Algebra Qualifying Exam Solutions. Thomas Goller


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1 Algebra Qualifying Exam Solutions Thomas Goller September 4, 2
2 Contents Spring Fall Spring Fall Spring Fall 28 25
3 Chapter Spring 2. The claim as stated is false. The identity element is a problem, since = (2)(2) is a product of commuting 2cycles, but has order. The statement is correct if we exclude the identity: Proposition. Let p be prime. An element σ S n has order p if and only if σ is the product of (at least one) commuting pcycles. Proof. Suppose σ has order p. Let σ S n have (disjoint) cyclic decomposition σ = C... C r. The order of σ is the least common multiple of the sizes of the C i, namely p = σ = lcm{ C i }, whence C i = p for each i. Thus r and the C i are disjoint, hence they commute. Conversely, suppose σ = C... C r is a decomposition into (not necessarily disjoint) commuting pcycles, with r. Then by commutativity, σ p = C p... Cp r =, so σ has order dividing p. Since σ, this proves that σ has order p. If p is not prime, then the theorem fails. For instance, (2)(345) S 5 has order lcm{2, 3} = 6, but is not the product of commuting 6cycles. 2. First we compute the order of G = GL 2 (F p ). For the first row of the matrix, there are p 2 possibilities, since we cannot have both entries be. For the second row, there are p 2 p possibilities, since anything but a multiple of the first row ensures a nonzero determinant. Thus GL 2 (F p ) = (p 2 )(p 2 p) = p(p + )(p ) 2. By Sylow s theorem, the number n p of Sylowp subgroups of G satisfies n p (p + )(p ) 2 and n p (mod p). 2
4 So we have n p {, p +, ( (p ) ) 2, (p + )(p ) 2 }. Consider the element, which is of order p and generates a Sylow psubgroup P. Its transpose is also order p, so we can rule out n p =. Now, note that if a, d, then ( ) ( ) ( ) ( ) a b d b ad =, ad d a ( ) a b whence every such is in the normalizer N d P (G). There are p(p ) 2 such elements, so since n p = (G : N P (G)) by Sylow s theorem, we obtain the bound n p p +. Thus n p = p (a) We have G = p k for k 2. Let G act on itself by conjugation, with g,..., g r distinct representatives of the noncentral conjugacy classes of G. The class equation gives p k = G = Z(G) + r (G: C G (g i )). Since G = C G (g i ) (G: C G (g i )), 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 p b for each j k (the cases j =, 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 G := G/Z has order p k, whence G has normal subgroups H,..., H k 2 of orders p,..., p k 2, plus the identity subgroup H :=. Letting G π G denote the natural surjection, we claim that the subgroups H j := π ( H j ) for j k 2 are of order p j+ and normal in G. The order statement is clear, since cosets of Z in G have cardinality p. Now for any x H j and g G, we have ḡ xḡ H j, so that since π is a homomorphism, gxg H j. Thus gxg H j, so H j is normal in G. i= 4. We have A = 2 3 and wish to compute the rational and Jordan canonical forms (over Q and C, respectively). The characteristic polynomial is f = (x ) 2 (x 2) and the minimal polynomial is p = (x )(x 2), which is proved by showing that 3
5 p(a) =. Thus the invariant factors are p = p = (x )(x 2) = x 2 3x + 2 and p 2 = x (we must have p p 2 = f), so the rational form is 2 A 3 To confirm this, we note that A =, A =, A 2 = 2, 3 so that a basis that yields the rational form is,, 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 P =, with P =, whence P AP = 2 3. (Since B old v old = B new v new and B new = B old P, we see that v old = P v new, 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 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 and the other of size for the eigenvalue 2. Since the minimal polynomial has linear powers, each block is diagonalizable, so the result is A. 2 4
6 5. First, Z[i] is an integral domain since by definition (a + bi)(c + di) = ac bd + (ad + bc)i is if and only if a = b = or c = d =. We need to prove that there is a norm Z[i] N Z such that for any α = a + bi, β = c + di Z[i], where β, there are p + qi, γ Z[i] such that α = (p + qi)β + γ, with γ = 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 2. 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(γ) = N(β)N(θ) N(β), 2 whence γ 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 ) = ab( ), with as a generator. Moreover, there exist integers a, b such that am+bn = d, so d( ) = (am + bn)( ) = (am bn) =, whence the order of the cyclic group divides d. Now consider the map Z/mZ Z/nZ ϕ Z/dZ defined by (a mod m, b mod n) ab mod d. It is Zbilinear, and the induced linear map Z/mZ Z Z/nZ ϕ Z/dZ maps, an element of order d. Thus has order a multiple of d, whence the cyclic group has order at least d, and therefore exactly d. So ϕ is an isomorphism. 8. Skip. 9. (a) First, note that x 4 5 is irreducible over Q by Eisenstein. Letting α denote the positive real number 4 5, the roots of x 4 5 in C are 4 5, ζ 4 5, ζ 2 4 5, ζ 3 4 5, 5
7 where ζ = i is a primitive fourth root of unity. Thus the splitting field of E of x 4 5 is E = Q( 4 5, i), which has degree 8 over Q. Since i is a root of the irreducible polynomial x 2 +, the 8 elements of G = Gal(E/Q) are determined by 4 5 i k 4 5, i ±i, all of which are automorphisms of E. Letting σ : 4 5 i 4 5, i i and τ : , i i, we see that G = σ, τ = {, σ, σ 2, σ 3, τ, στ, σ 2 τ, σ 3 τ}, and that σ, τ satisfy the relations σ 4 = τ 2 = and τσ = σ 3 τ. Thus G D 4. The inverted lattice of subgroups of D 4 is (DF 69) τ σ 2 τ σ 2 στ σ 3 τ σ 2, τ σ σ 2, 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(( + i) 4 5) Q(( i) 4 5) Q( 5) Q(i) Q(i 5) Q (b) The Galois group of Q( 4 5, i) over Q( 5) is σ 2, τ, which is isomorphic to Z 2 Z 2, the Klein Viergruppe. 6
8 (c) The Galois group of Q( 4 5) over Q(i) is σ, isomorphic to Z 4.. The extension Q( 2)/Q is Galois of degree 2, with basis {, 2} over Q. We claim that 3 / Q( 2). For suppose (a + b 2) 2 = 3 with a, b Q. Then a 2 + 2b 2 + 2ab 2 = 3, whence a or b is, but 3 is not the square of a rational number. Since Q( 3)/Q is of degree 2, so is Q( 2, 3)/Q( 2), whence Q( 2, 3)/Q is of degree 4, with basis {, 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 (x 2 2)(x 2 3), hence a Galois extension of Q. Automorphisms of Q( 2, 3) fixing Q must map 2 ± 2, 3 ± 3. Since there are four such maps, and the degree of the extension is 4, each of these maps is an automorphism. Let σ : 2 2, 3 3 τ : 2 2, 3 3. Then G = Gal(Q( 2, 3)/Q) = σ, τ. Since σ 2 = τ 2 = and στ = τσ, G is isomorphic to Z 2 Z 2.. Recall that for polynomials f, g F[x, y], where F is a field, we have deg(fg) = deg f + deg g. Suppose x 2 + y 2 is reducible, factoring into the nonunits f, g. Then f, g must each be of degree, so we get an expression (ax + by + c)(a x + b y + c ) = x 2 + y 2. This implies aa = bb = cc =, 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, contradicting the former equality. So there is no such factorization, in either Q[x, y] or C[x, y]. 7
9 Chapter 2 Fall 2. We begin by computing the order of G := SL 2 (F 5 ). The order of H := GL 2 (F 5 ) is (5 2 )(5 2 5), since the first row of a matrix can be anything except, 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 )(5 2 5) 4 = = 2. By Sylow s theorem, the number n 5 of Sylow5 subgroups of G satisfies n 5 24 and n 5 (mod 5). So n 5 = or n 5 = 6. But ( ) and ( ) are elements of degree 5 that generate distinct subgroups of order 5, so n 5 = The derived subgroup G () of G consists of the uppertriangular matrices of the form ( ) a, a R. A simple computation shows that every commutator is of this form, and we have ( ) ( ) ( ) ( ) ( ) ( ) ( ) a a a a a = =, and elements of this form do not generate any new elements. Note that G () is abelian, so that G (2), the subgroup of commutators of G (), is trivial. Thus the 8
10 derived series shows that G is solvable. = G (2) G () G () = G 3. We compute σ 2 = τ 2 =, that στ τσ, and that (στ) 3 =, whence τσ = (στ) 2. Set a = στ and b = σ. Then the group contains the six elements {, a, a 2, b, ba, b 2 a} and is nonabelian, so it is isomorphic to D 3 S 3. One way of proving the isomorphism is to consider how σ, τ act on the x values {2, 2, }. 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 2cycles (2) and (3) generate S 3, we get a surjective homomorphism σ, τ S 3. 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 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 A 5 + 2A 3 + 3A 6I = if and only if U AU does as well, it suffices to classify real diagonal matrices A with eigenvalues in increasing order that satisfy A 5 + 2A 3 + 3A 6I =. Since A is diagonal, each entry on the diagonal must be a zero of the polynomial f(x) = x 5 + 2x 3 + 3x 6. One such zero is x =, but we have f (x) = 5x 4 + 6x 2 + 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 =. 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 A 3 2A 2 + A =. This means the minimum polynomial p of A must divide x(x ) 2, so the possibilities for p are x, x, x(x ), (x ) 2, and x(x ) 2. Since p divides the characteristic polynomial, the nine possible Jordan forms (up to reordering of blocks), which correspond to conjugacy classes, are: 9
11 Minimal Polynomial Characteristic Polynomials Jordan Forms x x 4 x (x ) 4 (x ) 2 (x ) 4, x(x ) x 3 (x ), x 2 (x ) 2 x(x ) 3 x(x ) 2 x 2 (x ) 2, x(x ) 3 7. Note that α = 7 /6. The field extensions Q(α 3 )/Q and Q(α 2 )/Q have degrees 2 and 3 respectively (x 2 7 and x 3 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, α,..., α 5 are linearly independent over Q, hence also over Z, and x 6 7 is
12 the minimal polynomial of α (or use Eisenstein on x 6 7 to conclude that it is irreducible). It follows that Q(α) Q[x]/(x 6 7), using the map α x, so that also Z[α] Z[x]/(x 6 7) as subrings. Moreover, Z[α]/(α 2 ) (Z[x]/(x 6 7))/( x 2 ) Z[x]/(x 2, 7) (Z/7Z)[x]/(x 2 ), so this ring consists of Z[α]/(α 2 ) = {a + bα : a, b Z/7Z}, which consists of 7 7 = 49 elements. The elements of the form a + α, a are units. Thus every ideal of Z[α]/(α 2 ) is principal since if it is proper and nonzero, then it must contain + bα for some b, whence it contains b 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 α 2 =, so α is a zero divisor. 8. (i) Let α denote the positive real number α = 3 /6. By the argument in the previous problem (or Eisenstein), Q(α)/Q is an extension of degree 6. Over C, the polynomial x 6 3 factors as x 6 3 = (x 3 3 /2 )(x /2 ) = (x α)(x 2 + αx + α 2 )(x + α)(x 2 αx + α 2 ), so that using the quadratic formula, we see that the roots are ±α, ±α ± α 3. 2 Thus the splitting field of x 6 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 2, 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 x 6 3 is irreducible...? I don t know very much about finite fields. 9. Skip.. x p 2 is irreducible over Q by Eisenstein, with distinct roots p 2, ζ p 2,..., ζ p p 2,
13 where ζ is a primitive pth root of unity. Thus the splitting field of x p 2 is Q(ζ, p 2). Since x p 2 is irreducible, Q( p 2) is a degree p extension of Q. Since ζ is a root of the irreducible polynomial (x p )/(x ) = x p +x p (one proves irreducibility by substituting x + 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 extension of Q. Since p and p are relatively prime, we see immediately that the degree of the extension Q(ζ, p 2) over Q is p(p ). The Galois group has order p(p ), so each of the p(p ) elements of the form p σ : 2 ζ n p 2, ζ ζ m, where n p and m p, is an automorphism. If p = 2 then the Galois group is clearly abelian, so suppose p > 2. Setting σ : p 2 ζ p 2, ζ ζ p τ : p 2 2, ζ ζ 2, we see that (στ)( p 2) = ζ p 2, but (τσ)( p 2) = τ(ζ p 2) = ζ 2 p 2, so the Galois group is not abelian. 2
14 Chapter 3 Spring 2. For σ A n, let σ = C... C r be the (disjoint) cyclic decomposition. Suppose σ has order 2. Then 2 = σ = lcm{ C i } implies that each C i is a 2cycle. Since σ is an even permutation, r is even. The product (ab)(cd) of two disjoint 2cycles are the square of the 4cycle (acbd). Thus grouping the C i into pairs and taking the corresponding 4cycles gives an element τ S n of order 4 whose square is σ. 2. We may assume G = p k for k 2. We first prove the result when G is not abelian. Consider the homomorphism where G σ Aut(G), g σ g ; σ g := x gxg, x G. The kernel of σ is Z(G), the center of G, which is a subgroup of G and therefore a pgroup. The image of σ is a subgroup of Aut(G), which is also a pgroup since it is isomorphic to G/ker σ. The image is nontrivial since G Z(G) by assumption, so we are done by Lagrange s theorem. If G is abelian, then G = Z p k Z p k 2 Z p kr, where k i k i+ for each i and k + + k r = k 2. Let x,..., x r denote generators of the factors of G. It suffices to construct an automorphism with order p. If k =, then r 2, and the automorphism x 2 x x 2 has order p. So we may assume k 2, in which case the homomorphism ϕ: x x pk + 3
15 is an automorphism since p k + is relatively prime to p k. We claim that ϕ has order p. To see this, note that p ( ) p (p k + ) p = p (p n)(k ), n n= so since p ( ) p n for each n p, every term in the sum is divisible by p k except the last term, which is ( p p) p =. Thus x x pk + x (pk +) 2 x (pk +) p = proves that ϕ p =. To see that no lower power of ϕ is, note that when s < p, s ( ) s (p k + ) s = p (s n)(k ) n n= has exactly two terms not divisible by p k, whose sum is ( ) s p k + = s p k +, s which cannot be congruent to modulo p k. 3. Choose nonzero x M, and note that Rx = M. Thus there is a surjective Rmodule homomorphism R ϕ M defined by r rx. The kernel I is a (twosided) ideal of R. Thus M R/I as Rmodules, 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. 4. Q/Z is a torsion Zmodule, hence not a submodule of a free Zmodule, so it is not projective. Another way to see this is to note that the short exact sequence Z Q Q/Z does not split, since Q does not contain a submodule isomorphic to Q/Z, and there are no nontrivial Zmodule homomorphisms Q Z. Q/Z is injective since a Zmodule M is injective if and only if it is divisible, namely if nm = M for all 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 2n does not remain injective after tensoring with Q/Z (look at the element ( 2 + Z) ). 5. If n p =, G has a normal Sylow psubgroup, so assume n p >. By Sylow s theorem n p (mod p) and n p q, 4
16 so we deduce that n p = q and n p = + kp for some k. If we also assume that n q >, then since n q (mod q) and n q p 2, we must have n q = p 2. But these results imply that G has q (p 2 ) elements of order p or p 2 and p 2 (q ) elements of order q, which is a contradiction since if we also count the identity, then + q(p 2 ) + p 2 (q ) = + p 2 q q + p 2 q p 2 + p 2 q = + G. Another method of proof is to use the second application of Sylow s theorem to deduce that q p 2, whence q p or q p + since q is prime. But the first application of Sylow s theorem implies q = + kp with k, so the only possibility is q = p +, 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 2 to get the result. 6. We have M a 5 5 real matrix satisfying (M I) 2 =, and wish to show the subspace of R 5 of vectors fixed by M has dimension at least 3. Since the minimal polynomial p of M is either x or (x ) 2, the characteristic polynomial of M is f = (x ) 5. Viewing M as a complex matrix, the possible Jordan forms for M are,,. These Jordan forms fix subspaces of dimensions 5, 4, and 3, respectively. (If the new basis is {e,..., e 5 }, then the respective fixed subspaces are V, e 2, e 3, e 4, e 5, and e 2, e 4, e 5.) 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 Rmodules 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 nontrivial Rmodule that is not free since multiplication by x kills every element. 8. I m assuming Z 7 denotes the finite field with 7 elements, not the 7adic 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 Z 7. Now it seems difficult to proceed... 5
17 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 α}, so E is the splitting field of the separable polynomial x n a, whence E/F is Galois. Any automorphism of E fixing F must be a homomorphism of the form α ζ r α, r n. Let d > be minimal such that α d F, whence d n since α n = a F. We claim that x d α d F [x] is the minimal polynomial of α over F. For if f(x) F [x] is the minimal polynomial, then f (x d α d ), hence every root of f is in the set of roots above. Thus the constant term of f is of the form ( ) s ζ d ζ ds α s, which must be in F, whence d s, where s = deg f, so we must have f = x d α d. It follows that x d α d is irreducible and E/F has degree d. If α ζ r α defines an automorphism, then α d ζ rd α d implies ζ rd =, 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 ) = {, α ζ n/d α,..., α ζ (d )n/d α}, which is cyclic with generator α ζ n/d α.. Theorem (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 a,..., a n A since A is Noetherian. Choose elements f,..., f n a so that each f i has leading coefficient a i. Let r i denote the degree of f i and set r := max i {r i }. Then if f a has degree r and leading coefficient a I, choose u i A such that a = u a + + u n a n. It follows that by multiplying u i f i by appropriate powers of x and taking the sum, we can kill off the leading term of f. Thus a = (a M) + (f,..., f m ), where M is the Amodule generated by, x,..., x r. Since M is finitely generated and A is Noetherian, M is Noetherian, hence a M M is finitely generated. Combining the generators with the f i gives a finite generating set for a. 6
18 Chapter 4 Fall 29. Every element of S 7 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 4cycles, and the combinations of 4cycles with disjoint 2cycles. The number of 4cycles is ( 7 4) 3! = 7 6 5, and the number of possibilities for 2cycles using the remaining 3 indices is ( 3 2) = 3. Thus the total number of elements of order 4 is 7 6 5( + 3) = Let G act on itself by conjugation. Let g,..., g r denote representatives of the conjugacy classes of noncentral elements. The class equation yields G = Z(G) + r (G : C G (g i )), i= where C G (g i ) is the centralizer of g i 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 nontrivial. 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 σ = G, i.e. G is cyclic. Since F/E is a proper extension, we can choose p prime such that G = p k n, with k and (n, p) =. We wish to prove that n =, so assume for contradiction that n >. Then σ pk and σ n are proper subgroups of G, hence fix α. Since (p k, n) =, the Euclidean algorithm ensures we can find 7
19 a, b Z such that ap k + bn =. But then contradiction. σ(α) = (σ pk ) a (σ n ) b (α) = α, 4. If m n Q/Z is reduced, by which we mean < m < n and gcd{m, n} =, then m n has order n in Q/Z. Thus m n contains n elements, so that in particular we see that m n = n. Given n, n 2 Q/Z with n, n 2 2, let n = lcm{n, n 2 }. We claim that n,, n whence n, n 2 = n 2 n since the inclusion is obvious. For the claim, we use the Euclidean algorithm on n and n 2 to find a, b Z such that an + bn 2 = gcd{n, n 2 }, and then divide both sides by n n 2 to get b n + a n 2 = lcm{n, n 2 } = n. This proves the claim. Now we are ready to prove the result of the problem. Given m,..., m r Q/Z, n n r where each generator is reduced, set n = lcm(n,..., n r ). Then we have m,..., m r =. n n r n For the previous comments imply we can replace each mi n i by n i, and the above method for combining two generators, together with induction on the number of generators, give the result. 5. We compute 52 = Suppose G is simple. Then n 3 (mod 3) and n
20 implies n 3 = since n 3 >. Likewise n 5 = 2 3. So we have (3 ) elements of order 3 and (5 ) 2 3 elements of order 5. But contradiction > = = G, 6. (a) We have M a matrix satisfying M 2 + M + I = and wish to find the possible rational canonical forms over R. Since x 2 + x + = (x ζ)(x ζ) for ζ = +3i 2, we see that x 2 + x + is irreducible over R, hence is the minimal polynomial of M over R. The corresponding rational block is ( ), and M must be composed of 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 ζ, and (x ζ)(x ζ). In any case, the minimal polynomial factors linearly, so M is diagonalizable, with eigenvalues ζ and ζ. Thus the Jordan form of M is an n n diagonal matrix (n ) with some number r of ζs ( r n), then (n r) ζs on its diagonal. 7. Consider the ring of fractions R := S R. Prime ideals of R correspond to prime ideals of R that don t meet S via the ring homomorphism R π R, r r, 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 a of R, so a is prime, hence a = π (a ) is prime as well. 8. Skip. 9. First we state a general result. If ζ n is a primitive nth root of unity, then the Galois group of Q(ζ n )/Q is isomorphic to the multiplicative group (Z/nZ) of order φ(n). For ζ n is a root of the degree φ(n) irreducible cyclotomic polynomial Ω n (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(ζ n ) = ζ a n, and there are precisely φ(n) possibilities for a, every such map is an automorphism of Q(ζ n ), whence we obtain an isomorphism (Z/nZ) Gal(Q(ζ n )/Q) 9
21 a (ζ n ζ a n). Now we take n = above and let ζ be a primitive th root of unity. The Galois group G of the extension Q(ζ)/Q is isomorphic to the cyclic group (Z/Z) Z/Z 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 G G/H Z/5Z.. 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) = x n + a n x n + + a R[x] a monic polynomial such that each a i p, but a / p 2. 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 nonconstant polynomials (i.e. nonunits, since the only constants that divide are units). Reducing modulo p, we see that x n a(x) b(x) in R/p[x], which is an integral domain since R/p is an integral domain. Thus both ā and b must have constant term (otherwise the nonzero lowest degree terms of ā and b, which exist since neither ā nor b can be 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 p 2, contradiction. 2
22 Chapter 5 Spring 29. Elements of S n with the same cycle type are conjugate to each other. There are (p )! pcycles in S p, so the order of the orbit of σ under conjugation is (p )!. Thus C = S P /(p )! = p, so C = σ, which is abelian. Now let S p act on the set of its subgroups by conjugation. Since there are (p )! pcycles in S p, there are (p )!/(p ) = (p 2)! distinct subgroups of order p, all of which are in the orbit of σ since σ is conjugate to every other pcycle. So the order of N is p!/(p 2)! = p(p ). To see that N is not abelian, let τ N be an element satisfying Then τστ σ = σ. τστ = σ The derived subgroup G () of G consists of the p uppertriangular matrices of the form ( ) a, a Z/pZ. A simple computation shows that every commutator is of this form, and we have ( ) ( ) ( ) ( ) ( ) ( ) ( ) a a a a a = =, and elements of this form do not generate any new elements. Note that G () is abelian, so that G (2), the subgroup of commutators of G (), is trivial. Thus the derived series = G (2) G () G () = G shows that G is solvable. 2
23 3. We have a group G with G = 8 = 3 4 acting on a set X with X = 3. Since the stabilizer S x of x X is a subgroup of G, S x is a power of 3. The index (G : S x ) of S x in G also divides G, hence is a power of 3. Since the cosets of S x are in bijection with the elements of the orbit O x of x, (G : S x ) = O x, so O x 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 S x < 27 for each x X, so that O x > 3 for each x. This is impossible since no sum involving only 9 s and 27 s equals 3. 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, and suppose that for each prime p R, R p contains no nonzero nilpotents. Now if a R is nilpotent, then a n = 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 a R m is a nonzero nilpotent, yet ( a )n =, 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 = 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 b q r x 4 4x 3 + 3x 2 4x + 2 x 3 5x 2 + 6x 2 x + 2x 2 8x + 4 x 3 5x 2 + 6x 2 2x 2 8x x 2 Thus the gcd of the polynomials is 2x 2 8x+4, so the ideal a is (x 2 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
24 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 t 2. 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.. ζ = e 2πi/8 is a primitive 8th root of unity. Thus by the argument in 9 of Fall 29, Q(ζ)/Q has Galois group G (Z/8Z) Z 2 Z 2, generated by The inverted subgroup lattice is σ : ζ ζ 3 and τ : ζ ζ 5. σ στ τ σ, τ with corresponding lattice of fixed fields Q(ζ) Q(ζ + ζ 3 ) Q(ζ + ζ 7 ) Q(ζ 2 ) where the latter cannot be Q(ζ + ζ 5 ) since ζ + ζ 5 =! This is much clearer if we take a different approach. Write ζ = 2 ( + 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 ± 2, i ±i, Q 23
25 and since there are only 4 such possibilities, they must all be automorphisms. Let σ : 2 2, i i τ : 2 2, i 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) and we note that ζ + ζ 3 = i 2, ζ + ζ 7 = 2, and ζ 2 = i. Q 24
26 Chapter 6 Fall 28. By Sylow s theorem, n 5 (mod 5) and n ; n 7 (mod 7) and n The only possibilities are n 5 = n 7 =, so G has normal subgroups H, K of orders 5 2 and 7 2, respectively. Since 5, 7 are relatively prime, H K = {}. 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 p 2 are abelian, hence either isomorphic to Z p Z p or Z p 2. (Use the class equation to show G has nontrivial center; so if g 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, with possible isomorphism classes Z 5 2 Z 7 2, Z 5 2 Z 7 Z 7, Z 5 Z 5 Z 7 2, Z 5 Z 5 Z 7 Z 7. 25
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