Sameera Vemulapalli
5/16/2019
Start: 1:30pm
End: 3:30pm
general advice: everyone says this but my committee really was very nice! I was very nervous at the beginning but settled into a relaxed phase pretty quickly. It's also really hard to think while people are staring at you, and i'm pretty embarrassed about how long certain things took me to figure out, but I think this is normal and probably everyone does this.
Chair: Manjul Bhargava (B)
Other Member: Chris Skinner (S)
Other Member: Robert Gunning (G)
B, S, and G decide which blackboard to face, because there are two blackboards in B's office
they then debate between sitting on the couch and sitting on the chairs
me: this is very stressful
B: don't worry this is the hardest part of the exam (this was not true)
S: what do you want to start with
me: complex analysis
Complex analysis
G: what is the relationship between conformal mappings and holomorphic functions
me: conformal mappings are bijective holomorphic functions with a nonzero derivative
G: conformal mappings preserve angles. are all angle preserving maps holomorphic
me: i think so...
G: what about orientation
me: oh they're holomorphic if they preserve orientation
G: the most beautiful theorem in complex analysis is the riemann mapping theorem. State the riemann mapping theorem
me: stated it
G: why is the whole complex plane not conformally equivalent to the disc
me: Louisville's theorem
G: if a function holomorphic is on the plane except at the origin what can happen
me: there's three types of singularities, essential, removable, and pole. talked further
G: what's an example of an essential singularity at the origin
me: e^{1/z}
G: If you have a Laurent series on annulus with an infinite expansion, does it necessarily have essential singularity
me: no, take 1/z-1
G: state stone-weierstrass theorem
me: stated the locally compact version
G: what's your favorite function on the complex plane
me (not wanting to write down gamma or analytically continue it): the constant functions
committee: chuckles
thankfully they do not press further and we move on
Real analysis
G: what are different types of convergence
me: L1, convergence in measure, pointwise convergence, say that L1 implies convergence in measure implies a subsequence converges pointwise almost everywhere. I give counterexamples for L1 does not imply pointwise convergence and convergence in measure doesn't imply pointwise convergence, and for pointwise convergence doesn't imply convergence in measure.
G: ok i'm satisfied, do you guys want to ask anything else
me: relieved
S: no let's ask about Fourier analysis
me: internally screams because I thought they wouldn't ask me about Fourier and really did not study for it
S: What is the fourier transform of an L1 function
me: it's C0. Quickly prove that it vanishes at infinity and then fumble a little but eventually show that it is continuous
S: what is a Schwarz function
me: C^infinity function whose derivates all decays faster than all polynomials
S: what does fourier do to schwarz
me: isometry on Schwarz functions
S: prove it
me: fumbles for a few minutes while they stare at me.....i desperately wish i reviewed fourier analysis
S,B: its ok your not an analyst let's move on
me: ok but how do you do it
B: Plancherel theorem
B: there are number theory applications to Fourier analysis! Can you calculate sum of 1/n^2
me: I write down poisson summation formula and stare blankly at it
B: if you haven't seen it before it's ok! just work this out later today, it's cool
B: ok let's do algebra now
me: can i drink water first
B: it's good to rehydrate before algebra
i awkwardly drink as much water as possible so I can take a mental break
Algebra
B: classify finite integral domains of order 10
me: there are none because finite integral domains are finite fields
B: ok classify rings of order 10 without sylow. a ring for us means unital and commutative.
me (temporarily forgetting everything): Ok first the underlying abelian group must be Z/10Z because the 5-sylow is index 2 and thus normal and there are only two maps from Z/2 to Z/4 and one of them gives a dihedral group. (i used sylow after being explicitly told not to and also gave a reason for there being only one abelian group of order 10 yikes)
B: don't use sylow
me: i didn't I used that it's index 2
S: Prove without sylow that every group of order 10 has a 5-sylow
me, panicking: uhh if there's no 5-subgroup then all elements are 2 torsion
B: why can't a not necessarily abelian group of order 10 have all 2-torsion elements
me after mucking around for an embarrassingly long time: if ab!=ba then as a = a^-1 and b = b^-1 we have abab = (ab)^2 != 1, so groups whose elements all have order two are direct sums of Z/2 and thus have cardinality a power of 2
B: ok what are the possible ring structures on Z/10Z
me (after mucking around for an embarrassingly long time again): only the reasonable one is possible
S: what can you say about finitely generated modules over Z
me: we have structure theorem for finitely generated modules over PID and stated it
S: how do you prove this
me: gave map of relations to generators and put the matrix in smith normal form
B: how do you put it in smith normal form
me: explains this proof in detail, clear the rows and columns to get a diagonal matrix and then make the diagonals divide themselves
S: What can you say if its over something else, like a dedekind domain
me: state something similar to the structure theorem of finitely generated modules over a dedekind domain because I couldn't remember it exactly, like the module is isomorphic to I_1 \oplus \dots \oplus I_m \oplus R/J_1 \oplus \dots \oplus R/J_n for ideals J_i | J_{i+1} for integral ideals I_k and J_l.
S: I guess we're veering into number theory now. what's the uniqueness statement
me: uhhhhhhhhh, i think you can write J_i | J_{i+1} and this makes the torsion part is definitely unique.
S: is the free part unique
me (want to say yes but S's reaction tells me this is not true): probably not...uh ok if I_1 and I_2 are coprime then we have the exact sequence 0 -> I_1 I_2 -> I_1 \oplus I_2 -> I_1 + I_2 = R -> 0 so I_1 \oplus I_2 = R \oplus I_1 I_2
S: can we always assume coprime
me (after some more hints): uh yea we're only looking at ideal classes in the class group, so we can scale them so they're coprime......oh so the uniqueness is R^{n-1} \oplus I_1 \dots I_m, so we're just looking at rank (n) and an element of the class group for the non torsion part.
S: do you know Wedderburn's theorem
me: it says that the brauer group of finite field is trivial
S: laughs and says that's one statement of the theorem
B: how do you prove that
me: prove at finite level and take limits, let G be the galois group of a finite extension of finite fields, H2(G, Gm) = H^1(G, Gm) = 0 because Herbrand quotient of a finite cyclic group is trivial and the second is trivial by Hilbert 90.
S: classify conics over finite fields
me: they're all isomorphic because the Brauer group of finite field is trivial and conics over finite field are 2-torsion of the Brauer group
S: tell us more about Brauer group
me: its also division algebras up to equivalence and central simple algebras up to equivalence and twists of projective space
S: ok conics are quadratic forms in 3 variables, classify quadratic forms in n variables
me (forgetting everything about quadratic forms): diagonalize, just depends on the number of squares and nonsquares
B: so would x^2 + y^2 and ax^2 + by^2 be inequivalent for a, b nonsquares
me: uhhhhh
B: construct an explicit map between these two. you can choose a and b.
me (continuing to be very embarrassed): after many hints from B i realize that you can take the smallest nonsquare a in Fq and its a sum of squares, specifically 1 and (a-1). so then we can apply some linear transformation that makes the x^2 coefficient a. But because discriminant of a quadratic form is invariant, b must also be a nonsquare.
B: so what classifies quadratic forms over a finite field
me (regaining sanity): rank, discriminant
B: classify quadratic forms over Qp
me: rank, discriminant, hilbert symbol
S: define the Hilbert symbol, what else can you say about it
me: define Hilbert symbol and state hilbert reciprocity
B: ok let's do algebraic geometry now
Algebraic geometry
B: classify curves, genus 0 to genus 5
me: ok for genus 0 the anti-canonical bundle is very ample and embeds you as a conic in P2. if you have a rational point you're isomorphic to P1 because you have a degree 1 map to P1. genus 0 curves are classified by 2 torsion in Brauer group.
me (foolishly): for genus 1 you embed as a cubic in P2......but i forgot what causes this embedding
B: over what base field
me: uhhh........ok ill get back to this let's do the other genera first. for genus 2 i proved that the canonical bundle is base-point free, degree 2, and has 2 global sections. So it has a degree 2 cover of P1 and is hyperelliptic.
B: over what base field...
me: proved that if we have a non-hyperelliptic curve then the canonical bundle is very ample and thus for genus 3 we get a plane quartic
B: you proved that the canonical bundle is very ample over an algebraically closed field
me: very ampleness is preserved via base change
B: ok
S: what's the dimension of the moduli space of genus 3 curves
me: 6, you can see this from taking degree 4 monomials in P2 and subtracting a gl3
me: so for genus 4 you get complete intersection of quadric surface and cubic surface and you can get this by computing dimensions of cohomology.
S: i don't mind skipping genus 5
me (wanting to redeem myself): i know genus 5
B: ok let's do it
me: if its non-hyperelliptic the canonical bundle is very ample and the embedding is degree 8 in P^4 and it's either a complete intersection of 3 quadric surfaces or its a trigonal curve.
B: ok let's return the base field issues...does a hyperelliptic curve always have a double cover to P1
me: i guess sometimes its a double cover of a conic
B: ok and about elliptic curve..what do you need to embed it as a cubic in P2
me: a degree 3 line bundle
B: how do you get any line bundle
me: base change to algebraic closure, choose a point, take the finite extension in which this point is defined, take the galois conjugates, and then you have a line bundle. multiple it enough so it becomes very ample, then you have an embedding
B: how much do you need to multiply it
me: not very much, degree just has to be greater than 4 = 2g+2
B: ok so you have embeddings into projective space, how can you get one into P2
me: uhhhhhh
B: the answer is you can't always embed a genus 1 curve as a plane cubic. ok moving on what is a normal scheme
me (embarrassment returns...still can't believe i completely blanked on this definition): sorry i forgot what a normal scheme is
B: the stalks are integrally closed. when is an order a normal scheme
me: when its a ring of integers because the stalks are the localizations which must be maximal, i.e. the order must be maximal
B: what is normalization of a curve
me: a nonsingular thing birational to it
B: construct an isomorphism from x^2 + y^2 - z^2 to P1
me: after some fumbling, fix a point on the conic and let P1 be the slope and send that element of P1 to the second point intersected by a line of that slope
S: why does this work
me: Bezout
B: does this iso work over any field
me: yes because Bezout and the point [1:0:1] exists for any field
S: what's the Picard group
me: defines it
B: how does the class group of a ring of integers relate to the Picard group
me: the class group of a ring of integers is the Picard group of Spec(O_k)
B: what about an order
me: you have to take invertible ideals only, which correspond to invertible line bundles
B,S: some followup questions i don't remember on what exactly i mean
B: normalizing the spec of an order gives you the spec of a maximal order, so you should think about the spec of a maximal order as a smooth curve.
me: didn't realize this, thought this was very cool
B: let's move on to number theory
me: was hoping the slightly number theoretic questions in algebra were the number theory part but i guess i knew this is too optimistic
Algebraic number theory
S: speaking of class groups, how do you know the class group is finite
me: Minkowski bound gives you an upper bound on the norm of the ideal class and there's only finitely many ideals of bounded norm
S: what about for function fields
me: it's not finite
S: it's not *necessarily* finite
me: right, oops
S: give me an example of a function field with infinite Picard group
me elliptic curve with infinity many points
B: compute class group of Z adjoin sqrt -5
me: writes down minkowski bound, only need to check primes above 2 and 3 (call these p2, p3), 2 ramifies, 3 splits, and (1+sqrt-5) = p_2*p_3 so class group is Z/2
B: what's p2
me: write (2, 1+sqrt(-5))
S: what's p3
me: write (3, 1+sqrt(-5))
B: 3 splits, what's the other prime above 3
me: write (3, 1-sqrt(-5)) because it should be the conjugate
S: do you know how a general machine to write down generators of prime ideals?
me: no
S: ok let's work it out, what happens when you quotient the ring of integers by the ideal (2)
me: you get F_2[x]/(x^2 + 5) = F2[x]/(x+1)^2
S: what are the prime ideals here
me: (x+1)
S: ok we have an isomorphism O_k/(2) \cong F2[x]/(x+1)^2, what are the prime ideals on the left hand side
me: p2
S: ok so pulling back x+1 and lifting, what do we get
me: ohhh (2, 1+sqrt(-5))
B: ok if we do this for any p what happens
me: Fp[x]/(x^2 + 5). The polynomial x^2 + 5 is a square if p ramifies, is irreducible if p is inert, and splits into two different linear factors if p splits.
B: right, so lets do 41
me: i calculate and check that 41 splits. i factor x^2 + 5 mod 41 as (x-a)(x+a) but struggle to find a until B kindly points out that it is 6.
B: so what's the ideal it pulls back to
me: ohh, (41) = (41, 6 + \sqrt(-5))(41, 6 - sqrt(-5))
S: prove that any ideal is generated by at most 2 elements
me: uhh lets do it for prime ideals first
B: well what we did above basically shows it for prime ideals right
me: yea true
B: well in the case above the ring of integers was monogenic. what can we say if its not
me: well its fine just localize and then its monogenic
B: ok.
S: says something about Q(sqrt -15) but then decides to let me off the hook
me: thanks the heavens
S: compute hilbert class field of Q(sqrt -5)
me: just adjoin i
S: prove that Q(sqrt -5, i) is the Hilbert class field of Q(sqrt -5)
me: draws the lattice of subfields of Q(sqrt -5, i) and shows that the only possible ramification of the extension Q(sqrt -5, i)/Q(sqrt -5) can be at 2. However 2 is unramified in the sub extension Q(sqrt 5)/Q and is ramified in Q(sqrt -5)/Q and hence Q(sqrt -5, i)/Q(sqrt -5) is unramified.
B: Characterize all splitting behavior in Q(sqrt -5, i)/Q
me: shows that no primes are totally ramified, ramification of order 2 happens iff it happens in a quadratic sub extension, no primes are totally inert because the galois group is noncyclic, and a prime splits completely iff it splits completely in all 3 quadratic sub extensions
S: what if a prime splits completely in 2 quadratic sub extensions
me: the prime splits completely in the whole thing
B: what's the discriminant of Q(sqrt -5, i)/Q
me (yikes): -20 because Q(sqrt -5, i)/Q(sqrt -5) is unramified and the discriminant of Q(sqrt -5) is -20
B: what's the galois group of Q(sqrt -5, i) as a subgroup of S4
me: double transpositions
B: so is the galois group a subgroup of A4
me: yes
B: so what does that tell you about the discriminant
me: its a square
B: why
me: because Galois fixes \prod_{i