# SPRING 2006 PRELIMINARY EXAMINATION SOLUTIONS

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1 SPRING 006 PRELIMINARY EXAMINATION SOLUTIONS 1A. Let G be the subgroup of the free abelian group Z 4 consisting of all integer vectors (x, y, z, w) such that x + 3y + 5z + 7w = 0. (a) Determine a linearly independent subset of G which generates G as an abelian group. (b) Show that Z 4 /G is a free abelian group and determine its rank. Solution: (b) The linear map Z 4 Z, (x, y, z, w) x + 3y + 5z + 7w has kernel G, and is onto because and 3 are relatively prime. Hence Z 4 /G is isomorphic to the image Z, which is a free abelian group of rank 1. (a) There is a sequence of elementary column operations over Z (not involving divisions) that transforms the 1 4-matrix ( ) into ( ). For instance, subtract 3 times the first column from the fourth to get ( ), and then subtract appropriate multiples of the fourth from each of the first three columns to make them zero. The same sequence of operations applied to the 4 4 identity matrix eventually yields a matrix such that U = ( ) U = ( ). Because of the way U was constructed, it has an inverse U 1 with integer entries. The first three columns of U are in G, and we claim that they span G as an abelian group. Suppose v G. Then 0 = ( ) v = ( ) U 1 v, α so U 1 v = β γ for some α, β, γ Z. Thus 0 α v = U β γ, 0 which is an integer combination of the first three columns of U. Finally these first three columns of U are linearly independent, since U is invertible. 1

2 A. Find (with proof) all real numbers c such that the differential equation with boundary conditions f cf + 16f = 0, f(0) = f(1) = 1 has no solution. Solution: First suppose that the characteristic equation x cx + 16 = 0 has a repeated root. This happens when c = ±8. If c = 8, the repeated root is 4, and the general solution to the differential equation without boundary conditions has the form The boundary conditions impose f(t) = (at + b)e 4t. b = 1 (a + b)e 4 = 1, and this system has a solution. Similarly, there is a solution in the case c = 8. From now on, we suppose that the complex roots α, β of x cx + 16 = 0 are distinct. Then the general solution is f(t) = ae αt + be βt, where a, b C, and the boundary conditions impose (1) a + b = 1 ae α + be β = 1. This system is guaranteed to have a solution if e α e β. So assume e α = e β. Then α β = πik for some k Z. By interchanging α, β, we may assume k > 0. On the other hand, by the quadratic formula, (α β) = c 64. Thus 4π k = 64 c 64. The only possibility is k = 1, which leads to c = ± 64 4π. In this case e α = e β, but the common value is not 1, since e α e β = e α+β = e c e 0 = 1. So the system (1) has no solution. Thus the set of values c for which the differential equation with boundary conditions has no solution is {± 64 4π }. 3A. Let S = {(x 1,..., x n ) R n x 1 + +x n = 0}. Find (with justification) the n n matrix P of the orthogonal projection from R n onto S. That is, P has image S, and P = P = P T. Solution: The orthogonal complement of S is one-dimensional, and spanned by the unit vector w = 1 n (1,..., 1), because v S v, w = 0. So the orthogonal projection is given by P v = v v, w w = v v 1+ +v n (1,..., 1). Therefore n n n n n 1 n 1 1 P = Id w T w = n n n n 1 n n n 4A. Let D = {z C : z < 1}. Find all holomorphic functions f : D C such that f( 1 n + ie n ) is real for all integers n.

3 Solution: We show that the only such functions are the real constant functions. Let f(z) = a n z n be the Taylor series for f around 0. We first prove by contradiction that a k are real. Suppose that k is the smallest index so that Ima k 0. Then we must have Ima k = lim x 0,x R x k Imf(x) On the other hand, because there is a bound on f (z) in a closed disk containing all the numbers 1 n + ie n, as n. Hence Imf( 1 n ) = Imf( 1 n + ie n ) + O(e n ) = O(e n ) Im a k = lim n k Imf( 1 n n ) = 0, which is a contradiction. As a consequence, f( 1 ) must be real. n By bounding f (z) on a closed disk, we may write f( 1 n + ie n ) = f( 1 n ) + ie n f ( 1 n ) + O(e n ) Taking imaginary parts we get Ref ( 1 n ) = O(e n ) Arguing as above, the Taylor series at 0 for f (z) has purely imaginary coefficients. conclude that all a k s must vanish with the exception of a 0. We 5A. Consider the following four commutative rings: Z, Z[x], R[x], R[x, y]. Which of these rings contains a nonzero prime ideal that is not a maximal ideal? Solution: In the ring of integers Z the nonzero prime ideals are p, where p is a prime number. Each of these ideals is maximal since F p = Z/ p is a field. Hence every nonzero prime ideal in Z is maximal. The polynomial ring Z[x] in one variable x over the ring of integers Z is not a principal ideal domain. For instance,, x is not a principal ideal; it strictly contains the ideal, which is therefore not a maximal ideal. The ideal is a prime ideal, because Z[x]/ = F [x] is a polynomial ring over a field, and hence an integral domain. Hence is a nonzero prime ideal in Z[x] which is not maximal. The polynomial ring R[x] in one variable x over the field R is a principal ideal domain. Hence every nonzero ideal has the form f(x) where f(x) is a nonzero polynomial with real coefficients. The ideal is prime if and only if f(x) is an irreducible polynomial, i.e., if f(x) is a linear polynomial or f(x) is a quadratic polynomial with no real roots. In either case, the quotient R[x]/ f is a field, namely, either R or C, which means that f is a maximal ideal. Hence every nonzero prime ideal in R[x] is a maximal ideal. The polynomial ring R[x, y] in two variables x, y over R has many nonzero prime ideals which are not maximal ideals. For instance, x is a prime ideal, but it is not maximal since it is contained in the ideal x, y. 3

4 6A. Let u: R R be a function for which there exists B > 0 such that N 1 u(x k+1 ) u(x k ) B k=1 for all finite increasing sequences x 1 < x < < x N. Show that u has at most countably many discontinuities. Solution: Let A be the set of points of discontinuity for u. Then A = n 1 A n where A n = {x R : lim sup u(y) lim inf u(y) > 1 y x y x n } To prove that A is countable, we will prove that A n 4n B. If y 1 < y < < y N are in A n then we can choose a strictly increasing sequence (x k ) N k=1 such that x k 1 < y k < x k and u(x k ) u(x k 1 ) > 1 n for k = 1,..., N. Summing over k gives the inequality on the right in B N 1 k=1 u(x k+1 ) u(x k ) N u(x k ) u(x k 1 ) N k=1 Hence N 4n B, which concludes the proof. ( ) 1. n 7A. Recall that SL(, R) denotes the group of real matrices of determinant 1. Suppose that A SL(, R) does not have a real( eigenvalue. Show ) that there exists B SL(, R) such cos θ sin θ that BAB 1 equals a rotation matrix for some θ R. sin θ cos θ Solution: Since the eigenvalues of A are solutions to a real quadratic equation, they are complex conjugates of each other, call them λ and λ. Since det(a) = 1, it follows that λλ = 1, i.e. λ and λ are on the unit circle. Write λ = cos θ + i sin θ. Pick a nonzero eigenvector z C with Az = λz. Write z = v + iw with v, w R. Taking the real and imaginary parts of the equation Az = λz gives the equations Av = (cos θ)v (sin θ)w, Aw = (sin θ)v + (cos θ)w. Note also that A(v iw) = λ(v iw) and λ λ, so v + iw and v iw are linearly independent over C, so v and w are linearly independent over R. We can find B SL(, R) ( taking the basis ) {v, w} to a real multiple of the standard basis for R. cos θ sin θ Then BAB 1 =. This is of the desired form, with θ in place of θ. sin θ cos θ 8A. Let D = {z C : z < 1}. Let f : D C be holomorphic, and suppose that the restriction of f to D {0} is injective. Prove that f is injective. 4

5 Solution: Suppose on the contrary that there is a D {0} such that f(a) = f(0). Let α be the common value. Choose disjoint open disks D 0 and D a contained in D, centered at 0 and a, respectively. By the Open Mapping Theorem f(d 0 ) and f(d a ) are open subsets of C containing α. Hence G := f(d 0 ) f(d a ) is a nonempty open subset of C. Choose ξ G with ξ α. Then there exist z 0 D 0 and z a D a such that f(z 0 ) = f(z a ) = ξ. Since ξ α, neither z 0 nor z a is 0. This contradicts the injectivity of f restricted to D {0}. 9A. Let p be a prime. Let G be a finite non-cyclic group of order p m for some m. Prove that G has at least p + 3 subgroups. Solution: We will use the following two facts: (i) A nontrivial p-group has a nontrivial center Z (nontrivial conjugacy classes have size divisible by p, as does the whole group, so {1} cannot be the only trivial one). (ii) If G is a group with center Z, and G/Z is cyclic, then G is abelian (since if a G generates G/Z, every element of G is of the form a n z for some n Z and z Z). We use induction on m. Suppose m. Since G has order 1, p, or p, it is abelian (for order p, combine (i) and (ii) above). Since it is not cyclic, we have G Z/pZ Z/pZ. So G has one trivial subgroup, (p 1)/(p 1) = p + 1 subgroups of order p, and G itself. Thus G has exactly p + 3 subgroups. Now suppose m >. By (i), the center Z of G is nontrivial. Since G is a nontrivial p-group, it has a nontrivial center Z. If G/Z is non-cyclic, then by the inductive hypothesis it has p + 3 subgroups, and their inverse images in G are distinct subgroups of G. If G/Z is cyclic, then G is abelian by (ii); but G is not cyclic, so by the structure theory of finite abelian groups, it must contain Z/pZ Z/pZ, which already contains p + 3 subgroups. 1B. Let A 1 A be compact connected subsets of R n. Show that the set A = A m is connected. Solution: The intersection A is nonempty, since otherwise {A 1 A m } is a covering of A 1 (by sets open in A 1 ) with no finite subcover. Suppose that A is not connected. Then there exist sets B 0, C 0 open in A such that B 0 C 0 = A and B 0 C 0 =. Then B 0, C 0 are also closed in A, which (as an intersection of closed sets) is closed in R n, so B 0, C 0 are closed in R n. Hence we can find disjoint sets B, C open in A 1 such that B 0 B, C 0 C: for instance, we could let B be the set of points in A 1 that are strictly closer to B 0 than to C 0, and vice versa for C. Since A = B 0 C 0 B C, the sets B, C, and A 1 A m for m 1 form a cover of A 1 by sets open in A 1 ; thus there is a finite subcover consisting of B, C, and A 1 A m for m = 1,..., r. So r is such that A r B C. Since B, C are open, disjoint, and B A r B 0 A and C A r C 0 A, we have that A r is not connected, a contradiction. B. Let F be the field of elements. Let n be a prime. Show that there are exactly ( n )/n degree-n irreducible polynomials in F [x]. Solution: There is a unique field extension F n of degree n over F. It is Galois over F (this is because it is a splitting field for the separable polynomial x n x). If a F n F, then F (a) is a subfield of F n of degree dividing n but not equal to 1, so F (a) = F n. Hence 5

6 the minimal polynomial f a of a over F is an irreducible polynomial of degree n over F. Thus we have a map (F n F ) {degree-n irreducible polynomials in F [x]} a f a. On the other hand, if f F [x] is any degree-n irreducible polynomial, then f has a zero in F n (since F n is the unique degree-n extension of F ) and it follows that f has n distinct zeros in F n (since F n is Galois over F ). Moreover, f is automatically monic (the only nonzero element of F is 1) so it is the minimal polynomial of each of its zeros. Thus our map is n-to-1. Its domain has size n, so its range has size ( n )/n. 3B. Evaluate the integral for t > 0. e itx dx e x + e x Solution: The integral converges absolutely, since the numerator has absolute value 1, while the denominator decays exponentially in both directions. Use a rectangular contour C bounded by x = R, x = R, y = 0 and y = π. As R the integrals along the vertical parts of the contour tend to 0, since π 0 e it(r+iy) e R+iy + e R iy dy π The integral along the horizontal path y = π equals R R e it(x+πi) dx = e (x+πi) + e (x+πi) R Let I denote the integral we have to find. Then On the other hand, lim R C C R 0 1 π dy = e R e R e πt e itx dx = e x e πt e x e itz e z + e z dz = ( 1 + e πt) I. e itz dz = πi Res e z πi, + e z since the only singular point inside the contour is πi. Now so C Res πi = e πt i, e itz πt dz = e z πe + e z, πt e I = π 1 + e = π. πt + e πt 6 e πt e R e R. R R e itx dx. e x + e x

7 4B. Let n be a positive integer, and let GL n (R) be the group of invertible n n matrices. Let S be the set of A GL n (R) such that A I has rank. Prove that S generates GL n (R) as a group. Solution: By Gaussian elimination, GL n (R) is generated by the elementary matrices obtained from the identity matrix by interchanging two rows, by multiplying one row by a nonzero scalar, or by adding a multiple of one row to a different row. For each such matrix A, the matrix A I has at most two nonzero rows and hence has rank. 5B. Prove that there exists no continuous bijection from (0, 1) to [0, 1]. (Recall that a bijection is a map that is both one-to-one and onto.) Solution: Suppose on the contrary that there exists a continuous bijection f : (0, 1) [0, 1]. Then there exists x (0, 1) such that f(x) = 0. Let A = (0, x), B = (x, 1). We have A B = and since f is injective we have f(a) f(b) = f(a B) =. Since f is continuous and (0, x] is connected, f((0, x]) contains an interval [0, a) for some a > 0. Hence f(a) contains (0, a). Similarly, f(b) contains (0, b) for some b > 0. This gives f(a) f(b). Contradiction to ( ). 6B. Let A be the subring of R[t] consisting of polynomials f(t) such that f (0) = 0. Is A a principal ideal domain? Solution: No. Suppose A is a principal ideal domain. Then the A-ideal I generated by t and t 3 would be principal. Let p(t) be a generator of I. Then t = q(t)p(t) for some q(t) A, so p(t) divides t also in the unique factorization domain R[t]. Hence p(t) = ut m for some unit u of R[t] and some m {0, 1, }. The case m = 1 is impossible, since p(t) A. If m = 0, then p(t) is a unit also of A, and hence generates the unit ideal; this contradicts the fact that every element of I has constant term zero. If m =, then t 3 is not a multiple of p(t), since the element t 3 /p(t) R[t] is not in A. 7B. Let m be a fixed positive integer. (a) Show that if an entire function f : C C satisfies f(z) e z for all z C, then f (m) (0) m!em m m. (b) Prove that there exists an entire function f such that f(z) e z for all z and f (m) (0) = m!em m m. Solution: (a) Write f(z) = n 0 a nz n with a n C. Then a m is the coefficient of z 1 in the Laurent series of f(z)/z m+1, so a m = 1 πi z =R 7 f(z) z m dz z, ( )

8 for any R > 0, and we get a m 1 ( ) e R πr π R m R = er R m. Taking R = m (which calculus shows minimizes the right hand side) and multiplying by m! gives f (m) (0) = m!a m m!em m. m (b) Examining the proof of part (a) shows also that in order to have equality, f(z) must z m have constant modulus e m /m m and constant argument on the circle z = m. Thus we guess f(z) = em z m, and it remains to prove that f(z) e z for all z C. Equivalently, we must m m show that the minimum value of e x /x m on (0, ) is e m /m m. This can be seen by observing that the only zero of the derivative of log(e x /x m ) = x m log x is at x = m, while the second derivative is positive everywhere (it is m/x ). 8B. Let, be the standard Hermitian inner product on C n. Let A be an n n matrix with complex entries. Suppose x, Ax is real for all x C n. Prove that A is Hermitian. Solution: We have x, Ax = x H Ax = x H Ax (since x H Ax is real) = (x H Ax) H = x H A H x. Thus x H Ax = x H A H x. So x H (A A H )x = 0 for all x C n. Let B = A A H. We have x H Bx = 0 for all x C n and B H = A H A = B, so B is skew-hermitian (hence normal). Let x be an eigenvector of B with the eigenvalue λ, so Bx = λx. Then 0 = x H Bx (by ( )) = λx H x = λ x. This gives λ = 0. Thus all eigenvalues of B are zero. Being normal, B is diagonalizable, so B = 0. By definition of B, we get A = A H. Thus A is Hermitian. 9B. Find a bounded non-convergent sequence of real numbers (a n ) n 1 such that a n a n 1 a n+1 n for all n. Solution: We will let a n = f(n), where f(x) is a function similar to the sine function but with oscillations that slow down as x, so that f (x) 0. To be precise, we take f(x) := 1 sin(ln(x + 1)). This sequence is bounded. It also does not converge, since the spacing between values of ln n tends to zero, which means that the values of (ln n) mod (π) are dense in [0, π]. By Taylor s theorem with remainder (centered at n), ( ) f(n + 1) = f(n) + f (n) + 1 f (ξ + ) f(n 1) = f(n) f (n) + 1 f (ξ ) for some ξ + (n, n + 1), and for some ξ (n 1, n), so, f(n) f(n 1) f(n + 1) = 1 f (ξ + ) + f (ξ ) = f (ξ) for some ξ (ξ, ξ + ) (n 1, n + 1) 8

9 by the intermediate value theorem. We compute f 1 (x) = cos(ln(x + 1)) (x + 1) f 1 (x) = (cos(ln(x + 1)) + sin(ln(x + 1))), (x + 1) f 1 (x) (x + 1) f 1 (ξ) (ξ + 1) 1 n. so a n a n 1 a n+1 = f (ξ) n. 9

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