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, since1 = (12)(12) is a product of commuting 2-cycles, but has order 1. The statementis correct if we exclude the identity:

Proposition 1. Let p be prime. An element 1 6= σ ∈ Sn has order p if andonly if σ is the product of (at least one) commuting p-cycles.

Proof. Suppose σ has order p. Let σ ∈ Sn have (disjoint) cyclic decompositionσ = C1 . . . Cr. The order of σ is the least common multiple of the sizes of theCi, namely

p = |σ| = lcm|Ci|,whence |Ci| = p for each i. Thus r ≥ 1 and the Ci are disjoint, hence theycommute.

Conversely, suppose σ = C1 . . . Cr is a decomposition into (not necessarilydisjoint) commuting p-cycles, with r ≥ 1. Then by commutativity,

σp = Cp1 . . . Cpr = 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 hasorder lcm2, 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 thematrix, 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 multipleof 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 (mod p).

2

So we have np ∈ 1, p+ 1, (p− 1)2, (p+ 1)(p− 1)2.

Consider the element

(1 10 1

), which is of order p and generates a Sylow

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

ad

(a b0 d

)(1 10 1

)(d −b0 a

)=

(1 ad−1

0 1

),

whence every such

(a b0 d

)is in the normalizer NP (G). There are p(p − 1)2

such elements, so since np = (G : NP (G)) by Sylow’s theorem, we obtain thebound 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 ofG. The class equation gives

pk = |G| = |Z(G)|+r∑i=1

(G : CG(gi)).

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 orderpb for each 1 ≤ j ≤ k−1 (the cases j = 0, k are trivial). Since |Z(G)| is divisibleby 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 Zis contained in the center of G. Then G′ := G/Z has order pk−1, whence G′

has normal subgroups H1, . . . , Hk−2 of orders p1, . . . , pk−2, plus the identitysubgroup H0 := 1.

Letting Gπ // G′ denote the natural surjection, we claim that the sub-

groups Hj := π−1(Hj) for 0 ≤ j ≤ k−2 are of order pj+1 and normal in G. Theorder statement is clear, since cosets of Z in G have cardinality p. Now for anyx ∈ Hj and g ∈ G, we have gxg−1 ∈ Hj , so that since π is a homomorphism,

gxg−1 ∈ Hj . Thus gxg−1 ∈ Hj , so Hj is normal in G.

4. We have

A =

1 0 00 0 −20 1 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 theminimal 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+ 2and p2 = x− 1 (we must have p1p2 = f), so the rational form is

A ∼

0 −2 01 3 00 0 1

To confirm this, we note that

A

100

=

100

, A

010

=

001

, A2

010

=

0−23

,

so that a basis that yields the rational form is0

10

,

001

,

100

The associated change of basis matrix (expressing the new basis in terms of theold basis – just load in the vectors of the basis as the columns) is

P =

0 0 11 0 00 1 0

, with P−1 =

0 1 00 0 11 0 0

,

whence

P−1AP =

0 −2 01 3 00 0 1

.

(Since Boldvold = Bnewvnew and Bnew = BoldP , we see that vold = Pvnew, namelyP takes the old basis to the new basis, but takes vectors with coordinates in thenew basis to vectors with coordinates in the old basis. The new matrix shouldact on vectors expressed in the new basis. So take such a vector, convert to theold basis using P , then apply A, then use P−1 to take the result back to thenew basis.)

Now for the Jordan form. We have two Jordan blocks, one of size 2 associatedwith the eigenvalue 1 and the other of size 1 for the eigenvalue 2. Since theminimal polynomial has linear powers, each block is diagonalizable, so the resultis

A ∼

1 0 00 1 00 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.

We need to prove that there is a norm Z[i]N // 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 αβ = 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 ≤ 12 . 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 ofZ[i], we have

N(γ) = N(β)N(θ) ≤ N(β)

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/nZϕ // Z/dZ

defined by(a mod m, b mod n) 7→ ab mod d.

It is Z-bilinear, and the induced linear map Z/mZ⊗Z Z/nZϕ′ // Z/dZ maps

1⊗ 1 7→ 1, an element of order d. Thus 1⊗ 1 has order a multiple of d, whencethe cyclic group has order at least d, and therefore exactly d. So ϕ′ is an iso-morphism.

8. Skip.

9. (a) First, note that x4 − 5 is irreducible over Q by Eisenstein. Letting αdenote the positive real number 4

√5, the roots of x4 − 5 in C are

4√

5, ζ4√

5, ζ24√

5, ζ34√

5,

5

where ζ = i is a primitive fourth root of unity. Thus the splitting field of E ofx4 − 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 ofG = Gal(E/Q) are determined by

4√

5 7→ ik4√

5, i 7→ ±i,

all of which are automorphisms of E. Letting

σ :4√

5 7→ i4√

5, i 7→ i

andτ :

4√

5 7→ 4√

5, i 7→ −i,

we see that G = 〈σ, τ〉 = 1, σ, σ2, σ3, τ, στ, σ2τ, σ3τ, and that σ, τ satisfy therelations σ4 = τ2 = 1 and τσ = σ3τ . Thus G ' D4.

The inverted lattice of subgroups of D4 is (DF 69)

1

〈τ〉 〈σ2τ〉 〈σ2〉 〈στ〉 〈σ3τ〉

〈σ2, τ〉 〈σ〉 〈σ2, στ〉

〈σ, τ〉

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((1− i) 4√

5)

Q(√

5) Q(i) Q(i√

5)

Q

(b) The Galois group of Q( 4√

5, i) over Q(√

5) is 〈σ2, τ〉, which is isomorphicto Z2 × Z2, the Klein Viergruppe.

6

(c) The Galois group of Q( 4√

5) over Q(i) is 〈σ〉, isomorphic to Z4.

10. The extension Q(√

2)/Q is Galois of degree 2, with basis 1,√

2 overQ. We claim that

√3 /∈ Q(

√2). For suppose (a + b

√2)2 = 3 with a, b ∈ Q.

Then a2 + 2b2 + 2ab√

2 = 3, whence a or b is 0, but 3 is not the square of arational number. Since Q(

√3)/Q is of degree 2, so is Q(

√2,√

3)/Q(√

2), whenceQ(√

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 isthe splitting field of the separable polynomial (x2 − 2)(x2 − 3), hence a Galoisextension 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 ofthese maps is an automorphism. Let

σ :√

2 7→ −√

2,√

3 7→√

3

τ :√

2 7→√

2,√

3 7→ −√

3.

Then G = Gal(Q(√

2,√

3)/Q) = 〈σ, τ〉. Since σ2 = τ2 = 1 and στ = τσ, G isisomorphic to Z2 × Z2.

11. Recall that for polynomials f, g ∈ F[x, y], where F is a field, we havedeg(fg) = deg f + deg g. Suppose x2 + y2 − 1 is reducible, factoring into thenon-units f, g. Then f, g must each be of degree 1, so we get an expression

(ax+ by + c)(a′x+ b′y + c′) = x2 + y2 − 1.

This implies aa′ = bb′ = −cc′ = 1, so all the coefficients are nonzero. Moreover,we deduce ab′ = −a′b, ac′ = −a′c, and bc′ = −b′c, whence b′ = −a′b/a and−a′c/a = c′ = −b′c/b. But the latter equalities imply a′/a = b′/b, contradictingthe 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 except0, 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 ischosen for an element of H, there are 4 multiples of that row that yield elementsof H, but only one that gives an element of G. Thus

|G| = (52 − 1)(52 − 5)

4= 4 · 5 · 6 = 120.

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 10 1

)and

(1 01 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 matricesof the form (

1 a0 1

), a ∈ R.

A simple computation shows that every commutator is of this form, and wehave (

1 00 1

2

)(1 a0 1

)(1 00 2

)(1 −a0 1

)=

(1 a0 1

2

)(1 −a0 2

)=

(1 a0 1

),

and elements of this form do not generate any new elements. Note that G(1) isabelian, so that G(2), the subgroup of commutators of G(1), is trivial. Thus the

8

derived series1 = G(2) / G(1) / G(0) = G

shows that G is solvable.

3. We compute σ2 = τ2 = 1, that στ 6= τσ, and that (στ)3 = 1, whenceτσ = (στ)2. Set a = στ and b = σ. Then the group contains the six elements

1, a, a2, b, ba, b2a

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, 12 ,−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 〈σ, τ〉 → S3. Sincethe group contains six elements, the map is injective.

5. Hermitian matrices are normal, hence diagonalizable by unitary matri-ces, 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−1AU is diagonal withreal entries. Conversely, every real diagonal matrix is Hermitian, and real diag-onal matrices with different multiplicities of eigenvalues are not conjugate (bythe uniqueness of the Jordan form up to reordering of Jordan blocks). Since Asatisfies A5 + 2A3 + 3A− 6I = 0 if and only if U−1AU does as well, it sufficesto classify real diagonal matrices A with eigenvalues in increasing order thatsatisfy A5 + 2A3 + 3A− 6I = 0. Since A is diagonal, each entry on the diagonalmust be a zero of the polynomial f(x) = x5 + 2x3 + 3x − 6. One such zero isx = 1, but we have f ′(x) = 5x4 + 6x2 + 3, which is strictly positive in the realnumbers. Thus f is monotonic increasing on the reals, so the only real zero off is x = 1. Thus the only possibility for A is I, so the unique conjugacy class isI.

6. We wish to determine the number of conjugacy classes of 4× 4 complexmatrices A satisfying A3 − 2A2 +A = 0. This means the minimum polynomialp 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, thenine possible Jordan forms (up to reordering of blocks), which correspond toconjugacy classes, are:

9

Minimal Polynomial Characteristic Polynomials Jordan Forms

x x4

0

0

0

0

x− 1 (x− 1)4

1

1

1

1

(x− 1)2 (x− 1)4

1

1 1

1

1

,

1

1 1

1

1 1

x(x− 1) x3(x− 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,

0

0

1

1 1

x(x− 1)3

0

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 theyare of degree ≤ 3 and have no roots in Q), so both 2 and 3 divide the degreeof the extensions Q(α)/Q, whence this extension must have degree 6. Thus1, α, . . . , α5 are linearly independent over Q, hence also over Z, and x6 − 7 is

10

the minimal polynomial of α (or use Eisenstein on x6 − 7 to conclude that it isirreducible).

It follows that Q(α) ' Q[x]/(x6 − 7), using the map α 7→ x, so that alsoZ[α] ' Z[x]/(x6 − 7) as subrings. Moreover,

Z[α]/(α2) ' (Z[x]/(x6 − 7))/(x2) ' 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 contain0+ bα for some b 6= 0, whence it contains b−1 · bα = α and is the ideal (α). Notethat this ring is not a PID since it is not an integral domain, due to the factthat α2 = 0, so α is a zero divisor.

8. (i) Let α denote the positive real number α = 31/6. By the argumentin 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± α√−3

2.

Thus the splitting field of x6−3 is Q(√−3, α). Since Q(α) ⊆ R,

√−3 /∈ Q(α), so

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 verymuch about finite fields.

9. Skip.

10. xp − 2 is irreducible over Q by Eisenstein, with distinct roots

p√

2, ζp√

2, . . . , ζp−1p√

2,

11

where ζ is a primitive pth root of unity. Thus the splitting field of xp − 2 isQ(ζ, p

√2). Since xp−2 is irreducible, Q( p

√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 (oneproves irreducibility by substituting x+1 for x and then applying Eisenstein onthe 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, wesee 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 theform

σ :p√

2 7→ ζnp√

2, ζ 7→ ζm,

where 0 ≤ n ≤ p− 1 and 1 ≤ m ≤ p− 1, is an automorphism. If p = 2 then theGalois group is clearly abelian, so suppose p > 2. Setting

σ :p√

2 7→ ζp√

2, ζ 7→ ζ

τ :p√

2 7→ p√

2, ζ 7→ ζ2,

we see that(στ)(

p√

2) = ζp√

2,

but(τσ)(

p√

2) = τ(ζp√

2) = ζ2p√

2,

so the Galois group is not abelian.

12

Chapter 3

Spring 2010

1. For σ ∈ An, letσ = 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-cyclesgives 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 isnot abelian. Consider the homomorphism

Gσ // Aut(G) , g 7→ σg;

whereσg := x 7→ gxg−1, x ∈ G.

The kernel of σ is Z(G), the center of G, which is a subgroup of G and thereforea p-group. The image of σ is a subgroup of Aut(G), which is also a p-groupsince it is isomorphic to G/kerσ. The image is non-trivial since G 6= Z(G) byassumption, 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 denotegenerators of the factors of G. It suffices to construct an automorphism withorder p. If k1 = 1, then r ≥ 2, and the automorphism x2 7→ x1x2 has order p.

So we may assume k1 ≥ 2, in which case the homomorphism ϕ : x1 7→ xpk1−1+1

1

13

is an automorphism since pk1−1 + 1 is relatively prime to pk1 . We claim that ϕhas order p. To see this, note that

(pk1−1 + 1)p =

p∑n=0

(p

n

)p(p−n)(k1−1),

so since p |(pn

)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→ xpk1−1+1

1 7→ x(pk1−1+1)

2

1 7→ · · · 7→ x(pk1−1+1)

p

1 = 1

proves that ϕp = 1. To see that no lower power of ϕ is 1, note that when1 ≤ s < p,

(pk1−1 + 1)s =

s∑n=0

(s

n

)p(s−n)(k1−1)

has exactly two terms not divisible by pk1 , whose sum is(s

s− 1

)pk1−1 + 1 = s · pk1−1 + 1,

which cannot be congruent to 1 modulo pk1 .

3. Choose nonzero x ∈ M , and note that Rx = M . Thus there is a sur-

jective R-module homomorphism Rϕ // M defined by r 7→ rx. The kernel

I is a (two-sided) ideal of R. Thus M ' R/I as R-modules, whence M simpleimplies that R/I has no nonzero proper left ideals, namely I is a maximal leftideal. I’m not sure how to show that I is unique if R is commutative.

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 exactsequence

0 // Z // Q // Q/Z // 0

does not split, since Q does not contain a submodule isomorphic to Q/Z, andthere 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 notremain injective after tensoring with Q/Z (look at the element ( 1

2 + Z)⊗ 1).

5. If np = 1, G has a normal Sylow p-subgroup, so assume np > 1. BySylow’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 assumethat 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) elementsof order p or p2 and p2(q− 1) elements of order q, which is a contradiction sinceif we also count the identity, then

1 + q(p2 − 1) + p2(q − 1) = 1 + p2q − q + p2q − p2 ≥ 1 + p2q = 1 + |G|.

Another method of proof is to use the second application of Sylow’s theoremto deduce that q | p2 − 1, whence q | p − 1 or q | p + 1 since q is prime. Butthe first application of Sylow’s theorem implies q = 1 + kp with k ≥ 1, so theonly possibility is q = p+ 1, so that p = 2 and q = 3 is forced since there are noother consecutive primes. Now one uses the classification of groups of order 12to get the result.

6. We have M a 5 × 5 real matrix satisfying (M − I)2 = 0, and wish toshow the subspace of R5 of vectors fixed by M has dimension at least 3. Sincethe minimal polynomial p of M is either x − 1 or (x − 1)2, the characteristicpolynomial of M is f = (x− 1)5. Viewing M as a complex matrix, the possibleJordan forms for M are

11

11

1

,

11 1

11

1

,

11 1

11 1

1

.

These Jordan forms fix subspaces of dimensions 5, 4, and 3, respectively. (If thenew basis is e1, . . . , e5, then the respective fixed subspaces are V , 〈e2, e3, e4, e5〉,and 〈e2, e4, e5〉.) Since the dimension of a subspace is not changed by conjuga-tion (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 xkills every element.

8. I’m assuming Z7 denotes the finite field with 7 elements, not the 7-adicnumbers. Polynomials of degree 2 or 3 over a field are reducible if and only ifthey have roots in the field, so we need to find the number of degree 3 polyno-mials 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 αbe an nth root of a. Then E = F (α) contains all the nth roots of a, namely

α, ζα, . . . , ζn−1α,

so E is the splitting field of the separable polynomial xn − a, whence E/F isGalois.

Any automorphism of E fixing F must be a homomorphism of the form

α 7→ ζrα, 0 ≤ r ≤ n− 1.

Let d > 0 be minimal such that αd ∈ F , whence d|n since αn = a ∈ F . We claimthat 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 setof roots above. Thus the constant term of f is of the form (−1)sζd1 · · · ζdsαs,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→ ζrαdefines an automorphism, then αd 7→ ζrdαd implies ζrd = 1, so n|rd. Thus wemust 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→ ζn/dα, . . . , α 7→ ζ(d−1)n/dα,

which is cyclic with generator α 7→ ζn/dα.

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 finitelygenerated by some elements a1, . . . , an ∈ A since A is Noetherian. Chooseelements f1, . . . , fn ∈ a so that each fi has leading coefficient ai. Let ri denotethe degree of fi and set r := maxiri. Then if f ∈ a has degree ≥ r and leadingcoefficient a ∈ I, choose ui ∈ A such that a = u1a1 + · · ·+unan. It follows thatby multiplying uifi by appropriate powers of x and taking the sum, we can killoff 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 gen-erated and A is Noetherian, M is Noetherian, hence a ∩ M ⊆ M is finitelygenerated. Combining the generators with the fi gives a finite generating setfor a.

16

Chapter 4

Fall 2009

1. Every element of S7 has a unique (up to reordering) disjoint cycle decom-position, 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 of4-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 representativesof the conjugacy classes of non-central elements. The class equation yields

|G| = |Z(G)|+r∑i=1

(G : CG(gi)),

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 Econtained in F correspond bijectively to the subgroups of G = Gal(F/E) as thefixed subfields. Thus α is fixed by every proper subgroup of G, but not all ofG. It follows that if σ ∈ G is an element not fixing α, then 〈σ〉 = G, i.e. G iscyclic.

Since F/E is a proper extension, we can choose p prime such that |G| = pkn,with k ≥ 1 and (n, p) = 1. We wish to prove that n = 1, so assume for

contradiction that n > 1. Then 〈σpk〉 and 〈σn〉 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

σ(α) = (σpk

)a(σn)b(α) = α,

contradiction.

4. If 0 6= mn ∈ Q/Z is reduced, by which we mean 0 < m < n and

gcdm,n = 1, then mn has order n in Q/Z. Thus 〈mn 〉 contains n elements, so

that in particular we see that 〈mn 〉 = 〈 1n 〉.Given 1

n1, 1n2∈ Q/Z with n1, n2 ≥ 2, let n = lcmn1, n2. We claim that

1

n∈⟨

1

n1,

1

n2

⟩,

whence ⟨1

n1,

1

n2

⟩=

⟨1

n

⟩since the inclusion ⊆ is obvious. For the claim, we use the Euclidean algorithmon n1 and n2 to find a, b ∈ Z such that

an1 + bn2 = gcdn1, n2,

and then divide both sides by n1n2 to get

b1

n1+ a

1

n2= lcmn1, n2 = n.

This proves the claim.Now we are ready to prove the result of the problem. Given⟨

m1

n1, . . . ,

mr

nr

⟩≤ Q/Z,

where each generator is reduced, set n = lcm(n1, . . . , nr). Then we have⟨m1

n1, . . . ,

mr

nr

⟩=

⟨1

n

⟩.

For the previous comments imply we can replace each mi

niby 1

ni, and the above

method for combining two generators, together with induction on the numberof generators, give the result.

5. We compute 520 = 23 · 5 · 13. Suppose G is simple. Then

n13 ≡ 1 (mod 13) and n13 | 23 · 5

18

implies n13 = 23 ·5 since n13 > 1. Likewise n5 = 2 ·13. So we have (13−1) ·23 ·5elements 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 M2 + M + I = 0 and wish to findthe possible rational canonical forms over R. Since x2 + x+ 1 = (x− ζ)(x− ζ)for ζ = −1+3i

2 , we see that x2 +x+ 1 is irreducible over R, hence is the minimalpolynomial of M over R. The corresponding rational block is(

0 −11 −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. Thepossible minimal polynomials are x−ζ, x− ζ, and (x−ζ)(x− ζ). In any case, theminimal polynomial factors linearly, so M is diagonalizable, with eigenvalues ζand ζ. Thus the Jordan form of M is an n × n diagonal matrix (n ≥ 1) withsome number r of ζs (0 ≤ r ≤ n), then (n− r) ζs on its diagonal.

7. Consider the ring of fractions R′ := S−1R. Prime ideals of R′ correspondto prime ideals of R′ that don’t meet S via the ring homomorphism

Rπ // R′ , 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 correspondsto a maximal ideal a′ of R′, so a′ is prime, hence a = π−1(a′) 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). For ζn is a root of the degree φ(n) irreducible cyclotomicpolynomial Ωn(x) =

∏(x− ζa), where ζa ranges across the primitive nth roots

of unity, hence the degree of the extension is φ(n). This means the Galois grouphas order φ(n). Since any automorphism σ is determined by σ(ζn) = ζan, andthere are precisely φ(n) possibilities for a, every such map is an automorphismof Q(ζn), whence we obtain an isomorphism

(Z/nZ)× → Gal(Q(ζn)/Q)

19

a 7→ (ζn 7→ ζan).

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 toZ/2Z, which is normal since G is abelian. Let F be the fixed field of H. ThenF/Q is Galois by the fundamental theorem of Galois theory, with Galois groupG′ ' G/H ' Z/5Z.

10. We state and prove the Eisenstein criterion for irreducibility of poly-nomials.

Proposition 2 (Eisenstein criterion). Let R be an integral domain, p a primeideal of R, and p(x) = xn + an−1x

n−1 + · · · + a0 ∈ R[x] a monic polynomialsuch 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 arenon-constant polynomials (i.e. non-units, since the only constants that divide1 are units). Reducing modulo p, we see that xn ≡ a(x) b(x) in R/p[x], whichis an integral domain since R/p is an integral domain. Thus both a and b musthave 0 constant term (otherwise the non-zero lowest degree terms of a and b,which exist since neither a nor b can be 0 due to our equation, multiply tosomething nonzero of degree < n), i.e. the constant terms of both a and b arein 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. Thereare (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 = 〈σ〉, 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 oforder p, all of which are in the orbit of 〈σ〉 since σ is conjugate to every otherp-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

τστ−1 = σ2.

Then τστ−1σ−1 = σ 6= 1.

2. The derived subgroup G(1) of G consists of the p upper-triangularmatrices of the form (

1 a0 1

), a ∈ Z/pZ.

A simple computation shows that every commutator is of this form, and wehave (

1 00 1

2

)(1 a0 1

)(1 00 2

)(1 −a0 1

)=

(1 a0 1

2

)(1 −a0 2

)=

(1 a0 1

),

and elements of this form do not generate any new elements. Note that G(1) isabelian, so that G(2), the subgroup of commutators of G(1), is trivial. Thus thederived 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. Theindex (G : Sx) of Sx in G also divides |G|, hence is a power of 3. Since the cosetsof 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 sinceno 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 order27. If there is such an element, then every other element of the orbit is fixed byprecisely 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 thatfor each prime p / R, Rp contains no nonzero nilpotents. Now if 0 6= a ∈ Ris nilpotent, then an = 0 for some positive integer n. Then we can choose amaximal ideal m (Zorn’s lemma argument), which is prime and hence containsa. Then a

1 ∈ Rm is a nonzero nilpotent, yet (a1 )n = 01 , contradiction.

If we don’t assume that R has an identity, then there may not be maximalideals or even prime ideals (e.g. Q with trivial multiplication – ab = 0 for alla, b ∈ Q). I’m not sure what to do in this case.

6. Finitely generated ideals in a UFD are principal, generated by the gcdof the generators. So the first step is to compute the greatest common factor ofthe two polynomials. We do this by the Euclidean algorithm, which producesthe following output:

a b q r

x4 − 4x3 + 3x2 − 4x+ 2 x3 − 5x2 + 6x− 2 x+ 1 2x2 − 8x+ 4

x3 − 5x2 + 6x− 2 2x2 − 8x+ 4 12x−

12 0

Thus the gcd of the polynomials is 2x2−8x+4, so the ideal a is (x2−4x+2) =((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

and consider Q(t)/Q. Define Q(t)σ // 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, hencenot an automorphism of Q(t).

The statement is true with the additional assumption that L/K is an alge-braic extension. Then if α ∈ L−K, the fact that σ is a homomorphism impliesthat σ(α) must be another root of the minimal polynomial for α over K. Thismeans σ is injective. For surjectivity, suppose β ∈ L −K and let Σ ⊆ L −Kdenote the set of all roots of the minimal polynomial of β over K that are con-tained in L. Then β ∈ Σ, Σ is finite, and σ permutes the elements of Σ since itmaps 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 9of Fall 2009, Q(ζ)/Q has Galois group G ' (Z/8Z)

× ' Z2 × Z2, generated by

σ : ζ 7→ ζ3 and τ : ζ 7→ ζ5.

The inverted subgroup lattice is

1

〈σ〉 〈στ〉 〈τ〉

〈σ, τ〉

with corresponding lattice of fixed fields

Q(ζ)

Q(ζ + ζ3) Q(ζ + ζ7) Q(ζ2)

Q

where the latter cannot be Q(ζ + ζ5) since ζ + ζ5 = 0! This is much clearer ifwe take a different approach.

Write ζ = 1√2(1 + i). Then ζ2 = i and the remaining powers are easily

computed. Note that Q(ζ) = Q(√

2, i), which is clearly an extension of degree 4over Q since it decomposes into a pair of quadratic extensions. Automorphismsof 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 toσ, τ above. The lattice of fixed fields is then

Q(ζ)

Q(i√

2) Q(√

2) Q(i)

Q

and we note that ζ + ζ3 = i√

2, ζ + ζ7 =√

2, and ζ2 = 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 oforders 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 rightside 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 eitherisomorphic to Zp×Zp or Zp2 . (Use the class equation to show G has non-trivialcenter; so if g 6= 1 in the center, it either generates G or there is some h /∈ 〈g〉,whence G = 〈g, h〉, but g, h commute so G is abelian.) Thus G is abelian, withpossible isomorphism classes

Z52 × Z72 , Z52 × Z7 × Z7, Z5 × Z5 × Z72 , Z5 × Z5 × Z7 × Z7.

25