Algebra Qualifying Exam Solutions
Thomas Goller
September 4, 2011
Contents
1 Spring 2011
2
2 Fall 2010
8
3 Spring 2010
13
4 Fall 2009
17
5 Spring 2009
21
6 Fall 2008
25
1
Chapter 1
Spring 2011
1.
The claim as stated is false. The identity element is a problem, since
1 = (12)(12) is a product of commuting 2-cycles, but has order 1. The statement
is correct if we exclude the identity:
Proposition 1. Let p be prime. An element 1 6= σ ∈ Sn has order p if and
only if σ is the product of (at least one) commuting p-cycles.
Proof. Suppose σ has order p. Let σ ∈ Sn have (disjoint) cyclic decomposition
σ = C1 . . . Cr . The order of σ is the least common multiple of the sizes of the
Ci , namely
p = |σ| = lcm{|Ci |},
whence |Ci | = p for each i. Thus r ≥ 1 and the Ci are disjoint, hence they
commute.
Conversely, suppose σ = C1 . . . Cr is a decomposition into (not necessarily
disjoint) commuting p-cycles, with r ≥ 1. Then by commutativity,
σ p = C1p . . . Crp = 1,
so σ has order dividing p. Since σ 6= 1, this proves that σ has order p.
If p is not prime, then the theorem fails. For instance, (12)(345) ∈ S5 has
order lcm{2, 3} = 6, but is not the product of commuting 6-cycles.
2. First we compute the order of G = GL2 (Fp ). For the first row of the
matrix, there are p2 − 1 possibilities, since we cannot have both entries be 0.
For the second row, there are p2 − p possibilities, since anything but a multiple
of the first row ensures a nonzero determinant. Thus
|GL2 (Fp )| = (p2 − 1)(p2 − p) = p(p + 1)(p − 1)2 .
By Sylow’s theorem, the number np of Sylow-p subgroups of G satisfies
np |(p + 1)(p − 1)2
and np ≡ 1
2
(mod p).
So we have np ∈ {1, p + 1,
(p − 1)2 , (p + 1)(p − 1)2 }.
1 1
Consider the element
, which is of order p and generates a Sylow
0 1
p-subgroup P . Its transpose is also order p, so we can rule out np = 1. Now,
note that if a, d 6= 0, then
1 a b
1 1
d −b
1 ad−1
=
,
0 1
0 a
0
1
ad 0 d
a b
whence every such
is in the normalizer NP (G). There are p(p − 1)2
0 d
such elements, so since np = (G : NP (G)) by Sylow’s theorem, we obtain the
bound np ≤ p + 1. Thus np = p + 1.
3. (a) We have |G| = pk for k ≥ 2. Let G act on itself by conjugation,
with g1 , . . . , gr distinct representatives of the non-central conjugacy classes of
G. The class equation gives
pk = |G| = |Z(G)| +
r
X
(G : CG (gi )).
i=1
Since |G| = |CG (gi )| · (G : CG (gi )), each summand is divisible by p, so |Z(G)|
must be divisible by p and therefore nontrivial.
(b) Now, we prove by induction on k that G has normal subgroup of order
pb for each 1 ≤ j ≤ k − 1 (the cases j = 0, k are trivial). Since |Z(G)| is divisible
by p, |Z(G)| contains an element of order p by Cauchy’s theorem, whence Z(G)
contains an (abelian) subgroup Z or order p, which is normal in G since Z
is contained in the center of G. Then G0 := G/Z has order pk−1 , whence G0
has normal subgroups H̄1 , . . . , H̄k−2 of orders p1 , . . . , pk−2 , plus the identity
subgroup H̄0 := 1.
π /
G0 denote the natural surjection, we claim that the subLetting G
−1
groups Hj := π (H̄j ) for 0 ≤ j ≤ k − 2 are of order pj+1 and normal in G. The
order statement is clear, since cosets of Z in G have cardinality p. Now for any
x ∈ Hj and g ∈ G, we have ḡx̄ḡ −1 ∈ H̄j , so that since π is a homomorphism,
gxg −1 ∈ H̄j . Thus gxg −1 ∈ Hj , so Hj is normal in G.
4. We have

1 0
A = 0 0
0 1

0
−2
3
and wish to compute the rational and Jordan canonical forms (over Q and C,
respectively). The characteristic polynomial is f = (x − 1)2 (x − 2) and the
minimal polynomial is p = (x − 1)(x − 2), which is proved by showing that
3
p(A) = 0. Thus the invariant factors are p1 = p = (x − 1)(x − 2) = x2 − 3x + 2
and p2 = x − 1 (we must have p1 p2 = f ), so the rational form is


0 −2 0
A ∼ 1 3 0
0 0 1
To confirm this, we note that
   
   
1
1
0
0
A 0 = 0 ,
A 1 = 0 ,
0
0
0
1
   
0
0
A2 1 = −2 ,
0
3
so that a basis that yields the rational form is
     
0
1 
 0
1 , 0 , 0


0
1
0
The associated change of basis matrix (expressing the new basis in terms of the
old basis – just load in the vectors of the basis as the columns) is




0 1 0
0 0 1
with P −1 = 0 0 1 ,
P = 1 0 0 ,
1 0 0
0 1 0
whence

0
P −1 AP = 1
0
−2
3
0

0
0 .
1
(Since Bold vold = Bnew vnew and Bnew = Bold P , we see that vold = P vnew , namely
P takes the old basis to the new basis, but takes vectors with coordinates in the
new basis to vectors with coordinates in the old basis. The new matrix should
act on vectors expressed in the new basis. So take such a vector, convert to the
old basis using P , then apply A, then use P −1 to take the result back to the
new basis.)
Now for the Jordan form. We have two Jordan blocks, one of size 2 associated
with the eigenvalue 1 and the other of size 1 for the eigenvalue 2. Since the
minimal polynomial has linear powers, each block is diagonalizable, so the result
is


1 0 0
A ∼ 0 1 0 .
0 0 2
4
5. First, Z[i] is an integral domain since by definition
(a + bi)(c + di) = ac − bd + (ad + bc)i
is 0 if and only if a = b = 0 or c = d = 0.
N /
We need to prove that there is a norm Z[i]
Z≥0 such that for any
α = a + bi, β = c + di ∈ Z[i], where β =
6 0, there are p + qi, γ ∈ Z[i] such that
α = (p + qi)β + γ,
with γ = 0 or N (γ) < N (β).
In Q[i], we have α
β = r + si, with r, s ∈ Q. Let p be an integer closest to r
and q an integer closest to s, so that |r − p| and |s − q| are ≤ 21 . Then setting
θ = (r − p) + (s − q)i and γ = βθ, we have α = (p + qi)β + γ, so that γ ∈ Z[i].
Moreover, using the norm of Q[i] that is the natural extension of the norm of
Z[i], we have
N (β)
N (γ) = N (β)N (θ) ≤
,
2
whence γ 6= 0 implies N (γ) < N (β) since N (γ) is an integer.
6. Skip.
7. Z/mZ ⊗Z Z/nZ is cyclic since
a ⊗ b = b(a ⊗ 1) = ab(1 ⊗ 1),
with 1⊗1 as a generator. Moreover, there exist integers a, b such that am+bn =
d, so
d(1 ⊗ 1) = (am + bn)(1 ⊗ 1) = (am ⊗ bn) = 0,
whence the order of the cyclic group divides d. Now consider the map Z/mZ × Z/nZ
defined by
(a mod m, b mod n) 7→ ab mod d.
ϕ0
/ Z/dZ maps
It is Z-bilinear, and the induced linear map Z/mZ ⊗Z Z/nZ
1 ⊗ 1 7→ 1̄, an element of order d. Thus 1 ⊗ 1 has order a multiple of d, whence
the cyclic group has order at least d, and therefore exactly d. So ϕ0 is an isomorphism.
8. Skip.
9. (a) First, note that x4 − 5√is irreducible over Q by Eisenstein. Letting α
denote the positive real number 4 5, the roots of x4 − 5 in C are
√
√
√
√
4
4
4
4
5, ζ 5, ζ 2 5, ζ 3 5,
5
ϕ
/ Z/dZ
where ζ = i is a primitive
fourth root of unity. Thus the splitting field of E of
√
x4 − 5 is E = Q( 4 5, i), which has degree 8 over Q.
Since i is a root of the irreducible polynomial x2 + 1, the 8 elements of
G = Gal(E/Q) are determined by
√
√
4
4
5 7→ ik 5,
i 7→ ±i,
all of which are automorphisms of E. Letting
√
√
4
4
σ : 5 7→ i 5,
i 7→ i
and
τ:
√
4
5 7→
2
√
4
5,
i 7→ −i,
3
we see that G = hσ, τ i = {1, σ, σ , σ , τ, στ, σ 2 τ, σ 3 τ }, and that σ, τ satisfy the
relations σ 4 = τ 2 = 1 and τ σ = σ 3 τ . Thus G ' D4 .
The inverted lattice of subgroups of D4 is (DF 69)
1
hσ 2 τ i
hσ 2 i
hστ i
hσ 2 , τ i
hσi
hσ 2 , στ i
hτ i
hσ 3 τ i
hσ, τ i
The corresponding lattice of fixed fields is
√
Q( 4 5, i)
√
Q( 4 5)
√
Q(i 4 5)
√
Q( 5, i)
√
Q((1 + i) 4 5)
√
Q( 5)
Q(i)
√
Q(i 5)
√
Q((1 − i) 4 5)
Q
√
√
(b) The Galois group of Q( 4 5, i) over Q( 5) is hσ 2 , τ i, which is isomorphic
to Z2 × Z2 , the Klein Viergruppe.
6
√
(c) The Galois group of Q( 4 5) over Q(i) is hσi, isomorphic to Z4 .
√
√
10. The extension
is Galois of degree 2,√with basis {1, 2} over
√ Q( 2)/Q
√
Q. We claim that √
3 ∈
/ Q( 2). For suppose (a + b 2)2 = 3 with a, b ∈ Q.
2
2
Then a + 2b + 2ab 2 =√3, whence a or b is 0, but √
3 is √
not the√square of a
3)/Q
is
of
degree
2,
so
is
Q(
2,
3)/Q( 2), whence
rational
number.
Since
Q(
√ √
√ √ √
Q( 2, 3)/Q
is
of
degree
4,
with
basis
{1,
2,
3,
6}.
√
√
√ √
Since Q( 2, 3) is the smallest extension of Q containing ± 2, ± 3, it is
the splitting field of the separable polynomial (x2 − 2)(x2 − 3), hence a Galois
extension of Q.
√ √
Automorphisms of Q( 2, 3) fixing Q must map
√
√
√
√
2 7→ ± 2,
3 7→ ± 3.
Since there are four such maps, and the degree of the extension is 4, each of
these maps is an automorphism. Let
√
√
√
√
3 7→ 3
σ : 2 7→ − 2,
√
√
√
√
τ : 2 7→ 2,
3 7→ − 3.
√ √
Then G = Gal(Q( 2, 3)/Q) = hσ, τ i. Since σ 2 = τ 2 = 1 and στ = τ σ, G is
isomorphic to Z2 × Z2 .
11. Recall that for polynomials f, g ∈ F[x, y], where F is a field, we have
deg(f g) = deg f + deg g. Suppose x2 + y 2 − 1 is reducible, factoring into the
non-units f, g. Then f, g must each be of degree 1, so we get an expression
(ax + by + c)(a0 x + b0 y + c0 ) = x2 + y 2 − 1.
This implies aa0 = bb0 = −cc0 = 1, so all the coefficients are nonzero. Moreover,
we deduce ab0 = −a0 b, ac0 = −a0 c, and bc0 = −b0 c, whence b0 = −a0 b/a and
−a0 c/a = c0 = −b0 c/b. But the latter equalities imply a0 /a = b0 /b, contradicting
the former equality. So there is no such factorization, in either Q[x, y] or C[x, y].
7
Chapter 2
Fall 2010
1. We begin by computing the order of G := SL2 (F5 ). The order of H :=
GL2 (F5 ) is (52 −1)(52 −5), since the first row of a matrix can be anything except
0, and then the second row can be anything except a linear multiple of the first.
To modify this counting system for G, note simply that once the second row is
chosen for an element of H, there are 4 multiples of that row that yield elements
of H, but only one that gives an element of G. Thus
|G| =
(52 − 1)(52 − 5)
= 4 · 5 · 6 = 120.
4
By Sylow’s theorem, the number n5 of Sylow-5 subgroups of G satisfies
n5 |24
and n5 ≡ 1
(mod 5).
So n5 = 1 or n5 = 6. But
1
0
1
1
and
1
1
0
1
are elements of degree 5 that generate distinct subgroups of order 5, so n5 = 6.
2. The derived subgroup G(1) of G consists of the upper-triangular matrices
of the form
1 a
,
a ∈ R.
0 1
A simple computation shows that every commutator is of this form, and we
have
1 0
1 a
1 a
1 −a
1 0
1 −a
1 a
=
=
,
0 1
0 2
0 2
0 1
0 1
0 12
0 21
and elements of this form do not generate any new elements. Note that G(1) is
abelian, so that G(2) , the subgroup of commutators of G(1) , is trivial. Thus the
8
derived series
1 = G(2) / G(1) / G(0) = G
shows that G is solvable.
3. We compute σ 2 = τ 2 = 1, that στ 6= τ σ, and that (στ )3 = 1, whence
τ σ = (στ )2 . Set a = στ and b = σ. Then the group contains the six elements
{1, a, a2 , b, ba, b2 a}
and is non-abelian, so it is isomorphic to D3 ' S3 .
One way of proving the isomorphism is to consider how σ, τ act on the x
values {2, 21 , −1}. We see that σ swaps the first two, but leaves the third fixed,
while τ swaps the first and third, leaving the second fixed. Since the 2-cycles
(12) and (13) generate S3 , we get a surjective homomorphism hσ, τ i → S3 . Since
the group contains six elements, the map is injective.
5. Hermitian matrices are normal, hence diagonalizable by unitary matrices, and all their eigenvalues are real. In other words, if A is 5 × 5 Hermitian,
then there exists a 5 × 5 unitary matrix U such that U −1 AU is diagonal with
real entries. Conversely, every real diagonal matrix is Hermitian, and real diagonal matrices with different multiplicities of eigenvalues are not conjugate (by
the uniqueness of the Jordan form up to reordering of Jordan blocks). Since A
satisfies A5 + 2A3 + 3A − 6I = 0 if and only if U −1 AU does as well, it suffices
to classify real diagonal matrices A with eigenvalues in increasing order that
satisfy A5 + 2A3 + 3A − 6I = 0. Since A is diagonal, each entry on the diagonal
must be a zero of the polynomial f (x) = x5 + 2x3 + 3x − 6. One such zero is
x = 1, but we have f 0 (x) = 5x4 + 6x2 + 3, which is strictly positive in the real
numbers. Thus f is monotonic increasing on the reals, so the only real zero of
f is x = 1. Thus the only possibility for A is I, so the unique conjugacy class is
{I}.
6. We wish to determine the number of conjugacy classes of 4 × 4 complex
matrices A satisfying A3 − 2A2 + A = 0. This means the minimum polynomial
p of A must divide x(x − 1)2 , so the possibilities for p are x, x − 1, x(x − 1),
(x − 1)2 , and x(x − 1)2 . Since p divides the characteristic polynomial, the
nine possible Jordan forms (up to reordering of blocks), which correspond to
conjugacy classes, are:
9
Minimal Polynomial
Characteristic Polynomials
x
x4
x−1
(x − 1)4
(x − 1)2
x(x − 1)
(x − 1)4
x3 (x − 1),

1
1



Jordan Forms


0

 0






0
0


1

 1






1
1
 
1
 1 1
1
 
,
1
1  
1
1


0
 0







0
1
x2 (x − 1)2

0





0




1
1
x(x − 1)3

0





1




1
1
x(x − 1)2
x2 (x − 1)2 ,
x(x − 1)3

0





0





0
1
1




1

1
1




1
1
7. Note that α = 71/6 . The field extensions Q(α3 )/Q and Q(α2 )/Q have
degrees 2 and 3 respectively (x2 − 7 and x3 − 7 are irreducible over Q since they
are of degree ≤ 3 and have no roots in Q), so both 2 and 3 divide the degree
of the extensions Q(α)/Q, whence this extension must have degree 6. Thus
1, α, . . . , α5 are linearly independent over Q, hence also over Z, and x6 − 7 is
10





1
the minimal polynomial of α (or use Eisenstein on x6 − 7 to conclude that it is
irreducible).
It follows that Q(α) ' Q[x]/(x6 − 7), using the map α 7→ x̄, so that also
Z[α] ' Z[x]/(x6 − 7) as subrings. Moreover,
Z[α]/(α2 ) ' (Z[x]/(x6 − 7))/(x̄2 ) ' Z[x]/(x2 , 7) ' (Z/7Z)[x]/(x2 ),
so this ring consists of
Z[α]/(α2 ) = {a + bα : a, b ∈ Z/7Z},
which consists of 7 · 7 = 49 elements.
The elements of the form a + 0α, a 6= 0 are units. Thus every ideal of
Z[α]/(α2 ) is principal since if it is proper and non-zero, then it must contain
0 + bα for some b 6= 0, whence it contains b−1 · bα = α and is the ideal (α). Note
that this ring is not a PID since it is not an integral domain, due to the fact
that α2 = 0, so α is a zero divisor.
8. (i) Let α denote the positive real number α = 31/6 . By the argument
in the previous problem (or Eisenstein), Q(α)/Q is an extension of degree 6.
Over C, the polynomial x6 − 3 factors as
x6 − 3 = (x3 − 31/2 )(x3 + 31/2 ) = (x − α)(x2 + αx + α2 )(x + α)(x2 − αx + α2 ),
so that using the quadratic formula, we see that the roots are
√
±α ± α −3
±α,
.
2
√
√
Thus √
the splitting field of x6 −3 is Q( −3, α). Since Q(α)
/ Q(α), so
√ ⊆ R, −3 ∈
Q(α, −3)/Q(α) is an extension
of
degree
2.
Thus
Q(α,
−3)/Q
is
an
extension
√
√
√
of degree 12, so since Q( −3)/Q is degree 2, it follows that Q(α, −3)/Q( −3)
is degree 6.
√
Actually, a much easier method is to note that the element −3 has degree
2 over Q(α), and to use the multiplicativity of extension degrees over Q.
(ii) The first step is to show x6 − 3 is irreducible... ? I don’t know very
much about finite fields.
9. Skip.
10. xp − 2 is irreducible over Q by Eisenstein, with distinct roots
√
√
√
p
p
p
2, ζ 2, . . . , ζ p−1 2,
11
where√ζ is a primitive pth root of unity.
Thus the splitting field of xp − 2 is
√
p
p
p
Q(ζ, 2). Since x −2 is irreducible, Q( 2) is a degree p extension of Q. Since ζ
is a root of the irreducible polynomial (xp −1)/(x−1) = xp−1 +xp−2 +· · ·+1 (one
proves irreducibility by substituting x + 1 for x and then applying Eisenstein on
the prime p, or by showing that the cyclotomic polynomials are all irreducible),
Q(ζ) is a degree p − 1 extension of Q. Since p and p − 1√are relatively prime, we
see immediately that the degree of the extension Q(ζ, p 2) over Q is p(p − 1).
The Galois group has order p(p − 1), so each of the p(p − 1) elements of the
form
√
√
p
p
σ : 2 7→ ζ n 2, ζ 7→ ζ m ,
where 0 ≤ n ≤ p − 1 and 1 ≤ m ≤ p − 1, is an automorphism. If p = 2 then the
Galois group is clearly abelian, so suppose p > 2. Setting
√
√
p
p
σ : 2 7→ ζ 2, ζ 7→ ζ
√
√
p
p
τ : 2 7→ 2, ζ 7→ ζ 2 ,
we see that
but
√
√
p
p
(στ )( 2) = ζ 2,
√
√
√
p
p
p
(τ σ)( 2) = τ (ζ 2) = ζ 2 2,
so the Galois group is not abelian.
12
Chapter 3
Spring 2010
1. For σ ∈ An , let
σ = C1 . . . Cr
be the (disjoint) cyclic decomposition. Suppose σ has order 2. Then
2 = |σ| = lcm{|Ci |}
implies that each Ci is a 2-cycle. Since σ is an even permutation, r is even.
The product (ab)(cd) of two disjoint 2-cycles are the square of the 4-cycle
(acbd). Thus grouping the Ci into pairs and taking the corresponding 4-cycles
gives an element τ ∈ Sn of order 4 whose square is σ.
2. We may assume |G| = pk for k ≥ 2. We first prove the result when G is
not abelian. Consider the homomorphism
G
σ
/ Aut(G) ,
g 7→ σg ;
where
σg := x 7→ gxg −1 ,
x ∈ G.
The kernel of σ is Z(G), the center of G, which is a subgroup of G and therefore
a p-group. The image of σ is a subgroup of Aut(G), which is also a p-group
since it is isomorphic to G/ker σ. The image is non-trivial since G 6= Z(G) by
assumption, so we are done by Lagrange’s theorem.
If G is abelian, then
G = Zpk1 × Zpk2 × · · · × Zpkr ,
where ki ≥ ki+1 ≥ 1 for each i and k1 + · · · + kr = k ≥ 2. Let x1 , . . . , xr denote
generators of the factors of G. It suffices to construct an automorphism with
order p. If k1 = 1, then r ≥ 2, and the automorphism x2 7→ x1 x2 has order p.
k1 −1
+1
So we may assume k1 ≥ 2, in which case the homomorphism ϕ : x1 7→ x1p
13
is an automorphism since pk1 −1 + 1 is relatively prime to pk1 . We claim that ϕ
has order p. To see this, note that
p X
p (p−n)(k1 −1)
(pk1 −1 + 1)p =
p
,
n
n=0
so since p | np for each 1 ≤ n ≤ p − 1, every term in the sum is divisible by pk1
except the last term, which is pp p0 = 1. Thus
x1 7→ xp1
k1 −1
+1
(pk1 −1 +1)
2
7→ x1
(pk1 −1 +1)
7→ · · · 7→ x1
p
=1
proves that ϕp = 1. To see that no lower power of ϕ is 1, note that when
1 ≤ s < p,
s X
s (s−n)(k1 −1)
(pk1 −1 + 1)s =
p
n
n=0
has exactly two terms not divisible by pk1 , whose sum is
s
pk1 −1 + 1 = s · pk1 −1 + 1,
s−1
which cannot be congruent to 1 modulo pk1 .
Choose nonzero x ∈ M , and note that Rx = M . Thus there is a surϕ
/ M defined by r 7→ rx. The kernel
jective R-module homomorphism R
I is a (two-sided) ideal of R. Thus M ' R/I as R-modules, whence M simple
implies that R/I has no nonzero proper left ideals, namely I is a maximal left
ideal. I’m not sure how to show that I is unique if R is commutative.
3.
4. Q/Z is a torsion Z-module, hence not a submodule of a free Z-module,
so it is not projective. Another way to see this is to note that the short exact
sequence
/Z
/Q
/ Q/Z
/0
0
does not split, since Q does not contain a submodule isomorphic to Q/Z, and
there are no non-trivial Z-module homomorphisms Q → Z.
Q/Z is injective since a Z-module M is injective if and only if it is divisible,
namely if nM = M for all 0 6= n ∈ Z, and this latter condition is clear for Q/Z.
Q/Z is not flat since the injective map Z → Z defined by n 7→ 2n does not
remain injective after tensoring with Q/Z (look at the element ( 12 + Z) ⊗ 1).
5. If np = 1, G has a normal Sylow p-subgroup, so assume np > 1. By
Sylow’s theorem
np ≡ 1 (mod p) and np | q,
14
so we deduce that np = q and np = 1 + kp for some k ≥ 1. If we also assume
that nq > 1, then since
nq ≡ 1
(mod q)
and nq | p2 ,
we must have nq = p2 . But these results imply that G has q · (p2 − 1) elements
of order p or p2 and p2 (q − 1) elements of order q, which is a contradiction since
if we also count the identity, then
1 + q(p2 − 1) + p2 (q − 1) = 1 + p2 q − q + p2 q − p2 ≥ 1 + p2 q = 1 + |G|.
Another method of proof is to use the second application of Sylow’s theorem
to deduce that q | p2 − 1, whence q | p − 1 or q | p + 1 since q is prime. But
the first application of Sylow’s theorem implies q = 1 + kp with k ≥ 1, so the
only possibility is q = p + 1, so that p = 2 and q = 3 is forced since there are no
other consecutive primes. Now one uses the classification of groups of order 12
to get the result.
6. We have M a 5 × 5 real matrix satisfying (M − I)2 = 0, and wish to
show the subspace of R5 of vectors fixed by M has dimension at least 3. Since
the minimal polynomial p of M is either x − 1 or (x − 1)2 , the characteristic
polynomial of M is f = (x − 1)5 . Viewing M as a complex matrix, the possible
Jordan forms for M are

 
 

1
1
1

 1 1
 1 1
 1

 
 

.





1
1
1

,
,







1 1
1
1
1
1
1
These Jordan forms fix subspaces of dimensions 5, 4, and 3, respectively. (If the
new basis is {e1 , . . . , e5 }, then the respective fixed subspaces are V , he2 , e3 , e4 , e5 i,
and he2 , e4 , e5 i.) Since the dimension of a subspace is not changed by conjugation (change of basis), we get the desired result.
7. If R is a field, then R-modules are vector spaces over R, which are free.
Conversely, suppose R is not a field. Choose x ∈ R not a unit, so that Rx ⊂ R.
Then R/Rx is a non-trivial R-module that is not free since multiplication by x
kills every element.
8. I’m assuming Z7 denotes the finite field with 7 elements, not the 7-adic
numbers. Polynomials of degree 2 or 3 over a field are reducible if and only if
they have roots in the field, so we need to find the number of degree 3 polynomials without any roots in Z7 . Now it seems difficult to proceed...
15
9. Let ζ denote the primitive nth root of unity contained in F and let α
be an nth root of a. Then E = F (α) contains all the nth roots of a, namely
{α, ζα, . . . , ζ n−1 α},
so E is the splitting field of the separable polynomial xn − a, whence E/F is
Galois.
Any automorphism of E fixing F must be a homomorphism of the form
α 7→ ζ r α,
0 ≤ r ≤ n − 1.
d
Let d > 0 be minimal such that α ∈ F , whence d|n since αn = a ∈ F . We claim
that xd − αd ∈ F [x] is the minimal polynomial of α over F . For if f (x) ∈ F [x]
is the minimal polynomial, then f |(xd − αd ), hence every root of f is in the set
of roots above. Thus the constant term of f is of the form (−1)s ζ d1 · · · ζ ds αs ,
which must be in F , whence d|s, where s = deg f , so we must have f = xd − αd .
It follows that xd − αd is irreducible and E/F has degree d. If α 7→ ζ r α
defines an automorphism, then αd 7→ ζ rd αd implies ζ rd = 1, so n|rd. Thus we
must have r a multiple of n/d, and since there are precisely d such multiples,
we see that every such multiple must be an automorphism and
Gal(E/F ) = {1, α 7→ ζ n/d α, . . . , α 7→ ζ (d−1)n/d α},
which is cyclic with generator α 7→ ζ n/d α.
10.
Theorem 1 (Hilbert’s basis theorem). If A is a Noetherian ring (commutative,
with identity), then A[x] is Noetherian.
Proof. We prove the result by showing any ideal a / A[x] is finitely generated.
Let I / A be the ideal of leading coefficients of elements of a, which is finitely
generated by some elements a1 , . . . , an ∈ A since A is Noetherian. Choose
elements f1 , . . . , fn ∈ a so that each fi has leading coefficient ai . Let ri denote
the degree of fi and set r := maxi {ri }. Then if f ∈ a has degree ≥ r and leading
coefficient a ∈ I, choose ui ∈ A such that a = u1 a1 + · · · + un an . It follows that
by multiplying ui fi by appropriate powers of x and taking the sum, we can kill
off the leading term of f . Thus
a = (a ∩ M ) + (f1 , . . . , fm ),
where M is the A-module generated by 1, x, . . . , xr−1 . Since M is finitely generated and A is Noetherian, M is Noetherian, hence a ∩ M ⊆ M is finitely
generated. Combining the generators with the fi gives a finite generating set
for a.
16
Chapter 4
Fall 2009
1. Every element of S7 has a unique (up to reordering) disjoint cycle decomposition, and the order of the element is the lcm of the lengths of the cycles.
Thus the only elements of order 4 are the 4-cycles, and the
combinations of
4-cycles with disjoint 2-cycles. The number of 4-cycles is 74 · 3! = 7 · 6 · 5, and
the number of possibilities for 2-cycles using the remaining 3 indices is 32 = 3.
Thus the total number of elements of order 4 is
7 · 6 · 5(1 + 3) = 840.
2. Let G act on itself by conjugation. Let g1 , . . . , gr denote representatives
of the conjugacy classes of non-central elements. The class equation yields
|G| = |Z(G)| +
r
X
(G : CG (gi )),
i=1
where CG (gi ) is the centralizer of gi in G. The index of each centralizer divides
|G|, hence is divisible by p, so we must have p||Z(G)|, hence Z(G) is non-trivial.
3. By the fundamental theorem of Galois theory, the field extensions of E
contained in F correspond bijectively to the subgroups of G = Gal(F/E) as the
fixed subfields. Thus α is fixed by every proper subgroup of G, but not all of
G. It follows that if σ ∈ G is an element not fixing α, then hσi = G, i.e. G is
cyclic.
Since F/E is a proper extension, we can choose p prime such that |G| = pk n,
with k ≥ 1 and (n, p) = 1. We wish to prove that n = 1, so assume for
k
contradiction that n > 1. Then hσ p i and hσ n i are proper subgroups of G,
hence fix α. Since (pk , n) = 1, the Euclidean algorithm ensures we can find
17
a, b ∈ Z such that apk + bn = 1. But then
k
σ(α) = (σ p )a (σ n )b (α) = α,
contradiction.
4.
If 0 6= m
n ∈ Q/Z is reduced, by which we mean 0 < m < n and
m
gcd{m, n} = 1, then m
n has order n in Q/Z. Thus h n i contains n elements, so
1
that in particular we see that h m
i
=
h
i.
n
n
Given n11 , n12 ∈ Q/Z with n1 , n2 ≥ 2, let n = lcm{n1 , n2 }. We claim that
1
∈
n
whence
1 1
,
n1 n2
1 1
,
n1 n2
,
1
=
n
since the inclusion ⊆ is obvious. For the claim, we use the Euclidean algorithm
on n1 and n2 to find a, b ∈ Z such that
an1 + bn2 = gcd{n1 , n2 },
and then divide both sides by n1 n2 to get
b
1
1
+a
= lcm{n1 , n2 } = n.
n1
n2
This proves the claim.
Now we are ready to prove the result of the problem. Given
m1
mr
,...,
≤ Q/Z,
n1
nr
where each generator is reduced, set n = lcm(n1 , . . . , nr ). Then we have
m1
mr
1
,...,
=
.
n1
nr
n
1
i
For the previous comments imply we can replace each m
ni by ni , and the above
method for combining two generators, together with induction on the number
of generators, give the result.
5. We compute 520 = 23 · 5 · 13. Suppose G is simple. Then
n13 ≡ 1
(mod 13)
18
and
n13 | 23 · 5
implies n13 = 23 · 5 since n13 > 1. Likewise n5 = 2 · 13. So we have (13 − 1) · 23 · 5
elements of order 13 and (5 − 1) · 2 · 13 elements of order 5. But
12 · 23 · 5 + 4 · 2 · 13 > 12 · 23 · 5 + 23 · 5 = 13 · 23 · 5 = |G|,
contradiction.
6. (a) We have M a matrix satisfying M 2 + M + I = 0 and wish to find
the possible rational canonical forms over R. Since x2 + x + 1 = (x − ζ)(x − ζ̄)
, we see that x2 + x + 1 is irreducible over R, hence is the minimal
for ζ = −1+3i
2
polynomial of M over R. The corresponding rational block is
0 −1
,
1 −1
and M must be composed of ≥ 1 of these blocks on its diagonal.
(b) Now M is over C and we wish to find all possible Jordan forms. The
possible minimal polynomials are x−ζ, x− ζ̄, and (x−ζ)(x− ζ̄). In any case, the
minimal polynomial factors linearly, so M is diagonalizable, with eigenvalues ζ
and ζ̄. Thus the Jordan form of M is an n × n diagonal matrix (n ≥ 1) with
some number r of ζs (0 ≤ r ≤ n), then (n − r) ζ̄s on its diagonal.
7. Consider the ring of fractions R0 := S −1 R. Prime ideals of R0 correspond
to prime ideals of R0 that don’t meet S via the ring homomorphism
R
/ R0 ,
π
r 7→
r
,
1
since the preimage of a prime ideal under a ring homomorphism is a prime ideal.
An ideal a of R that is maximal among ideals not meeting S thus corresponds
to a maximal ideal a0 of R0 , so a0 is prime, hence a = π −1 (a0 ) is prime as well.
8. Skip.
9. First we state a general result. If ζn is a primitive nth root of unity,
then the Galois group of Q(ζn )/Q is isomorphic to the multiplicative group
(Z/nZ)× of order φ(n).
Q For ζn is a root of the degree φ(n) irreducible cyclotomic
polynomial Ωn (x) = (x − ζ a ), where ζ a ranges across the primitive nth roots
of unity, hence the degree of the extension is φ(n). This means the Galois group
has order φ(n). Since any automorphism σ is determined by σ(ζn ) = ζna , and
there are precisely φ(n) possibilities for a, every such map is an automorphism
of Q(ζn ), whence we obtain an isomorphism
×
(Z/nZ) → Gal(Q(ζn )/Q)
19
a 7→ (ζn 7→ ζna ).
Now we take n = 11 above and let ζ be a primitive 11th root of unity.
The Galois group G of the extension Q(ζ)/Q is isomorphic to the cyclic group
(Z/11Z)× ' Z/10Z ' Z/2Z × Z/5Z. Let H be a subgroup of G isomorphic to
Z/2Z, which is normal since G is abelian. Let F be the fixed field of H. Then
F/Q is Galois by the fundamental theorem of Galois theory, with Galois group
G0 ' G/H ' Z/5Z.
10. We state and prove the Eisenstein criterion for irreducibility of polynomials.
Proposition 2 (Eisenstein criterion). Let R be an integral domain, p a prime
ideal of R, and p(x) = xn + an−1 xn−1 + · · · + a0 ∈ R[x] a monic polynomial
such that each ai ∈ p, but a0 ∈
/ p2 . Then p is irreducible in R[x].
Proof. Suppose for contradiction that p(x) = a(x)b(x) in R[x], where a, b are
non-constant polynomials (i.e. non-units, since the only constants that divide
1 are units). Reducing modulo p, we see that xn ≡ a(x) b(x) in R/p[x], which
is an integral domain since R/p is an integral domain. Thus both ā and b̄ must
have 0 constant term (otherwise the non-zero lowest degree terms of ā and b̄,
which exist since neither ā nor b̄ can be 0 due to our equation, multiply to
something nonzero of degree < n), i.e. the constant terms of both a and b are
in p. But then the constant term of p is in p2 , contradiction.
20
Chapter 5
Spring 2009
1. Elements of Sn with the same cycle type are conjugate to each other. There
are (p − 1)! p-cycles in Sp , so the order of the orbit of σ under conjugation is
(p − 1)!. Thus |C| = |SP |/(p − 1)! = p, so C = hσi, which is abelian.
Now let Sp act on the set of its subgroups by conjugation. Since there are
(p − 1)! p-cycles in Sp , there are (p − 1)!/(p − 1) = (p − 2)! distinct subgroups of
order p, all of which are in the orbit of hσi since σ is conjugate to every other
p-cycle. So the order of N is p!/(p − 2)! = p(p − 1). To see that N is not abelian,
let τ ∈ N be an element satisfying
τ στ −1 = σ 2 .
Then τ στ −1 σ −1 = σ 6= 1.
2.
The derived subgroup G(1) of G consists of the p upper-triangular
matrices of the form
1 a
,
a ∈ Z/pZ.
0 1
A simple computation shows that every commutator is of this form, and we
have
1 0
1 a
1 a
1 0
1 −a
1 −a
1 a
=
=
,
0 1
0 2
0 1
0 2
0 1
0 12
0 21
and elements of this form do not generate any new elements. Note that G(1) is
abelian, so that G(2) , the subgroup of commutators of G(1) , is trivial. Thus the
derived series
1 = G(2) / G(1) / G(0) = G
shows that G is solvable.
21
3. We have a group G with |G| = 81 = 34 acting on a set X with |X| = 30.
Since the stabilizer Sx of x ∈ X is a subgroup of G, |Sx | is a power of 3. The
index (G : Sx ) of Sx in G also divides |G|, hence is a power of 3. Since the cosets
of Sx are in bijection with the elements of the orbit Ox of x, (G : Sx ) = |Ox |,
so |Ox | is a power of 3. Recall that the orbits partition X.
Suppose no element of X is fixed by at least 27 elements. This means that
|Sx | < 27 for each x ∈ X, so that |Ox | > 3 for each x. This is impossible since
no sum involving only 9’s and 27’s equals 30.
An element x ∈ X fixed by precisely 3 elements of G is in an orbit of order
27. If there is such an element, then every other element of the orbit is fixed by
precisely 3 elements as well. So there are 27 such elements.
4. Skip.
5. Suppose R is a commutative ring with identity 1 6= 0, and suppose that
for each prime p / R, Rp contains no nonzero nilpotents. Now if 0 6= a ∈ R
is nilpotent, then an = 0 for some positive integer n. Then we can choose a
maximal ideal m (Zorn’s lemma argument), which is prime and hence contains
a. Then a1 ∈ Rm is a nonzero nilpotent, yet ( a1 )n = 10 , contradiction.
If we don’t assume that R has an identity, then there may not be maximal
ideals or even prime ideals (e.g. Q with trivial multiplication – ab = 0 for all
a, b ∈ Q). I’m not sure what to do in this case.
6. Finitely generated ideals in a UFD are principal, generated by the gcd
of the generators. So the first step is to compute the greatest common factor of
the two polynomials. We do this by the Euclidean algorithm, which produces
the following output:
a
x − 4x + 3x2 − 4x + 2
x3 − 5x2 + 6x − 2
4
3
b
x − 5x + 6x − 2
2x2 − 8x + 4
3
2
q
x+1
1
1
2x − 2
r
2x − 8x + 4
0
2
Thus the
is 2x2 −8x+4, so the ideal a is (x2 −4x+2) =
√ gcd of the polynomials
√
((x − 2 + 2)(x − 2 − 2)). It follows that a is principal, but not prime, and
hence also not maximal.
7. Skip.
8. Skip.
22
9. No, σ is not necessarily an automorphism. Let t be an indeterminate
σ /
and consider Q(t)/Q. Define Q(t)
Q(t) by t 7→ t2 . Then σ fixes Q and
is easily shown to be a homomorphism of Q(t). But σ is not surjective, hence
not an automorphism of Q(t).
The statement is true with the additional assumption that L/K is an algebraic extension. Then if α ∈ L − K, the fact that σ is a homomorphism implies
that σ(α) must be another root of the minimal polynomial for α over K. This
means σ is injective. For surjectivity, suppose β ∈ L − K and let Σ ⊆ L − K
denote the set of all roots of the minimal polynomial of β over K that are contained in L. Then β ∈ Σ, Σ is finite, and σ permutes the elements of Σ since it
maps L into L, whence β must be in the image.
10. ζ = e2πi/8 is a primitive 8th root of unity. Thus by the argument in 9
×
of Fall 2009, Q(ζ)/Q has Galois group G ' (Z/8Z) ' Z2 × Z2 , generated by
σ : ζ 7→ ζ 3
and τ : ζ 7→ ζ 5 .
The inverted subgroup lattice is
1
hσi
hστ i
hτ i
hσ, τ i
with corresponding lattice of fixed fields
Q(ζ)
Q(ζ + ζ 3 )
Q(ζ + ζ 7 )
Q(ζ 2 )
Q
where the latter cannot be Q(ζ + ζ 5 ) since ζ + ζ 5 = 0! This is much clearer if
we take a different approach.
Write ζ = √12 (1 + i). Then ζ 2 = i and the remaining powers are easily
√
computed. Note that Q(ζ) = Q( 2, i), which is clearly an extension of degree 4
over Q
√ since it decomposes into a pair of quadratic extensions. Automorphisms
of Q( 2, i) fixing Q must map
√
√
2 7→ ± 2,
i 7→ ±i,
23
and since there are only 4 such possibilities, they must all be automorphisms.
Let
√
√
σ : 2 7→ − 2, i 7→ −i
√
√
τ : 2 7→ − 2, i 7→ i
be generators of the Galois group, where we have chosen them to correspond to
σ, τ above. The lattice of fixed fields is then
Q(ζ)
√
Q(i 2)
√
Q( 2)
Q(i)
Q
√
√
and we note that ζ + ζ 3 = i 2, ζ + ζ 7 = 2, and ζ 2 = i.
24
Chapter 6
Fall 2008
1. By Sylow’s theorem,
n5 ≡ 1
(mod 5)
and
n5 |72 ;
n7 ≡ 1
(mod 7)
and
n7 |52 .
The only possibilities are n5 = n7 = 1, so G has normal subgroups H, K of
orders 52 and 72 , respectively. Since 5, 7 are relatively prime, H ∩ K = {1}.
Thus HK ' H × K, and since the left side is a subgroup of G while the right
side has the same order as G, we see that G ' H × K.
Now, it is easy to show that groups of order p2 are abelian, hence either
isomorphic to Zp × Zp or Zp2 . (Use the class equation to show G has non-trivial
center; so if g 6= 1 in the center, it either generates G or there is some h ∈
/ hgi,
whence G = hg, hi, but g, h commute so G is abelian.) Thus G is abelian, with
possible isomorphism classes
Z52 × Z72 , Z52 × Z7 × Z7 , Z5 × Z5 × Z72 , Z5 × Z5 × Z7 × Z7 .
25