Practice Algebra Qualifying Exam Solutions

Practice Algebra Qualifying Exam Solutions 1. Let A be an n n matrix with complex coefficients. Define tr A to be the sum of the diagonal elements. Show that tr A is invariant under conjugation, i.e., tr A = tr PAP 1 for all invertible n n matrices P. Proof. Let P be an invertible matrix. Let p k be the k-th row of P, q j the j-th column of P 1, and a i the i-th column of A. The k-th row of the matrix PA is So the k, k-th entry of PAP 1 is Hence Now, the quantity tr(pap 1 ) = p k a 1,..., p k a n. p k a 1,..., p k a n q k. = = p k a 1,..., p k a n q k j=1 j=1 p k,i q j,k = a i,j p k,i q j,k a i,j n p k,i q j,k. { 1 if i = j 0 if i j since it represents the dot product of the i-th column of P and the j-th row of P 1. So n tr(pap 1 ) = p k,i q j,k = a j,j = tr(a), as desired. j=1 a i,j 2. Let K/F be a degree n Galois extension of fields with Galois group G. Let α K and A : K K be defined by x αx. We view K as a vector space over F so that A is an F-linear transformation. Choose an F-basis for K/F so that we may consider A as an n n matrix with coefficients in F. (a) Prove that N K F (α) := j=1 α σ = det A and tr K F (α) := α σ = tr A. Proof. Let m A (x) F[x] be the minimal polynomial for the transformation A, and m α (x) be the minimal polynomial for the algebraic element α. Note that m α (A)x = m α (α)x = 0. So m A (x) m α (x). But m α (x) is irreducible, so it follows that m A (x) = m α (x) =: m(x), moreover, the invariant factors of A are m(x), m(x),..., m(x)

2 with multiplicity k = n/ deg m(x) = H where H is the subgroup of G belonging to Q(α) by Galois correspondence. Since the determinant and trace are invariant under conjugation we might as well assume A is in JCF form. Considering the invariant factors of A, we get det A = /H(α σ ) k = α σ and as claimed. tr(a) = /H kα σ = (b) Let V be a cyclic subspace of K corresponding to an invariant factor of A. Is V necessarily a subfield of K? α σ, Proof. No. Let F = Q, k = Q(α) and K = k(β). Then V = Qβ Qαβ Qα deg m(x) 1 β is a cyclic subspace corresponding to an invariant factor of A, namely m(x). The field generated by V is K which, in general, has dimension over Q larger than V. 3. (a) Classify all groups, up to isomorphism, of order 10. Proof. Let G be a group of order 10. Let n 5 = Syl 5 (G), so n 5 1 mod 5 and n 5 2. Hence n 5, so there is a unique normal subgroup of G of order 5, say P. Let Q Syl 2 (G). Then PQ = {pq : p P, q Q} is a subgroup of G since P is normal, moreover, PQ = P Q / P Q 5. So G = PQ. For p i P and q j Q, note that p 1 q 1 p 2 q 2 = p 1 (q 1 p 2 q 1 1 )q 1q 2. Let ϕ : Q Aut(P) be defined by ϕ : q (φ q : p qpq 1 ). Since Q = 2, either ϕ is the trivial map or the non-identity element of Q maps to the inversion automorphism of P. Suppose ϕ is trivial. Then p 1 q 1 p 2 q 2 = p 1 p 2 q 1 q 2, hence G P Q. Since P and Q are cyclic and of co-prime order, it follows easily that G is cyclic of order 10. Suppose ϕ is not trivial. Let q Q be the non-identity element. Then for all p P we have qpq 1 = p 1, i.e., qp = p 1 q. Hence G p, q p 5 = q 2, qp = p 1 q D 10.

3 So there are, up to isomorphism, two groups of order 10: the cyclic group of order 10 and the dihedral group of order 10. (b) Suppose G is the Galois group of an irreducible degree 5 polynomial f Q[x] such that 0. Then G is non-abelian and f must have precisely 1 or 5 real roots. Proof. Suppose, for contradiction, that G is abelian, and let θ be a root of f(x). Let K be the splitting field of f. Since every subgroup of an abelian group is normal, it follows that Q(θ) K is normal. If this were the case, then f would split in Q(θ), so = 5 contrary to hypothesis. So G is not abelian. Let Θ = {θ 1,..., θ 5 } be the roots of f. The action of G on Θ induces an embedding ϕ : G S 5. The action of G on Θ is transitive. For each σ G, let fix(σ) = {θ i Θ : θ σ i = θ i }. Considering the cycle types in S 5 we get 0 if ord G (σ) = 5 1 if ord G (σ) = 2, ϕσ = (ab)(cd) fix(σ) = 3 if ord G (σ) = 2, ϕσ = (ab) 5 else. Recall the formula for Θ/G, the number of distinct G-orbits in Θ: Θ/G = 1 orb G (θ i ) stab G (θ i ) θ i Θ θ i Θ θ i Θ σ stab G (θ i ) θ σ i =θ i 1 fix(σ). In this case, we have 1 fix(σ) 1 (0 + 0 + 0 + 0 + 1 + 1 + 1 + 1 + 1 + 5), 10 10 the right hand side of the above inequality occuring when ϕg contains only 5-cycles and 2 2-cycles. It follows that ϕg must indeed only contain 5- cycles and 2 2-cycles. This means every non-identity element of G fixes, at most, one root θ i. So Q(θ i ) Q(θ j ) unless i = j. Moreover, complex conjugation must either fix every root (5 real roots) or one root (1 real root). 4. Prove [ or disprove: ] If G GL(2, F p ), then G contains an element conjugate to 1 1 A = if and only if p. 0 1 1

4 Proof. Suppose p. Let g G such that g p 1 = (g 1) p = 0. It follows that the minimal polynomial for g, say m g (x), divides (x 1) 2 (considering this is dimension 2). If m g (x) = x 1, then g does not have order p. So it must be that m g (x) = (x 1) 2 whence the A = JCF(g). So g is conjugate to A. Suppose G contains a matrix conjugate to A. Then there is a g G such that (g 1) 2 = 0. Let n = ord G (g), so g n 1 = 0. Since m g (x) = (x 1) 2, it follows that (x 1) 2 x n 1. Note that x n 1 has 1 as a repeated root if and only if d dx (xn 1) = 0, x=1 that is, if and only if n 0. Let n = pj, then g j has order p in G. 5. Let R be a unique factorization domain. Show that R is a principal ideal domain if and only if every prime ideal of R is a maximal ideal. Proof. Suppose R is a PID. Let p = (p) be a prime ideal. Then p m = (m) for some maximal ideal m. Let r R such that rm = p. Then either m p or r p. In the former case, we re done. In the latter, write r = ps for some s R. Then psm = p, hence, sm. This means m is a unit contrary to the assumption that m is a maximal ideal. Conversely, suppose every prime ideal of R is maximal. Let m be a maximal ideal of R. Let r m. Since R is a UFD, we factor r = p e 1 1 pe r r into irreducibles. Since maximal ideals are always prime, it follows that there exists some p i such that p i m (by repeated application of the definition of prime ideal). Since prime ideals are maximal, it must be that (p i ) = m. Now, let I be any ideal. Then I is contained in some maximal ideal m. By what has already been shown, we know that m = (p) for some irreducible p R. Let s I and factor s = p f 0 q f 1 1 qf t t into irreducibles. We have the containment of ideals (p f 0 q f 1 1 qf t t ) I (p) R. If I (s), choose s 1 I \ (s). Then gcd(s, s 1 ) factors gcd(s, s 1 ) = p f 0,1 q f 1,1 1 q f t,1 t, where f i,1 < f i for at least one i. If I (gcd(s, s 1 )), choose s 2 I \ (gcd(s, s 1 )), ad nauseam. Since f 0,j 1 for all j, this process must terminate. Hence R is principal. 6. Let F/Q be a finite Galois extension. Let R be a unique factorization domain such that Z R F, R is Galois invariant, and N F Q (R) Z. Let p Z be prime, and let p = p e 1 1 pe r r be the prime factorization of p into irreducibles p j R. (a) Show that r Gal(F/Q).

5 (b) Show that e 1 = e 2 = = e r and N F Q(p 1 ) = = N F Q(p r ) = p f, where efr = Gal(F/Q) Proof. Note that (p j ) Z is a non-trivial proper ideal (non-trivial since p (p j ) Z, and 1 (p j ) Z (p j ) implies p j is a unit, contrary to assumption). Since (p) (p j ) Z, it follows that (p) = (p j ) Z. Let G = Gal(F/Q), and note that N F Q(p j ) = p j (p j ) Z = (p), σ id p σ j whence p N F Q (p j). It s easy to see that if p is irreducible, then so is p σ for every σ G. With that in mind, note that for every σ G, we have (σp 1 ) e1 (σp r ) e r = σp = p. It follows that for every j there exists a unique k such that p σ j = p k up to a unit, moreover, if p k orb G (p j ) then e k = e j. We have a well-defined G action on the collection of ideals P := {(p 1 ),..., (p r )}, specifically, σ (p j ) := (p σ j ). Let p k, p j P be distinct. Consider the chain of divisibilities p k p N F Q(p j ) = p σ j. It follows that there exists σ G such that p σ j (p k) since (p k ) is a prime ideal, so (p σ j ) = (p k). Hence G acts transitively on P. From the remark at the end of the previous paragraph, we get that e 1 = e 2 = = e r. Also, by the orbit-stabilizer theorem we get that = orb G (p j ) stab G (p j ) = r stab G (p j ), hence r. Let p j P. Then, by the orbit-stabilizer theorem, we have N F Q(p j ) = (σp j ) /r = / stab G (p j ) p /r i Since p j was arbitrary we get the first statement of the claim. Now, using the fact that p = (p 1 p r ) e and taking norms of both sides, we see p = (N F Q(p i )) e. Since N F Q (p j) Z, it must be that N F Q (p j) = p f. The above equality then gives us that p = p ef = p efr, whence = efr.