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−  ###########################################################################
 +  {{SNES title = Zombies Ate My Neighbors 
−  1Q: What is the current status of Fermat's last theorem?
 +  image = [[Image:ZAMN Title.PNGcenter]] 
 +  name = ZOMBIES ATE MY NEIGHB 
 +  company = Konami 
 +  header = None 
 +  bank = LoROM 
 +  interleaved = No 
 +  sram = 0 Kb 
 +  type = Normal 
 +  rom = 8 Mb 
 +  country = USA 
 +  video = NTSC 
 +  romspeed = 200ns (SlowROM) 
 +  revision = 1.0 
 +  checksum = Good 0xF50C 
 +  crc32 = 7CFC0C7C 
 +  game = Zombies Ate My Neighbors 
 +  }} 
   
 +  ==Utilities== 
 +  * [http://www.romhacking.net/utils/363/ ZAMNEdit] 
 +  * [https://code.google.com/archive/p/zamneditor/ ZAMN Level Editor] ([http://www.romhacking.net/utilities/1178/ RHDN mirror]) 
   
−  and
 +  ==Hacks== 
 +  * [http://acmlm.noip.org/archive2/thread.php?id=9051 Ultimate Zombies Ate My Neighbors]  All 55 levels changed. ([http://www.romhacking.net/hacks/59/ RHDN mirror]) 
 +  * [http://www.romhacking.net/hacks/623/ Oh No! More Zombies Ate My Neighbors!] 
 +  * [http://www.romhacking.net/hacks/2745/ Sky's Zombies Ate My Neighbors Hack] 
   
−  Did Fermat prove this theorem?
 +  ==Translations== 
− 
 +  * None known. 
− 
 
−  Fermat's Last Theorem:
 
   
−  There are no positive integers x,y,z, and n > 2 such that x^n + y^n = z^n.
 +  ==Miscellaneous== 
 +  * No miscellaneous yet. 
   
−  I heard that <insert name here> claimed to have proved it but later
 +  ==Known Dumps== 
−  on the proof was found to be wrong. ...
 +  * Zombies (Beta) 
 +  * Zombies (E) 
 +  * Zombies Ate My Neighbors (U) [!] 
   
−  A: The status of FLT has remained remarkably constant. Every few
 +  ==External Links== 
−  years, someone claims to have a proof ... but oh, wait, not quite.
 +  * No external links yet. 
−   
−  UPDATE... UPDATE... UPDATE
 
−   
−  Andrew Wiles, a researcher at Princeton, Cambridge claims to have
 
−  found a proof.
 
−   
−  SECOND UPDATE...
 
−   
−  A mistake has been found. Wiles is working on it. People remain
 
−  mildly optimistic about his chances of fixing the error.
 
− 
 
−  The proposed proof goes like this:
 
−   
−  The proof was presented in Cambridge, UK during a three day seminar
 
−  to an audience including some of the leading experts in the field.
 
−   
−  The manuscript has been submitted to INVENTIONES MATHEMATICAE, and
 
−  is currently under review. Preprints are not available until the
 
−  proof checks out. Wiles is giving a full seminar on the proof this
 
−  spring.
 
−   
−  The proof is long and cumbersome, but here are some of the first
 
−  few details:
 
−   
−  *From Ken Ribet:
 
−   
−  Here is a brief summary of what Wiles said in his three lectures.
 
−   
−  The method of Wiles borrows results and techniques from lots and lots
 
−  of people. To mention a few: Mazur, Hida, Flach, Kolyvagin, yours
 
−  truly, Wiles himself (older papers by Wiles), Rubin... The way he does
 
−  it is roughly as follows. Start with a mod p representation of the
 
−  Galois group of Q which is known to be modular. You want to prove that
 
−  all its lifts with a certain property are modular. This means that the
 
−  canonical map from Mazur's universal deformation ring to its "maximal
 
−  Hecke algebra" quotient is an isomorphism. To prove a map like this is
 
−  an isomorphism, you can give some sufficient conditions based on
 
−  commutative algebra. Most notably, you have to bound the order of a
 
−  cohomology group which looks like a Selmer group for Sym^2 of the
 
−  representation attached to a modular form. The techniques for doing
 
−  this come from Flach; you also have to use Euler systems a la
 
−  Kolyvagin, except in some new geometric guise.
 
−   
−  CLARIFICATION: This step in Wiles' manuscript, the Selmer group
 
−  bound, is currently considered to be incomplete by the reviewers.
 
−  Yet the reviewers (or at least those who have gone public) have
 
−  confidence that Wiles will fill it in. (Note that such gaps are
 
−  quite common in long proofs. In this particular case, just such
 
−  a bound was expected to be provable using Kolyvagin's techniques,
 
−  independently of anyone thinking of modularity. In the worst of
 
−  cases, and the gap is for real, what remains has to be recast, but
 
−  it is still extremely important number theory breakthrough work.)
 
−   
−   
− 
 
−  If you take an elliptic curve over Q, you can look at the
 
−  representation of Gal on the 3division points of the curve. If you're
 
−  lucky, this will be known to be modular, because of results of Jerry
 
−  Tunnell (on base change). Thus, if you're lucky, the problem I
 
−  described above can be solved (there are most definitely some
 
−  hypotheses to check), and then the curve is modular. Basically, being
 
−  lucky means that the image of the representation of Galois on
 
−  3division points is GL(2,Z/3Z).
 
− 
 
−  Suppose that you are unlucky, i.e., that your curve E has a rational
 
−  subgroup of order 3. Basically by inspection, you can prove that if it
 
−  has a rational subgroup of order 5 as well, then it can't be
 
−  semistable. (You look at the four noncuspidal rational points of
 
−  X_0(15).) So you can assume that E[5] is "nice." Then the idea is to
 
−  find an E' with the same 5division structure, for which E'[3] is
 
−  modular. (Then E' is modular, so E'[5] = E[5] is modular.) You
 
−  consider the modular curve X which parameterizes elliptic curves whose
 
−  5division points look like E[5]. This is a "twist" of X(5). It's
 
−  therefore of genus 0, and it has a rational point (namely, E), so it's
 
−  a projective line. Over that you look at the irreducible covering
 
−  which corresponds to some desired 3division structure. You use
 
−  Hilbert irreducibility and the Cebotarev density theorem (in some way
 
−  that hasn't yet sunk in) to produce a noncuspidal rational point of X
 
−  over which the covering remains irreducible. You take E' to be the
 
−  curve corresponding to this chosen rational point of X.
 
− 
 
− 
 
−  *From the previous version of the FAQ:
 
− 
 
−  (b) conjectures arising from the study of elliptic curves and
 
−  modular forms.  The TaniyamaWeilShmimura conjecture.
 
− 
 
−  There is a very important and well known conjecture known as the
 
−  TaniyamaWeilShimura conjecture that concerns elliptic curves.
 
−  This conjecture has been shown by the work of Frey, Serre, Ribet,
 
−  et. al. to imply FLT uniformly, not just asymptotically as with the
 
−  ABC conj.
 
− 
 
−  The conjecture basically states that all elliptic curves can be
 
−  parameterized in terms of modular forms.
 
−   
−  There is new work on the arithmetic of elliptic curves. Sha, the
 
−  TateShafarevich group on elliptic curves of rank 0 or 1. By the way
 
−  an interesting aspect of this work is that there is a close
 
−  connection between Sha, and some of the classical work on FLT. For
 
−  example, there is a classical proof that uses infinite descent to
 
−  prove FLT for n = 4. It can be shown that there is an elliptic curve
 
−  associated with FLT and that for n=4, Sha is trivial. It can also be
 
−  shown that in the cases where Sha is nontrivial, that
 
−  infinitedescent arguments do not work; that in some sense 'Sha
 
−  blocks the descent'. Somewhat more technically, Sha is an
 
−  obstruction to the localglobal principle [e.g. the HasseMinkowski
 
−  theorem].
 
−   
−  *From Karl Rubin:
 
−   
−  Theorem. If E is a semistable elliptic curve defined over Q,
 
−  then E is modular.
 
−   
−  It has been known for some time, by work of Frey and Ribet, that
 
−  Fermat follows from this. If u^q + v^q + w^q = 0, then Frey had
 
−  the idea of looking at the (semistable) elliptic curve
 
−  y^2 = x(xa^q)(x+b^q). If this elliptic curve comes from a modular
 
−  form, then the work of Ribet on Serre's conjecture shows that there
 
−  would have to exist a modular form of weight 2 on Gamma_0(2). But
 
−  there are no such forms.
 
− 
 
−  To prove the Theorem, start with an elliptic curve E, a prime p and let
 
− 
 
−  rho_p : Gal(Q^bar/Q) > GL_2(Z/pZ)
 
− 
 
−  be the representation giving the action of Galois on the ptorsion
 
−  E[p]. We wish to show that a _certain_ lift of this representation
 
−  to GL_2(Z_p) (namely, the padic representation on the Tate module
 
−  T_p(E)) is attached to a modular form. We will do this by using
 
−  Mazur's theory of deformations, to show that _every_ lifting which
 
−  'looks modular' in a certain precise sense is attached to a modular form.
 
− 
 
−  Fix certain 'lifting data', such as the allowed ramification,
 
−  specified local behavior at p, etc. for the lift. This defines a
 
−  lifting problem, and Mazur proves that there is a universal
 
−  lift, i.e. a local ring R and a representation into GL_2(R) such
 
−  that every lift of the appropriate type factors through this one.
 
− 
 
−  Now suppose that rho_p is modular, i.e. there is _some_ lift
 
−  of rho_p which is attached to a modular form. Then there is
 
−  also a hecke ring T, which is the maximal quotient of R with the
 
−  property that all _modular_ lifts factor through T. It is a
 
−  conjecture of Mazur that R = T, and it would follow from this
 
−  that _every_ lift of rho_p which 'looks modular' (in particular the
 
−  one we are interested in) is attached to a modular form.
 
− 
 
−  Thus we need to know 2 things:
 
−   
−  (a) rho_p is modular
 
−  (b) R = T.
 
− 
 
−  It was proved by Tunnell that rho_3 is modular for every elliptic
 
−  curve. This is because PGL_2(Z/3Z) = S_4. So (a) will be satisfied
 
−  if we take p=3. This is crucial.
 
− 
 
−  Wiles uses (a) to prove (b) under some restrictions on rho_p. Using
 
−  (a) and some commutative algebra (using the fact that T is Gorenstein,
 
−  'basically due to Mazur') Wiles reduces the statement T = R to
 
−  checking an inequality between the sizes of 2 groups. One of these
 
−  is related to the Selmer group of the symmetric square of the given
 
−  modular lifting of rho_p, and the other is related (by work of Hida)
 
−  to an Lvalue. The required inequality, which everyone presumes is
 
−  an instance of the BlochKato conjecture, is what Wiles needs to verify.
 
− 
 
−  He does this using a Kolyvagintype Euler system argument. This is
 
−  the most technically difficult part of the proof, and is responsible
 
−  for most of the length of the manuscript. He uses modular
 
−  units to construct what he calls a 'geometric Euler system' of
 
−  cohomology classes. The inspiration for his construction comes
 
−  from work of Flach, who came up with what is essentially the
 
−  'bottom level' of this Euler system. But Wiles needed to go much
 
−  farther than Flach did. In the end, _under_certain_hypotheses_ on rho_p
 
−  he gets a workable Euler system and proves the desired inequality.
 
−  Among other things, it is necessary that rho_p is irreducible.
 
− 
 
−  Suppose now that E is semistable.
 
− 
 
−  Case 1. rho_3 is irreducible.
 
−  Take p=3. By Tunnell's theorem (a) above is true. Under these
 
−  hypotheses the argument above works for rho_3, so we conclude
 
−  that E is modular.
 
− 
 
−  Case 2. rho_3 is reducible.
 
−  Take p=5. In this case rho_5 must be irreducible, or else E
 
−  would correspond to a rational point on X_0(15). But X_0(15)
 
−  has only 4 noncuspidal rational points, and these correspond to
 
−  nonsemistable curves. _If_ we knew that rho_5 were modular,
 
−  then the computation above would apply and E would be modular.
 
− 
 
−  We will find a new semistable elliptic curve E' such that
 
−  rho_{E,5} = rho_{E',5} and rho_{E',3} is irreducible. Then
 
−  by Case I, E' is modular. Therefore rho_{E,5} = rho_{E',5}
 
−  does have a modular lifting and we will be done.
 
− 
 
−  We need to construct such an E'. Let X denote the modular
 
−  curve whose points correspond to pairs (A, C) where A is an
 
−  elliptic curve and C is a subgroup of A isomorphic to the group
 
−  scheme E[5]. (All such curves will have mod5 representation
 
−  equal to rho_E.) This X is genus 0, and has one rational point
 
−  corresponding to E, so it has infinitely many. Now Wiles uses a
 
−  Hilbert Irreducibility argument to show that not all rational
 
−  points can be images of rational points on modular curves
 
−  covering X, corresponding to degenerate level 3 structure
 
−  (i.e. im(rho_3) not GL_2(Z/3)). In other words, an E' of the
 
−  type we need exists. (To make sure E' is semistable, choose
 
−  it 5adically close to E. Then it is semistable at 5, and at
 
−  other primes because rho_{E',5} = rho_{E,5}.)
 
−   
− 
 
−  Referencesm:
 
−   
−   
−  American Mathematical Monthly
 
−  January 1994.
 
−   
−  Notices of the AMS, Februrary 1994.
 
−   
− 
 
−  ###########################################################################
 