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Department of Mathematics, University of California, Berkeley
STUDENT EXAM NUMBER
GRADUATE PRELIMINARY EXAMINATION, Part A Spring Semester 2014
- Please write your 1- or 2-digit student exam number on this cover sheet and on all problem sheets (even problems that you do not wish to be graded).
- Indicate below which six problems you wish to have graded. Cross out solutions you may have begun for the problems that you have not selected.
- Extra sheets should be stapled to the appropriate problem at the upper right corner. Do not put work for problem p on either side of the page for problem q if p 6 = q.
- No notes, books, or calculators may be used during the exam.
PROBLEM SELECTION
Part A: List the six problems you have chosen:
GRADE COMPUTATION
1A. 1B. Calculus 2A. 2B. Real analysis 3A. 3B. Real analysis 4A. 4B. Complex analysis 5A. 5B. Complex analysis 6A. 6B. Linear algebra 7A. 7B. Linear algebra 8A. 8B. Abstract algebra 9A. 9B. Abstract algebra
Part A Subtotal: Part B Subtotal: Grand Total:
Please cross out this problem if you do not wish it graded
Problem 1A. Score:
Find the sum of the series (^1) ×^12 × 3 + (^2) ×^13 × 4 + (^3) ×^14 × 5 + · · ·.
Solution: Use the partial fraction decomposition (^) (n−1)^1 n(n+1) = (^) n^1 /−^21 − (^1) n + (^) n^1 +1/^2 and rearrange
to find that the sum of the first m terms is 1/ 4 − 1 /2(m + 1)(m + 2) so the sum of all terms is 1/4.
Please cross out this problem if you do not wish it graded
Problem 3A. Score:
Find a real number c so that ∣ ∣ ∣ ∣ ∣ c −
− 1 / 2
exp(x) − 1 x
dx
Solution: The integral is given by integrating the power series term by term, so is (^20) ×^11 ×1! + 1 22 × 3 ×3! +^
1 24 × 5 ×5! +· · ·^. The sum of all but the first two terms is easily seen to be much less than 1 /100, so we can take c to be the sum of the first two terms which is 73/72 = 1. 0138888 · · ·. (A more accurate answer is 1.01399349965)
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Problem 4A. Score:
Let f be analytic on the closed unit disk, and assume that |f (z)| ≤ 1 for all z’s in this set. Suppose also that f (^12 ) = f ( 2 i ) = 0. Prove that |f (0)| ≤ 14.
Solution: Set
g(z) =
z − 2 2 z − 1
z − 2 i 2 z − i
f (z)
Then g is holomorphic inside the disc and |g(z)| = |f (z)| ≤ 1 for |z| = 1. Hence by the maximum principle we get |g| ≤ 1. The conclusion follows.
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Problem 6A. Score:
Let R be a finite ring (with 1) of characteristic p. For S a subring of R (not necessarily containing an identity element), S is a vector space over Fp. For a ∈ S let T (^) aS : S → S be the linear map T (^) aS (x) = ax. (a) Show: if 1 ∈ S then the minimal polynomial of T (^) aS = the minimal polynomial of T (^) aR. (b) Give an example of p, R, S, a where (a) is false.
Solution: (a) When 1 ∈ S, the map a → T (^) aS has kernel 0. So the minimal polynomial of T (^) aS is the minimal polynomial p(x) such that p(a) = 0 which depends only on a. (b) R = F 2 [a]/(a^2 ), S = { 0 , a}. min poly of T (^) aR = x^2 ; min poly of T (^) aS = x.
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Problem 7A. Score:
Let F be a finite field with q elements. A complete flag in the vector space F n^ is a nested sequence of linear subspaces V 1 ⊂ V 2 ⊂ · · · ⊂ V n−^1 of dimensions 1, 2 ,... , n−1 respectively. Let fn(q) be the number of complete flags in F n^ as a function of q. Find the limit of fn(q) as q tends to 1.
Solution: When the subpaces V 1 ⊂ · · · ⊂ V k^ are already selected, the subspace V k+1^ is determined by a 1-dimensional subspace in the quotient space F n/V k^ of dimension n−k. The numebr of such subspaces is equal to (qn−k^ − 1)/(q − 1) (non-zero vectors up to proportionality), which tends to n−k as q tends to 1. Thus the answer is the product of n−k over k = 0, 1 , 2 ,... , n−1, that is n!.
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Problem 9A. Score:
Let A be a finite abelian group (under +) and let R = End(A) be the ring of homomorphisms from A to A. Show there is a subring S of R such that A and S are isomorphic as abelian groups.
Solution: A is a product of cyclic groups Ci. For C finite cyclic, End(C) is isomorphic to C as groups. So S = the product of the End(Ci) works.
Department of Mathematics, University of California, Berkeley
STUDENT EXAM NUMBER
GRADUATE PRELIMINARY EXAMINATION, Part B Spring Semester 2014
- Please write your 1- or 2-digit student exam number on this cover sheet and on all problem sheets (even problems that you do not wish to be graded).
- Indicate below which six problems you wish to have graded. Cross out solutions you may have begun for the problems that you have not selected.
- Extra sheets should be stapled to the appropriate problem at the upper right corner. Do not put work for problem p on either side of the page for problem q if p 6 = q.
- No notes, books, or calculators may be used during the exam.
PROBLEM SELECTION
Part B: List the six problems you have chosen:
Please cross out this problem if you do not wish it graded
Problem 2B. Score:
Prove that a non-empty closed convex subset of the real vector space Rn^ with the usual Euclidean distance has a unique element of minimum norm (distance to the origin).
Solution: We first check uniqueness: if a and b are two distinct points of the same minimum norm then their midpoint has smaller norm, and is in the set as the set is convex: contradic- tion. To show existence, take a nonzero intersection with some closed ball with center the origin. This intersection is compact, so the norm attains a minimum value.
Please cross out this problem if you do not wish it graded
Problem 3B. Score:
Prove that there exists a constant C such that for every polynomial P of degree 2014
P (0) ≤ C
0
|P (x)| dx.
Solution:
In the space of polynomials of degree ≤ 2014, P 7 → P (0) is a linear function, and
P 7 → ‖P ‖ :=
0 |P^ (x)|dx^ is a continuous homogeneous function of degree 1 vanishing only at the origin. Thus, it suffices to take C to be the maximum of the linear function on the compact subset {P |‖P ‖ ≤ 1 }. To justify continuity of ‖ · ‖, consider a sequence Pn = P + ∆Pn such that ∆Pn → 0 coefficientwise. We have ∫ (^1)
0
ckxk|dx ≤
|ck|
0
xkdx =
∑ (^) |ck| k
and hence ‖∆Pn‖ → 0. Then by the triangle inequality
|‖Pn‖ − ‖P ‖| ≤
0
||Pn(x)| − |P (x)|| dx ≤
0
|∆Pn(x)| dx = ‖∆Pn‖
, and therefore ‖Pn‖ → ‖P ‖.
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Problem 5B. Score:
Find the integral
0
cos x 1+x^2 dx.
Solution: This is half the real part of
−∞
eix 1+x^2 dx. Taking the usual semicircular contour in the upper half plane (and checking that the integral over the curved bit tends to 0) we see that this latter integral is 2πi times the residue at i. The residue is e − 1 2 i , so the original integral is π/ 2 e.
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Problem 6B. Score:
Let n be an integer and let O(n) be the group of n × n orthogonal matrices. View O(n) as a topological group with the induced topology from the embedding O(n) ⊂ Rn 2 given by the entries. Show that O(n) is compact.
Solution: Let group O(n) is realized as the intersection of the compact closed unit ball in Rn 2 with the closed subsets characterized by the conditions vi · vj = 0 for i 6 = j, where vi denotes the i-th column vector.
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Problem 8B. Score:
Let B = C−^1 AC, where A and C are n×n-matrices with integer entries, such that det A = 1, and det C 6 = 0. Prove that there exists a positive integer m such that all entries of Bm^ are integers.
Solution: Let d = | det C|. Then dC−^1 is the adjoint matrix of C and thus has integer entries. Let m be the order of A in the finite group GLn(Z/dZ) of automorphisms of the abelian group (Z/dZ)n. Then Am^ = I + d A˜ where A˜ has integer entries. Therefore Bm^ = C−^1 AmC = I + dC−^1 AC˜ has integer entries.
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Problem 9B. Score:
Let G be a finite group acting on a finite set X with a single orbit. For an element g ∈ G let Fixg(X) denote the set {x ∈ X|g(x) = x}. (a) Show that #G =
g∈G
#Fixg(X).
Hint: Count the set S = {(x, g) ∈ X × G|gx = x} two ways. (b) Show that if X has more than 1 point then there exists an element g ∈ G fixing no points of X.
Solution: Summing over x we get that the size of S is equal to ∑
x∈X
Stab(x),
where Stab(x) is the stabilizer group of x. Summing over g first we get that the size of S is
∑
g∈G
#Fixg(X).
From the orbit stabilizer formula, and using the fact that we have just a single orbit, we have (^) ∑
g∈G
#Fixg(X) = #G
as desired in (a). For (b) note that since Fixe(X) = X we from (a) that if Fixg(X) 6 = ∅ for all g that #G ≥ #X + (#G − 1) > #G,
a contradiction.
