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1 Feb 2010: Find a conjugate harmonic functionv(x, y), that is, a function v such that u and v together satisfy the Cauchy-Riemann equations. ... Show that there is an analyticfunction H : C C such that h(z) = exp H(z) for all z.

27 Jan 2010: Use a suitableLagrange resolvent to find a Q(µ3) such that K = Q(ζ3, 3. ... Use (ii) to show that there exists b K such thatdet(σiσjQ1(b)) ̸= 0.

29 Jan 2010: The irreducibility of Sp(χ) for such a prime pin conjunction with a) implies that Φq,χ = 0. ... Since GẐ acts transitively on |MK |,. there exists a positive constant cn Q such that for every x |M|,.

12 Oct 2010: Suppose then that g is a. quartic with J = 0 such that z(g) = z(g). ... We now fix such a u and put αi = B(ei, u)where e1, e2, e3 K3 are the standard basis vectors.

22 Nov 2010: Let t = a! and d = t. By Ramsey’s theorem, there exists a positive integer b such that whenever[b](t) is ka-coloured, there exists a monochromatic subset of order ... d. Letu = b a 1 and v = u tr. By Ramsey’s theorem again, there exists a positive

18 Nov 2010: 14. Is there an uncountable family S PN such that A B is finite for all distinct A, B S? ... such that. (qn qn1)2 converges? A.G.Thomason@dpmms.cam.ac.uk - 1 - 18 November 2010.

29 Jan 2010: Therefore there are constants a, b, c k such that if v, v′ Spand w IndGB δ1/2, then. ... π′′ = I(1)0, then we can choose w such that HomGw (πw.

3 Nov 2010: Recall that the existence of such a mini-mizer was proved in lecture.). ... b) Choose numbers a1, b1 such that 0 < a1 < a < b < b1 < 1, and set, for x R and n = 1, 2, 3,. ,.

17 Dec 2010: such that. MIA A M/B. ll 1r MA oh_ MB. commuts up to natural isomorphism. ... For g : Y-, B in M/B we expect IIh 0 Y E M/A to be such that a fibre (II,, 0 Y)n is the modest set of sections

23 Feb 2010: 6. Let R be an integral domain. The highest common factor of non-zero elements a and b in R is an elementd in R such that d divides both a ... identified with the set Fq {} in such a way that GL2(Fq) acts on asthe group of Möbius transformations z 7 azb

29 May 2010: Let Γ Aut(X′) be a finite subgroup such that f factorsas X′ X′/Γ g X, where g is a radicial alteration. ... Then one knows that there exists a finite extension K′/K such that ρ(IK′) isunipotent.

19 Nov 2010: 5. (i) Suppose that f : R2 R is such that f/x is continuous in some open ball around (a,b),and f/y exists at (a,b). ... Show that f is differentiable at (a,b). (ii) Suppose that f : R2 R is such that f/x exists and is bounded near (a,b), and that fora

29 Jan 2010: W2 E = G. Q(1). and GrWj E = 0 for j > 0 or j < 2, such that. ... The reader who is sceptical of the general finiteness of groups such as IIIwill be reassured that we do not require equality in (b).).

28 Mar 2010: 4. Let F/K be a finite extension. Show that there is a unique intermediate fieldK L F such that L/K is separable and F/L is purely inseparable. ... Let a,b K such that the polynomials P(X) = Xn a and Q(X) = Xn b are irreducible.Show that P and Q have the

22 Jan 2010: a, b] be given such that for each polynomial P , the error. ... let fn : C befunctions such that fn converge uniformly on to a bounded function f : C.

17 Feb 2010: B(µp,νp) spanned by tensors pφ̂p such that. φ̂p belongs to Bs(µp,νp) for at least one p, and this is the unique invariant subspace of B(µ,ν) ... Then C is finite.Proof. Choose a prime p 7 such that GZp K.

19 Nov 2010: n}, let the non-negative numbers A(n)k and the. “nodes” x(n)k [a, b] be given such that for each polynomial P , the error. ... such that for every f C([0, 1]), there exists a unique series.

24 Nov 2010: Use a suitableLagrange resolvent to find a Q(µ3) such that K = Q(ζ3, 3. ... Use (ii) to show that there exists b K such thatdet(σiσjQ1(b)) ̸= 0.

18 Feb 2010: abelian category of “mixed motives”. The ultimate definition of such a category remains to be found. ... φ : IK T. Here T/K is a number field and there exist integers a, b with ab = w such that.

9 Nov 2010: Let a,b K such that the polynomials P(X) = Xn a and Q(X) = Xn b are irreducible. ... an i such that Gi = {1}.(iv) Let G be the group of invertible n n upper triangular matrices with entries in a.

15 Oct 2010: On the other hand, show that if fn does not converge uniformly to f, then we can find aconvergent sequence xm x in [a, b] such that fn(xn) does not ... Give an example of a sequence of uniformly continuousfunctions fn on R, such that fn converges

29 Apr 2010: c) Give a topology on R which is neither trivial nor discrete such that every open set isclosed (and vice-versa). ... Prove the following:. (a) G is homogeneous: given any x, y G there is a homeomorphism φ : G G such thatφ(x) = y.

8 Nov 2010: a linear func-tional θ V such that θ(x) 6= θ(y).(b) Suppose that V is finite dimensional. ... If no such subspaceexists, show that there exists a basis v, α(v),.

17 Dec 2010: such that. MIA A M/B. ll 1r MA oh_ MB. commuts up to natural isomorphism. ... For g : Y-, B in M/B we expect IIh 0 Y E M/A to be such that a fibre (II,, 0 Y)n is the modest set of sections

17 Dec 2010: clopen sets guxantecd by oil’s being separated. For each B c A such that V,, J: fl,. ... such that WAcJ,=fl where A(J) =; l_j {Ai 1 iE.7}. B! ( )

17 Dec 2010: clopen sets guxantecd by oil’s being separated. For each B c A such that V,, J: fl,. ... such that WAcJ,=fl where A(J) =; l_j {Ai 1 iE.7}. B! ( )

25 Nov 2010: 6 and 12, related by c34 c26 = 1728, such that(i) The ring of invariants K[Xn]. ... in K[c4,c6] such that. D(λ,µ) = D(λa µDb,λb µD. a)/D(a,b). c4(λ,µ) = c4(λa µDb ,λb µDa. ).

19 Oct 2010: Many peoplestill study them for practical purposes such as elliptic curve cryptography, but I. ... becomes just Q. Therefore refers to the fact that we replace this componentin the image with b Q, chosen such that NQ[θ].

29 Jan 2010: There exists such a basechange overwhich factors as a product of isogenies of prime degree. ... the smallest closed subset of V k[r] such thatV k[r]V k[r1] is smooth over S.