
G except the identity.
Proof. We first claim that the set of points fixed by any nontrivial isometry m ∈ G
has Lebesgue measure zero. We proceed by considering each kind of isometry in
Thm. 6.3.4. If m is a nontrivial translation, then there are no fixed points. If m is
a nontrivial rotation around a point p, then p is a fixed point. If m is a nontrivial
reflection about a line ℓ, then any point on ℓ is a fixed point. If m is a nontrivial glide,
then m has no fixed points. Thus, in all cases the set of fixed points has Lebesgue
measure zero in R2 .
Now for each m 6= e, let Fm be the set of fixed points of m; by the above, it
has Lebesgue measure zero. We claim there are only countably many m ∈ G. By
Thm. 6.3.2, any m is of the form tv ρθ or tv ρθ . By Thm. 6.5.5, there are only countably
many tv . By Prop. 6.5.10, there are only finitely many ρθ and ρθ r. Thus, there are
only countably many m ∈ G.
Finally, the set of all points fixed by some nontrivial element of G has Lebesgue
measure
!
[
X
µ
Fm ≤
µ(Fm ) = 0,
e6=m∈G
e6=m∈G
hence almost all points in R2 are not fixed by any nontrivial element of G.
16
6.6
Plane Crystallographic Groups
Exercise 6.6.2. Let G be the group of symmetries of an equilateral triangular lattice
L. Determine the index in G of the subgroup of translations in G.
Proof. The index of translations is given by [G : T ]. By Theorem 6.3.2, we see that
every isometry should be given by tv ρθ r; we therefore want the cardinality of the set
of ρθ r’s. But this is exactly the set D3 , and so [G : T ] = |D3 | = 6.
6.7
Abstract Symmetry: Group Operations
Exercise 6.7.1. Let G = D4 be the dihedral group of symmetries of the square.
(a) What is the stabilizer of a vertex? of an edge?
(b) G operates on the set of two elements consisting of the diagonal lines. What
is the stabilizer of a diagonal?
Solution for (a). If the vertex lies along the axis of reflection for y, {e, y} is the
stabilizer. If it does not, {e, x2 y} is the stabilizer. If the given edge lies immediately
to the +θ direction of the line of reflection, the stabilizer is {e, xy}. If it does not,
{e, x3 y} is the stabilizer.
Solution for (b). The stabilizer of the diagonal is {e, x2 , y, x2 y}.
Exercise 6.7.2. The group M of isometries of the plane operates on the set of lines
in the plane. Determine the stabilizer of a line.
Solution. It suffices to consider the classes of isometries in Thm. 6.3.4. The stabilizer
of a line ℓ consists of translations along ℓ, rotations of an angle π about a point p ∈ ℓ,
reflections about ℓ, and glide reflections about ℓ.
Exercise 6.7.8. Decompose the set C2×2 of 2×2 matrices into orbits for the following
operations of GL2 (C):
(a) left multiplication,
(b) conjugation.
Solution (a). Left multiplication by elements of GL2 (C) corresponds to products of
elementary row operations on matrices in C2×2 . By row reduction, we know that
applying elementary row operations on a given matrix in C2×2 gives a unique matrix
of one of the following forms:
0 0
0 1
1 a
1 0
.
,
,
,
0 0
0 0
0 0
0 1
17
Thus, C2×2 decomposes as
! a
1 a
0 0
0 1
1 0
.
GL2 (C) ·
∐
∐ GL2 (C) ·
∐
GL2 (C) ·
0 0
0 0
0 0
0 1
a∈C
Solution (b). Since every matrix is conjugate (similar) to a unique Jordan form (up
to ordering of Jordan blocks), we have that the orbits are generated by the different
Jordan forms, i.e., C2×2 decomposes as
!
!
a
a
λ 1
λ 0
GL2 (C) ∗
GL2 (C) ∗ 1
.
∐
0 λ
0 λ2
λ∈C
λ1 ≥λ2 ,λi ∈C
Exercise 6.7.11. Prove that the only subgroup of order 12 of the symmetric group
S4 is the alternating group A4 .
Proof. We first prove a lemma: If H is a subgroup of G, then H is normal if and
only if H is the union of conjugacy classes inSG. But H is normal if and only if it is
closed under conjugation if and only if H = h∈H G ∗ h.
Now suppose H 6 S4 has order 12. Then, by the counting formula (2.8.8),
|S4 | = 24 = 12 · [S4 : H] = |H|[S4 : H] implies [S4 : H] = 2, and so H ⊳ S4 by
Exercise 2.8.10. So, we use the classification of conjugacy classes in S4 from p. 201,
following Prop. 7.5.1:
Partition
Element No. in Conj. Class
1+1+1+1
e
1
4
2+1+1
(ab)
=6
2 1 4
2+2
(ab)(cd)
=3
2 2 4
3+1
(abc)
2· 3 =8
4
(abcd)
3! = 6
Now if H ⊳ S4 has order 12, it must arise from a union of conjugacy classes above
including {e} (since it is a subgroup), but the only way this is possible is the sum
1+3+8 = 12. The conjugacy classes associated with these terms are exactly the even
permutations in S4 , which form the subgroup A4 , and so A4 is the only subgroup of
order 12 in S4 .
6.8
The Operation on Cosets
Exercise 6.8.4. Let H be the stabilizer of the index 1 for the operation of the symmetric group G = Sn on the set of indices {1, . . . , n}. Describe the left cosets of H
in G and the map (6.8.4) in this case.
18
Solution. We see that H consists of all cycles that hold 1 fixed, i.e., permutations of
the remaining n − 1 elements, hence is isomorphic to Sn−1 . We claim
G/H = {(1i)H | i ∈ {1, . . . , n}}
Each (1i) gives a different coset since no (1i)H contains (1j) for i 6= j, and then by
(2.8.5). These are all the cosets since by Pop. 6.8.4, |G/H| = |O1 | = n. Thus, the
map ǫ : G/H → O1 is the map (1i)
i.
6.9
The Counting Formula
Exercise 6.9.4. Identify the group T ′ of all symmetries of a regular tetrahedron,
including orientation-reversing symmetries.
Solution. T ′ 6 O3 (R) operates transitively on the set F of faces of order 4, hence is
given by a permutation of the set {f1 , f2 , f3 , f4 }, and so there is a homomorphism
T ′ → S4 by Prop. 6.11.2. This is injective since if t ∈ T ′ fixes three faces, then it fixes
the three vectors defining the centers of each face, hence is the identity matrix in
O3 (R). We claim this homomorphism T ′ → S4 is surjective, hence an isomorphism;
it suffices to show |T ′ | = 24. But the stabilizer Gf of a given face f is the group
D3 generated by a rotation by 2π/3 about the center of f and a reflection about an
axis in f ; |D3 | = 6, and so the counting formula (6.9.2) gives |T ′ | = 6 · 4 = 24, hence
T ′ ≈ S4 .
6.10
Operations on Subsets
Exercise 6.10.1. Determine the orders of the orbits for left multiplication on the
set of subsets of order 3 of D3 .
Solution. We know that there are 63 = 20 subsets of order 3 of D3 ; we also know
by the counting formula (6.9.2) that there can only be orbits of order 1, 2, 3, 6 since
only these divide |D3 |. We see that only the subset {e, x, x2 } is a subgroup, which
produces the orbit {e, x, x2 }, {y, yx, yx2 }. The other 18 subsets form 3 orbits of order
6 of D3 . Letting H1 = {e, x, y}, we have
H1 = {e, x, y},
yH1 = {y, yx, e},
xH1 = {x, x2 , yx2 },
yxH1 = {yx, yx2 , x2 },
Letting H2 = {e, x, yx}, we have
H2 = {e, x, yx},
yH2 = {y, yx, x},
xH2 = {x, x2 , y},
yxH2 = {yx, yx2 , e},
19
x2 H1 = {x2 , e, yx},
yx2 H1 = {y, yx2 , x}.
x2 H2 = {x2 , e, yx2 },
yx2 H2 = {y, yx2 , x2 }.
Letting H3 = {e, x, yx2 }, we have
H3 = {e, x, yx2 },
yH3 = {y, yx, x2 },
xH3 = {x, x2 , yx},
yxH3 = {yx, yx2 , x},
x2 H3 = {x2 , e, y},
yx2 H3 = {y, yx2 , e}.
We therefore have one orbit of order 2 and three orbits of order 6; this means we
have 1 × 2 + 3 × 6 = 20 subsets in these orbits, and so we have found all of them.
6.11
Permutation Representations
Exercise 6.11.1. Describe all the ways in which S3 can operate on a set of four
elements.
Solution. By Cor. 6.11.3, it suffices to find all homomorphisms f : S3 → S4 . Recall
from p. 42 that S3 is generated by x = (123), y = (12); a homomorphism f is then
fully determined by specifying f (x), f (y). We moreover note that x3 = e, y 2 = e
implies f (x)3 = f (y)2 = e, hence f (x) = e or is one of the eight 3-cycles in S4 by the
classification in Exercise 6.7.11; similarly, f (y) = e or is one of the nine elements of
order 2 in the table from Exercise 6.7.11.
If f (x) = e, then there is no restriction on f (y), thus there are ten homomorphisms f of this kind.
If f (x) = (abc), then since yx has order 2, f (yx)2 = f (y)2 f (x)2 = e. Thus
f (y) 6= e. Now suppose f (y) acts nontrivially on the fourth element d unaffected by
f (x); then without loss of generality, f (y) interchanges a, d, and so f (yx) · d = a.
Hence f (yx) must map a
d as well, and so f (y) · b = d. But then f (y)2 · b =
a 6= b, contradicting that f (y) has order 2. Hence f (y) is a permutation of order
2 of the subset {a, b, c}, which are 2-cycles; assume without loss of generality that
f (y) = (ab). In the argument above, there are 4 choices for the fixed point d of f (x),
3 choices for the fixed point of f (y), and two choices for a = f (x) · c. Each choice
gives a homomorphism f by just renaming a, b, c as 1, 2, 3, hence realizing f as the
canonical operation of S3 on the subset {1, 2, 3} ⊂ {1, 2, 3, 4}. Thus, there are 24
homomorphisms f such that f (x) 6= e.
Exercise 6.11.5. A group G operates faithfully on a set S of five elements, and there
are two orbits, one of order 3 and one of order 2. What are the possible groups?
Hint: Map G to a product of symmetric groups.
Solution. Let S = {1, 2, 3, 4, 5} such that the two orbits are O3 = {1, 2, 3}, O2 =
{4, 5}, respectively. G operates on O3 , O2 separately, hence the action of G on
O3 , O2 respectively correspond to group homomorphisms f3 : G → S3 and f2 : G →
20
S2 . This defines a group homomorphism f = (f3 , f2 ) : G → S3 × S2 defined by
g
(f3 (g), f2 (g)). Since G acts faithfully on S, g is injective, hence G ≈ f (G) by
Cor. 2.12.11. Now the only group that acts transitively on O2 is S2 itself, and so
f2 (G) ≈ S2 . On the other hand, there are two groups that act transitively on O3 :
C3 and S3 . Hence G ≈ C3 × S2 or S3 × S2 .
Exercise 6.11.6. Let F = F3 . There are four one-dimensional subspaces of the space
of column vectors F 2 . List them. Left multiplication by an invertible matrix permutes
these subspaces. Prove that this operation defines a homomorphism ϕ : GL2 (F ) → S4 .
Determine the kernel and the image of this homomorphism.
Proof. We have the following one-dimensional subspaces, denoting F = {0, 1, 2}:
1
2
2
1
0
0
2
1
.
,
, V4 =
,
, V3 =
,
, V2 =
,
V1 =
2
1
2
1
2
1
0
0
Call ei the first listed vector in Vi ; note that Vi = {ei , 2ei }. These are all the subspaces
in F 2 since there are only 3 · 3 − 1 = 8 nontrivial vectors in F 2 .
Left multiplication by an invertible matrix defines an action on the vectors in F 2
since left multiplication by matrices in F is associative, and since the identity matrix
defines the trivial action; this descends to a well-defined action on the subspaces Vi
since the span of Aei is equal to the span of 2Aei for any A ∈ GL2 (F ). Hence, by
Cor. 6.11.3, this action defines a homomorphism ϕ : GL2 (F ) → S4 .
Now ker ϕ = {I, 2I} since ⊃ clearly holds, and for A ∈ GL2 (F ), Ae1 ∈ V1 implies
the first column of A is either e1 or 2e1 , Ae2 ∈ V2 implies the second column of A is
either e2 or 2e2 , and Ae3 ∈ V3 implies that the coefficients on ei in each column have
to be the same, hence A = I or 2I.
We claim im ϕ = S4 . By Cor. 2.8.13, we have |GL2 (F )| = |ker ϕ||im ϕ|, and since
|S4 | = 24 and |ker ϕ| = 2 from above, it suffices to show |GL2 (F )| = 48. Now every
matrix in GL2 (F ) consists of a pair of linearly independent vectors in F2 ; there are
8 · 6 = 48 of these pairs by the above decomposition of F2 into subspaces, hence
|GL2 (F )| = 48.
6.12
Finite Subgroups of the Rotation Group
Exercise 6.12.3. Let O be the group of rotations of a cube, and let S be the set of
four diagonal lines connecting opposite vertices. Determine the stabilizer of one of
the diagonals.
21
Solution. O acts on S, hence there is a homomorphism O → S4 by Prop. 6.11.2. We
first show this homomorphism is injective. If s ∈ O fixes all four diagonals, then
s either fixes or interchanges the two endpoints of each diagonal. If s 6= e, i.e., it
acts nontrivially on the vertexes, then we can pick three pairs of opposite vertexes
such that s interchanges one of the pairs of vertexes, or such that s interchanges all
three pairs of vertexes. In either case, give R3 a basis that points along the three
diagonals connecting these pairs of vertexes; then the matrix for s in this basis has
an odd number of −1’s on the diagonal, and so has determinant −1, contradicting
that O 6 SO3 (R). The homomorphism O → S4 is surjective, since looking at the
faces of the cube, O acts transitively on the set of faces, and each face has stabilizer
C4 , hence |O| = |Of ||O · f | = 4 · 6 = 24 by the counting formula (6.9.2).
Now we know O acts like S4 on the set of diagonals, hence the stabilizer of one
of the diagonals is the same as fixing one index in the set of indexes {1, 2, 3, 4} as
in Exercise 6.8.4, hence is equal to S3 .
Exercise 6.12.7. The 12 points (±1, ±α, 0)t , (0, ±1, ±α)t , (±α, 0, ±1)t form the
vertices of a regular icosahedron if α > 1 is chosen suitably. Verify this, and determine α.
Proof. We have that the 12 points form three categories of the form above. By the
distance formula, if they are in the same category, we have three possibilities for
square distances:
Vertex
Square Distance
Change sign of 1 Change sign of α
4
4α2
Change sign of both
4(α2 + 1)
If they are in different categories, then we have two possibilities for square distances.
If our coordinates are (x1 , x2 , x3 ) and (y1 , y2 , y3 ), then there is only one i such that
xi , yi 6= 0, and xi , yi must have different absolute values for each i, giving the table
Vertex
Square Distance
xi , yi have same sign
2(α2 − α + 1)
xi , yi have differing sign
2(α2 + α + 1)
where for a given vertexes, there are four vertexes each with square distance of each
form above.
We now find α. There must be five vertexes of shortest distance from v0 ; by
comparing the possible distances above, this shortest square distance must be equal
to both 4 and 2(α2 − α + 1). This gives the equation
√
1+ 5
2
2
2(α − α + 1) = 4 =⇒ α − α − 1 = 0 =⇒ α =
2
22
by choosing the root greater than 1. This shows each vertex has exactly five neighbors
of distance 2 away from it; we claim that the polyhedron formed by connecting
vertexes of distance 2 away from each other forms an icosahedron.
Now we show each face formed by the edges is a congruent equilateral triangle.
This is true since any face is formed by neighboring vertexes, which are distance 2
away from each other by the above. Each vertex moreover has the same number of
faces meeting there since every vertexes has exactly five neighbors by the above.
Finally, suppose we have two neighboring vertexes v, v ′ forming an edge; we claim
there are only two faces intersecting at that edge. It suffices to show there are only
two vertexes w that are of distance 2 from both. Suppose vi = vi′ = 0; in the
′
following, we consider subscripts mod 3. Then vi+1 = −vi+1
with absolute value
′
1, so wi+1 = 0, |wi+2 | = 1, |wi | = 3 by the table above. Next vi+2 = vi+2
, hence
′
wi+2 must have the same sign as vi+2 , vi+2 . Finally, wi can have either sign since
vi = vi′ = 0, hence there are only two vertexes that are of distance 2 from v, v ′ .
6.M
Miscellaneous Problems
Exercise 6.M.7. Let G be a finite group operating on a finite set S. For each element
g of G, let S g denote the subset of elements of S fixed by g: S g = {s ∈ S | gs = s},
and let Gs be the stabilizer of s.
(a) We may imagine a true-false table for the assertion that gs = s, say with rows
indexed by elements of G and columns indexed by elements of S. Construct
such a table for the P
action of theP
dihedral group D3 on the vertices of a triangle.
(b) Prove the formula s∈S |Gs | = g∈G |S g |.
P
(c) Prove Burnside’s Formula: |G| · (number of orbits) = g∈G |S g |.
Solution for (a). We construct the true-false table:
e
x
x2
y
yx
yx2
s1
True
False
False
True
False
False
s2
True
False
False
False
True
False
s3
True
False
False
False
False
True
where S = {s1 , s2 , s3 } is the set of the vertices of the triangle, x = (s1 s2 s3 ) rotations,
and y is the reflection with preferred vertex s1 .
23
Proof of (b). By summing over different sets, we have
X
X
|Gs | = |{(g, s) ∈ G × S | gs = s}| =
|S g |.
s∈S
g∈G
Proof of (c). By the orbit-stabilizer theorem (Prop. 6.8.4) we have that there is a
bijection between cosets G/Gs and orbits Os . By the counting formula (2.8.8), we
then have
|Gs | · |Os | = |Gs | · |G/Gs | = |G|.
Note in particular that |Gs′ | is equal for all s′ ∈ Os since Os = Os′ . Thus,
X
X
X
|G| · (number of orbits) =
|G| =
|Gs | · |Os | =
|Gs |,
orbits Os
orbits Os
s∈S
and so, combining (b),
|G| · (number of orbits) =
11
11.1
X
g∈G
|S g |.
Rings
Definition of a Ring
√
√
√
Exercise 11.1.1. Prove that 7 + 3 2 and 3 + −5 are algebraic numbers.
√
3
3
3
Proof.
7
+
21x2 + 147x − 345 = 0.
√
√ 2 is a root of (x2− 7) 2− 2 = x −
4
3 + −5 is a root of (x + 2) + 60 = x + 4x2 + 64 = 0.
Exercise 11.1.2. Prove that, for n 6= 0, cos(2π/n) is an algebraic number.
Proof. Suppose n > 0. Recall that the Chebyshev polynomials Tn (x) are defined by
the recurrence relations T0 (x) = 1, T1 (x) = x, and Tn (x) = 2x Tn−1 (x) − Tn−2 (x).
We claim Tn (cos θ) = cos nθ for any θ. This is clear for n = 0, 1. For arbitrary n,
Tn (cos θ) = 2 cos θ Tn−1 (cos θ) − Tn−2 (cos θ)
= 2 cos θ cos ((n − 1)θ) − cos ((n − 2)θ)
= cos nθ + cos ((n − 2)θ) − cos ((n − 2)θ) = cos nθ,
and so Tn (cos θ) = cos nθ for any θ as desired. Letting θ = 2π/n, we have that
Tn (cos(2π/n)) = cos 2π = 1, and so cos(2π/n) satisfies the polynomial equation
Tn (x) − 1 = 0.
If n < 0, then using the fact that cos(2π/n) = cos(−2π/n), we see cos(2π/n)
satisfies the polynomial equation T−n (x) − 1 = 0 by the above.
24
Exercise 11.1.3. Let Q[α, β] denote
√ the smallest
√ subring of C containing the rational
numbers Q and the elements α = 2 and β = 3. Let γ = α+β. Is Q[α, β] = Q[γ]?
Is Z[α, β] = Z[γ]?
Solution. Since γ ∈ Z[α, β], Q[α, β], the reverse inclusion ⊃ holds in both cases. We
claim that the inclusion ⊂ holds for Q, but not Z.
To show Q[α, β] ⊂ Q[γ], it suffices to show α, β ∈ Q[γ]. Now γ 2 = 5+2αβ ∈ Q[γ],
hence αβ ∈ Q[γ], so αβγ = 3α + 2β ∈ Q[γ]. Thus α = αβγ − 2γ ∈ Q[γ], and finally
β = γ − α ∈ Q[γ] as well.
To show Z[α, β] 6⊂ Z[γ], we claim αβ ∈
/ Z[γ].PIt suffices to show that γ k only has
k
even coefficients for αβ, for then, any element N
k=0 ak γ ∈ Q[γ] will have an even
coefficient for αβ, hence αβ cannot be expressed as such a sum.
To prove this, we claim
(
aα + bβ for a, b odd
when k odd
γk =
c + dαβ for c odd, d even when k even
This is clear for k = 0, k = 1. For arbitrary n, consider first when k is even. Then,
by inductive hypothesis
γ k = γ k−1 γ = (aα + bβ)(α + β) = 2a + 3b + (a + b)αβ,
for a, b odd, hence 2a + 3b is odd and a + b is even. Likewise, when k is odd,
γ k = γ k−1 γ = (c + dαβ)(α + β) = (3d + c)α + (c + 2d)β,
for c odd, d even, hence 3d + c, c + 2d are odd, and we are done.
11.2
Polynomial Rings
Exercise 11.2.1. For which positive integers n does x2 + x + 1 divide x4 + 3x3 +
x2 + 7x + 5 in [Z/(n)][x]?
Proof. We will perform the division algorithm on x4 + 3x3 + x2 + 7x + 5. We have
x2 + x + 1
x2 + 2x − 2
x4 + 3x3 + x2 + 7x + 5
− x4 − x3 − x2
2x3
+ 7x
2
2
−2x − 2x − 2x
−2x2 + 5x + 5
2x2 + 2x + 2
7x + 7
25
and so x4 + 3x3 + x2 + 7x + 5 = (x2 + 2x − 2)(x2 + x + 1) + (7x + 7), where the
remainder 7x + 7 ≡ 0 mod n if and only if n ∈ {1, 7}. Hence x2 + x + 1 divides
x4 + 3x3 + x2 + 7x + 5 in [Z/(n)][x] if and only if n ∈ {1, 7}.
Exercise 11.2.2. Let F be a field. The set of all formal power series p(t) = a0 +
a1 t + a2 t2 + · · · , with ai in F , forms a ring that is often denoted by F [[t]]. By formal
power series we mean that the coefficients form an arbitrary sequence of elements of
F . There is no requirement of convergence. Prove that F [[t]] is a ring, and determine
the units in this ring.
P
P
P
i
i
i
+ defined as
Proof. We denote
i bi t , r(t)
P p(t) = k i ai t , q(t) =
P =P i ci t . Let
k
p(t) + q(t) = k (ak + bk )t and × as p(t) × q(t) = k i+j=k ai bj t .
+ makes F [[t]] an abelian group because associativity and commutativity follow
since + is defined termwise, and because having ai = 0 for all i defines an identity
and letting q(t) such that biP
= −ai for all iPdefines an inverse for p(t).
× is commutative since i+j=k ai bj = i+j=k bi aj , and is associative since
(p(t) × q(t)) × r(t) =
X X
ℓ
i+j=ℓ
= p(t) ×
ai bj tℓ × r(t) =
X X
ℓ
j+k=ℓ
X X
ℓ
ai bj ck tℓ
i+j+k=ℓ
bj ck tℓ = p(t) × (q(t) × r(t)).
The identity is p(t) such that a0 = 1, ai = 0 for all i > 0.
It remains to show the distributive property:
X
X
X X
(p(t) + q(t)) × r(t) =
(ai + bi )ti ×
cj t j =
(ai + bi )cj tk
i
=
j
X X
k
ai cj tk +
i+j=k
k
X X
k
i+j=k
bi cj t k
i+j=k
= p(t) × r(t) + q(t) × r(t).
Finally, we claim that the p(t) such that a0 6= 0 are the units. Any unit must
have a0 6= 0, for p(t) × q(t) = 1 =⇒ a0 b0 = 1. In the other direction, suppose p(t)
is such that a0 6= 0. Define q(t) such that
b0 =
a−1
0 ,
bi =
−a−1
0
26
i
X
j=1
aj bi−j .
Then, a0 b0 = 1 but
k
X
i=0
ak−i bi = a0 bk + a1 bk−1 + a2 bk−2 + · · · + ak b0 = 0,
and so p(t) × q(t) = 1.
11.3
Homomorphisms and Ideals
Exercise 11.3.3. Find generators for the kernels of the following maps:
(a) R[x, y] → R defined by f (x, y)
f (0, 0),
(b) R[x] → C defined by f (x)
f (2 + √
i),
(c) Z[x] → R defined by f (x) √ f (1√+ 2),
2 + 3,
(d) Z[x] → C defined by x
(e) C[x, y, z] → C[t] defined by x
t, y
t2 , z
t3 .
Remark. We will denote each map as ϕ.
Solution for (a). We claim that (x, y) = ker ϕ. By the division algorithm, any polynomial f ∈ R[x, y] can be written g + a0 for g ∈ (x, y), and so ϕ(f ) = a0 = 0 if and
only if a0 = 0 if and only if f = g ∈ (x, y).
Solution for (b). We claim that (x2 − 4x + 5) = ker ϕ. x2 − 4x + 5 = (x − (2 + i))(x −
(2 − i)), hence (x2 − 4x + 5) ⊂ ker ϕ. Conversely, let f ∈ ker ϕ. By the division
algorithm, we can write f = g + r for g ∈ (x2 − 4x + 5), where r = a1 x + a0 for
ai ∈ R has degree less than 2. Then, ϕ(f ) = ϕ(g) + ϕ(r) = a1 (2 + i) + a0 , which is
zero only if a1 = a0 = 0, i.e., only if f = g ∈ (x2 − 4x + 5).
√
2
2
Solution
√ for (c). We2 claim that (x −2x−1) = ker ϕ. x −2x−1 = (x−(1+ 2))(x−
(1 − 2)), hence (x − 2x − 1) ⊂ ker ϕ. Conversely, let f ∈ ker ϕ. By the division
algorithm, we can write f = g + r for g ∈ (x2 − 4x + 5), where r = a1 x + a0 for
ai ∈ Z √
has degree less than 2, since x2 − 2x − 1 is monic. Then,
ϕ(f ) = ϕ(g) + ϕ(r) =
√
a1 (1 + 2) + a0 , which is zero only if a1 = a0 = 0 since 1, 2 are linearly independent
over Z, i.e., only if f = g ∈ (x2 − 2x − 1).
2
Solution for (d). We claim that (x√4 − 10x
+ 1) = ker
ϕ. √
We have (x4 − 10x2 + 1) ⊂
√
√
ker ϕ, since ϕ(x4 − 10x2 + 1) = ( 2 + 3)4 − 10( 2 + 3)2 + 1 = 0. Conversely,
27
let f ∈ ker ϕ. By the division algorithm, we can write f = g + r, where r =
a3 x3 + a2 x2 + a1 x + a0 for ai ∈ Z has degree less than 4. Then,
√
√
ϕ(f ) = ϕ(g + r) = r( 2 + 3)
√
√
√
√
√
√
= a3 ( 2 + 3)3 + a2 ( 2 + 3)2 + a1 ( 2 + 3) + a0
√
√
√
= (11a3 + a1 ) 2 + (9a3 + a1 ) 3 + (2a2 ) 6 + (5a2 + a0 ).
This gives rise to the system of equations represented by the matrices

   
1 0 5 0
a0
0
0 0 2 0  a1  0

   
0 1 0 9  a2  = 0 ,
0 1 0 11
a3
0
√ √ √
since 2, 3, 6, 1 are linearly independent over Z. But, this only has the trivial
solution a0 = a1 = a2 = a3 = 0, since
1
0
0
0
0
0
1
1
5 0
1 0 0
2 0
1 9
= −2 0 1 9 = −2
= −2(11 − 9) = −4 6= 0,
0 9
1 11
0 1 11
0 11
which implies f = g ∈ (x4 − 10x2 + 1).
Solution for (e). We claim that (x2 − y, x3 − z, y 3 − z 2 ) = ker ϕ. Clearly, (x2 − y, x3 −
z, y 3 − z 2 ) ⊂ ker ϕ, since
ϕ(x2 − y) = t2 − t2 = 0,
ϕ(x3 − z) = t3 − t3 = 0,
ϕ(y 3 − z 2 ) = t6 − t6 = 0.
Conversely, we first regard f as a polynomial in z whose coefficients are in x, y,
as in Corollary 11.3.8. We can apply the division algorithm to get f = g1 + r1 for
g1 ∈ (y 3 − z 2 ), where r1 is of degree less than 2 in z. If r1 = 0, then f ∈ (y 3 − z 2 ), and
so we are done. If not, then we can apply the division algorithm again with x2 − y,
this time in y, on r1 to have r1 = g2 + r2 for g2 ∈ (x2 − y), where r2 is of degree 0 in
y, and degree less than 2 in z. If r2 = 0, then f = g1 + g2 ∈ (y 3 − z 2 , x2 − y), and
so we are done. If not, then we can apply the division algorithm again with x3 − z,
this time in x, on r2 to have r2 = g3 + r3 for g3 ∈ (x3 − z), where r3 is of degree less
than 2 in x, degree 0 in y, and degree less than 2 in z. This means that we have
ϕ(r3 ) = ϕ(a101 xz + a001 z + a100 x + a000 ) = a101 t4 + a001 t3 + a100 t + a000 = 0,
and the linear independence of tk over C implies that r3 = 0, and so f = g1 +g2 +g3 ∈
(y 3 − z 2 , x2 − y, x3 − z).
28
Exercise 11.3.5. The derivative of a polynomial f with coefficients in a field F is
defined by the calculus formula (an xn + · · · + a1 x + a0 )′ = nan xn−1 + · · · + 1a1 . The
integer coefficients are interpreted in F using the unique homomorphism Z → F .
(a) Prove the product rule (f g)′ = f ′ g + f g ′ and the chain rule (f ◦ g)′ = (f ′ ◦ g)g ′ .
(b) Let α be an element of F . Prove that α is a multiple root of a polynomial f if
and only if it is a common root of f and of its derivative f ′ .
P
P
P
ProofP
of (a). Let f =
ai x i , g =
bj xj . Then by (11.2.7), f g =
ck xk , where
ck = i+j=k ai cj , and so
′
fg=
X
(i + 1)ai+1 x
i≥0
i
!
X
bj x
j
j≥0
!
=
!
=
X X
(i + 1)ai+1 bj xk ,
X X
(j + 1)ai bj+1 xk .
k≥0 i+j=k
and similarly
!
X
ai xi
f ′g + f g′ =
X
f g′ =
i≥0
X
(j + 1)bj+1 xj
j≥0
k≥0 i+j=k
Thus,
=
(i + 1)ai+1 bj + (j + 1)ai bj+1
k≥0
i+j=k
X
X
k≥0
=
X
X
k≥0
iai bj + jai bj
i+j=k+1
(k + 1)
X
!
ai bj xk =
i+j=k+1
!
xk
xk
X
(k + 1)ck+1 xk = (f g)′ .
k≥0
Proof of (b). Suppose f ∈ F [x]. Then, f = (x − α)k g for some k ≥ 0 and g ∈ F [x]
such that g(α) 6= 0. By (a), f ′ = k(x − α)k−1 g + (x − α)k g ′ if k ≥ 1, and f ′ = g ′ if
k = 0. Hence f (α) = f ′ (α) = 0 if and only if k ≥ 2, i.e., α is a multiple root of f if
and only if α is a common root of f, f ′ .
Exercise 11.3.7. Determine the automorphisms of the polynomial ring Z[x].
Proof. Suppose ϕ ∈ Aut(Z[x]). Then ϕ(1) = 1, hence
· · + 1}) = ϕ(n).
n = ϕ(1) + · · · + ϕ(1) = ϕ(1| + ·{z
{z
}
|
n times
n times
29
Pn
Pn
j
j
Thus, if f =
j=0 bj x ∈ Z[x], then ϕ(f ) =
j=0 bj ϕ(x) , and so ϕ is uniquely
determined by ϕ(x).
P
So let ϕ(x) = di=0 ai xi . Then,
!d
d
n
d
X
X
X
ϕ(f ) =
ai
bj x j
= ad bn xnd + · · · +
ai bdj .
(1)
i=0
j=0
i=0
hence deg(ϕ(f )) = nd. Suppose f ∈ Z[x] is the unique element such that ϕ(f ) = x.
Then, nd = 1 and ad bn = 1 imply that d = 1 and a1 ∈ {±1}. Thus ϕ(x) = ±x + a
for a ∈ Z, i.e., every ϕ ∈ Aut(Z[x]) must map x
±x+a for a ∈ Z. Finally, all such
ϕ give automorphisms of Z since they define ring homomorphisms by the equation
(1), and since Z[±x + a] = Z[x].
Exercise 11.3.8. Let R be a ring of prime characteristic p. Prove that the map
R → R defined by x
xp is a ring homomorphism. (It is called the Frobenius
map).
Proof. We first claim that for p prime, p | pi if 1 ≤ i ≤ p − 1. By definition,
p
p(p − 1) · · · (p − i + 1)
=
.
i
i(i − 1) · · · 2 · 1
1 ≥ i implies p appears in the
p does not appear
numerator, and i ≤ pp− 1 implies
p
p
in the denominator. Since i ∈ Z, this implies p | i , hence i = 0 ∈ R.
Now if x, y ∈ R, the binomial theorem gives
p−1 p X
X
p i p−i
p i p−i
p
p
p
xy
= xp + y p
xy
=x +y +
(x + y) =
i
i
i=1
i=0
Thus x
xp respects addition. Since trivially 1
1 and (xy)p = xp y p , we therefore
p
have that x
x defines a ring homomorphism R → R.
Exercise 11.3.9.
(a) An element x of a ring R is called nilpotent if some power is zero. Prove that
if x is nilpotent, then 1 + x is a unit.
(b) Suppose that R has prime characteristic p 6= 0. Prove that if a is nilpotent
then 1 + a is unipotent, that is, some power of 1 + a is equal to 1.
Proof of (a). Suppose xN = 0. Then
(1 + x)(1 − x + x2 − · · · + (−1)N −1 xN −1 ) = 1 − xN = 1,
hence 1 + x is a unit.
30
Proof of (b). Suppose aN = 0. By Exercise 11.3.8, x
xp is a homomorphism
R → R, so (1 + a)p = 1 + ap . Iterating this map n times such that pn ≥ N gives
n
n
n
(1 + a)p = 1 + ap = 1 + aN ap −N = 1.
Exercise 11.3.10. Determine all ideals of the ring F [[t]] of formal power series with
coefficients in a field F (see Exercise 11.2.2).
Solution. We claim all nonzero ideals are of the form (tn ) for some n. Let I be an
ideal and p ∈ I such the number n := min{i | ai 6= 0} is minimal. We claim I = (tn ).
First, p = tn q for some unit q, hence (tn ) ⊂ I. Conversely, any r ∈ I has first nonzero
coefficient at degree ≥ n, hence tn s for some s ∈ F [[t]], and so r ∈ (tn ).
Exercise 11.3.11. Let R be a ring, and let I be an ideal of the polynomial ring
R[x]. Let n be the lowest degree among nonzero elements of I. Prove or disprove: I
contains a monic polynomial of degree n if and only if it is a principal ideal.
Proof of ⇒. If I = 0, then it is principal so suppose not. Let f be a monic polynomial
of lowest degree n. We claim (f ) = I. f ∈ I hence (f ) ⊂ I. Now suppose g ∈ I;
then, by division with remainder we can write g = f q + r, where if r 6= 0, it has
degree lower than f . But then, f, g ∈ I, hence g − f q = r ∈ I, so r = 0, and
g ∈ (f ).
Counterexample for ⇐. Consider I = (2x) ⊂ Z[x]. Any element in I is obtained by
multiplying 2x by a polynomial of degree ≥ 1, in which case we get an element of
degree ≥ 2, or by multiplying by an element of Z. But then, 2 ∈
/ Z× , hence there is
no monic polynomial of degree 1 in I.
Exercise 11.3.12. Let I and J be ideals of a ring R. Prove that the set I + J of
elements of the form x + y, with x in I and y in J, is an ideal. This ideal is called
the sum of the ideals I and J.
Proof. Let (x + y), (x′ + y ′ ) ∈ I + J and s ∈ R, where x, x′ ∈ I, y, y ′ ∈ J. Then,
s(x + y) + (x′ + y ′ ) = (sx + x′ ) + (sy + y ′ ) ∈ I + J, hence I + J is an ideal.
Exercise 11.3.13. Let I and J be ideals of a ring R. Prove that the intersection
I ∩ J is an ideal. Show by example that the set P
of products {xy | x ∈ I, y ∈ J} need
not be an ideal, but that the set of finite sums
xν yν of products of elements of I
and J is an ideal. This ideal is called the product ideal, and is denoted by IJ. Is
there a relation between IJ and I ∩ J?
31
Proof. Let x, y ∈ I ∩ J and s ∈ R. Then, sx + y ∈ I and sx + y ∈ J since I, J are
ideals, hence sx + y ∈ I ∩ J and so I ∩ J is an ideal.
Now let R = Z[x], I = (2, x), J = (3, x). Then, 3x, 2x are in the set of products,
but their
− 2x = x is not, and so the set of products is not anP
ideal.
Pdifference
P ′ 3x
′
′
′
xi yi , xi yP
xi yi +
i ∈ IJ and s ∈ R, where xi , xi ∈ I, yi , yi ∈ J. Then, s
P Let
P
x′i yi′ =
sxi yi + x′i yi′ ∈ IJ, hence IJPis an ideal.
We have in general IJ ⊂ I ∩ J since
xi yi for xi ∈ I, yi ∈ J has xi yi ∈ I ∩ J.
IJ = I ∩ J if I + J = (1) by Exercise 11.6.8(a).
I ∩ J ⊂ IJ does not hold in general, for if R = Z, I = (m), J = (n), then
I ∩ J = (lcm(m, n)) but IJ = (mn).
11.4
Quotient Rings
Exercise 11.4.2. What does the Correspondence Theorem tell us about ideals of Z[x]
that contain x2 + 1?
Solution. By the Correspondence Theorem (Thm. 11.4.3), there is a bijective correspondence between the ideals of Z[x] that contain x2 + 1 and the ideals of Z[x]/(x2 +
1) ≈ Z[i], where these two rings are isomorphic as in Ex. 11.4.5. By Prop. 12.2.5(c),
Z[i] is a Euclidean domain, hence a principal ideal domain by Prop. 12.2.7. Thus,
every ideal in Z[i] is of the form (a + bi) for a, b ∈ Z. These correspond to ideals
(a + bx) in Z[x]/(x2 + 1) by the isomorphism i
x from above, hence the ideals in
2
2
Z[x] containing x + 1 are of the form (a + bx, x + 1).
Exercise 11.4.3. Identify the following rings: (a) Z[x]/(x2 − 3, 2x + 4), (b) Z[i]/(2 +
i), (c) Z[x]/(6, 2x − 1), (d) Z[x]/(2x2 − 4, 4x − 5), (e) Z[x]/(x2 + 3, 5).
Solution for (a). We see that 2(x2 − 3) − (x − 2)(2x + 4) = 2 ∈ (x2 − 3, 2x + 4), and
so (x2 − 3, 2x + 4) = (x2 − 3, 2x + 4, 2) = (x2 − 3, 2), since 2(x + 2) = 2x + 4. Then,
Z[x]/(x2 − 3, 2x + 4) = Z[x]/(x2 − 3, 2). But then, Z[x]/(x2 − 3, 2) ≈ F2 [x]/(x2 − 3) =
F2 [x]/(x2 + 1) = F2 [x]/(x + 1)2 since x2 + 1 = (x + 1)2 in F2 .
Solution for (b). First recall Z[i] ≈ Z[x]/(x2 + 1). Thus, Z[i]/(2 + i) ≈ Z[x]/(x2 +
1, 2 + x). We first consider the quotient Z[x]/(2 + x). Since (2 + x) is the kernel of
the homomorphism Z[x] → Z[−2], f (x)
f (−2), we see that this is isomorphic to
Z, as in Example 11.4.5. Then, we have that the residue of g = x2 + 1 is 5, and so
we have Z[i]/(2 + i) ≈ Z/5Z = F5 .
Solution for (c). We first note 6x − 3(2x − 1) = 3 ∈ (6, 2x − 1), and so (6, 2x − 1) =
(3, 2x − 1). Also, 3x − (2x − 1) = x + 1 ∈ (3, 2x − 1), and so (3, 2x − 1) = (3, x + 1).
Thus, Z[x]/(6, 2x − 1) = Z[x]/(3, x + 1) ≈ F3 [x]/(x + 1) = F3 .
32
Solution for (d). Since (4x + 5)(4x − 5) − 8(2x2 − 4) = 7 ∈ (2x2 − 4, 4x − 5), hence
(2x2 − 4, 4x − 5) = (7, 2x2 − 4, 4x − 5). Hence Z[x]/(2x2 − 4, 4x − 5) = Z[x]/(7, 2x2 −
4, 4x−5) = F7 [x]/(2x2 −4, 4x−5). In F7 [x], 4(2x2 −4) = x2 +5 and 2(4x−5) = x+4,
hence (2x2 −4, 4x−5) = (x2 +5, x+4). But x+4 | x2 +5, hence (x2 +5, x+4) = (x+4).
Finally, F7 [x]/(2x2 − 4, 4x − 5) = F7 [x]/(x2 + 5, x + 4) = F7 [x]/(x + 4) ≈ F7 .
Solution for (e). We see Z[x]/(x2 + 3, 5) ≈ F5 [x]/(x2 + 3). But since x2 + 3 6= 0 for
any x ∈ F5 , √
we see that we cannot reduce x2 + 3. But in F5 , x2 + 3 = x2 − 2, and so
we have F5 [ 2].
Exercise 11.4.4. Are the rings Z[x]/(x2 + 7) and Z[x]/(2x2 + 7) isomorphic?
Solution. We claim they are not. Let R = Z[x]/(x2 + 7) and S = Z[x]/(2x2 + 7).
Asume α : R → S is an isomorphism. α(1) = 1, hence α(2) = 2, and so if α
is an isomorphism, then R/(2) ≈ S/(2). We claim this is a contradiction. For,
R/(2) ≈ F2 [x]/(x2 + 1) 6= 0, whereas S/(2) ≈ F2 [x]/(1) = 0.
11.5
Adjoining Elements
Exercise 11.5.1. Let f = x4 +x3 +x2 +x+1 and let α denote the residue of x in the
ring R = Z[x]/(f ). Express (α3 + α2 + α)(α5 + 1) in terms of the basis (1, α, α2 , α3 )
of R.
Proof. We first have (x3 + x2 + x)(x5 + 1) = x8 + x7 + x6 + x3 + x2 + x. We then
perform the division algorithm:
x4 + x3 + x2 + x + 1
x4
−x
x8 + x7 + x6
+ x3 + x2 + x
− x8 − x7 − x6 − x5 − x4
− x5 − x4 + x3 + x2 + x
x5 + x4 + x3 + x2 + x
2x3 + 2x2 + 2x
and so (α3 + α2 + α)(α5 + 1) = 2α3 + 2α2 + 2α ∈ Z[x]/(f ).
Exercise 11.5.3. Describe the ring obtained from Z/12Z by adjoining an inverse of
2.
Solution. Let R = (Z/12Z)[x]/(2x − 1) be our ring; it is isomorphic to Z[x]/(12, 2x −
1). (12, 2x − 1) = (3, 2x − 1) since ⊂ clearly holds and 12x2 − (3 + 6x)(2x − 1) =
3 ∈ (12, 2x − 1). Thus R ≈ Z[x]/(3, 2x − 1) ≈ F3 [x]/(2x − 1). But 2 ∈ F×
3 , hence
F3 [x]/(2x − 1) ≈ F3 , so R ≈ F3 .
33
Exercise 11.5.4. Determine the structure of the ring R′ obtained from Z by adjoining an element α satisfying each set of relations.
(a) 2α = 6, 6α = 15, (b) 2α − 6 = 0, α − 10 = 0, (c) α3 + α2 + 1 = 0, α2 + α = 0.
Solution for (a). Z[α] = Z[x]/(2x−6, 6x−15). But (6x−15)−3(2x−6) = 3 ∈ (2x−
6, 6x−15), and 3(x−2)−(2x−6) = x ∈ (2x−6, 6x−15) imply (2x−6, 6x−15) = (x, 3).
Thus, Z[α] ≈ Z[x]/(3, x) ≈ Z/3Z = F3 .
Solution for (b). Z[α] = Z[x]/(2x − 6, x − 10). But (2x − 6) − 2(x − 10) = 14 ∈ (2x −
6, x−10) and (x−10)+14 = x+4 ∈ (2x−6, x−10) imply (2x−6, x−10) = (x+4, 14),
since 2x − 6 = 2x(x + 4) − 14. Thus, Z[α] ≈ Z[x]/(x + 4, 14) ≈ Z/14Z.
Solution for (c). Z[α] = Z[x]/(x3 + x2 + 1, x2 + x). But x3 + x2 + 1 − x(x2 + x) =
1 ∈ (x3 + x2 + 1, x2 + x), and so Z[α] ≈ Z[x]/(1) = 0, the zero ring.
Exercise 11.5.6. Let a be an element of a ring R, and let R′ be the ring R[x]/(ax−1)
obtained by adjoining an inverse of a to R. Let α denote the residue of x (the inverse
of a in R′ ).
(a) Show that every element β of R′ can be written in the form β = αk b, with b
in R.
(b) Prove that the kernel of the map R → R′ is the set of elements b of R such
that an b = 0 for some n > 0.
(c) Prove that R′ is the zero ring if and only if a is nilpotent (see Exercise 11.3.9).
P
Proof of (a). Any element β ∈ R′ can be written as a finiteP
sum
bk αk . Letting K
K−k
be the largest k such that bk 6= 0, we see that defining b =
a
bk , αK b = β.
Proof of (b). Let Γa (R) = {b ∈ R | an b = 0 for some n > 0}, and call the map
defined ϕ; it suffices to show Γa (R) = ϕ−1 ((ax − 1)), where (ax − 1)P⊂ R[x]. But
b ∈ R has ϕ(b) ∈ (ax − 1) if and only if (ax − 1)g = b for some g =
gi xi ∈ R[x].
Solving recursively, we must have gi = −ai b for all i. Such a g exists if and only if
b ∈ Γa (R), for otherwise g would be an infinite sum.
Proof of (c). a is nilpotent if and only if 1 ∈ Γa (R). By the proof of (b), this holds
if and only if 1 ∈ (ax − 1), and since R[x] → R′ is surjective, this holds if and only
if R′ = 0 by the first isomorphism theorem (Thm. 11.4.2(b)).
34
11.6
Product Rings
Exercise 11.6.1. Let ϕ : R[x] → C×C be the homomorphism defined by ϕ(x) = (1, i)
and ϕ(r) = (r, r) for r ∈ R. Determine the kernel and the image of ϕ.
Solution. ϕ is the evaluation map f
(f (1), f (i)). So f ∈ ker ϕ ⇐⇒ f (1) =
2
f (i) = 0 ⇐⇒ f ∈ ((x − 1)(x + 1)), i.e., ker ϕ = ((x − 1)(x2 + 1)).
We now claim im ϕ = R × C. ⊂ clearly holds. Now let (c, a + bi) ∈ R × C. Then,
if f = a3 x3 + a2 x2 + a1 x + a0 ∈ R[x] is of degree 3,
ϕ(f ) = (a3 + a2 + a1 + a0 , −ia3 − a2 + a1 i + a0 )
= (a3 + a2 + a1 + a0 , (a0 − a2 ) + (a1 − a3 )i),
and the condition f
(c, a + bi) gives the system of equations


 a0 + a1 + a2 + a3 = c
a0
− a2
=a


a1
− a3 = b
which has a solution by linear algebra, hence ⊃ also holds, and im ϕ = R × C.
Exercise 11.6.2. Is Z/(6) isomorphic to the product ring Z/(2) × Z/(3)? Is Z/(8)
isomorphic to Z/(2) × Z/(4)?
Solution. Letting R = Z/(6), I = (2), J = (3) in Exercise 11.6.8(c), we have Z/(6) ≈
Z/(2) × Z/(3), since IJ = 0 in R.
Z/(8) 6≈ Z/(2)×Z/(4), for Z/(8) has an additive element of order 8 while Z/(2)×
Z/(4) does not.
Exercise 11.6.3. Classify rings of order 10.
Proof. Let R be a ring of order 10. By the Sylow Theorems (Thms. 7.7.2, 7.7.4, 7.7.6),
the abelian group (R, +) contains exactly one normal subgroup each of orders 2 and 5,
with trivial intersection, hence by Prop. 2.11.4(d), (R, +) ≈ Z/(2) × Z/(5) ≈ Z/(10).
It remains to show R ≈ Z/(10) also as a ring. Let a ∈ R be an element of order 10;
a then generates R, and so 1 ∈ R implies na = 1 for some 1 ≤ n ≤ 9. If n = 2, then
5a = 2a · 5a = 10a2 = 0, contradicting that a has order 10; similarly, n 6= 5. Thus,
na = 1 has order 10, and generates R, hence R ≈ Z/(10).
Exercise 11.6.4. In each case, describe the ring obtained from the field F2 by adjoining an element α satisfying the given relation:
(a) α2 + α + 1 = 0, (b) α2 + 1 = 0, (c) α2 + α = 0.
35
Remark. Since each equation is of degree 2, we can write F2 [α] = {0, 1, α, α + 1} by
Prop. 11.5.5(a) in each case.
Solution for (a). Since α(α + 1) = α2 + α = 1, every element of F2 [α] has an inverse.
Thus, we have a field of order 4, i.e., F2 [α] = F4 .
Solution for (b). F2 [α] = F2 [x]/(x2 + 1) = F2 [i], the ring of Gauss integers mod 2.
We see (α + 1)2 = α2 + 1 = 0, and so α2 = 1, α(α + 1) = α2 + α = α + 1. This is
not a field, or even an integral domain, since (α + 1)2 = 0.
Solution for (c). We have α2 = α, and so by Prop. 11.6.2 we have F2 [α] ≈ αF2 [α] ×
(α+1)F2 [α]. Since αF2 [α] = {0, α} and (α+1)F2 [α] = {0, α+1} are both isomorphic
to F2 , we see that F2 [α] ≈ F2 × F2 .
Exercise 11.6.5. Suppose we adjoin an element α satisfying the relation α2 = 1 to
the real numbers R. Prove that the resulting ring is isomorphic to the product R × R.
Proof. The resulting ring is R = R[x]/(x2 − 1). Let ϕ : R[x] → R × R be defined
by f
(f (1), f (−1)). ker ϕ = (x2 − 1) = ((x + 1)(x − 1)), and so by the first
isomorphism theorem (Thm. 11.4.2(b)) it suffices to show ϕ is surjective, inducing
an isomorphism R → R × R. Letting f = ax + b ∈ R[x] of degree 1, we have
ϕ(ax + b) = (a + b, a − b),
which can be solved for arbitrary (a + b, a − b) = (x, y) by linear algebra.
Exercise 11.6.6. Describe the ring obtained from the product ring R×R by inverting
the element (2, 0).
Solution. Using the isomorphism in Exercise 11.6.5, since x + 1
(2, 0), the ring is
R[x, (x + 1)−1 ]
R[x, (x + 1)−1 ]
R[x, (x + 1)−1 ]
R[x, y]
≈
=
=
≈ R,
(x2 − 1, (x + 1)y − 1)
(x2 − 1)
((x + 1)(x − 1))
(x − 1)
since x + 1 has residue 2 in R[x]/(x − 1), which already has an inverse in R.
Exercise 11.6.7. Prove that in the ring Z[x], the intersection (2) ∩ (x) of the
principal ideals (2) and (x) is the principal ideal (2x), and that the quotient ring
R = Z[x]/(2x) is isomorphic to the subring of the product ring F2 [x] × Z of pairs
(f (x), n) such that f (0) ≡ n modulo 2.
36
Proof. Clearly (2x) ⊂ (2) ∩ (x). Conversely, if f ∈ (2) ∩ (x), then f = 2g = xh for
some g, h ∈ Z[x]. But this implies g ∈ (x), hence f ∈ (2x).
P
Now
consider
the
ring
homomorphism
ϕ
:
Z[x]
→
F
[x]
×
Z
where
ai xi
2
P
(ai xi , a0 ), where ai is the residue of ai in F2 . We claim ϕ(Z[x]) = S := {(f (x), n) ∈
F2 [x] × Z | f (0) ≡ n mod 2}. By
ϕ(Z[x]) ⊂ S; conversely, any
P construction,
ai xi where a0 = P
n, and so choosing repre(f (x), n) ∈ S must have f (x) =
ai xi 7→ (f (x), n), hence
sentatives ai of ai in Z such that a0 = n, we see that
ϕ(Z[x]) ⊃ S.
Finally we show ker ϕ = (2x). (2x) ⊂ ker ϕ by construction. Conversely, ϕ(f ) =
(0, 0) =⇒ a0 = 0 and 2 | ai for i 6= 0, hence f ∈ (2) ∩ (x) = (2x) by the above.
Thus, R ≈ S by the first isomorphism theorem (Thm. 11.4.2(b)).
Exercise 11.6.8. Let I and J be ideals of a ring R such that I + J = R.
(a) Prove that IJ = I ∩ J (see Exercise 11.3.13).
(b) Prove the Chinese Remainder Theorem: For any pair a, b of elements of R,
there is an element x such that x ≡ a modulo I and x ≡ b modulo J. (The
notation x ≡ a modulo I means that x − a ∈ I.)
(c) Prove that if IJ = 0, then R is isomorphic to the product ring (R/I) × (R/J).
(d) Describe the idempotents corresponding to the product decomposition in (c).
Proof of (a). Clearly IJ ⊂ I ∩J. In the other direction, let u+v = 1 for u ∈ I, v ∈ J.
Then, if x ∈ I ∩ J, x = x(u + v) = xu + xv ∈ IJ.
Proof of (b). It suffices to show the ring homomorphism ϕ : R → R/I × R/J defined
by x 7→ (x+I, x+J) is surjective. Let (a+I, b+J) ∈ R/I ×R/J. Since I +J = R, we
have u + v = 1 for some u ∈ I, v ∈ J. Let x = bu + av. Then, ϕ(x) = (x + I, x + J) =
(av + I, bu + J) = (a(1 − u) + I, b(1 − v) + J) = (a + I, b + J).
Proof of (c). By definition in (b) ker ϕ = I ∩ J, and by (a), ker ϕ = IJ = 0, and so
we are done by the first isomorphism theorem (Thm. 11.4.2(b)).
Solution for (d). By (b), if u + v = 1 for u ∈ I, v ∈ J, then ϕ(v) = (1, 0) and
ϕ(u) = (0, 1). Hence the idempotents corresponding to the product decomposition
in (c) are the images of u, v from partitions of unity u + v = 1 for u ∈ I, v ∈ J.
11.7
Fractions
Exercise 11.7.1. Prove that a domain of finite order is a field.
37
Proof. Suppose R is a finite domain, and consider R× := R \ {0}. It suffices to show
any r ∈ R× is a unit. If |R× | = n, then r · R× ⊂ R× since R is a domain, and
|r · R× | = n by the cancellation law (11.7.1). Hence 1 ∈ r · R× , and so rr′ = 1 for
some r′ ∈ R× .
Exercise 11.7.2. Let R be a domain. Prove that the polynomial ring R[x] is a
domain, and identify the units in R[x].
Proof. Let f, g ∈ R[x] be nonzero, and suppose f g = 0. If deg f = d, deg g = d′ ,
then deg f g = d + d′ since the product ad bd′ of their leading coefficients is nonzero,
contradicting that deg f g is undefined. Hence R[x] is a domain.
If f g = 1 then deg f g = d + d′ = 0, so f and g are constant polynomials. Viewing
constant polynomials as elements of R, the units of R[x] are then the units of R.
Exercise 11.7.3. Is there a domain that contains exactly 15 elements?
Solution. Suppose R is such a domain; it is a field by Exercise 11.7.1. It suffices to
show any finite field F has order pn for some prime power pn , since 15 cannot be
written in this way. By Lem. 3.2.10, F has characteristic p for some prime p, and
so contains Fp . Then, F is of finite dimension n as a vector space over Fp , and so
contains pn elements.
Exercise 11.7.4. Prove that the field of fractions of the formal power series F [[x]]
over a field F can be obtained by inverting the element x. Find a neat description of
the elements of that field (see Exercise 11.2.2).
P
i
Proof. Take an element fg in the field of fractions of F [[x]] and suppose g = ∞
i=n ai x
with n ≥ 0 and an 6= 0. Then, g = xn g0 , where g0 is a unit of F [[x]] by Exercise 11.2.2,
f g −1
and so fg = x0n . Thus, the field of fractions of F [[x]] is contained in F [[x]][x−1 ], and
the reverse inclusion is clear, and so the field of fractions
is F [[x]][x−1 ].
P of F [[x]]
i
This is the ring of infinite series of the form h = i≥n ai x for n ∈ Z, which is
called the ring of formal Laurent series F ((x)).
11.8
Maximal Ideals
Exercise 11.8.1. Which principal ideals in Z[x] are maximal ideals?
Solution. We claim that none are. Suppose (f ) ⊂ Z[x] is maximal. Then, deg f > 0,
for otherwise (f ) ( (f, x) ( Z[x], contradicting maximality of (f ). Now choose p
such that p ∤ ai for any coefficient ai of f . Then, (f ) ( (p, f ) since p ∈
/ (f ), and
(p, f ) ( Z[x] since Z[x]/(p, f ) = Fp [x]/(f ) 6= 0. Thus, (f ) cannot be maximal.
38
Exercise 11.8.2. Determine the maximal ideals of each of the following rings:
(a) R × R, (b) R[x]/(x2 ), (c) R[x]/(x2 − 3x + 2), (d) R[x]/(x2 + x + 1).
Solution for (a). The units in R × R are elements (a, b) for a, b 6= 0, and so any
maximal ideal in R × R cannot contain a pair (a, 0) and (0, b) for a, b 6= 0. Hence
every element in a maximal ideal of R × R must have the same coordinate equal to
zero, and so it is of the form R × 0 or 0 × R since 0 × 0 is not maximal; these are
maximal by Prop. 11.8.2(b) since taking quotients gives R, a field.
Solution for (b). The ideals in R[x]/(x2 ) are the ideals in R[x] containing (x2 ) by the
correspondence theorem (Thm. 11.4.3). Since R[x] is a PID by Props. 12.2.5, 12.2.7,
the only proper ideals containing (x2 ) are (x) and (x2 ), since if x2 ∈ (f ), then x2 = f g
implies f | x2 . We have (x2 ) ( (x), and so (x) is the only maximal ideal.
Solution for (c). x2 − 3x + 2 = (x − 2)(x − 1) and (x − 2) + (x − 1) = R[x] implies
R[x]/(x2 − 3x + 2) ≈ R[x]/(x − 1) × R[x]/(x − 2) ≈ R × R by Exercise 11.6.8(c).
The maximal ideals of R × R are R × 0 and 0 × R as in (a), which correspond to
(x − 2), (x − 1), respectively, in R[x]/(x2 − 3x + 2).
Solution for (d). We note that x2 + x + 1 has no real roots, and so is irreducible in
R[x]. Thus, (x2 +x+1) maximal and R[x]/(x2 +x+1) is a field with unique maximal
ideal (0) by Prop. 11.8.2.
Exercise 11.8.3. Prove that the ring F2 [x]/(x3 +x+1) is a field, but that F3 [x]/(x3 +
x + 1) is not a field.
Proof. By Prop. 11.8.2(b), it suffices to show (x3 + x + 1) is maximal in F2 [x] but not
in F3 [x]. But this is true since x3 +x+1 has no roots in F2 (03 +0+1 = 1 = 13 +1+1),
while x3 + x + 1 has the root 1 in F3 (13 + 1 + 1 = 0), hence (x3 + x + 1) ( (x − 1).
Exercise 11.8.4. Establish a bijective correspondence between maximal ideals of R[x]
and points in the upper half plane.
Proof. R[x] is a PID by Props. 12.2.5, 12.2.7, and so any maximal ideal is of the
form (f ). But (f ) is maximal if and only if f is an irreducible nonunit in R[x].
Every polynomial in R[x] of degree at least 3 has a real root, and therefore is not
irreducible. So, every maximal ideal is of the form (f ) for f a linear polynomial or
an irreducible quadratic polynomial.
Now recall we can identify the upper half plane with the subset of C of elements z
with Im(z) ≥ 0. Therefore, we can define the function H that sends every maximal
ideal (f ) to the root of f in the upper half plane. Notice that the function is
39
well-defined: if f is linear, then it has only one solution, a real number, and if f is an
irreducible quadratic polynomial, then z and z are the roots of f , for some z ∈ C \ R,
so only one of the roots is in the upper half-plane.
We need to show H is a bijection. We define the inverse H −1 sending z to the
ideal generated by x − z if z ∈ R, and (x − z)(x − z) otherwise. H −1 is clearly a
two-sided inverse of H since (f ) = (cf ) for any c ∈ R, and so we are done.
11.9
Algebraic Geometry
Exercise 11.9.4. Let U and V be varieties in Cn . Prove that the union U ∪ V and
the intersection U ∩ V are varieties. What does the statement U ∩ V = ∅ mean
algebraically? What about the statement U ∪ V = Cn ?
Proof. Let U be defined by {f1 , . . . , fr }, and V by {g1 , . . . , gs }. U ∩ V is defined by
{f1 , . . . , fr , g1 , . . . , gs } since x ∈ U ∩ V if and only if fi (x) = 0 for all i and gj (x) = 0
for all j. We claim U ∪ V is equal to the variety W defined by {fi gj }i,j . U ∪ V ⊂ W
since if x ∈ U (resp. V ), then fi = 0 for all i (resp. gj = 0 for all j), hence fi gj = 0
for all i, j. Conversely, suppose x ∈ W \ U ∪ V . Then, there exists i0 , j0 such that
fi0 (x), gj0 (x) 6= 0, and so fi0 gj0 6= 0, a contradiction.
Now by Thm. 11.9.1 and Cor. 11.9.3, U ∩ V = ∅ if and only if the quotient ring
C[x1 , . . . , xn ]/(f1 , . . . , fr , g1 , . . . , gs ) = 0, i.e., (f1 , . . . , fr , g1 , . . . , gs ) = (1).
Now U ∪ V = Cn if and only if any point x ∈ Cn is a common zero of all fi gj .
But there are only finitely many fi gj with finitely many zeros each, and so this holds
if and only if all the fi gj are equal to 0, which is true if and only if all the fi or all
the gj are 0, if and only if U or V equals Cn .
Exercise 11.9.5. Prove that the variety of zeros of a set {f1 , . . . , fr } of polynomials
depends only on the ideal they generate.
Proof. For sets of polynomials {f1 , . . . , fr } and {g1 , . . . , gs }, let V, W be the varieties
formed by their zero sets and let I, J be the ideals they generate, respectively. We
claim I ⊃ J implies V ⊂ W . Every gj can then be written as
Pa linear combination
of the fi , hence if x ∈ V , fi (x) = 0 for all i, and so gj (x) =
ai fi (x) = 0 for all j
as well, i.e., V ⊂ W .
Finally, if I = J, then V ⊂ W and W ⊂ V , and so V = W . Thus, the variety
defined by a set {f1 , . . . , fr } depends only on the ideal (f1 , . . . , fr ).
Exercise 11.9.6. Prove that every variety in C2 is the union of finitely many points
and algebraic curves.
40
Proof. Let U be defined by {f1 , . . . , fr }, and let g = gcd(f1 , . . . , fr ); g exists since
C[x, y] is a UFD (Thm. 12.3.10). Then, U = V ∪ W , where V is defined by g and
W by {f1 /g, . . . , fr /g} as in Exercise 11.9.4. W is the union of finitely many points
by Thm. 11.9.10 since gcd(gcd(f1 /g, . . . , fr−1 /g), fr /g) = Q
gcd(f1 /g, . . . , fr /g) = 1.
V is the union of finitely many algebraic curves since g = gi for some irreducible
polynomials gi , hence V is the union of curves defined by gi as in Exercise 11.9.4.
Exercise 11.9.11. Let C1 and C2 be the zeros of quadratic polynomials f1 and f2
respectively that don’t have a common linear factor.
(a) Let p and q be distinct points of intersection of C1 and C2 , and let L be the
(complex) line through p and q. Prove that there are constants c1 and c2 , not
both zero, so that g = c1 f1 + c2 f2 vanishes identically on L. Prove also that g
is the product of linear polynomials.
(b) Prove that C1 and C2 have at most 4 points in common.
Proof of (a). Let λ(t) = (1 − t)p + tq parametrize L ⊂ C2 . Then, f1 (λ(t)), f2 (λ(t))
are quadratic in t, and so c1 f1 (λ(t)) + c2 f2 (λ(t)) is at most quadratic in t for any
ci ∈ C. Let t0 ∈ C \ {0, 1}, and choose ci such that c1 f1 (λ(t0 )) + c2 f2 (λ(t0 )) = 0;
we can moreover choose ci to be nonzero. Then, h(t) = c1 f1 (λ(t)) + c2 f2 (λ(t)) is at
most quadratic in t hence has at most two zeros if h 6= 0 by the fundamental theorem
of algebra (Thm. 15.10.1); but h(0) = h(1) = h(t0 ) = 0 implies h(t) = 0. Hence
g = c1 f1 + c2 f2 vanishes identically on L.
Now since g vanishes on L, by unique factorization (Thm. 12.3.10) g = gL g ′ for
some linear polynomial gL that defines L. g is at most quadratic, hence g ′ is either
linear or constant. In either case, g is then a product of linear polynomials.
Proof of (b). Suppose C1 ∩ C2 contains more than four points. Let g as in (a). Then,
g = c1 f1 + c2 f2 implies that C1 ∩ C2 ⊂ {g = 0}. g is either the product of one
or two linear polynomials as in (a), and so C1 ∩ C2 is contained in either one or
two lines. In either case, this implies at least three points in C1 ∩ C2 lie on one line
parametrized by η(t). Then, f1 (η(t)), f2 (η(t)) are quadratic and both vanish at three
values of t, and so f1 (η(t)) = f2 (η(t)) = 0 by the fundamental theorem of algebra
(Thm. 15.10.1). Thus, f1 , f2 have infinitely many common zeros in C2 , hence have a
common linear factor by Thm. 11.9.10, a contradiction.
Exercise 11.9.12. Prove in two ways that the three polynomials f1 = t2 + x2 − 2,
f2 = tx−1, f3 = t3 +5tx2 +1 generate the unit ideal in C[x, t] in two ways: by showing
that they have no common zeros, and also by writing 1 as a linear combination of
f1 , f2 , and f3 with polynomial coefficients.
41
Proof. We first show that there are no common zeros. If f2 (x, t) = 0, then x, t 6= 0
and t = x−1 . Thus,
f1 (x, x−1 ) = x2 − 2 + x−2 = x−2 (x4 − 2x2 + 1) = x−2 (x2 − 1)2 ,
hence if f (x, t) = 0 as well, then (x, t) = ±(1, 1). But f3 (1, 1) = 7, f3 (−1, −1) = −5,
hence f1 , f2 , f3 have no common zeros, and (f1 , f2 , f3 ) = (1) by Exercise 11.9.4.
Now we write 1 as a linear combination of f1 , f2 , f3 . 1 can be expressed as


(−6t2 x − 24tx2
− 142t2 − 352tx
+ 53t − 36x
)f1
1 
+ ( 12t2 x − 24tx2 − 36x3 + 284t2 − 568tx + 12x2 + 36t + 140x − 1278)f2 
1225
+(
6tx + 12x2 + 142t + 68x − 53)f3
which we found by using Macaulay2.
Exercise 11.9.13. Let ϕ : C[x, y] → C[t] be a homomorphism that is the identity on
C and sends x
x(t), y
y(t) and such that x(t) and y(t) are not both constant.
Prove that the kernel of ϕ is a principal ideal.
Proof. We claim ker ϕ is principal. If not, then ker ϕ contains two elements f, g
that do not have a common factor. We claim they do not have a common factor in
C(x)[y]. For, suppose h ∈ C(x)[y] is a common factor; then, h = a−1 h0 for some
a ∈ C[x], h0 ∈ C[x, y] by clearing denominator. Assuming without loss of generality
that nothing of the form x − α divides h0 , then h0 divides f, g in C(x)[y], hence also
does in C[x, y] by Prop. 11.9.9, contradicting that f, g do not have a common factor.
Therefore, there exist r0 , s0 ∈ C(x)[y] such that r0 f + s0 g = 1, and clearing
denominators we get rf +sg = q ∈ C[x] for some r, s ∈ C[x]. This implies ker ϕ∩C[x]
is nontrivial, but this is a contradiction for any g ∈ ker ϕ∩C[x] must satisfy g(x(t)) =
0 for all t, hence g = 0.
11.M
Miscellaneous Problems
Exercise 11.M.3. Let R denote the set of sequences a = (a1 , a2 , a3 , . . .) of real
numbers that are eventually constant: an = an+1 = . . . for sufficiently large n.
Addition and multiplication are componentwise, that is, addition is vector addition
and multiplication is defined by ab = (a1 b1 , a2 b2 , . . .). Prove that R is a ring, and
determine its maximal ideals.
Proof. Denote a to be the real number a converges to. R is a ring since sums and
products of eventually constant sequences are eventually constant, commutativity
42
and associativity of +, × and the distributive property follow since addition and
multiplication are defined componentwise, and R has additive identity (0, 0, . . .),
additive inverses −a = (−a1 , −a2 , . . .) for a, and multiplicative identity (1, 1, . . .).
We now want to find its maximal ideals. Consider for each i ∈ Z>0 the map
ϕi : R → R defined by a
ai ; since +, × are defined termwise, and 0
0, this
defines a ring homomorphism. It is moreover surjective, hence mi := ker ϕi = {a |
ai = 0} is maximal by Prop. 11.8.2(a). Now consider the map ϕ∞ : R → R defined by
mapping a to the real number it converges to; since 0
0 and a + b = a + b, ab = ab,
this defines a ring homomorphism. It is moreover surjective, hence m∞ := ker ϕ∞ =
{a | a = 0} is maximal by Prop. 11.8.2(a).
We claim these are all the maximal ideals. It suffices to show if m ⊂ R is maximal
and not equal to mi , it is equal to m∞ . So suppose not; then, there exists a ∈ m such
that a 6= 0. There are therefore finitely many i such that ai = 0. Since m 6= mi , for
each i there exists a sequence ai with aii 6= 0; we can moreover assume aij = 0 for all
j 6= i by multiplying by the sequence in R with 1 in the ith index and 0 otherwise.
X
b=a+
ai ∈ m
{i|ai =0}
is then a unit in m, since bi 6= 0 for all i, contradicting maximality of m.
Exercise 11.M.4.
(a) Classify rings R that contain C and have dimension 2 as a vector space over
C.
(b) Do the same for rings that have dimension 3.
Solution for (a). Let {1, r} be a basis for R, and let ϕ : C[x] → R be defined by
1
1, x
r. ϕ is then surjective, and ker ϕ = (f ) for some f ∈ C[x] since C[x] is
a PID (Props. 12.2.5, 12.2.7), giving R ≈ C[x]/(f ) by the first isomorphism theorem
(Thm. 11.4.2(b)). deg f > 1 since otherwise {1, r} would be linearly dependent; since
r2 = ar + b ∈ R for some a, b ∈ C, we then have f = x2 − ax − b = (x − ζ1 )(x − ζ2 )
for some ζ1 , ζ2 ∈ C. If ζ1 = ζ2 =: ζ, then
R ≈ C[x]/(f ) ≈ C[x]/(x2 )
by composing with the isomorphism defined by x
(x − ζ2 ) = R as ideals, hence
x + ζ. If ζ1 6= ζ2 , then (x − ζ1 ) +
R ≈ C[x]/(f ) ≈ C[x]/(x − ζ1 ) × C[x]/(x − ζ2 ) ≈ C × C
by Exercise 11.6.8(c). Hence the two possibilities are R ≈ C[x]/(x2 ) or C × C.
43
Solution for (b). Suppose there exists r ∈ R such that {1, r, r2 } is a basis for R.
Then, defining the map ϕ : C[x] → R by x
r, we again have that ker ϕ = (f ) for
deg f > 2, since if deg f ≤ 2 then {1, r, r2 } would not be linearly independent. Also,
r3 = ar2 + br + c for some a, b, c ∈ C, and so f = x3 − ax2 − bx − c. If f has one
triple root ζ, then
R ≈ C[x]/((x − ζ)3 ) ≈ C[x]/(x3 ).
If f has a double root ζ and a simple root ζ ′ , then since ((x − ζ)2 ) + (x − ζ ′ ) = R as
ideals,
R ≈ C[x]/((x − ζ)2 )(x − ζ ′ ))
≈ C[x]/((x − ζ)2 ) × C[x]/(x − ζ ′ )
≈ C[x]/(x2 ) × C
by the same argument as before. Finally, if f has three distinct roots, then
R ≈ C[x]/((x − ζ1 )(x − ζ2 )(x − ζ3 ))
≈ C[x]/(x − ζ1 ) × C[x]/(x − ζ2 ) × C[x]/(x − ζ3 )
≈C×C×C
by repeated applications of Exercise 11.6.8(c) since, for example, ((x − ζ1 )(x − ζ2 )) +
(x − ζ3 ) = R as ideals.
Now suppose no r ∈ R exists such that {1, r, r2 } is a basis for R. Thus, if {1, r, s}
is a basis for R, then
r2 = a1 r + c1 ,
s 2 = b2 s + c 2 ,
rs = a3 r + b3 s + c3
for some ai , bi , ci ∈ C. We claim we can assume c1 = c2 = 0. For, changing
coordinates r
r + α gives
(r + α)2 = a1 (r + α) + c1 =⇒ r2 = (a1 − 2α)r − (α2 + a1 α − c1 ),
and so letting α be such that α2 + a1 α − c1 = 0, we have r2 = a1 r. Similarly for s,
we have s2 = b2 s. We then have
(r + zs)2 = r2 + 2zrs + z 2 s2
= (a1 + 2a3 z)r + (b2 z 2 + 2b3 z)s + 2c3 z
= Az (r + s) + Bz
for some Az , Bz ∈ C since {1, r + zs, (r + zs)2 } is linearly dependent. Thus,
Az = a1 + 2a3 z = b2 z 2 + 2b3 z,
44
Bz = 2c3 z.
Since the first equation must hold for all z, we have a1 = b2 = 0, hence r2 = s2 = 0,
and also a3 = b3 . This implies (rs)2 = r2 s2 = 0, but since also
(rs)2 = (a3 r + a3 s + c3 )2
= 2a23 rs + 2a3 c3 r + 2a3 c3 s + c23
= 2a3 (a23 + c3 )r + 2a3 (a23 + c3 )s + c3 (2a23 + c3 ),
we have a3 (a23 + c3 ) = c3 (2a23 + c3 ) = 0. If one of a3 , c3 is nonzero, then the other is
also. But this is impossible, and so we have a3 = b3 = c3 = 0.
Finally, let ϕ : C[x, y] → R be defined by 1
1, x
r, y
s. ϕ is then surjective, giving R ≈ C[x, y]/ker ϕ by the first isomorphism theorem (Thm. 11.4.2(b)).
We know I := (x2 , y 2 , xy) ⊂ ker ϕ, and the reverse inclusion holds since for any
f ∈ C[x, y], f ≡ αr + βs + γ mod I for some α, β, γ ∈ C, and f
0 in the composition C[x, y]/I → C[x, y]/ker ϕ → R if and only if α = β = γ = 0 since {1, r, s} is a
basis for R, i.e., if and only if f ∈ I. Thus, in this case R ≈ C[x, y]/(x2 , y 2 , xy).
In summary, there are four possibilities for R:
C[x, y]/(x2 , y 2 , xy),
12
12.1
C[x]/(x3 ),
C[x]/(x2 ) × C,
C × C × C.
Factoring
Factoring Integers
Exercise 12.1.4. Solve the following simultaneous congruences:
(a) x ≡ 3 modulo 8, x ≡ 2 modulo 5,
(b) x ≡ 3 modulo 15, x ≡ 5 modulo 8, x ≡ 2 modulo 7,
(c) x ≡ 13 modulo 43, x ≡ 7 modulo 71.
Solution for (a). x = 27 = 8 · 3 + 3 = 5 · 5 + 2.
Solution for (b). x = 93 = 15 · 6 + 3 = 8 · 11 + 5 = 7 · 13 + 2.
Solution for (c). x = 2421 = 56 · 34 + 13 = 34 · 71 + 7.
Exercise 12.1.5. Let a and b be relatively prime integers. Prove that there are
integers m and n such that am + bn = 1 modulo ab.
Proof. Since a, b are relatively prime, am + bn = 1 mod ab if and only if am + bn =
1 mod a and am + bn = 1 mod b. Note am + bn = bn mod a, and am + bn = am mod b.
45
Now consider a, a2 , a3 , . . . mod b. There are only b equivalence classes 0, 1, . . . , b−
1 mod b, hence for some i < j we have ai ≡ aj mod b. Then, ai (aj−i − 1) ≡ 0 mod b,
and so b | ai (aj−i − 1). But a, b are relatively prime, hence b | aj−i − 1. Thus, letting
m := j − i, we have am ≡ 1 mod b.
By the same argument working mod a, there exists n such that bn ≡ 1 mod a,
and so am + bn ≡ 1 mod a and mod b, hence am + bn ≡ 1 mod ab.
12.2
Unique Factorization Domains
Exercise 12.2.1. Factor the following polynomials into irreducible factors in Fp [x].
(a) x3 + x2 + x + 1, p = 2, (b) x2 − 3x − 3, p = 5, (c) x2 + 1, p = 7
Solution for (a). Let f (x) = x3 + x2 + x + 1. Since f (1) = 0, we can then factor out
x + 1 to get f (x) = (x + 1)(x2 + 1). Now since 12 + 1 = 0, x2 + 1 can be factored
into (x + 1)(x + 1), and so we have f (x) = (x + 1)3 .
Solution for (b). Let f (x) = x2 − 3x − 3. Since f (1) = f (2) = 0, we claim f (x) =
(x−1)(x−2); this follows since (x−1)(x−2) = x2 −3x+2 ≡ x2 −3x−3 mod 5.
Solution for (c). Since f (x) = x2 + 1 6= 0 for all x ∈ F7 (f (0) = 1, f (1) = 2, f (2) =
5, f (3) = 3, f (4) = 3, f (5) = 5, f (6) = 2), we know it is irreducible in F7 [x].
Exercise 12.2.2. Compute the greatest common divisor of the polynomials x6 + x4 +
x3 + x2 + x + 1 and x5 + 2x3 + x2 + x + 1 in Q[x].
Solution. We perform the Euclidean algorithm (p. 45):
x6 + x4 + x3 + x2 + x + 1 = x(x5 + 2x3 + x2 + x + 1) + (−x4 + 1)
x5 + 2x3 + x2 + x + 1 = −x(−x4 + 1) + (2x3 + x2 + 2x + 1)
3 2 3
x 1
3
2
4
(2x + x + 2x + 1) +
x +
−x + 1 = − +
2 4
4
4
3 2 3 8
2x3 + x2 + 2x + 1 =
x +
x+0
4
4 3
Since the last nonzero remainder is 34 x2 + 34 , the greatest common divisor is x2 +1.
Exercise 12.2.5 (partial fractions for polynomials).
(a) Prove that every element of C(x) can be written as a sum of a polynomial and
a linear combination of functions of the form 1/(x − a)i .
(b) Exhibit a basis for the field C(x) of rational functions as vector space over C.
46
Proof of (a). We first show that f (x)/(x − a)i ∈ C(x) can be so written, where
(x − a)i ∤ f (x) ∈ C[x]. We apply the Euclidean algorithm (p. 45) considering f (x)
as an element in C[x] i times to obtain
f (x) = q1 (x)(x − a) + r1
= (q2 (x)(x − a) + r2 )(x − a) + r1
..
.
= qi (x)(x − a)i + ri−1 (x − a)i−1 + · · · + r2 (x − a) + r1 ,
for some rj ∈ C, and qi (x) ∈ C(x). Dividing this out by (x − a)i , we have
f (x)
ri−1
r2
r1
= qi (x) +
+ ··· +
+
.
i
i−1
(x − a)
(x − a)
(x − a)
(x − a)i
Now suppose we have arbitrary f (x)/g(x) ∈ C(x), where we can
Qnassume without
loss of generality that gcd(f, g) = 1. We can first write g(x) = i=1 (x − ai )αi for
ai ∈ C, αi ∈ N, by the fundamental theorem of algebra (Thm. 15.10.1) (and moving
any unit factor into f (x)). If n = 1, we are done by the above, and so we proceed by
induction on n. If g(x) has n distinct roots, we can write g(x) = (x−a)j h(x) for some
a ∈ C, j ∈ N, h(x) ∈ C[x] where h(x) has n − 1 distinct roots. Since (x − a)j , h(x)
share no roots, they are relatively prime and so 1 = r(x)(x − a)j + s(x)h(x) for some
r(x), s(x) ∈ C[x]. Multiplying throughout by f (x)/g(x) gives
r(x)
s(x)
f (x)
,
=
+
g(x)
h(x) (x − a)j
where the first term can be written as in the statement by inductive hypothesis, and
the second by the above paragraph.
Proof of (b). We claim that
i
B = {x : i ∈ N}
[
1
: a ∈ C, i ∈ N
(x − a)i
is a basis for C(x) over C. We see that this set spans C(x) by (a); it suffices to show
that any finite subset of this set is linearly independent to show this is a basis. Let
S ( B be finite, and suppose it is linearly dependent. Then,
X
j∈J
cj xj +
X
i∈I
bi
= 0,
(x − ai )αi
47
where J indexes the xj ∈ S and I the 1/(x − ai )αi ∈ S. Let D(x) be the product of
all the denominators appearing in the sum on the right; then,
!
X
X
b
i
cj xj +
D(x)
= 0,
(x − ai )αi
j∈J
i∈I
and so cj = 0 for all j since the powers of x that come from the sum over J have
higher degree than those that come from the sum over I. Thus, we have
D(x)
X
i∈I
bi
= 0.
(x − ai )αi
But then, all the bi must equal zero, for suppose not. Then, we can write
D(x)
X
{i∈I|a1 6=ai }
bi
= −D(x)
(x − ai )αi
X
{i∈I|a1 =ai }
bi
(x − ai )αi
Note that the left side has all terms divisible by x − a1 , while the right side has no
terms divisible by x − a1 . Suppose the right side is zero; then the left side is nonzero
mod x − a′ for any a′ 6= a1 , while the right side is zero mod x − a′ . If the right side
is nonzero, then the left side is zero mod x − a1 and the right side is nonzero mod
x − a1 . In either case this is a contradiction, and so cj = bi = 0 for all i, j, and B is
linearly independent.
Exercise 12.2.6. Prove that
√ the following are Euclidean domains:
2πi/3
(a) Z[ω], ω = e
; (b) Z[ −2].
√
Proof. Let α = ω or −2. In either case Z[α] = {x + yα | x, y ∈ Z}. We claim that
σ(x + yα) := |x + yα|2 is an appropriate size function. This simplifies to
σ(x + yα) = |x + yα|2 = (x + yα)(x + yα) = x2 + xt(α + α) + y 2 |α|2 .
(2)
We claim this is a nonnegative integer in both cases. If α = ω, then ω + ω = −1 and
|α|2 = 1, and so
If α =
√
σ(x + yα) = x2 − xy + y 2 ≥ x2 − 2xy + y 2 = (x − y)2 ∈ Z≥0 .
−2, then
√
−2 +
√
√
−2 = 0 and | −2|2 = 2, and so
σ(x + yα) = x2 + 2y 2 ∈ Z≥0 .
48
Now suppose a, b ∈ Z[α] are given; we want to show the property in (12.2.4) holds.
First, we have
b
ba
= 2 = s + tω,
a
|a|
for some p, q ∈ Q, since ba ∈ Z[α] and |a|2 ∈ Z by (2). We can then find q ∈ Z[α]
close to s + tα, by letting
1
1
where |x − s| ≤ , |y − t| ≤ ,
2
2
which is possible since every rational lies within 1/2 of an integer. By (2), if α = ω,
q = x + yα,
b
−q
a
2
= |(s − x) + (t − y)ω|2 = (s − x)2 − (s − x)(t − y) + (t − y)2 ≤
and if α =
√
−2,
b
−q
a
2
√
= (s − x) + (t − y) −2
2
= (s − x)2 + 2(t − y)2 ≤
3
< 1,
4
3
< 1.
4
Multiplying throughout by |a|2 and letting r = b − aq, we either get r = 0 or
σ(r) = |b − aq|2 < |a|2 = σ(a),
and so our function is an appropriate size function.
Exercise 12.2.9. Let F be a field. Prove that the ring F [x, x−1 ] of Laurent polynomials (Chapter 11, Exercise 5.7) is a principal ideal domain.
Proof. It suffices by Prop. 12.2.7 to show that F [x, x−1 ] is a Euclidean domain. We
define the size function σ(f ) as deg+ (f )−deg− (f ), where deg+ (f ) denotes the largest
power of x in f and deg− (f ) denotes the smallest power of x in f , both including
negative powers. σ(f ) ∈ Z≥0 by considering possible values for the degree function.
Now consider the Euclidean algorithm on f where we divide by g. Let p = x− deg− (f )
and q = x− deg− (g) ; note they are both units in F [x, x−1 ]. Then, f p, gq have all
non-negative exponents and
deg(gq) = deg+ (gq) = deg+ (g) − deg− (g) = σ(g).
Thus, f p, gq ∈ F [x]. We can then apply the Euclidean algorithm (p. 45) on f p, gq
in F [x] to get (f p) = r(gq) + s, where r, s ∈ F [x], and s = 0 or deg(s) < deg(gq)
by the Euclidean algorithm in F [x]. Then, dividing by p throughout, we get f =
g(rqp−1 ) + sp−1 , and (12.2.4) holds since either sp−1 = 0, or
σ(sp−1 ) = σ(s) = deg(s) < deg(gq) = σ(g).
49
12.3
Gauss’s Lemma
Exercise 12.3.1.
√ Let ϕ denote the√homomorphism Z[x] → R defined by
(a) ϕ(x) = 1 + 2, (b) ϕ(x) = 12 + 2.
Is the kernel of ϕ a principal ideal? If so, find a generator.
√
Solution for (a). We want to find the polynomials f (x) such that f (1 +
√ 2) = 0.
Using the canonical embedding of Z[x] into R[x], we see that√(x − (1 + 2))√would
generate the kernel of ϕ in R[x]. But then, since (x − (1 + 2))(x − (1 − 2)) =
x2 − 2x − 1, and (x2 − 2x − 1) ⊆ Z[x], we obtain (x2 − 2x − 1) ⊆ ker ϕ, as an ideal
in Z[x]. But
x2 − 2x − 1 is primitive and irreducible in Q[x], since its roots
√ since
are (1 ± 2), x2 − 2x − 1 is irreducible in Z[x] by Prop. 12.3.7(a). Now suppose
that f (x) ∈ ker ϕ \ (x2 − 2x − 1). By the Euclidean algorithm dividing f (x) by
x2 − 2x − 1, √
we can assume without loss of generality that f (x) = ax + b. But then,
ϕ(f ) = a+a 2+b 6= 0 unless a = b = 0, and so we see that (x2 −2x−1) = ker ϕ.
Solution for (b). We proceed similarly
√ to (a). Considering the embedding of Z[x]
into R[x], we first see that (x − ( 12 + 2)) would generate the kernel in R[x], which
√
√
contains (x − ( 12 + 2))(x − ( 21 − 2)) = x2 − x − 47 . x2 − x − 47 is in Q[x], and is
irreducible since the roots are in theextension R[x] as in (a). But then since Q is
a field, (4x2 − 4x − 7) = x2 − x − 47 as ideals. This former ideal is generated by a
primitive and irreducible element in Q[x], which as an element of Z[x] is irreducible
as well by Prop. 12.3.7(a). We thus have (4x2 − 4x − 7) ⊆ ker ϕ. Now suppose that
f (x) ∈ ker ϕ \ (4x2 − 4x − 7). But taking the canonical embedding of Z[x] into Q[x],
and seeing that the latter is a principal ideal domain since Q is a field, we obtain that
f (x) ∈ (4x2 −4x−7) as an ideal in Q[x]. But then, 4x2 −4x−7 | f (x) in Q[x], and by
Thm. 12.3.6, this implies 4x2 − 4x − 7 | f (x) in Z[x]; thus (4x2 − 4x − 7) = ker ϕ.
Exercise 12.3.2. Prove that two integer polynomials are relatively prime elements
of Q[x] if and only if the ideal they generate in Z[x] contains an integer.
Proof. f, g ∈ Z[x] are relatively prime in Q[x] if and only if af + bg = 1 for some
a, b ∈ Q[x]. By clearing denominators of a, b, this is true if and only if a′ f + b′ g = d
for some a′ , b′ ∈ Z[x], d ∈ Z, which is equivalent to saying (f, g) ∩ Z 6= ∅.
Exercise 12.3.4. Let x, y, z, w be variables. Prove that xy − zw, the determinant of
a variable 2 × 2 matrix, is an irreducible element of the polynomial ring C[x, y, z, w].
Proof. Suppose xy − zw factors as pq for p, q ∈
/ C, where p = a0 + a1 + a2 and q =
b0 + b1 + b2 for ai , bi homogeneous of degree i or zero. We claim p, q are homogeneous
of degree one; we proceed by comparing degrees. a0 b0 = 0 hence without loss of
50
generality a0 = 0. a1 b0 = 0 as well; if a1 = 0, then b1 = b2 = 0, contradicting that
q ∈
/ C, hence a1 6= 0, b0 = 0. This implies b2 = 0, hence p, q are homogeneous of
degree one.
Thus, since xy − zw = x2 (y/x − z/x · w/x) = x2 (p/x)(q/x), xy − zw is reducible
only if y/x − z/x · w/x is reducible in C[y/x, z/x, w/x]. Relabeling variables and
taking the contrapositive, it suffices to show t − rs is irreducible in C[r, s, t]. But
C[r, s, t]/(t − rs) ≈ C[r, s], a domain, hence t − rs is prime (Prop. 13.5.1(a)) and
thus irreducible since C[r, s, t] is a domain (Lem. 12.2.10).
Exercise 12.3.6. Let α be a complex number. Prove that the kernel of the substitution map Z[x] → C that sends x
α is a principal ideal, and describe its generator.
Proof. Let ϕ be this map. By the mapping property of fractions (Thm. 11.7.2(c),
which is still true if ϕ is not injective), we have the commutative diagram
Z[x]
ϕ
C
ψ
Q[x]
Since Q[x] is a Euclidean domain, ker ψ is principal. We claim it is generated by the
polynomial f of minimal degree which has α as a root, or zero if α is transcendental.
For, in the former case, if g ∈ ker ϕ, we can write g = f q + r by the Euclidean
algorithm, and r 6= 0 implies deg r < deg f , contradicting the minimality of f .
We claim if f = cf0 as in Lem. 12.3.5 for f0 primitive, then ker ϕ = (f0 ).
(f0 ) ⊂ ker ϕ by the commutativity of the diagram, so it suffices to show the opposite inclusion. Let g ∈ ker ϕ. Then, f | g in Q[x] by the above, hence f0 | g in Z[x]
by Thm. 12.3.6(a). Thus ker ϕ is generated by f0 , the primitive polynomial obtained
from f as above, or is zero if α is transcendental.
12.4
Factoring Integer Polynomials
Exercise 12.4.4. Factor the integer polynomial x5 + 2x4 + 3x3 + 3x + 5 mod 2,
mod 3, and over Q.
Remark. We denote f (x) = x5 + 2x4 + 3x3 + 3x + 5.
Solution for F2 . f (x) = x5 + x3 + x + 1 over F2 . Then, f (1) = 0, hence f (x) = (x +
1)(x4 +x3 +1). Now x4 +x3 +1 has no roots in F2 hence has no linear factors; x2 +x+1
is the only irreducible quadratic in F2 [x] and has (x2 +x+1)2 = x4 +x2 +1 6= x4 +x3 +1,
hence we cannot factor further.
51
Solution for F3 . f (x) = x5 + 2x4 + 2 over F3 . Then, f (2) = 0, hence f (x) =
(x + 1)(x4 + x3 + 2x2 + x + 2). But 2 is also a root of x4 + x3 + 2x2 + x + 2,
hence f (x) = (x + 1)2 (x3 + 2x + 2), and we cannot factor further since x3 + 2x + 2
has no roots in F3 .
Solution for Q. From (a), (b), we suspect that −1 is a root. Indeed, f (x) = (x +
1)(x4 + x3 + 2x2 − 2x + 5). The residue x4 + x3 + 1 of the second factor in F2 [x] is
irreducible as in (a), and so x4 +x3 +2x2 −2x+5 is irreducible in Q[x] by Prop. 12.4.3.
Thus we cannot factor further.
Exercise 12.4.6. Factor x5 + 5x + 5 into irreducible factors in Q[x] and in F2 [x].
Solution for Q[x]. x5 + 5x + 5 is irreducible in Q[x] by the Eisenstein criterion
(Prop. 12.4.6) since 5 doesn’t divide the leading coefficient 1 while it divides all
other coefficients, and 52 = 25 does not divide the constant term 5.
Solution for F2 [x]. The residue in F2 [x] is x5 + x + 1, which is reducible in F2 [x] since
x5 + x + 1 = (x3 + x2 + 1)(x2 + x + 1); each of these factors have no roots in F2 hence
we cannot factor further.
Exercise 12.4.7. Factor x3 + x + 1 in Fp [x], when p = 2, 3, and 5.
Remark. We denote f (x) = x3 + x + 1.
Solution for p = 2. Since f (0) = 1, f (1) = 1, we see that there are no roots in F2 ,
and so x3 + x + 1 is irreducible.
Solution for p = 3. Since f (1) = 0, we can factor out (x + 2) to get f (x) = (x +
2)(x2 + x + 2). We cannot factor further since x2 + x + 2 has no roots in F3 .
Solution for p = 5. Since f (0) = 1, f (1) = 3, f (2) = 1, f (3) = 1, f (4) = 4, we see
that there are no factors of f (x) in F5 and so it is irreducible.
Exercise
(a) the
(b) the
(c) the
12.4.12. Determine:
monic irreducible polynomials of degree 3 over F3 ,
monic irreducible polynomials of degree 2 over F5 ,
number of irreducible polynomials of degree 3 over the field F5 .
Remark. It suffices to make sure our polynomials have no roots, for any factorization
of a quadratic or a cubic will produce a linear factor.
52
Solution for (a). We consider all polynomials of the form x3 + ax2 + bx + c in F3 .
c = 1, 2 since otherwise x = 0 would be a root. Moreover, 1 + a + b + c 6≡ 0 mod 3
and 2 + a + 2b + c 6≡ 0 mod 3 must be true for irreducibility. Suppose c = 1. Then,
1 + a + b + 1 6≡ 0 implies a + b 6≡ 1. Likewise, 2 + a + 2b + 1 ≡ a + 2b 6≡ 0, and so the
only ordered triples that work for c = 1 are (0, 2, 1), (1, 2, 1), (2, 0, 1), (2, 1, 1), i.e.,
x3 + 2x + 1,
x3 + x2 + 2x + 1,
x3 + 2x2 + 1,
x3 + 2x2 + x + 1
are monic irreducible polynomials of degree 3 over F3 .
Suppose c = 2. Then, 1 + a + b + 2 ≡ a + b 6≡ 0, and likewise, 2 + a + 2b + 2 6≡
0 implies a + 2b 6≡ 2. Thus, the only ordered triples that work for c = 2 are
(0, 2, 2), (1, 0, 2), (1, 1, 2), (2, 2, 2), i.e.,
x3 + 2x + 2,
x3 + x2 + 2,
x3 + x2 + x + 2,
x3 + 2x2 + 2x + 2
are monic irreducible polynomials of degree 3 over F3 .
Solution for (b). We consider polynomials of the form x2 + ax + b in F5 . b = 1, 2, 3, 4
since otherwise x = 0 would be a root. Moreover, we have the system of equations

1 + a + b 6≡ 0


(

 4 + 2a + b 6≡ 0
a, 4a 6≡ 4 − b,
=⇒

4 + 3a + b 6≡ 0
2a, 3a 6≡ 1 − b.



1 + 4a + b 6≡ 0
We can consider this system for each b value. For b = 1, we have
(
a, 4a 6≡ 3
=⇒ a = 1, 4.
2a, 3a 6≡ 0
For b = 2, we have
(
a, 4a 6≡ 2
=⇒ a = 0, 1, 4.
2a, 3a 6≡ 4
(
a, 4a 6≡ 1
=⇒ a = 0, 2, 3.
2a, 3a 6≡ 3
For b = 3, we have
53
For b = 4, we have
(
a, 4a 6≡ 0
=⇒ a = 2, 3.
2a, 3a 6≡ 2
Thus, our monic irreducible polynomials of degree 2 over F5 are
x2 + x + 1, x2 + 4x + 1, x2 + 2,
x2 + 3, x2 + 2x + 3, x2 + 3x + 3,
x2 + x + 2, x2 + 4x + 2,
x2 + 2x + 4, x2 + 3x + 4.
Solution for (c). There are 5 monic polynomials of degree 1, and from (b) there are
10 monic irreducible polynomials of degree 2. The irreducible monic polynomials of
degree 3 are those that cannot be obtained by multiplying together polynomials of
lower degree. We see that any choice of factors gives a unique polynomial of degree
3 by Prop. 12.2.14(c), and so it suffices to count all monic polynomials of degree 3 in
F5 [x], and subtract those that can be obtained by multiplying together polynomials
of lower degree.
We see that reducible polynomials of degree 3 must be a product of 3 polynomials
of degree 1 or a product of 2 polynomials of degree 1 and 2 respectively. The number
of polynomials in the former category is
5
5 4
5
+
+
= 35,
3
1 1
1
where the terms describe all different, 2 same, and all same factors. The number
of polynomials in the latter category is 5 × 10 = 50. Since the number of degree 3
monic polynomials is 53 = 125, we see that there are 125 − (35 + 50) = 40 degree 3
monic irreducible polynomials in F5 .
Since we want to find the number of (not necessarily monic) irreducible polynomials of degree 3 in F5 [x], we see that we can multiply our 40 monic polynomials by
any element of F5 \ {0}; this results in 160 irreducible polynomials of degree 3.
Exercise 12.4.13 (Lagrange interpolation formula).
(a) Let a0 , . . . , an be distinct complex numbers. Determine a polynomial p(x) of
degree n, which has a1 , . . . , an as roots, and such that p(a0 ) = 1.
(b) Let a0 , . . . , ad and b0 , . . . , bd be complex numbers, and suppose that the ai are
distinct. There is a unique polynomial g of degree ≤ d such that g(ai ) = bi for
each i = 0, . . . , d. Determine the polynomial g explicitly in terms of ai and bi .
54
Solution for (a). We let
pi (x) =
Y x − aj
.
a
i − aj
j6=i
Any aj with j 6= i is a root since pi (x) = 0 ⇐⇒ x − aj = 0 for some j, and p(ai ) = 1
since each term in the product would equal 1. Letting p(x) = p0 (x), we are done.
Solution for (b). We let
g(x) =
d
X
bi pi (x) =
i=0
d
X
i=0
bi
Y x − aj
,
a
i − aj
j6=i
where pi (x) is as in (a). g(ai ) = bi for all i since pi (aj ) = δij . deg g ≤ d by the fact
that each term in the sum has degree d.
We now prove uniqueness. Suppose h(x) is another such polynomial. Then,
f (x) = g(x) − h(x) has degree ≤ d and therefore has ≤ d roots; however, h(ai ) =
g(ai ) − h(ai ) = bi − bi = 0 for all 0 ≤ i ≤ d. But then, f (x) has d + 1 distinct roots,
which contradicts Prop. 12.2.20 unless g(x) = h(x).
Exercise 12.4.15. With reference to the Eisenstein criterion, what can one say
when
(a) f¯ is constant, (b) f¯ = xn + b̄xn−1 ?
Claim (a). Let f (x) = an xn + · · · + a0 ∈ Z[x] and let p ∈ Z be prime. Then f is
irreducible in Q[x] if (i) p ∤ a0 ; (ii) p | ai for 1 ≤ i ≤ n; (iii) p2 ∤ an . Note that (i)
and (ii) hold if and only if f¯ is a nonzero constant.
Proof of Claim (a). Suppose not; then f = gh for g = br xr + · · · + b0 , h = cs xs +
· · · + c0 , r + s = n. Since then f¯ = ḡ h̄ = a0 6= 0 by (i) and (ii), we see that ḡ = b0 and
h̄ = c0 since they must divide a0 . But then p | br and p | cs imply p2 | br cs = an .
Claim (b). Let f (x) = an xn + · · · + a0 ∈ Z[x] and let p ∈ Z be prime. Then f is
irreducible in Q[x] if (i) an ≡ 1 mod p; (ii) p ∤ an−1 ; (iii) p | ai for 0 ≤ i ≤ n − 2;
(iv) p ∤ a0 . Note that (i), (ii), and (iii) hold if and only if f¯ = xn + b̄xn−1 .
Proof of Claim (b). Suppose not; then f = gh for g = br xr + · · · + b0 , h = cs xs +
· · · + c0 , r + s = n. Since then f¯ = ḡ h̄ = xn + an−1 xn−1 = xn−1 (x + an−1 ) by (i), (ii),
and (iii), without loss of generality x + an−1 | ḡ, and so ḡ = br xr−1 (x + an−1 ) and
h̄ = cs xs . But then, p | c0 implies p | b0 c0 = a0 .
Exercise 12.4.19. Factor x5 − x4 − x2 − 1 mod 2, mod 16, and over Q.
55
Remark. We denote f (x) = x5 − x4 − x2 − 1.
Solution for F2 . f (x) = x5 + x4 + x2 + 1 over F2 . Then, f (1) = 0, and so f (x) =
(x + 1)(x4 + x + 1). x4 + x + 1 has no roots in F2 , and the only irreducible quadratic
in F2 [x] is x2 + x + 1 which has (x2 + x + 1)2 = x4 + x2 + 1 6= x4 + x + 1, hence we
cannot factor further.
Solution for Z/16Z. f (−5) ≡ 0 mod 16, and so f (x) = (x+5)(x4 −6x3 −2x2 +9x+3).
This does not factor further, for a factorization of x4 − 6x3 − 2x2 + 9x + 3 over Z/16Z
would produce a factorization of x4 + x + 1 over F2 , contradicting (a).
Solution for Q. The only possible factorization is as in (b). But the constant term of
the linear factor in (b) has constant term 5 6≡ ±1 mod 16, which would be necessary
for f to have constant term −1, and so f is irreducible.
12.5
Gauss Primes
Exercise 12.5.2. Find the greatest common divisor in Z[i] of (a) 11 + 7i, 4 + 7i, (b)
11 + 7i, 8 + i, (c) 3 + 4i, 18 − i.
Proof of (a). First, (11+7i)(11−7i) = 112 +72 = 170 = 2×5×17 = (1+i)(1−i)(2+
i)(2 − i)(4 + i)(4 − i), and so 11 + 7i = (1 + i)(2 − i)(4 + i) is a prime factorization.
Likewise, (4 + 7i)(4 − 7i) = 42 + 72 = 65 = 5 × 13 = (2 + i)(2 − i)(3 + 2i)(3 − 2i), and
so 4 + 7i = (2 + i)(3 + 2i) is a prime factorization. Thus, gcd(11 + 7i, 4 + 7i) = 1.
Proof of (b). 11 + 7i = (1 + i)(2 − i)(4 + i) is a prime factorization by (a). Then,
(8 + i)(8 − i) = 82 + 12 = 65 = 5 × 13 = (2 + i)(2 − i)(3 + 2i)(3 − 2i), and so
8 + i = (2 − i)(3 + 2i). Thus, gcd(11 + 7i, 8 + i) = 2 − i.
Proof of (c). Since (18 − i)/(3 + 4i) = 2 − 3i, gcd(3 + 4i, 18 − i) = 3 + 4i.
Exercise 12.5.3. Find a generator for the ideal of Z[i] generated by 3+4i and 4+7i.
Proof. Z[i] is a PID (Props. 12.2.5(c), 12.2.7), so it suffices to find gcd(3 + 4i, 4 +
7i). Since (3 + 4i)(3 − 4i) = 25 = (2 + i)2 (2 − i)2 , (3 + 4i) = (2 + i)(2 + i) is a
prime factorization, and from Exercise 12.5.2(a), 4 + 7i = (2 + i)(3 + 2i) is a prime
factorization. Thus, 2 + i is a generator of the ideal.
Exercise 12.5.5. Let π be a Gauss prime. Prove that π and π are associates if and
only if π is an associate of an integer prime, or ππ = 2.
56
Proof. Suppose π and π are associates. Then, e(a + bi) = a − bi for some unit
e ∈ {±1, ±i}. Thus, ππ = e(a + bi)2 = e(a2 − b2 + 2bi). First suppose e = ±1. Then,
b = 0 since ππ ∈ Z, and so ππ = ea2 = a2 , since ππ > 0. By Theorem 12.5.2(a), π
prime implies a is an integer prime, and so π must be an associate of a, an integer
prime. Now suppose e = ±i. Then, ππ = ±i(a2 − b2 + 2bi) = ∓2b ± i(a2 − b2 ).
Then, a2 − b2 = 0 since ππ ∈ Z, and so ππ = ∓2b. By Theorem 12.5.2(a), we see
b ∈ {±1, ±2}. But since a2 − b2 = 0 and b = ±2 imply a = ±2, we have that 2 | π,
which contradicts its primeness. Thus, b = ±1, and so ππ = ∓2b = 2.
Now if π is an associate of an integer prime, then π = ep for some e ∈ {±1, ±i},
and so π = ep; since then π/π ∈ {±1, ±i}, π and π are associates. Now suppose
ππ = 2. Then, (a − bi)(a + bi) = a2 + b2 = 2 = (1 + i)(1 − i). By the fact that Z[i] is
a UFD, this prime factorization is unique (up to a unit), and so since 1 + i = i(1 − i),
we see that π and π are associates.
√
Exercise 12.5.6. Let R be the ring Z[ −3]. Prove that an integer prime p is a
prime element of R if and only if the polynomial x2 + 3 is irreducible in Fp [x].
√
Proof. We consider the ring R = Z[ −3]/(p). Since R ≈ Z[x]/(x2 + 3), we can take
the quotient with respect to (p) and (x2 + 3) in different orders:
Z[x]
kill
p
kill
x2 + 3
kill
x2 + 3
R
Fp [x]
kill
p
R
Now R is a domain if and only if p is prime in R by Prop. 13.5.1. But R is a domain
if and only if x2 + 3 is prime in Fp [x] as well, and since Fp [x] is a domain, this is true
if and only if x2 + 3 is irreducible in Fp [x] by Lem. 12.2.10.
Exercise 12.5.7. Describe the residue ring Z[i]/(p) for each prime p.
Solution. Recall that Z[i] ≈ Z[x]/(x2 + 1), and so Z[i]/(p) ≈ (Z[x]/(p))/(x2 + 1) ≈
Fp [x]/(x2 + 1).
When p ≡ 3 mod 4, p is a Gauss prime and so Fp [x]/(x2 + 1) ≈ Fp [i] is a field by
Lem. 12.5.3, and so is isomorphic to Fp2 by Thm. 15.7.3(d).
When p ≡ 1 mod 4 or p = 2, p is not a Gauss prime and so x2 + 1 is reducible in
Fp [x] with roots α, α−1 by Lem 12.5.3. Then, (x − α) + (x − α−1 ) = Fp [x] as ideals,
57
and (x − α)(x − α−1 ) = (x2 + 1) = 0 as ideals in Fp [x]/(x2 + 1), hence by Exercise
11.6.8(c) we have Fp [x]/(x2 + 1) ≈ Fp [x]/(x − α) × Fp [x]/(x − α−1 ) ≈ Fp × Fp .
Exercise 12.5.9. Let R = Z[ω] with ω = e2πi/3 . Let p be a prime integer 6= 3. Adapt
the proof of Theorem 12.5.2 to prove the following:
(a) The polynomial x2 + x + 1 has a root in Fp if and only if p = 1 modulo 3.
(b) (p) is a maximal ideal of R if and only if p = −1 modulo 3.
(c) p factors in R if and only if it can be written in the form p = a2 + ab + b2 , for
some integers a, b.
Proof of (a). x2 + x + 1 has a root in Fp if and only if x3 − 1 has a nontrivial root
×
in Fp , i.e., F×
p has an element of order 3. This is true if and only if 3 | |Fp | =
p − 1 by corollaries to Lagrange’s theorem (Cor. 2.8.10) and the first Sylow theorem
(Cor. 7.7.3). This is equivalent to having p ≡ 1 mod 3.
Proof of (b). We consider the ring R = Z[ω]/(p). Since R ≈ Z[x]/(x2 + x + 1), we
can take the quotient with respect to (p) and (x2 + x + 1) in different orders:
Z[x]
kill
p
kill
x2 + x + 1
kill
x2 + x + 1
R
Fp [x]
kill
p
R
Now R is a field if and only if (p) is maximal in R by Prop. 11.8.2(b). But R is a
field if and only if (x2 + x + 1) is maximal in Fp [x] as well, and since Fp [x] is a PID
(Props. 12.2.5(b), 12.2.7), this is true if and only if x2 + x + 1 is irreducible in Fp [x]
by Cor. 12.2.9(C). But this holds if and only if p ≡ −1 mod 3 by (a) since all prime
integers 6= 3 are equivalent to ±1 mod 3.
Proof of (c). If p = a2 + ab + b2 , then p factors as (a − bω)(a − bω 2 ) in R.
Conversely, suppose p factors in R. p is not a unit in R hence is divisible by a
prime π ∈ R. Then, π | p = p, so ππ | p2 in R and Z, and so ππ = p or p2 . The
latter case is impossible for otherwise p would be an associate of π, hence irreducible.
Thus, p = (a − bω)(a − bω)(a − bω)(a − bω 2 ) = a2 + ab + b2 for some a, b ∈ Z.
Exercise 12.5.10.
(a) Let α be a Gauss integer. Assume that α has no integer factor, and that αα
is a square integer. Prove that α is a square in Z[i].
58
(b) Let a, b, c be integers such that a and b are relatively prime and a2 + b2 = c2 .
Prove that there are integers m and n such that a = m2 − n2 , b = 2mn, and
c = m2 + n2 .
Proof of (a). We can decompose α = uπ1k1 · · · πnkn for πj prime and u a unit, such
that πi , πj are not associates for any i 6= j; we want to show 2 | kj for each j. Then,
αα = u2 (π1 π1 )k1 · · · (πn πn )kn = c2 for some c ∈ Z.
By Theorem 12.5.2(a), each πj πj is an integer prime or a square of an integer prime.
For each j, πj πj integer prime implies 2 | kj , since otherwise c2 would not be a
square integer. Now suppose πj πj is a square of an integer prime for some j. Then,
p2 = πj πj implies πj is an associate of p, since this prime decomposition is unique up
to units; this implies p | α, which contradicts that α has no integer factor. Finally,
c2 > 0 hence u2 = 1, and so u = ±1. In either case, α is a square since i2 = −1.
Proof of (b). Not both of a, b are odd: otherwise, c2 = a2 + b2 ≡ 1 + 1 = 2 mod 4,
a contradiction since all even square integers must be 0 mod 4. So assume without
loss of generality that a is odd. Let α = a + bi. Then, αα = a2 + b2 = c2 , hence by
(a) letting α = (m + ni)2 we have a = m2 − n2 , b = 2mn, and c = m2 + n2 .
12.M
Miscellaneous Problems
Exercise 12.M.5. For which integers n does the circle x2 + y 2 = n contain a point
with integer coordinates?
Solution. We claim that x2 + y 2 = n has integer solutions if and only if every prime
≡ 3 mod 4 has even exponent in the prime factorization of n.
⇐ The prime factorization can be written n = ab2 , where a is square free. Then,
by Q
Thm. 12.5.2(d), every p that divides a can be written p = πp πp . Letting x + iy =
b · p|a πp , we have x2 + y 2 = (x − iy)(x + iy) = ab2 = n.
⇒ Let n = x2 + y 2 = (x − iy)(x + iy). If p ≡ 3 mod 4 divides a, it is a Gauss
prime by Thm. 12.5.2(b), and so p | x − iy or x + iy. But then, p | x, y, hence
p | x ± iy. Thus, if pk is the power of p that appears in the prime factorization
for x − iy, then p2k is the power of p that appears in the prime factorization for
(x − iy)(x + iy) = n.
Exercise 12.M.6. Let R be a domain and let I be an ideal that is the product of
distinct maximal ideals in two ways, say I = P1 · · · Pr = Q1 · · · Qs . Prove that the
two factorizations are the same, except for the ordering of the terms.
59
Proof. Suppose there is an i such that Qj 6⊂ Pi for all j. Let ϕi : R → R/Pi be the
canonical quotient map. Then, for each j there is qj ∈ Qj \ Pi , and so
!
Y
Y
ϕi (qj ) = ϕi
qj 6= 0 ∈ R/Pi ,
j
j
Q
which is a contradiction since j qj ∈ Q1 · · · Qs ⊂ Pi . Thus, for every i, there is a j
such that Qj ⊂ Pi ; since both ideals are maximal, we moreover have Qj = Pi . Thus,
{P1 , . . . , Pr } ⊂ {Q1 , . . . , Qs } as sets; interchanging the role of the Pi and Qj gives
the opposite inclusion, so the two factorizations are equal.
Exercise 12.M.7. Let R = Z[x].
(a) Prove that every maximal ideal in R has the form (p, f ), where p is an integer
prime and f is a primitive integer polynomial that is irreducible modulo p.
(b) Let I be an ideal of R generated by two polynomials f and g that have no
common factors other than ±1. Prove that R/I is finite.
Proof of (a). Let M ⊂ R be a maximal ideal. It is not principal by Exercise 11.8.1,
and so there exist f1 , f2 ∈ M that do not share a common factor. f1 , f2 do not share
a common factor in Q[x], either by Thm. 12.3.6(b). Thus, r0 f1 + s0 f2 = 1 for some
r, s ∈ Q[x], and so clearing denominators, rf1 + sf2 = q ∈ Q for r, s ∈ Z[x], i.e.,
M ∩ Z 6= (0).
Now M ∩ Z is prime, since if ab ∈ M ∩ Z, then, say, a ∈ M and moreover
a ∈ Z for otherwise ab ∈
/ Z. Thus, M ∩ Z = (0) or (p) for some prime p; the
former case is impossible by the above. Now consider the image M ′ of M in Fp [x];
then, Z[x]/M ≈ Fp [x]/M ′ is a field, and so M ′ is maximal by Prop. 11.8.2(b), and
is generated by some irreducible f0 ∈ Fp [x] by Cor. 12.2.9(c) since Fp [x] is a PID by
Props. 12.2.5(b), 12.2.7. Now if f ∈ Z[x] is a lift of f0 , then (p, f ) ⊂ M since f = 0 in
Fp [x]/M ′ ; M ⊂ (p, f ) since if g ∈ M \ (p, f ), then g 6= 0 ∈ Fp [x]/M ′ ≈ Z[x]/M .
Proof of (b). As in (a), if I = (f, g) and f, g have no common factors, then I ∩ Z 6=
(0); since Z is a PID, M ∩ Z = (n) for some
Z such that n 6= ±1, for otherwise
Q nk∈
i
R/I = 0 and we are done. So, letting n = pi be a prime factorization, we claim
Q
(Z/pki i Z)[x] Y (Z/pki i Z)[x]
(Z/nZ)[x]
R
≈
≈ i
≈
.
(3)
I
(f, g)
(f, g)
(f, g)
i
The first isomorphism is clear, and the second follows by applying Exercise 11.6.8(c)
repeatedly. For the third, consider the canonical surjection
Y
Y (Z/pki Z)[x]
ki
i
π:
(Z/pi Z)[x] →
.
(f, g)
i
i
60
ker π = (f, g) since an element on the left maps to zero on the right if and only
if each direct factor is in (f, g), and so (3) holds by the first isomorphism theorem
(Thm. 11.4.2(b)). Let each direct factor on the right be called Ri .
It suffices to show each Ri is finite. Since f, g do not share a common factor,
without loss of generality pi ∤ f . Then, separating out the terms in f , we can write
f = f1 − f2 where pi ∤ f1 while pi | f2 . Then, pki i | f2ki , hence in (Z/pki i Z)[x],
f1ki = f1ki − f2ki = (f1 − f2 )h = f h ∈ (f, g)
for some h ∈ (Z/pki i Z)[x]. If a is the leading coefficient of f1 , then f1ki has leading
coefficient aki . This is a unit since if not, then (aki ) is contained in a maximal ideal
of Z/pki i Z by Thm. 11.9.2, but (pi ) is the unique maximal ideal of this ring by the
correspondence theorem (Thm. 11.4.3), contradicting that pi ∤ a. Thus, m = a−ki f1ki
is a monic polynomial in (Z/pki i Z)[x], hence each Ri is finite since any polynomial
with degree ≥ deg m can be reduced to a polynomial of lower degree using that
m = 0, and there are only finitely many polynomials in (Z/pki i Z)[x] with degree
< deg m since Z/pki i Z is finite.
Exercise 12.M.8. Let u and v be relatively prime integers, and let R′ be the ring
obtained from Z byadjoining
an element
α with the relation vα = u. Prove that R′
is isomorphic to Z uv and also to Z v1 .
Proof. We see that R′ = Z[x]/(vx − u) ≈ Z[α]. Now since vα
u = u by construction,
u
we see that having
α
gives
an
isomorphism
Z[α]
≈
Z
, and so R′ ≈ Z uv .
v
v
To show R′ ≈ Z v1 , we first see that Z v1 ⊂ Z uv as a subring, since u · v1 = uv .
Since u, v are relatively prime, au +
= 1 for some
a, b ∈
Z.
Then, dividing out by
bv
v gives a · uv + b = v1 . Thus, v1 ∈ Z uv , and so Z v1 = Z uv .
13
13.1
Quadratic Number Fields
Algebraic Integers
√
√
Exercise 13.1.4. Let d and d′ be integers. When are the fields Q( d) and Q( d′ )
distinct?
2
′
′2 ′
′
Solution.
Write
√ d = s e and d = s ′ e such that e, e are square-free. We claim
√
Q( d) ≈ Q( d′ ) if and only if e = e . ⇐ clearly holds by p. 384, so it suffices to
′
show ⇒. So
e 6= 1. √We claim there
√ suppose e 6= e , and without loss of generality
√
′
is no x + y e with x, y ∈ Q such that e = (x + y e′ )2 = x2 + 2xy e′ + y 2 e′ . This
implies 2xy = 0 and x2 + y 2 e′ = e. If x = 0, then y 2 e′ = e, so if y = a/b for (a, b) = 1,
61
we have a2 e = b2 e′ ; since e, e′ are square-free, then a = b = 1, so e = e′ . On the
other hand, if y = 0, then x2 = e, contradicting that e =
6 1 is square-free.
13.2
Factoring Algebraic Integers
Exercise
√ 13.2.2. For which negative integers d = 2 modulo 4 is the ring of integers
in Q[ d] a unique factorization domain?
√
√
Solution. Suppose d ≡ 2 mod 4. The integers R in Q[ d] are of the form a +√b d
for a, b ∈ Z (Prop. 13.1.6). We claim R is a UFD if and only if d = −2. ⇐. Z[ −2]
is a Euclidean domain (Exercise 12.2.6), hence a UFD by Props.
12.2.7,
√
√ 12.2.14(b).
⇒. Suppose d 6= −2. Consider 2(2 − d/2) = 4 − d = (2 − d)(2 + √
d). We claim
2 is irreducible
in
R;
as
on
p.
386
it
suffices
to
show
there
is
no
a
+
b
d such that
√
2
2
2
N (a + b d) = a − b d = 2. For, this would imply 2 | d hence 2 | a , and so 2 | a,
which implies a = 0, b = 1, d = −2 is the only
Thus, for
√ a contradiction.
√
√ possibility,
/ R when
R to be a UFD we must have either 2 | 2 − d or 2 + d, but 1 ± d/2 ∈
d ≡ 2 mod 4 by Prop. 13.1.6.
13.3
√
Ideals in Z[ −5]
√
Exercise 13.3.2. Let δ := −5. Decide whether or not the lattice of integer
combinations of the given vectors is an ideal: (a) (5, 1 + δ), (b) (7, 1 + δ), (c)
(4 − 2δ, 2 + 2δ, 6 + 4δ).
Remark. We denote A to be the lattice spanned by the given vectors.
Solution for (a). A is not an ideal. For, suppose it were; then, δ(1+δ) = −5+δ ∈ A,
and so −5 + δ = 5a + (1 + δ)b for some a, b ∈ Z. Thus, b = 1, but −5 = 5a + 1 has
no solution a ∈ Z.
Solution for (b). A is not an ideal, for if it were, then as in (a), δ(1 + δ) = −5 + δ =
7a + (1 + δ)b ∈ A implies b = 1, but −5 = 7a + 1 has no solution a ∈ Z.
Solution for (c). We claim A is an ideal. A is closed under addition since it is a
lattice, and so it suffices to show A is closed under external multiplication. By the
distributive property, it is moreover sufficient to show δ times any of the generators
is in A, for A is closed under integer combinations. But this follows since
δ(4 − 2δ) = 10 + 4δ =
(4 − 2δ) + 3(2 + 2δ)
δ(2 + 2δ) = −10 + 2δ = −2(4 − 2δ) − (2 + 2δ)
δ(6 + 4δ) = −20 + 6δ = −4(4 − 2δ) + (2 + 2δ) − (6 + 4δ)
62
Exercise 13.3.3. Let A be an ideal of the ring of integers R in an imaginary
quadratic field. Prove that there is a lattice basis for A, one of whose elements
is an ordinary positive integer.
Proof. Let a′ ∈ A; then a′ a′ ∈ A is a positive integer hence A∩N is nonempty. Choose
a ∈ A ∩ N that is minimal, and choose b′ ∈ A which is Z-linearly independent from
a′ . Consider the parallelogram Π(a, b′ ) = {ra + sb′ | 0 ≤ r, s ≤ 1}. A ∩ Π(a, b′ ) is
finite, and so we can choose b ∈ A ∩ Π(a, b′ ) with minimal positive imaginary part.
We claim a, b form a lattice basis for A. First, A ∩ Π(a, b) = {0, a, b, a + b}, for if
c ∈ (A ∩ Π(a, b)) \ {0, a, b, a + b}, then either c ∈ Z but c < a, violating minimality of
a, or 0 < Im(c) < Im(b), violating minimality of b. Finally, we can move Π(a, b) by
Z-linear combinations of a, b, and none of these parallelograms contain anything in
A other than linear combinations of a, b for otherwise we can move that point into
the parallelogram Π(a, b) with an appropriate Z-linear combination of a, b. Thus,
a, b form a lattice basis for A.
13.4
Ideal Multiplication
√
Exercise 13.4.3. Let R be the ring Z[δ], where δ = −5, and let A and B be ideals
of the form A = (α, 12 (α + αδ)), B = (β, 21 (β + βδ)). Prove that AB is a principal
ideal by finding its generator.
Solution. By Prop. 13.1.6, we know 2 | α and 2 | β. By Prop. 13.4.3(c), we have
A = α2 (2, 1 + δ) and B = β2 (2, 1 + δ). Thus, by (13.4.2) and Prop. 13.4.3(a), we get
αβ
αβ
αβ
αβ
2
.
(2, 1 + δ) =
(4, 2 + 2δ, −4 + 2δ) =
(2) =
AB =
4
4
4
2
13.5
Factoring Ideals
√
Exercise 13.5.2. Let δ = −3 and R = Z[δ]. This is not the ring of integers in
the imaginary quadratic number field Q[δ]. Let A be the ideal (2, 1 + δ).
(a) Prove that A is a maximal ideal, and identify the quotient ring R/A.
(b) Prove that AA is not a principal ideal, and that the Main Lemma is not true
for this ring.
(c) Prove that A contains the principal ideal (2) but that A does not divide (2).
Proof of (a). It suffices to show R/A is a field by Prop. 11.8.2(b). But R/A =
Z[δ]/(2, 1 + δ) ≈ Z/(2) = F2 .
63
Proof of (b). First, we see
AA = (2, 1 − δ) · (2, 1 + δ) = (4, 2 + 2δ, 2 − 2δ, 4) = (4, 2 + 2δ) = 2A.
Now if the Main Lemma holds and AA is a principal ideal, then by the cancellation
law (Cor. 13.4.9(a)) we have A = (2), a contradiction since 1 + δ ∈ A \ (2).
Proof of (c). (2) ⊂ A since 2 is a generator of A. Now suppose AB = (2) for some
ideal B ( R. B 6⊂ A for otherwise
(2) = AB ⊂ A2 = (2, 1 + δ)2 = (4, 2 + 2δ, 2 − 2δ) = 2A,
which implies 1 ∈ A, contradicting that A is maximal from (a). Then, 2A = AAB =
2AB by (b), and so A = AB. Now let C be a maximal ideal containing B as exists
by Thm. 11.9.2; note C 6= A since B 6⊂ A. Then, A = AB ⊂ AC ⊂ A, and so we
have equalities throughout. But this contradicts Exercise 12.M.6, for A and AC are
two distinct factorizations of A into a product of distinct maximal ideals.
14
14.1
Linear Algebra in a Ring
Modules
Exercise 14.1.3. Let R = Z[α] be the ring generated over Z by an algebraic integer
α. Prove that for any integer m, R/mR is finite.
Proof. Let f (x) ∈ Z[x] be the (monic) irreducible polynomial for α. Then, I = (m, f )
is an ideal in Z[x] generated by two polynomials that have no common factors, and
so R/mR ≈ Z[x]/(m, f ) is finite by Exercise 12.M.7(b).
Exercise 14.1.4. A module is called simple if it is not the zero module and if it
has no proper submodule.
(a) Prove that any simple R module is isomorphic to an R module of the form
R/M where M is a maximal ideal.
(b) Prove Schur’s Lemma: Let ϕ : S → S ′ be a homomorphism of simple modules.
Prove that ϕ is either zero, or an isomorphism.
Proof of (a). Let S be simple and let s ∈ S be nonzero. Define ψ : R → S by r 7→ rs;
this is a homomorphism since S is a module. Then, im ϕ is a submodule of S, hence
equal to S since S is simple, and so ψ is surjective. By the first isomorphism theorem
(Thm. 14.1.6(c)), R/ker ϕ ≈ S. Now M := ker ϕ is a submodule of R, then by the
correspondence theorem (Thm. 14.1.6(d)) M is a maximal submodule of R, hence a
maximal ideal of R by Prop. 14.1.3.
64
Proof of (b). ker ϕ is a submodule of S, and im ϕ is a submodule of S ′ . Since S ′ is
simple, im ϕ is 0 or S ′ . If im ϕ = 0, then ϕ = 0. If im ϕ = S ′ , then ker ϕ ( S ′ , hence
equals 0, and so S ≈ S ′ by the first isomorphism theorem (Thm. 14.1.6(c)).
14.2
Free Modules
Exercise 14.2.3. Let A be the matrix of a homomorphism ϕ : Zn → Zm of free
Z-modules.
(a) Prove that ϕ is injective if and only if the rank of A, as a real matrix, is n.
(b) Prove that ϕ is surjective if and only if the greatest common divisor of the
determinants of the m × m minors of A is 1.
Remark. By Thm. 14.4.6, there exist matrices P ∈ GLn (Z), Q ∈ GLm (Z) such that
A′ = Q−1 AP is diagonal of the form


d1


..


.
(4)




dk
0
with di ∈ Z>0 and d1 | d2 | · · · | dk . Note A defines an injective (resp. surjective) map
ϕ : Zn → Zm if and only if A′ defines an injective (resp. surjective) map ϕ′ : Zn → Zm
since P, Q are invertible as integer matrices.
Proof of (a). ϕ is injective if and only if A′ above has k = n ≤ m, and this is
equivalent to A having rank n as a real matrix since rank is preserved by column
and row operations P, Q.
Proof of (b). ϕ is surjective if and only if A′ above has d1 = · · · = dk = 1 and
k = m ≥ n, i.e., the greatest commmon divisor of the determinants of the m × m
minors of A′ is 1. It suffices to show that this property is preserved by column and
row operations P, Q.
Recall (14.4.2): the operations are (i) adding an integer multiple of a row (resp.
column) to another, (ii) interchanging two rows (resp. columns), and (iii) multiplying a row (resp. column) by −1. But (i) simply adds integer multiples of some
determinants to others, (ii) switches some pairs of determinants and multiplies some
by −1, and (iii) multiplies some determinants by −1. So we are done.
65
14.4
Diagonalizing Integer Matrices
Exercise 14.4.6. Let ϕ : Zk → Zk be a homomorphism given by multiplication by
an integer matrix A. Show that the image of ϕ is of finite index if and only if A is
non-singular that it if so, then the index is equal to |det A|.
Proof. Since Zk is abelian, the index of im ϕ 6 Zk is given by |Zk / im ϕ|. By
Thm. 14.4.6, there exist P, Q ∈ GLk (Z) such that A′ = Q−1 AP is diagonal of the
form (4), with diagonal entries d1 | d2 | · · · | dr followed by k − r zeros, and so
k
k
k
k
′
k
Z / im ϕ = Z /AZ ≈ Z /A Z ≈
r
Y
i=1
(Z/d1 Z) × Zk−r .
(5)
Thus, |Zk / im ϕ| < ∞ if and only if r = k. But this is true if and only if
′
det A = det Q det A det P = det Q det P
k
Y
i=1
di 6= 0,
i.e., det A is nonsingular. If |Zk / im ϕ| < ∞, then (5) gives |Zk / im ϕ| =
which equals |det A| by (6).
14.5
(6)
Qk
i=1
di ,
Generators and Relations
Exercise 14.5.1. Let R = Z[δ] where δ =
as an R-module for the ideal (2, 1 + δ).
√
−5. Determine a presentation matrix
Proof. Let ϕ : R2 → I that sends (x, y)
2x + (1 + δ)y. Now
1+δ
1+δ
2
2
y ↔ y∈R
y∈R
ker ϕ = (x, y) ∈ R
x=−
2
2
where ↔ denotes bijection. But y = 2, 1−δ are the elements in R with smallest norm
satisfying this condition, and moreover the corresponding vectors (−3, 1 − δ), (−1 −
δ, 2) ∈ R are independent over R since 2r 6= 1 − δ for any r ∈ R. Thus, ker ϕ is
generated by these two vectors, and by pp. 424–425 we have a presentation matrix
−3 −1 − δ
A=
.
1−δ
2
66
14.7
Structure of Abelian Groups
Exercise 14.7.7. Let R = Z[i] and let V be the R-module given by the elements v1
and v2 with relations (1 + i)v1 + (2 − i)v2 = 0 and 3v1 + 5iv2 = 0. Write this module
as a direct sum of cyclic modules.
Proof. We use the relations to find a presentation matrix A and then diagonalize it
following the proof of Thm. 14.4.6:
1+i 3
2 − i 5i
1 0
−1+i 1 ·
i 0 ·
1 0
1+i
3
−i −3 + 8i
1
−8 − 3i
1+i
3
0 1 ·
1 0
1 0
−1−i 1 ·
−i −3 + 8i
1+i
3
1 −8 − 3i
0 8 + 11i
This reduces to the 1×1 matrix (8+11i) by Prop. 14.5.7(iv), and so V ≈ R/(8+11i).
To decompose further, we need to factor 8 + 11i. But
N (8 + 11i) = 64 + 121 = 185 = 5 · 37 = (2 + i)(2 − i)(6 + i)(6 − i),
and i(2 − i)(6 − i) = i(12 − 8i − 1) = 8 + 11i, hence V ≈ R/(8 + 11i) ≈ R/(2 − i) ⊕
R/(6 − i) by Exercise 11.6.8(c), where we note the map ϕ in Exercise 11.6.8 is also
an R-module isomorphism.
14.8
Applications to Linear Operators
Exercise 14.8.2. Let M be a C[t]-module of the form C[t]/(t − α)n . Show that there
is a C-basis for M , such that the matrix of the corresponding linear operator is a
Jordan block.
Proof. By Prop. 11.5.5, 1, t, . . . , tn−1 is a basis for M over C, and following p. 435,
we have the following matrix for multiplication by t in this basis:


0
−a0
1 0
−a1 

n
.. 
..


i
.
.  , an−i = (−α)
T = 1


i
..

. 0 −an−2 
1 −an−1
67
by using the binomial formula. This has characteristic polynomial f (t) = (t − α)n as
on p. 435, hence by the Jordan normal form it suffices to show dim ker(αI − T ) = 1.
By rank-nullity it suffices to show rk(αI −T ) ≥ n−1, for an eigenvalue has geometric
multiplicity at least one. For α = 0, we are done, so suppose α 6= 0. Then, the matrix


α
a0
−1 α
a1 


..
..


.
−1
.
αI − T = 



.
.. α

an−2 
−1 α + an−1
turns into the matrix

1
 1


..

.


1

a0 /α
a1 /α + a0 /α2
..
.
an−2 /α + · · · + a0 /αn−1
1 + an−1 /α + an−2 /α2 + · · · + a0 /αn
by Gaussian elimination, and so rk(αI − T ) ≥ n − 1.






Exercise 14.8.4. Let V be an F [t]-module, and let B = (v1 , . . . , vn ) be a basis for
V as F -vector space. Let B be the matrix of T with respect to this basis. Prove that
A = tI − B is a presentation matrix for the module.
Proof. We want to show that im A = ker(p : F [t]n → V ) where p is given by (fi )ni=1
P
n
i=1 fi vi . The inclusion ⊂ is clear, for B is multiplication by
Pnt on V .
n
For
i=1 fi vi = 0. If each
P the j other direction, suppose (fi )i=1 ∈ ker p, i.e.,
fi =
aij t , we then have
n
X
f i vi =
i=1
n X
X
aij tj vi
i=1 j≥0
=
n
X
i=1
"
ai0 vi +
=
i=1
"
aij t
j−1
j≥1
= (p ◦ A)
n
X
X
X
aij tj−1
j≥0
ai0 vi +
((t − B)vi + Bvi )
!n
i=1
X
+
n
X
i=1
#
aij tj−1 Bvi ,
j≥1
68
"
ai0 vi +
#
X
j≥1
aij tj−1 Bvi
#
since p ◦ A = 0 by the above. Then, repeating this process, we get
n
X
i=1
f i vi =
n X
X
j
aij B vi =
i=1 j≥0
n
X
bi v i = 0
i=1
for some coefficients bi ∈ F , i.e., each contribution from the fi above could be put in
the form (p ◦ A)(w) for some w ∈ F [t]n , and so im A ⊃ ker p.
14.M
Miscellaneous Problems
Exercise 14.M.10.
(a) Prove that the multiplicative group Q× of rational numbers is isomorphic to the
direct sum of a cyclic group of order 2 and a free abelian group with countably
many generators.
(b) Prove that the additive group Q+ of rational numbers is not a direct sum of
two proper subgroups.
(c) Prove that the quotient group Q+ /Z+ is not a direct sum of cyclic groups.
Proof of (a). Let h−1i = {−1, 1} with the obvious group structure making it isomorphic to Z/2Z. Let hpi = {pk | k ∈ Z} with the obvious group structure making
it isomorphic to Z. Then, define the map
M
hpi → Q×
h−1i ⊕
p prime
σ⊕
L
M
p kp
σ
Y
p kp
p prime
p prime
Note that hpi is a free abelian group. This
is a group
since
Q kmap
Q kp homomorphism
Q ℓp
p +ℓp
(σ, k2 , . . .) · (τ, ℓ2 , . . .) gets mapped to στ p
= σ p · τ p . This map
is clearly surjective since any rational number r/s has prime decompositions for r
and s, which we use to find a preimage on the left. Similarly, this is injective since
something on the right is equal to 1 if and only if σ = 1 and kp = 0 for all p.
Proof of (b). Let G, H be proper nontrivial subgroups of Q+ ; it suffices to show they
have nontrivial intersection by Prop. 2.11.4(d). If p/q ∈ G, s/t ∈ H for some p, s 6= 0,
then qs(p/q) = pt(s/t) = ps ∈ G ∩ H, while ps 6= 0 by assumption.
L
Proof of (c). Suppose Q+ /Z+ ≈ k Hk for Hk cyclic. Let (the image of) pk /qk +
Z+ ∈ Hk generate Hk for (pk , qk ) = 1. Let spk + tqk = 1 for some integers s, t; then,
1
spk tqk
pk
+ Z+ =
+
+ Z + = s + Z + ∈ Hk ,
qk
qk
qk
qk
69
hence we can assume that (the image of) 1/qk generates Hk . Now consider the
element 1/q12 + Z+ ∈ Q+ /Z+ . 1/q12 + Z+ then has a decomposition
X rk
1
+
+ Z+ ,
+
Z
=
q12
q
k
k
where qk | rk for all but finitely many values of k. Adding it to itself q1 times gives
X q1 r k
1
+ Z+ =
+ Z+
q1
q
k
k
and since we had a direct sum decomposition, qk | q1 rk for all k 6= 1, and q1 | q1 r1 − 1.
But this last property implies q1 | 1, a contradiction.
15
15.2
Fields
Algebraic and Transcendental Elements
Exercise 15.2.1. Let α be a complex root of the polynomial x3 − 3x + 4. Find the
inverse of α2 + α + 1 in the form a + bα + cα2 with a, b, c ∈ Q
Proof. Suppose 1 = (a + bα + cα2 )(1 + α + α2 ). We can rewrite this as
1 = a + (a + b)α + (a + b + c)α2 + (b + c)α3 + cα4
We then note that α3 = 3α − 4 and α4 = 3α2 − 4α, and so we get the equation
1 = a + (a + b)α + (a + b + c)α2 + (b + c)(3α − 4) + c(3α2 − 4α)
= (a − 4b − 4c) + (a + 4b − c)α + (a + b + 4c)α2
Since 1, α, and α2 are linearly independent over Q, we get a system of linear equations, with solution
 

 
−1  
  
1
1
17 12 20
17
1 −4 −4
a
 b  = 1 4 −1 0 = 1 −5 8 −3 0 = 1 −5
49
49
0
0
−3 −5 8
−3
1 1
4
c
and so (1 + α + α2 )−1 =
1
(17
49
− 5α − 3α2 ).
√
Exercise 15.2.3. Let β = ω 3 2 where ω = e2πi/3 , and let K = Q(β). Prove that the
equation x21 + · · · + x2k = −1 has no solution with xi in K.
√
3
2 = ω 2 β, β, and ωβ. Thus,
Proof. β has minimal polynomial x3 −2, which has roots
√
2
2
by Prop. 15.2.8, there is an isomorphism Q(β) ≈ Q( 3 2). But√then, if x√
1 + · · · + xk =
3
3
−1 has a solution in Q(β), it has a solution xi = ai + bi 2 ∈ Q( 2) ⊂ R by
Prop. 15.2.10, contradicting that this equation has no real solutions.
70
15.3
The Degree of a Field Extension
Exercise
15.3.2. Prove that the polynomial x4 + 3x + 3 is irreducible over the field
√
3
Q[ 2].
Proof. Since f := x4 + 3x + 3 ≡ x4 + x + 1 mod 2 is irreducible over F2 by Exercise
12.4.19, f is irreducible over Q as well by Prop. 12.4.3.
√
3
Now
let
α
be
a
root
of
f
;
then
[Q(α)
:
Q]
=
4
while
[Q(
2) :√Q] = 3, and so
√
√
√
3
3
3
3
2)] = 12
by
Cor.
15.3.8.
But
12
=
[Q(α,
2)
:
Q(
2)][Q(
2) : Q] implies
[Q(α, √
√
3
3
property of the degree (Thm. 15.3.5).
[Q(α, 2) : Q( 2)] = 4 by the multiplicative
√
3
Thus, the minimal
polynomial of α over Q( 2) has degree 4, and so x4 + 3x + 3 has
√
no factors in Q( 3 2)[x] by Lem. 15.3.2(b).
Exercise 15.3.5. Determine the values of n such that ζn has degree at most 3 over
Q.
Proof. If p is a prime dividing n, then ζpn = 1, hence Q(ζn ) ⊃ Q(ζp ). Now [Q(ζp ) :
Q] = p − 1 by Thm. 12.4.9, and so [Q(ζn ) : Q] ≤ 3 implies that if p | n, then
p ∈ {2, 3}. Thus, n = 2i 3j .
We claim i ≤ 2, j ≤ 1. [Q(ζ2 ) : Q] = 1 and [Q(ζ4 ) : Q] = 2 because the minimal
polynomial of ζ4 is x2 + 1. Similarly, [Q(ζ3 ) : Q] = 2. But any extension of Q(ζ4 ) or
Q(ζ3 ) will have degree ≥ 4 over Q by Thm. 15.3.5, hence i ≤ 2, j ≤ 1.
By the above, n ∈ {1, 2, 3, 4, 6, 12}, and also [Q(ζn ) : Q] ≤ 3 for n ∈ {1, 2, 3, 4},
hence it suffices to check whether n ∈ {6, 12} are also possible values of n. ζ6 = −ζ3
hence Q(ζ6 ) = Q(ζ3 ). On the other hand, Q(ζ12 ) ⊃ Q(ζ3 , ζ4 ), which has degree ≥ 6
over Q by Thm. 15.3.5, and so n 6= 12. Thus, n ∈ {1, 2, 3, 4, 5, }.
√
√
√
Exercise 15.3.7. (a) Is i in the field Q( 4 −2)? (b) Is 3 5 in the field Q( 3 2)?
√
4
Proof
of
(a).
Suppose
i
∈
Q(
−2). Then, there exists a, b ∈ Q such that (a +
√
b 4 −2)2 = −1. But then,
√
√
√
(a + b 4 −2)2 = a2 + 2ab 4 −2 + b2 2 −2 = −1,
√
√
and since {1, 4 −2, 2 −2} are linearly independent over Q by Prop. 15.2.7, a2 = −1,
which is impossible since a ∈ Q.
√
√
√
√
√
Proof of (b). If 3 5 ∈ Q( 3 2), then α := 3 2 + 3 5 ∈ Q( 3 2) must have degree at
most 3 over Q. Let f (x) = x9 − 21x6 − 123x3 − 343; then, f (α) = 0. But f (x) is
irreducible by the Eisenstein criterion (Prop. 12.4.6) since 3 | 21, 123 while 9 ∤ 343,
hence f is irreducible and so α has degree 9 over Q, a contradiction.
71
Exercise 15.3.9. Let α and β be complex roots of irreducible polynomials f (x) and
g(x) in Q[x]. Let K = Q(α) and L = Q(β). Prove that f (x) is irreducible in L[x]
iff g(x) is irreducible in K[x].
Proof. Let m = deg f, n = deg g. By the multiplicative property of the degree
(Thm. 15.3.5), we have [Q(α, β) : Q] = [L(α) : L][L : Q] = [K(β) : K][K : Q].
Hence, f (x) is irreducible in L[x] if and only if [Q(α, β) : Q] = mn if and only if g(x)
is irreducible in K[x].
Exercise 15.3.10. A field extension K/F is an algebraic extension if every element
of K is algebraic over F . Let K/F and L/K be algebraic field extensions. Prove
that L/F is an algebraic extension.
P
Proof. Let α ∈ L; we want to show α is algebraic over F . Let α = m
i=1
Pbni αi for
bi ∈ K, αi ∈ L linearly independent and algebraic over K. Write αi = j=1 cj βj
for cj ∈ F , βj ∈ K linearly independent and algebraic over F . Thus, we have α ∈
F (β1 , . . . , βn )(α1 , . . . , αm ). Now if α is transcendental over F , then by Thm. 15.3.5,
[F (β1 , . . . , βn )(α1 , . . . , αm ) : F (β1 , . . . , βn )][F (β1 , . . . , βn ) : F ] ≥ [F (α) : F ] = ∞,
which contradicts Cor. 15.3.6(c).
15.4
Finding the Irreducible Polynomial
Exercise 15.4.1. Let K = Q(α), where α is a root of x3 − x − 1. Determine the
irreducible polynomial of γ = 1 + α2 over Q.
Proof. First, γ satisfies (x − 1)(x − 2)2 − 1 = x3 − 5x2 + 8x − 5 = 0 since
(γ − 1)(γ − 2)2 − 1 = α2 (α2 − 1)2 − 1 = (α3 − α)2 − 1 = 12 − 1 = 0.
Now Thm. 15.3.5 implies [K : Q] = [K : Q(γ)][Q(γ) : Q] = 3 and so [Q(γ) : Q] = 3
since γ ∈
/ Q. Thus, x3 − 5x2 + 8x − 5 is the irreducible polynomial of γ over Q.
√
√
Exercise 15.4.2. Determine √
the irreducible
polynomial
for
α
=
3
+
5 over the
√
√
following fields. (a) Q, (b) Q( 5), (c) Q( 10), (d) Q( 15).
Solution for (a). Let f (x) = (x2 − 8)2 − 60 = x4 − 16x2 + 4. Then, f (α) = 0; we
claim f is irreducible. f cannot have linear factors since a root must be an integer
dividing 4 by the rational root test, and these are not roots. Now suppose f had
quadratic factors; then by Prop. 12.3.7(a), for some a, b, c, d ∈ Z we have
f (x) = (x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (b + ac + d)x2 + (ad + bc)x + bd.
72
Now a + c = 0 implies ad + bc = a(d − b) = 0. a 6= 0 since otherwise b + d = −16 and
bd = 4, a contradiction. So, d = b = 2, but then 2 + ac + 2 = −16 implies ac = −20,
which contradicts a + c = 0. Hence f is the irreducible polynomial for α over Q.
√
√
2
Solution for (b). Let f (x) =√(x − √ 5)2 − 3 = x√
− 2 5 +√2; then √
f (α) = 0. f splits
3 are in Q( 5). But 3 ∈
/√Q( 5), for otherwise
if
if the two roots 5 ± √
√ and only √
√
2
2
3 = a + b 5 implies 3 = a + 2ab 5 + 5b , and solving for 5 gives
that
5 ∈ Q,
√
a contradiction. Hence f is the irreducible polynomial for α over Q( 5).
√
Solution for (c). Let f (x) = x4 − 16x2 + 4. By Cor. 15.3.8 and (a), Q(α, 10) must
have degree either
√ 4 or 8 over Q, hence by the multiplicative
√ property of the degree
(Thm. 15.3.5), 10 has degree 2 or 1 over Q(α). To show 10 does not have degree
1 over Q(α),
it suffices to show it is not in Q(α) by Lem. 15.3.2(b). But this is clear
√
since if 10 = a + bα for a, b ∈ Q, then
√
√
√
10 = a2 + 2abα + b2 α2 = a2 + 8b2 + 2ab 3 + 2ab 5 + 2b2 15
√
√
/ Q. Now α has degree 4 over Q( 10) since
implies b = 0, contradicting that 10 ∈
√
√
√
√
√
[Q(α, 10) : Q( 10)][Q( 10) : Q] = [Q(α, 10) : Q( 10)] · 2 = 8
√
and so f is the irreducible polynomial for α over Q( 10).
√
Solution for (d). Let f (x) = x2 − 8 − 2 15; then f (α) = 0. Now
√
√
√
√
4 = [Q(α, 15) : Q] = [Q(α, 15) : Q( 15)][Q( 15) : Q]
√
√
α
has
degree
2
over
Q(
15), and so f
by (a) since 15 = (α2 − 8)/2 ∈ Q(α), hence
√
is the irreducible polynomial for α over Q( 15).
15.6
Adjoining Roots
Exercise 15.6.1. Let F be a field of characteristic 0, let f ′ be the derivative of
f ∈ F [x], and let g be an irreducible polynomial that is a common divisor of f and
f ′ . Prove that g 2 divides f .
Proof. f = gh for some h ∈ F [x], hence f ′ = gh′ + g ′ h by Exercise 11.3.5(a). If
g | f ′ , then g | g ′ h. g is prime since it is irreducible, hence g | g ′ or g | h. But g ∤ g ′
since deg g ′ < deg g, hence h = gr for some r ∈ F [x]. Thus, f = g 2 r, i.e., g 2 | f .
73
15.7
Finite Fields
Exercise 15.7.4. Determine the number of irreducible polynomials of degree 3 over
F3 and over F5 .
3
Proof. By Thm. 15.7.3(b), for prime p the irreducible factors of xp − x are the monic
irreducible polynomials in Fp [x] with degree 1 or 3. x, x − 1, x − 2, . . . , x − (p − 1) are
the linear irreducible polynomials, hence there are p(p−1)(p+1)/3 monic irreducible
polynomials of degree 3 in Fp [x]. Since by Thm. 15.7.3(c) there are p − 1 units in Fp ,
we have that there are p(p − 1)2 (p + 1)/3 irreducible polynomials of degree 3 over Fp .
Thus, for F3 we have 16, and for F5 we have 160 irreducible polynomials of degree
3. Note this matches our result from Exercise 12.4.12.
Exercise 15.7.6. Factor the polynomial x16 − x over the fields F4 and F8 .
Solution for F4 . We claim x16 − x factors as the product of all monic irreducible
polynomials over F4 of degree 1 and 2. First, x4 − x | x16 − x by Thm. 15.7.3(a), (e),
hence every monic irreducible polynomial of degree 1, x − a | x16 − x for a ∈ F4
divides x16 − x. Now let g be a monic irreducible polynomial of degree 2 over F4 . If
β is a root of g in F16 then [F4 (β) : F4 ] = 2, hence F4 (β) ≈ F16 by Thm. 15.7.3(d).
Thus, x − β | x16 − x, so g | x16 − x by considering the other root of g and proceeding
in the same way. Thus, x16 − x is divisible by every monic irreducible polynomial
over F4 of degree
1 or 2. Now there are 4 irreducible monic polynomials of degree 1
4
2
and 4 − 2 − 4 = 6 irreducible monic polynomials of degree 2 over F4 , which means
the total degree of their product is 16 as desired. Thus, we have that
x16 − x = x(x − 1)(x − α)(x − α − 1)(x2 + x + α)(x2 + x + (α + 1))
(x2 + αx + 1)(x2 + αx + α)(x2 + (α + 1)x + 1)(x2 + (α + 1)x + (α + 1)).
Solution for F8 . We claim x16 − x factors in the same way as over F2 , i.e.,
x16 − x = x(x − 1)(x2 + x + 1)(x4 + x + 1)(x4 + x3 + 1)(x4 + x3 + x2 + x + 1)
It suffices to show the degree 4 factors are irreducible over F8 . Since there are no
intermediate fields between F2 and F8 by Thm. 15.7.3(e), we know there are no new
linear factors of x16 − x, since its roots are the elements in F16 . Let α ∈ F8 such that
F8 ≈ F2 (α). Suppose one of the degree 4 factors has a degree 2 factor. Let β be one
of its roots; by Thm. 15.7.3(a) it is an element of F16 , and F16 ≈ F2 (β) as in the
solution for F4 . Then, [F8 (β) : F8 ] = 2, hence F8 (β) ≈ F64 by Thm. 15.7.3(d). But
F64 ≈ F8 (β) ≈ F2 (α, β) ≈ F2r where r = 12 by Cor. 15.3.8, a contradiction.
74
Exercise 15.7.7. Let K be a finite field. Prove that the product of the nonzero
elements of K is −1.
Proof. For every nonzero a ∈ K there is a−1 ∈ K such that aa−1 = 1 since K is a
field. In the product Π of nonzero elements of K, we can pair off each a, a−1 such
that a 6= a−1 so that Π is the product of all nonzero elements of K such that a−1 = a.
We claim this is true if and only if a = ±1. If a = a−1 , then a2 = 1 and so a is a
root of x2 − 1 = (x + 1)(x − 1). Since K[x] is a UFD (Prop. 12.2.14(c)), this implies
a = ±1. Thus, Π = 1 · (−1) = −1.
Exercise 15.7.8. The polynomials f (x) = x3 + x + 1 and g(x) = x3 + x2 + 1 are
irreducible over F2 . Let K be the field extension obtained by adjoining a root of f ,
and let L be the extension obtained by adjoining a root of g. Describe explicitly an
isomorphism from K to L, and determine the number of such isomorphisms.
Proof. Let α be a root of f , and β one of g. Then, α3 = α + 1 and β 3 = β 2 + 1. Any
field homomorphism ϕ : K → L is specified by ϕ(α), satisfying that ϕ(α3 ) = ϕ(α+1)
by the mapping property of quotient rings (Thm. 11.4.2). So let ϕ(α) = a + bβ + cβ 2 .
ϕ(α3 ) = (a2 + b2 β 2 + c2 β 4 )(a + bβ + cβ 2 )
= (a3 + ac2 + b3 + b2 c + bc2 ) + (a2 b + ac2 + b2 c + bc2 + c3 )β
+ (a2 c + ab2 + ac2 + b3 + b2 c + c3 )β 2
must then be equal to ϕ(α + 1).
We claim ϕ is an isomorphism if and only if (a, b, c) ∈ {(1, 1, 0), (1, 0, 1), (0, 1, 1)}.
By Prop. 11.8.4(b), it suffices to show ϕ is surjective if and only if this is true. Suppose
ϕ is surjective; then not both b, c are zero by the above. Matching 1, β, β 2 terms,
we have that if b = 0, then a = c = 1; if c = 0, then a = b = 1; and if b = c = 1,
then a = 0. Conversely, since ϕ(α + 1) = β in the first case, ϕ(α2 + α + 1) = β in
the second case, and ϕ(α2 + 1) = β in the third case, we see in each case that ϕ is
surjective by using that ϕ is an homomorphism.
Exercise 15.7.10. Let F be a finite field, and let f (x) be a nonconstant polynomial
whose derivative is the zero polynomial. Prove that f cannot be irreducible over F .
Lemma 15.7.10a. In a finite field of order pr , every element is a pth power.
Proof of Lemma 15.7.10a. Let a ∈ F ; if a = 0 we are done so suppose not. |F × | =
r
r
r−1
satisfies
pr − 1 by Thm. 15.7.3(c), hence ap −1 = 1 and ap = a. Thus, b := ap
p
b = a.
75
Pn
Pn
i
′
i−1
Main Proof. Let f (x) =
= 0, we have
i=0 ai x . Then, if f (x) =
i=1 iai x
iai = 0 for all iP
≥ 1. Since F is a domain, this implies either ai = 0 or i ≡ 0 mod p.
pj
Thus, f (x) = m
for some aj ∈ F where p = char F . There exist bj such
j=0 aj x
P
p
j
that bj = aj by Lemma 15.7.10a. Thus, f (x) = g(x)p where g(x) = m
j=0 bj x by
Exercise 11.3.8, and so f (x) is not irreducible.
Exercise 15.7.11. Let f = ax2 + bx + c with a, b, c in a ring R. Show that the ideal
of the polynomial ring R[x] that is generated by f and f ′ contains the discriminant,
the constant polynomial b2 − 4ac.
Proof. f ′ = 2ax + b. Proceeding as in the Euclidean algorithm (p. 45), we have
2ax2 + 2bx + 2c = x(2ax + b) + (bx + 2c),
b(2ax + b) = 2a(bx + 2c) + (b2 − 4ac),
hence b2 − 4ac = −4af + f ′2 ∈ (f, f ′ ) ⊂ R[x].
Exercise 15.7.13. Prove that a finite subgroup of the multiplicative group of any
field F is a cyclic group.
Proof. A finite subgroup H 6 F × is a finite abelian group. Thus, H ≈ C1 ⊕ · · · ⊕ Cn
where the Ci are cyclic groups, and |Ci | divides |Ci+1 | by Thm. 14.7.3. Let d := |Cn |.
Then, xd = 1 for every x ∈ H. This means that every x ∈ H is a root of xd − 1.
Now, xd − 1 has at most d roots, so d ≥ |H|. On the other hand, d ≤ |H| by the
decomposition above, so d = |H|, and n = 1, i.e., H is cyclic.
15.8
Primitive Elements
√ √
Exercise 15.8.2. Determine all primitive elements for the extension K = Q( 2, 3)
of Q.
√
√
√
Proof. We claim γ = a + b 2 + c 3 + d 6 is a primitive element for the extension
K/Q if and only if at least two of b, c, d are nonzero. Recall that
√ √
√ √
√
√
√
√ √
Q( 2, 3) = Q( 2, 6) = Q( 3, 6) = {a + b 2 + c 3 + d 6 | a, b, c, d ∈ Q}.
We can assume without loss of generality that a = 0 since
a). ⇒ is
√ Q(γ)√= Q(γ −√
clear since if only one of b, c, d is nonzero, then K = Q( 2), Q( 3), or Q( 6).
Conversely, suppose at least two of b, c, d are nonzero.
√ √ If
√ exactly two of b, c, d
are nonzero, then relabel γ = kα + jβ for α, β ∈ { 2, 3, 6}; we can moreover
assume k = 1 by dividing by k. Note K = Q(α, β) as above, and so by Lem. 15.8.2
γ = α + jβ is a primitive element for K/Q for all but finitely many j. In particular,
76
as in the proof of Lem. 15.8.2, γ fails to be primitive when at least two of ±α ± jβ
are equal, but this occurs if and only if j = 0, a contradiction.
Now suppose that all three of b, c, d are nonzero. Letting
√
√
√
γ ′ = γ 2 − 2b2 − 3c2 − 6d2 = 6cd 2 + 4bd 3 + 2bc 6,
we have that
√
√
bγ ′ − 6cdγ = 2d(2b2 − 3c2 ) 3 + 2c(b2 − 3d2 ) 6
√
√
cγ ′ − 4bdγ = 2d(3c2 − 2b2 ) 2 + 2b(c2 − 2d2 ) 6
√
√
dγ ′ − 2bcγ = 2c(3d2 − b2 ) 2 + 2b(2d2 − c2 ) 3
are all in Q(γ). But each of their coefficients have no nonzero solutions over Q by
clearing denominators and checking for integer solutions, hence any three of these
produces a primitive element for K/Q by the paragraph above, and so γ is a primitive
element for K/Q as well.
15.9
Function Fields
Exercise 15.9.1. Let f (x) be a polynomial with coefficients in a field F . Prove that
if there is a rational function r(x) such that r2 = f , then r is a polynomial.
Proof. Suppose f 6= 0 for otherwise r is necessarily 0. Let r = p(x)/q(x) for p, q
coprime; then p2 = f q 2 . Now using that F [x] is a UFD (Prop. 12.2.14(c)), since p ∤ q
we have p2 | f , and so f = p2 s for some s ∈ F [x]. But then, f q 2 = p2 sq 2 = p2 , and
so sq 2 = 1 and s, q ∈ F [x]; since the units in F [x] are the constant polynomials, we
have s, q ∈ F and so r ∈ F [x].
15.10
The Fundamental Theorem of Algebra
Exercise 15.10.1. Prove that A ⊂ C is algebraically closed, where A is the subset
of C consisting of the algebraic numbers.
Remark. We prove that a root of a polynomial with coefficients that are algebraic
over a field F is also algebraic.
Proof. Let f ∈ A[x] be non-constant, and let α be a root of f . It suffices to show α
is algebraic over F . Suppose not, and let a0 , . . . , an be the coefficients of f . Then,
[F (α) : F ] ≤ [F (α, a0 , . . . , an ) : F ] ≤ n ·
77
n
Y
i=0
mi < ∞
by Cor. 15.3.8, where mQ
i is the degree of ai over F . Thus, there exists a polynomial
in F [x] of degree ≤ n · ni=0 mi with α as a root.
Exercise 15.10.2. Construct an algebraically closed field that contains the prime
field Fp .
Remark. We prove that any field K is contained in an algebraically closed field L.
Proof. We first claim the monic irreducible polynomials of degree ≥ 1 in K[x] can be
well-ordered. This is a consequence of the well-ordering theorem, but in the case of
Fp (or any countable field) it is possible to well-order these polynomials without the
axiom of choice since there are only countably many of them. Then, letting (fi ) be
a well-ordering of these polynomials, and letting L0 = K and Li be the extension
S of
:=
Li−1 such that fi splits completely which exists by Prop. 15.6.3, we claim L
i Li
is an algebraically closed field containing K. It suffices to show it is algebraically
closed, but this is true since if f (x) ∈ L[x] has a root α, then the proof of Exercise
15.10.1 shows α is algebraic over K, hence in L by construction.
15.M
Miscellaneous Problems
Exercise 15.M.1. Let K = F (α) be a field extension generated by a transcendental
element α, and let β be an element of K that is not in F . Prove that α is algebraic
over F (β).
Proof. Suppose β ∈ K \ F ; then β = p(α)/q(α) for p(x), q(x) ∈ F [x], and so
βq(α) − p(α) = 0. Thus, α is a root of the polynomial βq(x) − p(x) ∈ F (β)[x],
i.e., α is algebraic over F (β).
Exercise 15.M.2. Factor x7 + x + 1 in F7 [x].
Proof. First substitute x
x + 3, giving x7 + x. Then, x7 + x = x(x6 + 1). Now
x6 + 1 has solutions x2 = 3, 5, 6 ∈ F7 , hence x7 + x = x(x2 + 1)(x2 + 2)(x2 + 4). Now
1, 2, 4 are the squares in F7 , none of which are 3, 5, 6, hence we cannot factor further.
Substituting back x
x − 3, since (x − 3)2 = x2 + x + 2,
x7 + x + 1 = (x − 3)(x2 + x + 3)(x2 + x + 4)(x2 + x + 6).
Exercise 15.M.3. Let f (x) be an irreducible polynomial of degree 6 over a field F ,
and let K be a quadratic extension of F . What can be said about the degrees of the
irreducible factors of f in K[x]?
78
Proof. K = F (α), where α is the root of some irreducible quadratic polynomial
g ∈ F [x]. Let β be a root of f ; then, 2 = [K : F ] and 6 = [F (β) : F ] divide
[K(β) : F ] ≤ 12 by Cor. 15.3.8, and so [K(β) : F ] ∈ {6, 12}. In either case, 3 divides
[K(β) : K], and so the minimal polynomial for β over K is of degree 3 or 6. In the
former case, f splits into a product of polynomials of degree 3 over K. In the latter,
f remains irreducible over K. These are the only possibilities for the irreducible
factors of f in K, for the only possibility is that the factors of degree 3 factor further
to include linear factors, which is impossible since β is of degree 6 over F .
Exercise 15.M.4.
(a) Let p be an odd prime. Prove that exactly half of the elements of F×
p are squares
and that if α and β are nonsquares, then αβ is a square.
(b) Prove the same assertion for any finite field of odd order.
(c) Prove that in a finite field of even order, every element
√ is a square.
√
(d) Prove that the irreducible polynomial for γ = 2 + 3 over Q is reducible
modulo p for every prime p.
Remark. (b) implies (a) and Lemma 15.7.10a implies (c), so it suffices to show (b)
and (d).
Proof of (b). Let F be our finite field of odd order, and consider the group homomorphism ϕ : F × → F × defined by α
α2 . If α ∈ ker ϕ then α2 = 1, hence
2
(x − α) | x − 1 in F [x]. Since F [x] is a UFD (Prop. 12.2.14(c)), this implies α = ±1.
Thus |ker ϕ| = 2 since char F 6= 2 implies 1 6= −1, and so the set of squares im ϕ in
F × has order 12 |F × | by the counting formula (Cor. 2.8.13).
Now let β be a nonsquare. Consider the map µ : F × → F × of sets defined by
α
αβ. This is a bijection, and so to show µ sends non-squares to squares, it suffices
to show µ sends squares to non-squares. But this is true since if α is a square and
αβ is also a square, then α−1 is also a square, hence β is also, a contradiction.
Proof of (d). First, f := x4 − 10x2 + 1 is the irreducible polynomial for γ since
f (γ) = 0 and Q(γ) = Q(α, β) by Exercise 11.1.3, and so [Q(γ) : Q] = 4. In this field,
we have the factorization
√
√
√
√
√
√
√
√
f = (x − ( 2 + 3))(x − ( 2 − 3))(x − (− 2 + 3))(x − (− 2 − 3)).
If f reduces over Fp , then it must reduce to quadratic factors since γ ∈
/ Fp ; thus, f
can factor in one of the three following ways by pairing up the factors above:
√
√
√
√
(x2 − 1 − 2 2)(x2 − 1 + 2 2), (x2 + 1 − 2 3)(x2 + 1 + 2 3),
√
√
(x2 − 5 − 2 6)(x2 − 5 + 2 6).
79
The first factorization can occur if 2 is a square mod p, the second can occur if 3 is
a square mod p, and the last can occur if 6 is a square mod p. By (a), (c), at least
one of 2, 3, 6 is a square in Fp , so at least one of these factorizations is possible and
f is reducible mod p for all p.
Exercise 15.M.6.
(a) Prove that a rational function f (t) that generates the field C(t) of all rational
functions defines a bijective map T ′ → T ′ .
(b) Prove a rational function f (x) generates the field of rational functions C(x) if
and only if it is of the form (ax + b)/(cx + d), with ad − bc 6= 0.
(c) Identify the group of automorphisms of C(x) that are the identity on C.
Proof of (a). Let f (t) ∈ F = C(t), and f = p/q for some coprime p, q ∈ C[t].
C[t](f ) is then isomorphic to F , which is isomorphic to F [x]/(p − qx); note this
is still isomorphic to F . Likewise, consider F [y]/(y); this is also isomorphic to F .
By Prop. 15.9.5, we have isomorphic field extensions, hence the Riemann surfaces
{p − qx = 0} and {y = 0} are isomorphic branched coverings of T ′ . Since {y = 0} is
the complex t-plane T ′ , the isomorphism of coverings is an isomorphism T ′ → T ′ .
Proof of (b). Any function (ax + b)/(cx + d) with ad − bc 6= 0 generates C(x) by
Exercise 12.2.5, since any element in C(x) can be expressed as a C-linear combination
of elements 1/(cx + d)i , and then by using the Euclidean algorithm (p. 45) to get
q(cx + d) + r
r
ax + b
=
=q+
, q, r ∈ C
cx + d
cx + d
cx + d
which implies 1/(cx + d)i for any i can be expressed as a C-linear combination of
(ax + b)i /(cx + d)i . Conversely, suppose a rational function f (x) generates C(x).
By (a), it induces an automorphism of the Riemann surface T ′ ; it suffices to show
that the automorphisms of T ′ are of the stated form. Suppose f = p/q defines
an automorphism of T ′ ; then, if either p, q had degree larger than 1, the induced
morphism would not be bijective since then p has two zeros; likewise, if q had degree
larger than 1, considering 1/f gives us the same argument. Moreover, one of p, q
must be non-constant also to induce a bijection. ad − bc 6= 0 corresponds to having
p, q coprime, which was necessary in (a).
Solution for (c). By (a), (b), Aut(C(x)) consists of maps x
(ax + b)/(cx + d) for
ad − bc 6= 0, which is a group under composition. We claim there is a surjective
homomorphism ϕ : GL2 (C) → Aut(C(x)) defined by ( ac db )
(ax + b)/(cx + d). This
1
0
map clearly maps ( 0 1 )
x, is surjective, and is a homomorphism since
′
′ ′ aa + b′ c a′ b + b′ d
a b
a b
,
=
ac′ + cd′ bc′ + dd′
c d
c′ d ′
80
and both sides get mapped to
′
a x + b′
a′ (ax + b) + b′ (cx + d)
ax + b
(aa′ + b′ c)x + (a′ b + b′ d)
◦
=
=
.
c′ x + d ′
cx + d
c′ (ax + b) + d′ (cx + d)
(ac′ + cd′ )x + (bc′ + dd′ )
It remains to find the kernel of ϕ. (ax + b)/(cx + d) = x ∈ C(x) if and only if
ax + b = x(cx + d) if and only if c = b = 0 and a = d. Hence ker ϕ = C · ( 10 01 ), and so
by the first isomorphism theorem (Thm. 11.4.2(b)), Aut(C(x)) ≈ GL2 (C)/ ker ϕ =:
PGL2 (C), the projective general linear group of order 2.
Exercise 15.M.7. Prove that the homomorphism SL2 (Z) → SL2 (Fp ) obtained by
reducing the matrix entries modulo p is surjective.
Proof. Let π : Z → Fp be the quotient map reducing mod p. We first show that if
u, v ∈ Fp are not both zero, then there exist c, d ∈ Z such that π(c) = u, π(d) = v,
and gcd(c, d) = 1. So suppose v 6= 0, and let 0 ≤ ũ, ṽ < p be lifts of u, v in Z. Now
since p ∤ ṽ, there exist x, y ∈ Z such that xṽ + yp = 1 since (p) is maximal in Z, and
so let c = xũṽ + yp and d = ṽ. Then, π(d) = v, π(c) = u since xṽ ≡ 1 mod p, and
gcd(c, d) = 1 since 1 = c + xd(1 − ũ). If v = 0, then necessarily u 6= 0, and switching
the roles of u, v gives the desired result.
Now suppose we have ( us vt ) ∈ SL2 (Fp ); let 0 ≤ s̃, t̃ < p be lifts of s, t in Z. Then,
letting c, d as constructed above, we have s̃d − t̃c = 1 + N p for some N ∈ Z. So,
letting a = s̃ + mp and b = t̃ + np where m, n ∈ Z such that N = cn − dm, which is
possible since gcd(c, d) = 1,
ad − bc = (s̃ + mp)d − (t̃ + np)c
= s̃d − t̃c + (dm − cn)p = 1 + (N + dm − cn)p = 1,
hence ( ac db ) ∈ SL2 (Z) maps to ( us vt ) ∈ SL2 (Fp ).
16
16.1
Galois Theory
Symmetric Functions
Exercise 16.1.1. Determine the orbit of the polynomial below. If the polynomial is
symmetric, write it in terms of the elementary symmetric functions.
(a) u21 u2 + u22 u3 + u23 u1 (n = 3),
(b) (u1 + u2 )(u2 + u3 )(u1 + u3 ) (n = 3),
(c) (u1 − u2 )(u2 − u3 )(u1 − u3 ) (n = 3),
81
(d) u31 u2 + u32 u3 + u33 u1 − u1 u32 − u2 u33 − u3 u31
(e) u31 + u32 + · · · + u3n .
(n = 3),
Solution for (a). The orbit consists of u21 u2 + u22 u3 + u23 u1 and u23 u2 + u22 u1 + u21 u3 ,
corresponding to odd and even permutations respectively in S3 .
Solution for (b). The orbit consists of only (u1 + u2 )(u2 + u3 )(u1 + u3 ), i.e., this
polynomial is symmetric. Since this polynomial is homogeneous of degree 3, write
(u1 + u2 )(u2 + u3 )(u1 + u3 ) = c1 s31 + c2 s1 s2 + c3 s3 .
Substituting (1, 0, 0), we have c1 = 0. Substituting (1, 1, 0), we have 2 = 2c2 hence
c2 = 1. Lastly, substituting (1, 1, 1), we have 8 = 9 + c3 hence c3 = −1. Thus,
(u1 + u2 )(u2 + u3 )(u1 + u3 ) = s1 s2 − s3 .
Solution for (c). The orbit consists of (u1 −u2 )(u2 −u3 )(u1 −u3 ) and −(u1 −u2 )(u2 −
u3 )(u1 − u3 ), corresponding to odd and even permutations respectively in S3 .
Solution for (d). The orbit consists of u31 u2 + u32 u3 + u33 u1 − u1 u32 − u2 u33 − u3 u31 and
u33 u2 + u32 u1 + u31 u3 − u3 u32 − u2 u31 − u1 u33 , corresponding to odd and even permutations
respectively in S3 .
Solution for (e). The orbit consists of only u31 + u32 + · · · + u3n , i.e., this polynomial
is symmetric. Since this polynomial is homogeneous of degree 3, write
u31 + u32 + · · · + u3n = c1 s31 + c2 s1 s2 + c3 s3 .
Substituting (1, 0, . . . , 0), we have 1 = c1 . Substituting (1, 1, 0, . . . , 0), we have 2 =
8 + 2c2 hence c2 = −3. Lastly, substituting (1, 1, 1, 0 . . . , 0), we have 3 = 27 − 27 + c3 ,
hence c3 = 3. Thus,
u31 + u32 + · · · + u3n = s31 − 3s1 s2 + 3s3 .
Exercise 16.1.3. Let wk = uk1 + · · · + ukn .
(a) Prove Newton’s identities: wk − s1 wk−1 + · · · ± sk−1 w1 ∓ ksk = 0.
(b) Do w1 , . . . , wn generate the ring of symmetric functions?
Proof of (a). For 1 < i ≤ k, let ri be the sum of all distinct monomials of degree k
where each monomial is the product of one variable raised to the power i and k − i
distinct other variables; note rk = wk . We claim sk−i wi = ri + ri+1 for 1 < i < k.
This is true since each product of terms with distinct variables on the left contributes
82
to ri , while each product which has the term from wi occurring in the term from
sk−i contributes to ri+1 , and since all terms on the right are obtained exactly once
in this way. For i = k we recall s0 = 1, hence s0 wk = wk = rk . Finally, for i = 1 we
have sk−1 w1 = ksk + r2 , since the terms contributing to r2 arise in the same way as
before, while the remaining terms produce k times each monomial in sk . The desired
identity is formed by taking the alternating sum of these equations.
Claim (b). w1 , . . . , wn generates the ring of symmetric functions if and only if m ∈
R× for all 1 ≤ m ≤ n.
Proof of Claim (b). ⇐. By Thm. 16.1.6, it suffices to show sk ∈ R[w1 , . . . , wn ] for
all k; we proceed by induction. If k = 1 then we are done, for s1 = w1 . Then, if
sj ∈ R[w1 , . . . , wn ] for all j < k, by (a) we have
sk =
1
(wk − s1 wk−1 + · · · ± sk−1 w1 ) ∈ R[w1 , . . . , wn ].
±k
⇒. We show the contrapositive. Suppose m ∈
/ R× for some 0 < m ≤ n; we claim
sm ∈
/ R[w1 , . . . , wn ]. Suppose not; then, since sm is homogeneous of degree m, write
Y
X
wi , rI ∈ R.
sm =
rI
I⊂{1,...,n}
P
i∈I i=m
i∈I
Then, substituting u1 = · · · = um = 1, um+1 = · · · = un = 0, we get wi = m for all i,
while sm = 1. But this is a contradiction, for the right side is in the ideal mR while
the left side = 1 ∈
/ mR.
16.2
The Discriminant
Exercise 16.2.2.
(a) Prove that the discriminant of a real cubic is non-negative if and only if the
cubic has three real roots.
(b) Suppose that a real quartic polynomial has positive discriminant. What can
you say about the number of real roots?
Proof of (a). Recall that the discriminant D is the product of squares of the differences of roots. So, if f has 3 roots in R, then the discriminant is trivially non-negative.
Conversely, if f does not have 3 roots in R, we know it must have one root a ∈ R,
and the other two roots must be a conjugate pair z, z since otherwise the constant
term in f would not be real. Then, we have
2
D = (a − z)2 (a − z)2 (z − z)2 = (a − z)2 (a − z) (z − z)2 .
83
Now (a − z)(a − z) > 0, and so it suffices to show (z − z)2 < 0. But this is true since
z − z = 2 Im z.
Solution for (b). We claim that a real quartic polynomial with positive discriminant
can have 0 or 4 real roots; note in particular this implies all roots are distinct. The
discriminant is given by
D = (α1 − α2 )2 (α1 − α3 )2 (α1 − α4 )2 (α2 − α3 )2 (α2 − α4 )2 (α3 − α4 )2 ,
where αi are our roots; recall that complex roots must come in conjugate pairs, and
so we can have 0, 2, or 4 real roots in general. If all the roots are real D is clearly
non-negative. If none of the roots are real and α2 = α1 , α4 = α3 , we get
D = (α1 − α1 )2 (α1 − α3 )2 (α1 − α3 )2 (α1 − α3 )2 (α1 − α3 )2 (α3 − α3 )2
2
2
= (2 Im α1 )2 (2 Im α3 )2 (α1 − α3 )2 (α1 − α3 ) (α1 − α3 )2 (α1 − α3 ) ,
which is positive since each consecutive pair of factors multiply to be positive. However, 2 real roots is impossible since if α2 = α1 and α3 , α4 are real,
D = (2 Im α1 )2 (α1 − α3 )2 (α1 − α4 )2 (α1 − α3 )2 (α1 − α4 )2 (α3 − α4 )2
2
2
= (2 Im α1 )2 (α1 − α3 )2 (α1 − α3 ) (α1 − α4 )2 (α1 − α4 ) (α3 − α4 )2 ,
which is negative since the first factor is negative and the rest of the product is
positive.
Exercise 16.2.4. Use undetermined coefficients to determine the discriminant of
the polynomial
(a) x3 + px + q, (b) x4 + px + q, (c) x5 + px + q.
Lemma 16.2.4a. The discriminant Dn of xn + px + q for n ≥ 2 is given by
(−1)
(n−1)(n−2)
2
(n − 1)n−1 pn + (−1)
n(n−1)
2
nn q n−1 .
Proof of Lemma 16.2.4a. Let Dn (p, q) be the determinant of xn +px+q. If u1 , . . . , un
are the roots of xn + px + q, by the first equation in §16.2 we have sn (u1 , . . . , un ) =
(−1)n q, sn−1 (u1 , . . . , un ) = (−1)n−1 p, and sk (u1 , . . . , un ) = 0 for all 0 < k < n − 1.
Note the discriminant is a homogeneous symmetric polynomial of degree n(n − 1);
we claim the only monomials in the si appearing in Dn (p, q) are pn , q n−1 . For, by the
above sn−1 , sn are the only nonzero elementary symmetric polynomials that would
appear in Dn (p, q), and moreover if pk q j had degree n(n − 1), then the equation
84
k(n − 1) + jn = n(n − 1) must hold, for which k = 0, j = n − 1 and k = n, j = 0 are
the only non-negative solutions. Thus, we have Dn (p, q) = r(n)pn + s(n)q n−1 .
Now consider the discriminant Dn (p, 0) for xn + px; since this has the same roots
as xn−1 + p in addition to the root un = 0, we have
Dn (p, 0) = Dn−1 (0, p)
n−1
Y
u2i = Dn−1 (0, p)p2
i=1
and so r(n)pn = s(n−1)pn , i.e., r(n) = s(n−1), and Dn (p, q) = s(n−1)pn +s(n)q n−1 .
It remains to find s(n). Calculating the discriminant Dn (0, −1) for f := xn − 1,
Dn (0, −1) = (−1)
n−1
g(n) = (−1)
n(n−1)
2
n Y
Y
i=1 i6=j
(ξi − ξj )
Q
where ξi are the roots of xn − 1. Then, we have the equation xn − 1 =
i (x − ξi ),
Q
and so deriving both sides and substituting in x = ξi , we get nξin−1 = i6=j (ξi − ξj ),
n(n−1) Qn
n−1
. Since ξi satisfies f , ξin−1 = ξi−1 , and so
and so Dn (0, −1) = (−1) 2
i=1 nξi
Dn (0, −1) = (−1)
hence s(n) = (−1)
n(n−1)
2
(−1)
n(n−1)
2
n
n
n
Y
ξi−1 = (−1)
n(n−1)
2
(−1)n−1 nn ,
i=1
nn , and finally
(n−1)(n−2)
2
(n − 1)n−1 pn + (−1)
n(n−1)
2
nn q n−1 .
Solution for (a). D3 = −4p3 − 27q 2 by Lemma 16.2.4a.
Solution for (b). D4 = −27p4 + 256q 3 by Lemma 16.2.4a.
Solution for (c). D5 = 256p5 + 3125q 4 by Lemma 16.2.4a.
Exercise
16.2.7. There are n variables. Let m = u1 u22 u33 · · · un−1
n−1 and let p(u) =
X
σ(m). The Sn -orbit of p(u) contains two elements, p and another polynomial q.
σ∈An
Prove that (p − q)2 = D(u).
Q
Proof. Let R[u1 , . . . , un ] be our polynomial ring, and let δ(u) = i<j (ui − uj ). We
first claim ui − uj | p − q for all i < j. This is equivalent to showing p − q = 0 if
ui − uj = 0.
85
So suppose ui = uj for i < j, and let τ = (ij) ∈ Sn . Then, p(u) = p(τ (u)) and
q(u) = q(τ (u)), and we have
X
X
p(τ (u)) =
σ(τ (m)) =
σ ′ (m) = q(u).
σ∈An
σ ′ ∈An ·τ
Similarly, q(τ (u)) = p(u), since An has exactly two cosets in Sn . Thus, p(u) − q(u) =
q(u) − p(u), and so p(u) − q(u) = 0.
Now since δ(u) is homogeneous
of degree n(n − 1)/2, as are p, q, ui − uj | p − q
Q
(u
for all i < j implies
p
−
q
=
a
i − uj ) for some a ∈ R. But a = ±1 since in the
i<j
Q
in the expression
equation
Qp − q = σ∈Sn sgn(σ)σ(m), the coefficient on m is 1, while
2
δ(u) = i<j (ui − uj ), the coefficient on m is ±1. Thus, (p − q) = δ(u)2 = D(u).
16.3
Splitting Fields
Exercise 16.3.1. Let f be a polynomial of degree n with coefficients in F and let K
be a splitting field for f over F . Prove that [K : F ] divides n!.
Proof. We use induction on the degree of f . The claim is trivial if n = 1, so suppose
n > 1. Suppose f is reducible, and let g be an irreducible factor of f of degree
m. Then, we can choose a subfield F1 ⊂ K such that g splits completely, hence
[F1 : F ] | m! by inductive hypothesis. Similarly, [K : F1 ] | (n − m)! by inductive
hypothesis since K is a splitting field for f /g over F1 . Hence by the multiplicative
property of the degree (Thm. 15.3.5), [K : F ] | m!(n − m)! | n!.
Now suppose f is irreducible. Then, F1 := F [x]/(f ) is a field extension of degree
n such that f has a root α. Thus, [K : F1 ] | (n − 1)! by inductive hypothesis since K
is a splitting field for f /(x − α) over F1 , and so [K : F ] | n(n − 1)! = n! as above.
16.4
Isomorphisms of Field Extensions
Exercise 16.4.1.
√
√
(a) Determine all automorphisms of the field Q( 3 2), and of the field Q( 3 2, ω),
where ω = e2πi/3 .
(b) Let K be the splitting field over Q of f (x) = (x2 − 2x − 1)(x2 − 2x − 7).
Determine all automorphisms of K.
Solution for (a). We first claim that any automorphism ϕ of either field must leave
Q fixed. Any integer is held fixed by Exercise 2.6.2 and since ϕ(1) = 1. Any rational
p/q must then be fixed since multiplicative inverses are preserved through ϕ.
86
√
√
Now let ϕ ∈ Aut Q(√3 2). ϕ is√then determined by ϕ( 3 2), and it is necessary
that
√
√
3
3
3
3
j 3
ϕ( 2) = 2. Thus, ϕ( 2) = ω 2 for some j ∈ {0, 1, 2}; however, Q( 2) ⊂ R, and
ω ∈ C \ R, hence j = 0 √
and ϕ = id.
√
2) and ϕ(ω), and it is
Now let ϕ ∈ Aut
Q( 3 2, ω). ϕ is then determined√by ϕ( 3 √
√
3
3
3
3
j 3
necessary that ϕ( 2) = 2 and ϕ(ω) = 1. Thus, ϕ( 2) = ω 2 and ϕ(ω) = ω k for
some j, k ∈ {0, 1, 2}. k 6= 0 for otherwise √
ϕ is not surjective. Any of the remaining
3
choices gives an automorphism, hence Q( 2, ω) has six automorphisms.
√
√
Solution√for (b). The roots of f are 1 ± 2, 1 ± 2 2 by the quadratic formula, hence
K
√ = Q( √2), which only has the trivial automorphism and the automorphism sending
2
− 2 as in Ex. 16.4.1.
16.5
Fixed Fields
t+i
it − i
and τ (t) =
of
t−i
t+1
C(t) generate a group isomorphic to the alternating group A4 , and determine the
fixed field of this group.
Exercise 16.5.2. Show that the automorphisms σ(t) =
Solution. We first calculate products of σ, τ :
1
1
τσ = − ,
στ 2 σ = ,
t
t
t+i
t−i
it + 1
−it + 1
2
2
2
σ=
,
σ τ=
,
τσ =
, τ =
,
t−i
t+i
−it + 1
it + 1
t+1
it − i
t+1
−it + 1
σ2 =
,
τ=
, τ 2σ =
,
στ 2 =
,
−it + i
t+1
it − i
t+1
id = t,
στ = −t,
(7)
where we have used the relations σ 3 = τ 3 = στ στ = id, hence there are 12 elements
in this group G. Now consider the four pairs of 3 points defining circles in CP1 :
{{0, −i, 1}, {−1, i, ∞}},
{{0, i, 1}, {−1, −i, ∞}},
{{0, −1, i}, {−i, 1, ∞}},
{{0, −1, −i}, {1, i, ∞}}.
Labeling these 1, 2, 3, 4 respectively, we have that σ permutes these pairs as (123)
and τ does as (234), hence G is isomorphic to a subgroup of S4 . By the above, this
has order 12; by Exercise 6.7.11 we then know that G ≈ A4 .
To find the fixed field, the first row in (7) gives a subgroup of A4 isomorphic to
the Klein 4-group C2 × C2 . We then note by Thm. 16.5.2(a) that the irreducible
polynomial for t over the fixed field is the polynomial whose roots form its orbit:
(x − t)(x + t)(x − 1/t)(x + 1/t) = (x2 − t2 )(x2 − 1/t2 ) = x4 − (t2 + t−2 )x2 + 1.
87
Thus letting u = t2 + t−2 , we have that [C(t) : C(u)] ≤ 4. Note u is fixed by C2 × C2 ,
hence C(u) ⊂ C(t)C2 ×C2 , and that since by the fixed field theorem (Thm. 16.5.4),
[C(t) : C(t)C2 ×C2 ] = 4, it follows that C(u) = C(t)C2 ×C2 .
We now want to find the subfield of C(u) that is fixed by all of A4 ; by (7) since σ
and C2 × C2 generate A4 , it suffices to find the subfield of C(u) fixed by σ. We have
2 2
1
2(u − 6)
t+i
2(t4 − 6t2 + 1)
t−i
σ(u) = σ(t) +
=
=
+
=
σ(t)2
t−i
t+i
(t2 + 1)2
u+2
2 2
4
2
1
t+1
−2(u + 6)
t−1
−2(t + 6t + 1)
σ 2 (u) = σ 2 (t)2 + 2 2 = i
=
+ i
=
2
2
σ (t)
t−1
t+1
(t − 1)
u−2
2
hence
(x − u)(x − σ(u))(x − σ 2 (u)) = x3 −
4u(u2 − 36)
u(u2 − 36) 2
x
−
36x
+
.
u2 − 4
u2 − 4
2
−36)
Now letting v = u(uu2 −4
, v is fixed by σ and [C(u) : C(v)] ≤ 3, hence by the fixed
field theorem (Thm. 16.5.4) C(v) = C(u)hσi = C(t)A4 . Explicitly,
v=
16.6
(t4 + 1)(t8 − 34t4 + 1)
.
t2 (t4 − 1)2
Galois Extensions
Exercise 16.6.1. Let α be a complex root√of x3 + x + 1 over Q, and let K be a
splitting field of this polynomial over Q. Is −31 in the field Q(α)? Is it in K?
√
Solution. By the multiplicative
property of degree (Thm. 15.3.5),
−31 ∈
/ Q(α)
√
√
since α has degree 3 while −31 has degree 2 over Q. However, −31 ∈ K because
the discriminant of x3 + x + 1 is −31 by Exercise 16.2.4(a), and the square root of
the discriminant is a product of differences of elements in K.
16.7
The Main Theorem
√ √ √
Exercise 16.7.4. Let F = Q, and let K = Q( 2, 3, 5). Determine all intermediate fields.
Proof. K is the splitting field of the polynomial (x2 −2)(x2 −3)(x2 −5), and so F ⊂ K
is a Galois extension of order 8 with Galois group G =
2 × C2 = hσ2 , σ3 , σ5 i,
√ C2 × C√
− k.
where σk is the field automorphism over Q such that k
88
The lattice diagram for subgroups of C2 × C2 × C2 is given by
{e}
hσ5 i
hσ2 i
hσ2 σ3 i
hσ2 σ3 σ5 i
hσ3 σ5 i
hσ2 σ5 i
hσ3 i
hσ3 , σ5 i
hσ2 , σ5 i
hσ2 , σ3 i
hσ5 , σ2 σ3 i hσ2 , σ3 σ5 i hσ2 σ3 , σ2 σ5 i hσ3 , σ2 σ5 i
C2 × C2 × C2
which corresponds to the lattice diagram of intermediate fields
√ √ √
Q( 2, 3, 5)
√ √
√ √
√ √
√ √
√ √
√ √
√ √
Q( 2, 3) Q( 3, 5) Q( 5, 6) Q( 6, 10) Q( 2, 15) Q( 3, 10) Q( 2, 5)
√
Q( 2)
√
Q( 3)
√
Q( 5)
√
Q( 6)
√
Q( 15)
√
Q( 30)
√
Q( 10)
Q
by computing fixed fields according to Thm. 16.7.1.
Exercise 16.7.6. Let K/F be a Galois extension whose Galois group is S3 . Is K
the splitting field of an irreducible cubic polynomial over F ?
89
Solution. We claim it is. Let S2 6 S3 be the subgroup fixing 3, and consider the
chain F ⊂ K S2 ⊂ K. By Thm. 16.7.5, the first extension in this chain is not Galois
since S2 6 S3 is not normal, while the second extension is Galois with Galois group
S2 . By Cor. 16.7.2(c), [K S2 : F ] = [S3 : S2 ] = 3. Now if α ∈ K S2 \ F , let f ∈ F [x] be
its irreducible polynomial; note that deg f = 3, and that K S2 = F (α). Since K S2 /F
is not Galois, f does not split in K S2 by Thm. 16.6.4, but it does split in K by the
splitting theorem (Thm. 16.3.2). Since [K : K S2 ] = 2, there are no intermediate
fields between K S2 , K, and so K/F must be the splitting field of f .
Exercise 16.7.10. Let K/F be a Galois extension with Galois group G, and let H
be a subgroup of G. Prove that there exists an element β ∈ K whose stabilizer is
equal to H.
Proof. Recall from p. 484 that we assume K, F have characteristic zero, and that
K/F is a finite extension. So consider the chain of extensions F ⊂ K H ⊂ K;
by the primitive element theorem (Thm. 15.8.1) there exists β ∈ K H such that
K H = F (β). We claim Gβ = H. Clearly H ⊂ Gβ , so suppose σ ∈ Gβ \ H. Then,
′
H ′ := hH, σi ) H, and K H ⊂ K H by Cor. 16.7.2(a). Every element in K H is
′
fixed by H ′ , hence K H = K H . But then [K : K H ] = |H| = |H ′ | by Cor. 16.7.2(c),
contradicting that H ( H ′ .
√
√
Exercise 16.7.11. Let α = 3 2, β = 3, and γ = α + β. Let L be the field Q(α, β),
and let K be the splitting field of the polynomial (x3 − 2)(x2 − 3) over Q.
(a) Determine the irreducible polynomial f for γ over Q, and its roots in C.
(b) Determine the Galois group of K/Q.
Solution for (a). First, (γ − β)3 = γ 3 − 3βγ 2 + 9γ − 3β = 2, hence
β=
γ 3 + 9γ − 2
,
3γ 2 + 3
and so Q(γ) = Q(α, β). Moreover, since α has degree 3 and β has degree 2 over Q,
by Cor. 15.3.8 [L : Q] = 6. Thus, the irreducible polynomial for γ has degree 6. Now
letting f (x) = ((x − β)3 − 2)((x + β)3 − 2), f (γ) = 0 by construction, and
f (x) = (x3 + 9x − 2 − 3β(x2 + 1))(x3 + 9x − 2 + 3β(x2 + 1))
= (x3 + 9x − 2)2 − 27(x2 + 1)2 = x6 − 9x4 − 4x3 + 27x2 − 36x − 23.
Since deg f (x) = 6, it is then the irreducible polynomial for γ. Now z ∈ C is a root
of f (x) if and only if z ± β = ω j α where ω = e2πi/3 and j ∈ {0, 1, 2}, hence the roots
of f (x) are ±β + ω j α.
90
Solution for (b). Denote G = G(K/Q). f (x) from (a) splits completely in K by the
splitting theorem (Thm. 16.3.2); similarly, (x3 − 2)(x2 − 3) splits completely in the
splitting field for f (x) since α, β can be obtained as linear combinations of the roots
of f (x). Thus, K is the splitting field for f (x). If j ∈ {1, 2},
±β + ω j α = ±β +
−1 − (−1)j iβ
1
(−1)j
α = ±β − α −
iβ,
2
2
2
hence K = Q(γ, i) = Q(α, β, i), and so [K : Q] = [K : Q(γ)][Q(γ) : Q] = 12 by the
multiplicative property of the degree (Thm. 15.3.5).
Now let K1 be the splitting field for x3 − 2 and let K2 be the splitting field for
2
x − 3. We have the lattice diagram of intermediate fields
splits x2 − 3
K
splits x3 − 2
K1
K2
splits x3 − 2
Q
splits x2 − 3
Each field extension is Galois by Thm. 16.6.4(c), and so in particular the subgroups
G1 := G(K/K1 ) and G2 := G(K/K2 ) of G fixing K1 , K2 respectively are normal
subgroups of G by Thm. 16.7.5. The extension K/K1 is quadratic, hence G1 ≈ C2 ;
the extension K/K2 is cubic with [K : K2 ] = 6, hence G2 ≈ S3 as on p. 492.
We therefore claim G ≈ G1 × G2 ≈ C2 × S3 . Consider the multiplication map
µ : G1 × G2 → G. We first claim µ is injective. By Thm. 16.6.6(b), it suffices to show
any σ ∈ G1 ∩ G2 operates as the identity on ±β + ω j α. But ω j α ∈ K1 and ±β ∈ K2
by definition, hence G1 ∩ G2 = {1}. Now since also |G1 × G2 | = 12 = |G|, µ is also
surjective, and is therefore an isomorphism.
16.8
Cubic Equations
Exercise 16.8.4. Let K = Q(α), where α is a root of the polynomial x3 + 2x + 1,
and let g(x) = x3 + x + 1. Does g(x) have a root in K?
Solution. Let f (x) = x3 + 2x + 1. By the rational root test, f (x), g(x) are both
irreducible in Q[x]. By (16.2.8), their discriminants are −59, −31 respectively, so
since both are not squares, by Thm. 16.8.5 their splitting fields Lf , Lg have degree 6
over Q, with Galois group S3 . Now if g(x) has a root in K, then it splits completely
91
in K by the splitting theorem (Thm. 16.3.2), hence Lf = Lg since both are of degree
6 over Q. Since the Galois group is √
S3 , there should
√ be one intermediate field of
degree 2 over Q, but we have two, Q( −59) and Q( −31), a contradiction.
16.9
Quartic Equations
Exercise 16.9.6. Compute the discriminant of the quartic polynomial x4 + 1, and
determine its Galois group over Q.
Solution. By Exercise 16.2.4(c), the discriminant is 256. By Prop. 16.9.5, this implies
the Galois group G is A4 or D2 . Now the roots of x4 + 1 are
1
1
α1 = √ (1 + i), α2 = − √ (1 + i),
2
2
1
α3 = √ (1 − i),
2
1
α4 = − √ (1 − i).
2
β1 = 0, β2 = 2, β3 = −2 gives g(x) = x(x − 2)(x + 2), so G = D2 by Prop. 16.9.8.
Exercise 16.9.13. Let K be the splitting field over Q of the polynomial x4 − 2x2 − 1.
Determine the Galois group G of K/Q, find all intermediate fields, and match them
up with the subgroups of G.
Solution. We find
the roots of x4 − 2x2 − 1. We first apply the quadratic equation
√
to get x2 = 1 ± 2; then,
q
q
√
√
α1 = 1 + 2, α2 = 1 − 2, α3 = −α1 , α4 = −α2 .
β1 = 2i, β2 = −2, β3 = −2i, so g(x) = (x + 2)(x2 + 4). As in Ex. 16.9.2(a), G is a
subgroup of D4 = hσ, τ i where σ = (1234), τ = (24). By Prop. 16.9.8, G = D4 or C4 .
As in Ex. 16.9.2(a), ρ = σ 2 = (13)(24) corresponds to an automorphism
K/Q; call
√
2
N the normal subgroup of√order 2 generated by ρ. Now α1 = 1+ 2√and α1 α2 = i are
both fixed
⊂ K N . The chain of fields Q ⊂ Q( 2, i) ⊂ K N ⊂ K
√ by ρ, hence Q( 2, i)
N
has [Q( 2, i) : Q] = 4, [K : K ] = 2 by Thm. 16.5.4, and so [K : Q] = 8, so G = D4 .
The lattice diagram for subgroups of D4 is as given in Exercise 6.4.2:
{e}
{e, τ }
{e, τ σ 2 }
{e, σ 2 }
{e, τ σ}
{e, σ 2 , τ, τ σ 2 }
{e, σ, σ 2 , σ 3 }
{e, σ 2 , τ σ, τ σ 3 }
D4
92
{e, τ σ 3 }
which corresponds to the lattice diagram of intermediate fields
K
Q(α1 )
Q(α2 )
√
Q( 2, i)
Q(i)
√
Q(i 2)
Q(α1 − α2 )
Q(α1 + α2 )
√
Q( 2)
Q
by computing fixed fields according to Thm. 16.7.1.
2πi/3
Exercise 16.9.14. Let F = Q(ω),
where ω = ep
. Determine the Galois group
p
√
√
3
over F of the splitting field of (a) 2 + 2, (b) 2 + 3 2.
Remark. We note there are analogues of results from §§12.3–12.4 for checking irreducibility of polynomials in F [x]: Gauss’s lemma (Prop. 12.3.4(b)) only needs F
to be the fraction field of a UFD, and Z[ω] is a UFD by Exercise 12.2.6(a) and
Props. 12.2.7 and 12.2.14(b). Moreover, Eisensten’s criterion generalizes to Z[ω][x]
by looking at prime elements p ∈ Z[ω].
p
√
3
Solution for (a). Let α = 2 + 2, with splitting field K/F . α satisfies the polynomial f (x) = (x3 − 2)2 − 2 = p
x6 − 4x3 + 2. The quadratic equation gives that
√
√
3
x3 = 2 ± 2, hence writing α′ = 2 − 2, we can write the roots as
α1 = α, α2 = α′ ,
α3 = ωα,
α4 = ωα′ ,
α5 = ω 2 α,
α6 = ω 2 α ′ .
Thus if α1
αi , then α3
ωαi , α5
ω 2 αi . The permutations with this property
are generated by σ = (123456), τ = (246) in S6 ; these form what is called the
semidirect product C6 ⋊ C3 , and so G(K/F ) 6 C6 ⋊ C3 .
Now f is irreducible over F by Eisenstein’s criterion
using p = 2 which is prime
√
by Exercise 12.5.9(b), hence [F (α) : F ] = 6. Also, 2 ∈ F (α), hence α′ has degree 3
or 1 over F (α). Thus, [K : F ] = 6 or 18.
We claim [K : F ] = 18. Consider σ 2 = (135)(246); it is in all subgroups of C6 ⋊C3
of order 6, hence extends to an F -automorphism of K. Let N be
of
q the subgroup
√
3
2
N
2 ′
order 3 generated by σ . The fixed field K contains α α = 2(2 + 2); let L
be the field generated by this element over F . The chain F ⊂ L ⊂ K N ⊂ K has
[K : K N ] = 3 by the fixed field theorem (Thm. 16.5.4), and [L : F ] ≥ 3, hence
[K : F ] = 18, and so G(K/F ) = C6 ⋊ C3 .
93
p
√
Solution for (b). Let α = 2 + 3 2, with splitting field K/F . α satisfies the polynomial f (x) = (x2 − 2)3 − 2 = x6 − 6x4 + 12x2 − 10. The roots of f (x) are
q
q
√
√
3
3
α1 = α, α2 = 2 + ω 2, α2 = 2 + ω 2 2, α4 = −α1 , α5 = −α2 , α6 = −α3 .
Thus if αi
αj , then αi+3
αj+3 , where we treat subscripts mod 6. The group
G of permutations satisfying these is generated by the subgroup of S6 isomorphic to
S3 permuting {{1, 4}, {2, 5}, {3, 6}} isomorphic to S3 and C2 ≈ h(14)(25)(36)i 6 S6 .
Now S3 ∩ C2 = {1}, and if σ ∈ S3 , τ ∈ C2 , στ = τ σ, hence S3 , C2 are normal in G,
G ≈ S3 × C2 by Prop. 2.11.4(d), and G(K/F ) 6 G.
Now f is irreducible over F √
by Eisenstein’s criterion, again using that 2 is prime,
j 3
hence [F (α) : F ] = 6. Also, ω 2 ∈ F (α) for j ∈ {0, 1, 2}, hence α2 , α3 have degree
2 or 1 over F (α). Thus, [K : F ] ∈ {6, 12, 24}. But |G| = 12, hence [K : F ] = 24 is
impossible.
We claim [K : F ] = 12. Consider ρ = ({1, 4}{2, 5}{3, 6}) ∈ G; it is contained in
every subgroup of G of order 6, hence extends to an F -automorphism of K. Let N
√ be
N
the subgroup of order 3 generated by ρ. The fixed field K contains α1 α2 α3 = 10;
let L be the field generated by this element over F . The chain F ⊂ L ⊂ K N ⊂ K
has [K : K N ] = 3 by the fixed field theorem (Thm. 16.5.4), and [L : F ] = 2.
But [K N : L] > 1 since α1 + α2 + α3 ∈ K N \ L, hence [K : F ] = 12, and so
G(K/F ) = S3 × C2 .
16.10
Roots of Unity
Exercise 16.10.3. Let ζ = ζ7 . Determine the degree of the following elements over
Q.
(a) ζ + ζ 5 , (b) ζ 3 + ζ 4 , (c) ζ 3 + ζ 5 + ζ 6 .
Remark. Suppose we want to find the degree of α ∈ Q(ζ). By Prop. 16.10.2,
G(Q(ζ)/Q) = C6 consisting of σi given by σi (ζ) = ζ i for 1 ≤ i ≤ 6. If H 6 C6
is the subgroup fixing α, then by the main theorem (Thm. 16.7.1) Q(α) = Q(ζ)H .
Thus, the multiplicative property of degree (Thm. 15.3.5) gives us [Q(α) : Q] =
[Q(ζ) : Q]/[Q(ζ) : Q(ζ)H ] = 6/|H| by the fixed field theorem (Thm. 16.5.4).
Solution for (a). {σ1 } fixes ζ + ζ 5 , hence ζ + ζ 5 has degree 6.
Solution for (b). {σ1 , σ6 } fixes ζ 3 + ζ 4 , hence ζ 3 + ζ 4 has degree 3.
Solution for (c). {σ1 , σ2 , σ4 } fixes ζ 3 + ζ 5 + ζ 6 , hence ζ 3 + ζ 5 + ζ 6 has degree 2.
94
16.11
Kummer Extensions
Exercise 16.11.5.
(a) How does Cardano’s formula (16.11.5) express the roots of the polynomials
x3 + 3x, x3 + 2 and x3 − 3x + 2?
(b) What are the correct choices of roots in Cardano’s formula?
p
p
√
√
3
3
3
Solution for (a). For xp
+ 3x, we getp 0 + 1 + 0 − 1.
√
√
3
3
For x3 + 2 we get −1p
+ 1 + −1p
− 1.
√
√
√
√
3
3
For x3 − 3x + 2 we get −1 + 0 + −1 − 0 = 3 −1 + 3 −1.
√
√
Solution for (b). For x3 + 3x, the solutions are 0, ±i 3. To get√0, we choose 1 to
3
2πik/3
and
have
the same sign to be ±1 in both
√ terms, and then
√ choose ±1 = ±e
√
3
2πik/3
∓1 = p
∓e
for kp∈ {1, 2}. For i 3, we choose 1 to have different signs ±1, ∓1
√
√
√
√
√
3
3
so that 0 + 1 + 0 − 1 = 2 3 ±1 = ±2 3 1, and then choose 3 1√= e2πik/3 for
k = 1 if we chose ± to be +, and k = 2 if we chose ± to be −. For −i 3 we would
switch our choice for k.
p
p
√
√
3
3
3
−1
+
1
+
−1
−
1=
For
x
+
2,
we
choose
the
same
square
roots,
giving
√
3
−2, and each choice of cube root gives us a solution.
For x3 − 3x + 2, x3 − 3x + 2 = (x − 1)2 (x + 2), so 1, −2 are the solutions. −2 is
obtained by choosing −1 for both. −1 is obtained by choosing a primitive cube root
of unity α for one root, and α for the other.
16.12
Quintic Equations
Exercise 16.12.7. Find a polynomial of degree 7 over Q whose Galois group is S7 .
Solution. We first claim Sp is generated by a p-cycle and a transposition for p prime.
Let (ab · · · cd), (ef ) be our p-cycle and transposition. By renaming e = 1, we can assume they are (ab · · · 1 · · · cd) = (1 · · · cdab · · · ), (1f ). Now (1 · · · cdab · · · )k = (1f · · · )
for some 1 ≤ k ≤ p − 1 since p is prime, hence we can assume our generators are
(1f · · · ), (1f ); by renaming f = 2 and the rest of the elements in (1f · · · ) accordingly,
we can assume our generators are α = (12 · · · p), β = (12). Now
α−1 βα = (1p), α−1 (1p)α = ((p − 1)p),
α−1 ((p − 1)p)α = ((p − 2)(p − 1)),
···
so we can generate all permutations of the form ((k − 1)k). These generate Sp .
Now let f (x) = (x3 −2)(x2 −4)(x2 −32)+2 = x7 −36x5 −2x4 +128x3 +72x2 −254.
This is irreducible by the Eisenstein criterion; also it has 5 real roots and 2 complex
roots by looking at its graph. The only permutations of its roots that fix the real
95
roots is conjugation, which corresponds to a transposition in G; moreover, G operates
transitively on the roots, hence contains a 7-cycle, and the Galois group is therefore
S7 by the above.
16.M
Miscellaneous Problems
Exercise 16.M.4.
(a) The non-negative real numbers are those having a real square root. Use this
fact to prove that the field R has no automorphism except the identity.
(b) Prove that C has no continuous automorphisms other than complex conjugation and the identity.
Proof of (a). Any automorphism ϕ is clearly the identity on the integers and rationals, since ϕ(n) = ϕ(1+· · ·+1) = ϕ(1)+· · ·+ϕ(1) = n, and then qϕ(p/q) = ϕ(p) = p.
Let a be such that ϕ(a) = b 6= 0. Then b < a after possibly multiplying by −1.
Subtracting a suitable rational from each, we get b < 0 < a. But then a = α2 , so
ϕ(α)2 = ϕ(α2 ) = ϕ(a) = b, contradicting b < 0.
Proof of (b). By (a), any automorphism ϕ is the identity on Q. Now letting i be
such that i2 = −1, we see ϕ(i)2 = −1, but since only ±i square to get −1 in C, on
Q[i] we have that ϕ is either the identity or complex conjugation.
Finally, Q[i] is dense in C, hence any automorphism of C must also be either the
identity or complex conjugation by continuity.
Exercise 16.M.7. A polynomial f in F [u1 , . . . , un ] is 21 -symmetric if f (uσ1 , . . . , uσn )
= f (u1 , . . . , un ) for every even permutation σ, and skew-symmetric if f (uσ1 , . . . , uσn )
= (sign σ)f (u1 , . . . , un ) for every permutation σ.
Q
(a) Prove that the square root of the discriminant δ := i<j (xi − xj ) is skewsymmetric.
(b) Prove that every 12 -symmetric polynomial has the form f + gδ, where f , g are
symmetric polynomials.
Proof of (a). It suffices to show that for any permutation of two elements contributes
a negative sign, for any permutation is generated by permutations of two elements,
and the number of permutation of two elements needed to generate a given permutation is equal to its sign.
So suppose we are interchanging ui , uj ; assume without loss of generality that
i < j. Then the terms in δ involving ui , uj have the product
(ui −uj )
i−1
Y
k=1
(uk −ui )
j−1
Y
k=1
(uk −uj )
j−1
Y
k=i+1
(uk −uj )
96
j−1
Y
k=i+1
(ui −uk )
n
Y
k=j+1
(ui −uk )
n
Y
k=j+1
(uj −uk )
Now interchanging ui , uj causes a sign change in each factor; the sign change in
ui − uj is the only one that remains since each two adjacent factors in the rest of the
product have canceling sign changes. Hence δ also changes sign when interchanging
ui and uj , i.e., δ is skew-symmetric.
Proof of (b). Let h be our 21 -symmetric polynomial. Assume that char F 6= 2. If h is
symmetric, we are done by letting f = h, g = 0 so suppose not. The action of S = Sn
on h has orbit {h, h′ } for some other polynomial h′ , since by the orbit-stabilizer
theorem |Sh ||Sh| = |S|, but Sh = An so |Sh| = 2. This implies f = 12 (h + h′ ) is
symmetric, for h, h′ will just be interchanged. g̃ = h − f = 21 (h − h′ ) is antisymmetric
for the same reason.
It now suffices to show that any antisymmetric polynomial g̃ is divisible by δ,
i.e., any binomial ui − uj divides it. For, if ϕ is the substitution map letting uj = ui ,
then ϕ(g̃) = −ϕ(g̃) implies ϕ(g̃) = 0. So letting g = g̃/δ suffices, since g cannot be
anti-symmetric otherwise g̃ would be symmetric by (a).
Exercise 16.M.10. Let K be a finite extension of a field F , and let f (x) be in K[x].
Prove that there is a nonzero g(x) in K[x] such that the product f (x)g(x) is in F [x].
Proof. We can assume f (x) is irreducible by working with irreducible factors individually; suppose moreover that it is monic. Any root α of f (x) is algebraic over K
hence algebraic over F by Exercise 15.10.1. Thus we have h(x) ∈ F [x] the minimal
polynomial for α over F . Then f | h since f is defined over K; thus h = f g for some
g ∈ K[x].
97
List of Solved Exercises
2
6
Groups
2.1 Laws of Composition . . . . .
Exercise 2.1.2 . . . . . . . . .
Exercise 2.1.3 . . . . . . . . .
2.2 Groups and Subgroups . . . .
Exercise 2.2.3 . . . . . . . . .
Exercise 2.2.4 . . . . . . . . .
2.4 Cyclic Groups . . . . . . . . .
Exercise 2.4.1 . . . . . . . . .
Exercise 2.4.3 . . . . . . . . .
Exercise 2.4.6 . . . . . . . . .
Exercise 2.4.9 . . . . . . . . .
Exercise 2.4.10 . . . . . . . .
2.6 Isomorphisms . . . . . . . . .
Exercise 2.6.2 . . . . . . . . .
Exercise 2.6.3 . . . . . . . . .
Exercise 2.6.6 . . . . . . . . .
2.8 Cosets . . . . . . . . . . . . .
Exercise 2.8.4 . . . . . . . . .
Exercise 2.8.8 . . . . . . . . .
Exercise 2.8.10 . . . . . . . .
2.10 The Correspondence Theorem
Exercise 2.10.3 . . . . . . . .
2.11 Product Groups . . . . . . . .
Exercise 2.11.1 . . . . . . . .
Exercise 2.11.3 . . . . . . . .
2.12 Quotient Groups . . . . . . .
Exercise 2.12.2 . . . . . . . .
Exercise 2.12.4 . . . . . . . .
Exercise 2.12.5 . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Symmetry
6.3 Isometries of the Plane . . . . .
Exercise 6.3.2 . . . . . . . . . .
Exercise 6.3.6 . . . . . . . . . .
6.4 Finite Groups of Orthogonal
Operators on the Plane . . . . .
Exercise 6.4.2 . . . . . . . . . .
Exercise 6.4.3 . . . . . . . . . .
6.5 Discrete Groups of Isometries .
Exercise 6.5.5 . . . . . . . . . .
Exercise 6.5.9 . . . . . . . . . .
6.6 Plane Crystallographic Groups
Exercise 6.6.2 . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
4
6.7 Abstract Symmetry: Group Operations . . . . . . . . . . . . .
4
Exercise 6.7.1 . . . . . . . . . .
4
Exercise 6.7.2 . . . . . . . . . .
4
Exercise 6.7.8 . . . . . . . . . .
4
Exercise 6.7.11 . . . . . . . . .
4
6.8 The Operation on Cosets . . . .
5
Exercise 6.8.4 . . . . . . . . . .
5
6.9 The Counting Formula . . . . .
5
Exercise 6.9.4 . . . . . . . . . .
6
6.10 Operations on Subsets . . . . .
6
Exercise 6.10.1 . . . . . . . . .
6
6.11 Permutation Representations .
6
Exercise 6.11.1 . . . . . . . . .
7
Exercise 6.11.5 . . . . . . . . .
7
Exercise 6.11.6 . . . . . . . . .
7
6.12 Finite Subgroups of the Rota8
tion Group . . . . . . . . . . . .
8
Exercise 6.12.3 . . . . . . . . .
8
Exercise 6.12.7 . . . . . . . . .
9
6.M
Miscellaneous Problems . . . .
9
Exercise
6.M.7 . . . . . . . . . .
9
9
9 11 Rings
11.1 Definition of a Ring . . . . . . .
9
Exercise 11.1.1 . . . . . . . . .
10
Exercise 11.1.2 . . . . . . . . .
10
Exercise 11.1.3 . . . . . . . . .
10
11.2 Polynomial Rings . . . . . . . .
11
Exercise 11.2.1 . . . . . . . . .
11
Exercise 11.2.2 . . . . . . . . .
11.3 Homomorphisms and Ideals . .
13
Exercise 11.3.3 . . . . . . . . .
13
Exercise 11.3.5 . . . . . . . . .
13
Exercise 11.3.7 . . . . . . . . .
13
Exercise 11.3.8 . . . . . . . . .
Exercise 11.3.9 . . . . . . . . .
14
Exercise 11.3.10 . . . . . . . . .
14
Exercise 11.3.11 . . . . . . . . .
15
Exercise 11.3.12 . . . . . . . . .
16
Exercise 11.3.13 . . . . . . . . .
16
11.4 Quotient Rings . . . . . . . . .
16
Exercise 11.4.2 . . . . . . . . .
17
Exercise 11.4.3 . . . . . . . . .
17
98
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
17
17
17
17
18
18
18
19
19
19
19
20
20
20
21
.
.
.
.
.
21
21
22
23
23
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
24
24
24
24
25
25
25
26
27
27
29
29
30
30
31
31
31
31
32
32
32
Exercise 11.4.4 . . . . .
11.5 Adjoining Elements . . .
Exercise 11.5.1 . . . . .
Exercise 11.5.3 . . . . .
Exercise 11.5.4 . . . . .
Exercise 11.5.6 . . . . .
11.6 Product Rings . . . . . .
Exercise 11.6.1 . . . . .
Exercise 11.6.2 . . . . .
Exercise 11.6.3 . . . . .
Exercise 11.6.4 . . . . .
Exercise 11.6.5 . . . . .
Exercise 11.6.6 . . . . .
Exercise 11.6.7 . . . . .
Exercise 11.6.8 . . . . .
11.7 Fractions . . . . . . . . .
Exercise 11.7.1 . . . . .
Exercise 11.7.2 . . . . .
Exercise 11.7.3 . . . . .
Exercise 11.7.4 . . . . .
11.8 Maximal Ideals . . . . .
Exercise 11.8.1 . . . . .
Exercise 11.8.2 . . . . .
Exercise 11.8.3 . . . . .
Exercise 11.8.4 . . . . .
11.9 Algebraic Geometry . .
Exercise 11.9.4 . . . . .
Exercise 11.9.5 . . . . .
Exercise 11.9.6 . . . . .
Exercise 11.9.11 . . . . .
Exercise 11.9.12 . . . . .
Exercise 11.9.13 . . . . .
11.M Miscellaneous Problems
Exercise 11.M.3 . . . . .
Exercise 11.M.4 . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
33
33
33
33
34
34
35
35
35
35
35
36
36
36
37
37
37
38
38
38
38
38
39
39
39
40
40
40
40
41
41
42
42
42
43
12 Factoring
12.1 Factoring Integers . . . . . . .
Exercise 12.1.4 . . . . . . . .
Exercise 12.1.5 . . . . . . . .
12.2 Unique Factorization Domains
Exercise 12.2.1 . . . . . . . .
Exercise 12.2.2 . . . . . . . .
Exercise 12.2.5 . . . . . . . .
Exercise 12.2.6 . . . . . . . .
Exercise 12.2.9 . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
45
45
45
45
46
46
46
46
48
49
12.3 Gauss’s Lemma . . . . . . . .
Exercise 12.3.1 . . . . . . . .
Exercise 12.3.2 . . . . . . . .
Exercise 12.3.4 . . . . . . . .
Exercise 12.3.6 . . . . . . . .
12.4 Factoring Integer Polynomials
Exercise 12.4.4 . . . . . . . .
Exercise 12.4.6 . . . . . . . .
Exercise 12.4.7 . . . . . . . .
Exercise 12.4.12 . . . . . . . .
Exercise 12.4.13 . . . . . . . .
Exercise 12.4.15 . . . . . . . .
Exercise 12.4.19 . . . . . . . .
12.5 Gauss Primes . . . . . . . . .
Exercise 12.5.2 . . . . . . . .
Exercise 12.5.3 . . . . . . . .
Exercise 12.5.5 . . . . . . . .
Exercise 12.5.6 . . . . . . . .
Exercise 12.5.7 . . . . . . . .
Exercise 12.5.9 . . . . . . . .
Exercise 12.5.10 . . . . . . . .
12.M Miscellaneous Problems . . .
Exercise 12.M.5 . . . . . . . .
Exercise 12.M.6 . . . . . . . .
Exercise 12.M.7 . . . . . . . .
Exercise 12.M.8 . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
50
50
50
50
51
51
51
52
52
52
54
55
55
56
56
56
56
57
57
58
58
59
59
59
60
61
13 Quadratic Number Fields
13.1 Algebraic Integers . . . . . .
Exercise 13.1.4 . . . . . . .
13.2 Factoring Algebraic Integers
Exercise 13.2.2
. . . . . . .
√
13.3 Ideals in Z[ −5] . . . . . .
Exercise 13.3.2 . . . . . . .
Exercise 13.3.3 . . . . . . .
13.4 Ideal Multiplication . . . . .
Exercise 13.4.3 . . . . . . .
13.5 Factoring Ideals . . . . . . .
Exercise 13.5.2 . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
61
61
61
62
62
62
62
63
63
63
63
63
14 Linear Algebra in a
14.1 Modules . . . .
Exercise 14.1.3
Exercise 14.1.4
14.2 Free Modules .
Exercise 14.2.3
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
64
64
64
64
65
65
99
Ring
. . . .
. . . .
. . . .
. . . .
. . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
14.4 Diagonalizing Integer Matrices . .
Exercise 14.4.6 . . . . . . . . . .
14.5 Generators and Relations . . . .
Exercise 14.5.1 . . . . . . . . . .
14.7 Structure of Abelian Groups . . .
Exercise 14.7.7 . . . . . . . . . .
14.8 Applications to Linear Operators
Exercise 14.8.2 . . . . . . . . . .
Exercise 14.8.4 . . . . . . . . . .
14.M Miscellaneous Problems . . . . .
Exercise 14.M.10 . . . . . . . . .
66
66
66
66
67
67
67
67
68
69
69
15 Fields
15.2 Algebraic and Transcendental
Elements . . . . . . . . . . . . . .
Exercise 15.2.1 . . . . . . . . . .
Exercise 15.2.3 . . . . . . . . . .
15.3 The Degree of a Field Extension
Exercise 15.3.2 . . . . . . . . . .
Exercise 15.3.5 . . . . . . . . . .
Exercise 15.3.7 . . . . . . . . . .
Exercise 15.3.9 . . . . . . . . . .
Exercise 15.3.10 . . . . . . . . . .
15.4 Finding the Irreducible Polynomial . . . . . . . . . . . . . . . .
Exercise 15.4.1 . . . . . . . . . .
Exercise 15.4.2 . . . . . . . . . .
15.6 Adjoining Roots . . . . . . . . . .
Exercise 15.6.1 . . . . . . . . . .
15.7 Finite Fields . . . . . . . . . . . .
Exercise 15.7.4 . . . . . . . . . .
Exercise 15.7.6 . . . . . . . . . .
Exercise 15.7.7 . . . . . . . . . .
Exercise 15.7.8 . . . . . . . . . .
Exercise 15.7.10 . . . . . . . . . .
Exercise 15.7.11 . . . . . . . . . .
Exercise 15.7.13 . . . . . . . . . .
15.8 Primitive Elements . . . . . . . .
Exercise 15.8.2 . . . . . . . . . .
15.9 Function Fields . . . . . . . . . .
Exercise 15.9.1 . . . . . . . . . .
15.10 The Fundamental Theorem of
Algebra . . . . . . . . . . . . . .
Exercise 15.10.1 . . . . . . . . . .
Exercise 15.10.2 . . . . . . . . . .
70
70
70
70
71
71
71
71
72
72
72
72
72
73
73
74
74
74
75
75
75
76
76
76
76
77
77
77
77
78
15.M Miscellaneous Problems
Exercise 15.M.1 . . . . .
Exercise 15.M.2 . . . . .
Exercise 15.M.3 . . . . .
Exercise 15.M.4 . . . . .
Exercise 15.M.6 . . . . .
Exercise 15.M.7 . . . . .
.
.
.
.
.
.
.
78
78
78
78
79
80
81
16 Galois Theory
16.1 Symmetric Functions . . . . . . .
Exercise 16.1.1 . . . . . . . . . .
Exercise 16.1.3 . . . . . . . . . .
16.2 The Discriminant . . . . . . . . .
Exercise 16.2.2 . . . . . . . . . .
Exercise 16.2.4 . . . . . . . . . .
Exercise 16.2.7 . . . . . . . . . .
16.3 Splitting Fields . . . . . . . . . .
Exercise 16.3.1 . . . . . . . . . .
16.4 Isomorphisms of Field Extensions
Exercise 16.4.1 . . . . . . . . . .
16.5 Fixed Fields . . . . . . . . . . . .
Exercise 16.5.2 . . . . . . . . . .
16.6 Galois Extensions . . . . . . . . .
Exercise 16.6.1 . . . . . . . . . .
16.7 The Main Theorem . . . . . . . .
Exercise 16.7.4 . . . . . . . . . .
Exercise 16.7.6 . . . . . . . . . .
Exercise 16.7.10 . . . . . . . . . .
Exercise 16.7.11 . . . . . . . . . .
16.8 Cubic Equations . . . . . . . . .
Exercise 16.8.4 . . . . . . . . . .
16.9 Quartic Equations . . . . . . . .
Exercise 16.9.6 . . . . . . . . . .
Exercise 16.9.13 . . . . . . . . . .
Exercise 16.9.14 . . . . . . . . . .
16.10 Roots of Unity . . . . . . . . . .
Exercise 16.10.3 . . . . . . . . . .
16.11 Kummer Extensions . . . . . . .
Exercise 16.11.5 . . . . . . . . . .
16.12 Quintic Equations . . . . . . . . .
Exercise 16.12.7 . . . . . . . . . .
16.M Miscellaneous Problems . . . . .
Exercise 16.M.4 . . . . . . . . . .
Exercise 16.M.7 . . . . . . . . . .
Exercise 16.M.10 . . . . . . . . .
81
81
81
82
83
83
84
85
86
86
86
86
87
87
88
88
88
88
89
90
90
91
91
92
92
92
93
94
94
95
95
95
95
96
96
96
97
100
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.