Solutions to the January 2009 Qualifying Examination

Any student with questions regarding the solutions is encouraged to contact the Chair of the Qualifying Examination Committee. Arrange- ments will then be made for the student to meet with a faculty member who can answer their questions.

1 Qualifying Examination Solutions (January 2009)

Abstract Algebra

1. Let H,K be subgroups of a group G. Put HK = {hk | h ∈ H, k ∈ K}.

(a) Prove that H ∩ K is a subgroup of G.

Solution: Note that H ∩ K 6= ∅ since 1 ∈ H ∩ K. As H ∩K is a subset of the group G, it suffices to show that H ∩K is closed under multiplication and inversion. Since H and K are both closed under multiplication, we have g, g1 ∈ H ∩ K =⇒ g, g1 ∈ H and g, g1 ∈ K

=⇒ gg1 ∈ H and gg1 ∈ K =⇒ gg1 ∈ H ∩ K. Similarly, since H and K are both closed under inversion, we have g ∈ H ∩ K =⇒ g ∈ H and g ∈ K =⇒ g−1 ∈ H and g−1 ∈ K =⇒ g−1 ∈ H ∩ K.

|H| · |K| (b) Let G be finite. Prove that |HK| = . |H ∩ K| Solution: Let ϕ : H × K → HK be the map given by ϕ(h, k) = hk for all h ∈ H, k ∈ K. As ϕ is onto, we can express H × K as a disjoint union: [ H × K = ϕ−1(g). (1) g∈HK Since

ϕ(h, k) = ϕ(h1, k1) ⇐⇒ hk = h1k1 −1 −1 ⇐⇒ h h1 = kk1 = m for some m ∈ H ∩ K −1 ⇐⇒ (h1, k1) = (hm, m k) for some m ∈ H ∩ K, it follows that ϕ−1(g) = (hm, m−1k) | m ∈ H ∩ K} if g = hk for some h ∈ H, k ∈ K. Thus, |ϕ−1(g)| = |H ∩ K| for all g ∈ HK. In view of (1) we derive that X X |H| |K| = |H × K| = |ϕ−1(g)| = |H ∩ K| = |HK| |H ∩ K|, g∈HK g∈HK and the result follows.

(c) Let G be finite. Assume that the indexes [G : H] and [G : K] are relatively prime. Prove that HK = G. Solution: Let m = [G : H], n = [G : K], r = [H : H ∩ K], s = [K : H ∩ K]. By hypothesis, gcd(m, n) = 1. Observe that mr = [G : H][H : H ∩ K] = [G : H ∩ K] = [G : K][K : H ∩ K] = ns. In particular, n divides mr, and since gcd(m, n) = 1, it follows that n divides r; thus, n ≤ r. On the other hand, using the result of part (b), we see that r ≤ n since |H| |HK| |G| r = [H : H ∩ K] = = ≤ = [G : K] = n. |H ∩ K| |K| |K| Therefore, r = n, and the inequality (≤) above becomes an equality (=). This shows that |G| = |HK|, or G = HK.

2. Let R be a commutative ring with multiplicative identity 1 6= 0.

(a) State the definition of ideal in R.

Solution: An ideal is a nonempty subset I ⊆ R such that (i) a + b ∈ I for all a, b ∈ I; (ii) ra ∈ I for all r ∈ R, a ∈ I.

(b) Prove that R is a field if and only if the only ideals in R are 0 and R.

Solution: First, suppose that R is a field, so that every nonzero element of R is invertible. Let I be a nonzero ideal in R, fix a nonzero element a ∈ I, and let b ∈ R be the inverse of a (that is, ba = 1). Then, for every element r ∈ R, since a ∈ I and rb ∈ R, we have by property (ii) from part (a): r = r · 1 = r · ba = rb · a ∈ I. This shows that R ⊆ I, and therefore I = R. Now suppose that the only ideals in R are 0 and R. Let r be a nonzero element of R. It is easily seen that the set I = {ra | a ∈ R} is a nonzero ideal in R; therefore, I = R, and since 1 ∈ R we have ra = 1 for some a ∈ R. This argument shows that every nonzero element of R is invertible, hence R is a field.

3. For an integral domain R, let R[x] be the polynomial ring over R in the vari- able x. For r ∈ R, let φr : R[x] → R be the evaluation homomorphism defined by
φr f(x) = f(r).

(a) Prove that if I is a proper ideal of C[x], then there exists r ∈ C with φr(I) 6= C.

Solution: Since C is a field, C[x] is a principal ideal domain. For any r ∈ C, the principal ideal hx − ri is maximal. On the other hand, since C is algebraically closed, the only irreducible polynomials are the linear polynomials, hence every maximal ideal in C[x] has the form hx − ri for some r ∈ C. If I is a proper ideal of C[x], it is contained in a maximal ideal hx − ri for some r ∈ C. Then, every f(x) ∈ I can be factored in the form f(x) = (x − r)g(x) with some g(x) ∈ C[x], and we have φr(f(x)) = f(r) = (r − r)g(r) = 0. Therefore, φr(I) = h0i= 6 C.

(b) Let I be the ideal of Z[x] generated by 3 and x2 +1. Prove that I is a proper ideal of Z[x], and that φr(I) = Z for every r ∈ Z.

Solution: Let F3 = Z/3Z, and let ρ1 be the natural mapping of Z[x] onto F3[x]. Let 2 ρ2 be the natural mapping of F3[x] onto F3[x]/hx + 1i. The composite mapping 2 ρ = ρ2 ◦ρ1 maps Z[x] onto F3[x]/hx +1i, the latter being a set with nine elements. 2 On the other hand, ρ does not map I onto F3[x]/hx + 1i (and thus, I is a proper ideal of Z[x]). Indeed, since 3 and x2 + 1 both lie in the kernel of ρ, we have ρ(I) = h0i. Fix an arbitrary r ∈ Z. Since r2 +1 ≡ 1 or 2 (mod 3), we have gcd(3, r2 +1) = 1. Hence, there exist integers m and n such that 3m + (r2 + 1)n = 1. For any k ∈ Z 2 let fk(x) = k(3m + (x + 1)n); then fk(x) is contained in the ideal I, and we have 2 φr(fk(x)) = fk(r) = k(3m + (r + 1)n) = k.

This shows that φr(I) = Z.

4. Let F be a field and suppose f(x) ∈ F [x] is irreducible. Fix a positive integer n. Let g(x) = f(xn), and let h(x) be any irreducible factor of g(x) in F [x].

(a) Identify F as a subfield of the field K = F [x]/hhi. Prove that the degree [K : F ] for the field extension F ⊂ K is equal to the degree of h(x).

Solution: More generally, if p(x) is any irreducible polynomial in F [x], then hpi is a maximal ideal in F [x], and L = F [x]/hpi is an algebraic extension of F , and we have [L : F ] = deg p. To prove this, let φ be the canonical epimorphism of F [x] onto L = F [x]/hpi, let α = φ(x), and suppose that deg p = r. It suffices to show that the set B = {1, α, . . . , αr−1} is a basis for L over F . r−1 First, suppose that a0 + a1α + ··· + ar−1α = 0. Then φ(q(x)) = 0, where r−1 q(x) = a0 +a1x+···+ar−1x , hence p(x) | q(x). Since deg q ≤ r −1 < r = deg p, it must be the case that q(x) = 0, i.e., a0 = a1 = ··· = ar−1 = 0. This shows that B is a linearly independent set. Since φ is onto, every element of L has the form u(α) = φ(u(x)) for some u(x) ∈ F [x]. Writing u(x) = p(x)q(x) + t(x), where deg t ≤ r − 1, and taking into account the fact that p(α) = φ(p(x)) = 0, it follows that u(α) = t(α). As t(α) clearly lies in the span of B, we conclude that B spans L, and we are done.

(b) Show that the degree of h(x) is a multiple of the degree of f(x).

Solution: As above, let α be the image of x under the canonical epimorphism of F [x] onto K = F [x]/hhi. So α ∈ K. Consider the subfield F (αn) of K generated by αn. The field extensions F ⊂ F (αn) ⊂ K yields the identity: [K : F ] = [K : F (αn)] · [F (αn): F ]. It follows from (a) that the degree of h(x) is a multiple of [F (αn): F ]. It remains to show that [F (αn): F ] is equal to the degree of f(x). To see this, define a ring homomorphism φ : F [x] → F [αn] = F (αn) by mapping u(x) ∈ F [x] to u(αn) ∈ F [αn]. Since φ is surjective, we obtain F (αn) ∼= F [x]/Ker(φ). Note that f(αn) = g(α) = 0 since h(α) = 0 and g(x) is a multiple of h(x). So f(x) ∈ Ker(φ). Since f(x) is irreducible, we must have Ker(φ) = hfi and F (αn) ∼= F [x]/hfi. By (a) again, [F (αn): F ] is equal to the degree of f(x). Linear Algebra

A. Let  1 0   1 1   1 1   0 1  β = , , , 1 0 1 0 0 1 0 1 which is a basis of R2×2, the vector space of 2 × 2 matrices with coefficients in K. Let 2×2 2×2 L : R → R be the linear map defined by  x y   x − y z + y  L = . z w y z + w (You can assume that β is a basis and that L is a linear map. You are not asked to prove these statements).

(a) Find the matrix A of L with respect to the basis β.(A is the matrix such that

A · (α)β = (L(α))β 2×2 whenever α ∈ R . Here (α)β is the coordinate vector, a column vector, of α with respect to the basis β).

Solution: Let  1 0   1 1   1 1   0 1  β = , β = , β = , β = . 1 1 0 2 1 0 3 0 1 4 0 1 If  e f   a + b + c b + c + d  = aβ + bβ + cβ + dβ = , g h 1 2 3 4 a + b c + d then a = −f + g + h, b = f − h, c = e − g, d = −e + g + h. In particular,

 0   0   1 1  0  0 2  1 (L(β )) = =   , (L(β )) = =   , 1 β 0 1  1  2 β 1 1  −1  β   β   0 2  1   1   0 1  0  −1 1  0 (L(β )) = =   , (L(β )) = =   . 3 β 1 1  −1  4 β 1 1  −2  β   β   2 3 Therefore,  0 0 1 1   0 1 0 0  A =   .  1 −1 −1 −2  0 2 2 3

(b) Suppose that  1   2  (L(α))β =   .  3  4 What is α?

Solution: We have  6 9  L(α) = β + 2β + 3β + 4β = 1 2 3 4 3 7 and if  x y  α = , z w then we have  x y   x − y z + y  L(α) = L = . z w y z + w Solving for x, y, z, w we deduce that  9 3  α = . 6 1

B. (a) An n × n matrix A is idempotent if A2 = A. Show that if λ is an eigenvalue of an idempotent matrix, then λ must be either 0 or 1.

Solution: Let v 6= 0 be an eigenvector corresponding to a fixed eigenvalue λ of the idempotent matrix A. Then, λv = Av = A2v = A(Av) = A(λv) = λ(Av) = λ(λv) = λ2v. That is, (λ − λ2)v = 0, which implies that λ − λ2 = 0, whence λ = 0 or 1.

(b) A matrix A is nilpotent if An = 0 for some positive integer n. If A is nilpotent and not the zero matrix, show that A is not diagonalizable. Solution: Let D = diag(d1, . . . , dr) be a diagonal matrix D with diagonal entries n n n d1, . . . , dr. Since D = diag(d1 , . . . , dr ) for each integer n ≥ 1, we see that n n D = 0 ⇐⇒ dj = 0 for each j ⇐⇒ dj = 0 for each j ⇐⇒ D = 0 Now suppose that A is nilpotent (An = 0) and diagonalizable (B−1AB = D for some invertible matrix B and diagonal matrix D). We have Dn = (B−1AB)n = B−1AnB = B−1 · 0 · B = 0, hence D = 0, and A = BDB−1 = 0. The result follows. C. (a) Let A, B be square matrices of the same size. Show that the eigenvalues of AB are the same as the eigenvalues of BA.

Solution: By symmetry, it suffices to show that every eigenvalue of AB is an eigenvalue of BA. If λ is an eigenvalue of AB, then for some vector v 6= 0 we have ABv = λv. (2) Multiplying by B and using the fact that λ is a scalar, it follows that BA Bv = λ Bv. If Bv 6= 0, this shows that λ is also an eigenvalue of BA (with eigenvector Bv), and we are done. On the other hand, if Bv = 0, then from (2) it follows that λ = 0. If λ = 0 is an eigenvalue of AB, it is a solution to det(AB − λI) = 0. Then, det(AB) = 0. Since det(BA) = det(B) det(A) = det(A) det(B) = det(AB) = 0, and we see that λ = 0 is a solution to det(BA − λI) = 0. In other words, λ = 0 is also an eigenvalue of BA. This completes the proof.

(b) Let I denote the n × n identity matrix in Rn×n. Prove that there do not exist n × n matrices A and B in Rn×n satisfying AB − BA = I. Solution: Using properties of the trace, for any n × n matrices A and B we have trace(AB − BA) = trace(AB) − trace(BA) = 0, whereas trace(I) = n 6= 0 if I is the n × n identity matrix. Thus, AB − BA 6= I. D. (a) Suppose that V is an n-dimensional (finite dimensional) vector space over a field K, and v1, . . . , vn ∈ V . Show that v1, . . . , vn are linearly dependent if and only if there exists a nonzero linear map L : V → K such that L(vi) = 0 for 1 ≤ i ≤ n.

Solution: Let W denote the span of the vectors v1, . . . , vn in V . If the vectors v1, . . . , vn are linearly independent, then {v1, . . . , vn} is a basis for W , and we have dim(W ) = n = dim(V ); this implies that W = V . In the case, if L(vi) = 0 for 1 ≤ i ≤ n, then L also vanishes on the span of the vectors v1, . . . , vn, which is V ; hence, L : V → K is the zero map. Conversely, if the vectors v1, . . . , vn are linearly dependent, let {w1, . . . , wr} be a maximal set of linearly independent vectors selected from the set {v1, . . . , vn}, where r < n. Then {w1, . . . , wr} is a basis for W , which can be extended to a basis {w1, . . . , wr, u1, . . . , us} for V , where r + s = n. If we define L to be (say) the map

L(a1w1 + ··· + arwr + b1u1 + ··· + bsus) = b1 + ··· + bs for all a1, . . . , ar, b1, . . . , bs ∈ K, then L : V → K is a nonzero linear map which vanishes on W ; thus, L(vi) = 0 for 1 ≤ i ≤ n.

(b) Let V be a finite dimensional vector space over C. Let A : V → V be a linear operator. Show that A can be written in the form A = D + N where D is a diagonalizable operator, N is a nilpotent operator, and DN = ND.

Solution: Fixing a basis {v1, . . . , vn} for V over C, every linear transformation V → V can be identified with an n × n matrix with complex entries. Thus, it suffices to prove the following statement: For every n × n complex matrix A, there is a diagonalizable matrix D and a nilpotent matrix N such that A = D + N and DN = ND. Note that if the conclusion holds for a particular matrix A, then it also holds for every conjugate B = P −1AP , where P is invertible, since E = P −1DP is diagonalizable, M = P −1NP is nilpotent, B = P −1AP = P −1(D + N)P = P −1DP + P −1NP = E + M, and EM = P −1DP · P −1NP = P −1DNP = P −1NDP = P −1NP · P −1DP = ME. Hence, it suffices to prove the statement for one matrix in each conjugacy class. Every n × n complex matrix A is conjugate to a block matrix

 J1  . J =  ..  , Jm where each block Ji has the form   λi 1  ...   λi  Ji =  .  .  .. 1  λi This is the Jordan canonical form of the matrix A. Let D and N denote the block matrices

 D1   N1  . . D =  ..  and N =  ..  , Dm Nm

where the blocks Di and Ni are given by    λi  0 1 . λi  ..     0  Di =  ..  and Ni =  .  .  .   .. 1  λi 0 Then D is diagonal, N is nilpotent, J = D + N, and DN = ND.

E.

(a) Let A = (aij) be a nonsingular n × n matrix with nonzero cofactor Ann i+j (The cofactor Aij is (−1) times the determinant of the (n − 1) × (n − 1) submatrix of A obtained by deleting the i-th row and j-th column. A is nonsingular if Det(A) 6= 0). Show that if one subtracts Det(A)/Ann from ann, then the resulting matrix is singular.

0 0 Solution: Let A = (aij) be the resulting matrix. Note that  0 aij if i 6= n or j 6= n; aij = ann − Det(A)/Ann if i = j = n. 0 0 Also, since A and A differ only in the last column, the cofactors Ain and Ain along 0 the last column agree: Ain = Ain for 1 ≤ i ≤ n. Now, n n−1 X X det(A) = ainAin = ainAin + annAnn, i=1 i=1 and therefore, n n−1 0 X 0 0 X 0 0 det(A ) = ainAin = ainAin + annAnn = (det(A) − annAnn) + annAnn = 0, i=1 i=1 0 where we have used the fact that ann − ann = − det(a)/Ann in the last step. This shows that A0 is singular. (b) Let A = (aij) be an n × n matrix. The trace of A is

trace(A) = a11 + a22 + ··· + ann. Let d1 dk f(x) = (x − c1) ··· (x − ck) be the characteristic polynomial of A. Show that

c1d1 + ··· + ckdk = trace(A). Solution: The characteristic polynomial of an arbitrary matrix A can be found by computing the determinant f(x) = det(xI − A). In its expansion according to the first column, the determinant contains a diagonal term (x − a11) ··· (x − ann) that gives a contribution to the coefficient of xn−1 equal to

−(a11 + ··· + ann) = −trace(A). No other term in the expansion will give a contribution to the coefficient of xn−1 since the power of x occurring in another term is at most xn−2. Therefore, trace(A) is the negative of the coefficient of xn−1 in the characteristic polynomial of A. In the case that d1 dk (x − c1) ··· (x − ck) , n−1 the negative of the coefficient of x is c1d1 + ··· + ckdk, thus we only the stated result.