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| | {{Bob |
| 1Q: What is the current status of Fermat's last theorem?
| | | bobscreen= ZAMN Title.PNG |
| | | title= Zombies Ate My Neighbors |
| | | aka= Zombies (EU/AU) |
| | | developer= LucasArts |
| | | publisher= Konami |
| | | system= SNES |
| | | usa= {{date|1993|July|19}} |
| | | europe= {{date|1994|January|27|nocat=true}} |
| | | tcrf=Zombies Ate My Neighbors (SNES) |
| | }} |
| | {{SNES| title = Zombies Ate My Neighbors |
| | |image = [[File:ZAMN Title.PNG|center]] |
| | |name = ZOMBIES ATE MY NEIGHB |
| | |company = Konami |
| | |header = None |
| | |bank = LoROM |
| | |interleaved = No |
| | |sram = 0 KiB |
| | |type = Normal |
| | |rom = 1 MiB |
| | |country = USA |
| | |video = NTSC |
| | |romspeed = 200ns (SlowROM) |
| | |revision = 1.0 |
| | |checksum = Good 0xF50C |
| | |crc32 = 7CFC0C7C |
| | |game = Zombies Ate My Neighbors |
| | }} |
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| |
|
| | ==Sub-pages== |
| | {{namespaces}} |
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|
| and
| | ==Utilities== |
| | * [http://www.romhacking.net/utils/363/ ZAMN-Edit] |
| | * [https://code.google.com/archive/p/zamn-editor/ ZAMN Level Editor] ([http://www.romhacking.net/utilities/1178/ RHDN mirror]) |
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|
| Did Fermat prove this theorem?
| | ==Hacks== |
|
| | * [http://acmlm.no-ip.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!] |
| Fermat's Last Theorem:
| | * [http://www.romhacking.net/hacks/2745/ Sky's Zombies Ate My Neighbors Hack] |
|
| |
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| There are no positive integers x,y,z, and n > 2 such that x^n + y^n = z^n.
| | ==Translations== |
| | * None known. |
|
| |
|
| I heard that <insert name here> claimed to have proved it but later
| | ==Miscellaneous== |
| on the proof was found to be wrong. ...
| | * No miscellaneous yet. |
|
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| A: The status of FLT has remained remarkably constant. Every few
| | ==Known Dumps== |
| years, someone claims to have a proof ... but oh, wait, not quite.
| | * Zombies (Beta) |
| | * Zombies (E) |
| | * Zombies Ate My Neighbors (U) [!] |
|
| |
|
| UPDATE... UPDATE... UPDATE
| | ==External links== |
| | | * No external links yet. |
| Andrew Wiles, a researcher at Princeton, Cambridge claims to have
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| found a proof.
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| | |
| SECOND UPDATE...
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| | |
| A mistake has been found. Wiles is working on it. People remain
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| mildly optimistic about his chances of fixing the error.
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|
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| The proposed proof goes like this:
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| The proof was presented in Cambridge, UK during a three day seminar
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| to an audience including some of the leading experts in the field.
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| The manuscript has been submitted to INVENTIONES MATHEMATICAE, and
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| is currently under review. Preprints are not available until the
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| proof checks out. Wiles is giving a full seminar on the proof this
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| spring.
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| The proof is long and cumbersome, but here are some of the first
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| few details:
| |
| | |
| *From Ken Ribet:
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| | |
| Here is a brief summary of what Wiles said in his three lectures.
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| | |
| The method of Wiles borrows results and techniques from lots and lots
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| of people. To mention a few: Mazur, Hida, Flach, Kolyvagin, yours
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| truly, Wiles himself (older papers by Wiles), Rubin... The way he does
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| it is roughly as follows. Start with a mod p representation of the
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| Galois group of Q which is known to be modular. You want to prove that
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| all its lifts with a certain property are modular. This means that the
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| canonical map from Mazur's universal deformation ring to its "maximal
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| Hecke algebra" quotient is an isomorphism. To prove a map like this is
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| an isomorphism, you can give some sufficient conditions based on
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| commutative algebra. Most notably, you have to bound the order of a
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| cohomology group which looks like a Selmer group for Sym^2 of the
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| representation attached to a modular form. The techniques for doing
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| this come from Flach; you also have to use Euler systems a la
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| Kolyvagin, except in some new geometric guise.
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| | |
| CLARIFICATION: This step in Wiles' manuscript, the Selmer group
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| bound, is currently considered to be incomplete by the reviewers.
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| Yet the reviewers (or at least those who have gone public) have
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| confidence that Wiles will fill it in. (Note that such gaps are
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| quite common in long proofs. In this particular case, just such
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| a bound was expected to be provable using Kolyvagin's techniques,
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| independently of anyone thinking of modularity. In the worst of
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| cases, and the gap is for real, what remains has to be recast, but
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| it is still extremely important number theory breakthrough work.)
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|
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| If you take an elliptic curve over Q, you can look at the
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| representation of Gal on the 3-division points of the curve. If you're
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| lucky, this will be known to be modular, because of results of Jerry
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| Tunnell (on base change). Thus, if you're lucky, the problem I
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| described above can be solved (there are most definitely some
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| hypotheses to check), and then the curve is modular. Basically, being
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| lucky means that the image of the representation of Galois on
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| 3-division points is GL(2,Z/3Z).
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|
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| Suppose that you are unlucky, i.e., that your curve E has a rational
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| subgroup of order 3. Basically by inspection, you can prove that if it
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| has a rational subgroup of order 5 as well, then it can't be
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| semistable. (You look at the four non-cuspidal rational points of
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| X_0(15).) So you can assume that E[5] is "nice." Then the idea is to
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| find an E' with the same 5-division structure, for which E'[3] is
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| modular. (Then E' is modular, so E'[5] = E[5] is modular.) You
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| consider the modular curve X which parameterizes elliptic curves whose
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| 5-division points look like E[5]. This is a "twist" of X(5). It's
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| therefore of genus 0, and it has a rational point (namely, E), so it's
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| a projective line. Over that you look at the irreducible covering
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| which corresponds to some desired 3-division structure. You use
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| Hilbert irreducibility and the Cebotarev density theorem (in some way
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| that hasn't yet sunk in) to produce a non-cuspidal rational point of X
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| over which the covering remains irreducible. You take E' to be the
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| curve corresponding to this chosen rational point of X.
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|
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|
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| *From the previous version of the FAQ:
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|
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| (b) conjectures arising from the study of elliptic curves and
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| modular forms. -- The Taniyama-Weil-Shmimura conjecture.
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|
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| There is a very important and well known conjecture known as the
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| Taniyama-Weil-Shimura conjecture that concerns elliptic curves.
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| This conjecture has been shown by the work of Frey, Serre, Ribet,
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| et. al. to imply FLT uniformly, not just asymptotically as with the
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| ABC conj.
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|
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| The conjecture basically states that all elliptic curves can be
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| parameterized in terms of modular forms.
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| | |
| There is new work on the arithmetic of elliptic curves. Sha, the
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| Tate-Shafarevich group on elliptic curves of rank 0 or 1. By the way
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| an interesting aspect of this work is that there is a close
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| connection between Sha, and some of the classical work on FLT. For
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| example, there is a classical proof that uses infinite descent to
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| prove FLT for n = 4. It can be shown that there is an elliptic curve
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| associated with FLT and that for n=4, Sha is trivial. It can also be
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| shown that in the cases where Sha is non-trivial, that
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| infinite-descent arguments do not work; that in some sense 'Sha
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| blocks the descent'. Somewhat more technically, Sha is an
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| obstruction to the local-global principle [e.g. the Hasse-Minkowski
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| theorem].
| |
| | |
| *From Karl Rubin:
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| | |
| Theorem. If E is a semistable elliptic curve defined over Q,
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| then E is modular.
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| | |
| It has been known for some time, by work of Frey and Ribet, that
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| Fermat follows from this. If u^q + v^q + w^q = 0, then Frey had
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| the idea of looking at the (semistable) elliptic curve
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| y^2 = x(x-a^q)(x+b^q). If this elliptic curve comes from a modular
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| form, then the work of Ribet on Serre's conjecture shows that there
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| would have to exist a modular form of weight 2 on Gamma_0(2). But
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| there are no such forms.
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|
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| To prove the Theorem, start with an elliptic curve E, a prime p and let
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|
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| rho_p : Gal(Q^bar/Q) -> GL_2(Z/pZ)
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|
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| be the representation giving the action of Galois on the p-torsion
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| E[p]. We wish to show that a _certain_ lift of this representation
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| to GL_2(Z_p) (namely, the p-adic representation on the Tate module
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| T_p(E)) is attached to a modular form. We will do this by using
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| Mazur's theory of deformations, to show that _every_ lifting which
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| 'looks modular' in a certain precise sense is attached to a modular form.
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|
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| Fix certain 'lifting data', such as the allowed ramification,
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| specified local behavior at p, etc. for the lift. This defines a
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| lifting problem, and Mazur proves that there is a universal
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| lift, i.e. a local ring R and a representation into GL_2(R) such
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| that every lift of the appropriate type factors through this one.
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|
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| Now suppose that rho_p is modular, i.e. there is _some_ lift
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| of rho_p which is attached to a modular form. Then there is
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| also a hecke ring T, which is the maximal quotient of R with the
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| property that all _modular_ lifts factor through T. It is a
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| conjecture of Mazur that R = T, and it would follow from this
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| that _every_ lift of rho_p which 'looks modular' (in particular the
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| one we are interested in) is attached to a modular form.
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|
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| Thus we need to know 2 things:
| |
| | |
| (a) rho_p is modular
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| (b) R = T.
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|
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| It was proved by Tunnell that rho_3 is modular for every elliptic
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| curve. This is because PGL_2(Z/3Z) = S_4. So (a) will be satisfied
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| if we take p=3. This is crucial.
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|
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| Wiles uses (a) to prove (b) under some restrictions on rho_p. Using
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| (a) and some commutative algebra (using the fact that T is Gorenstein,
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| 'basically due to Mazur') Wiles reduces the statement T = R to
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| checking an inequality between the sizes of 2 groups. One of these
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| is related to the Selmer group of the symmetric square of the given
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| modular lifting of rho_p, and the other is related (by work of Hida)
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| to an L-value. The required inequality, which everyone presumes is
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| an instance of the Bloch-Kato conjecture, is what Wiles needs to verify.
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|
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| He does this using a Kolyvagin-type Euler system argument. This is
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| the most technically difficult part of the proof, and is responsible
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| for most of the length of the manuscript. He uses modular
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| units to construct what he calls a 'geometric Euler system' of
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| cohomology classes. The inspiration for his construction comes
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| from work of Flach, who came up with what is essentially the
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| 'bottom level' of this Euler system. But Wiles needed to go much
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| farther than Flach did. In the end, _under_certain_hypotheses_ on rho_p
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| he gets a workable Euler system and proves the desired inequality.
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| Among other things, it is necessary that rho_p is irreducible.
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|
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| Suppose now that E is semistable.
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|
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| Case 1. rho_3 is irreducible.
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| Take p=3. By Tunnell's theorem (a) above is true. Under these
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| hypotheses the argument above works for rho_3, so we conclude
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| that E is modular.
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|
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| Case 2. rho_3 is reducible.
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| Take p=5. In this case rho_5 must be irreducible, or else E
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| would correspond to a rational point on X_0(15). But X_0(15)
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| has only 4 noncuspidal rational points, and these correspond to
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| non-semistable curves. _If_ we knew that rho_5 were modular,
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| then the computation above would apply and E would be modular.
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|
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| We will find a new semistable elliptic curve E' such that
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| rho_{E,5} = rho_{E',5} and rho_{E',3} is irreducible. Then
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| by Case I, E' is modular. Therefore rho_{E,5} = rho_{E',5}
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| does have a modular lifting and we will be done.
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|
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| We need to construct such an E'. Let X denote the modular
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| curve whose points correspond to pairs (A, C) where A is an
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| elliptic curve and C is a subgroup of A isomorphic to the group
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| scheme E[5]. (All such curves will have mod-5 representation
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| equal to rho_E.) This X is genus 0, and has one rational point
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| corresponding to E, so it has infinitely many. Now Wiles uses a
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| Hilbert Irreducibility argument to show that not all rational
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| points can be images of rational points on modular curves
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| covering X, corresponding to degenerate level 3 structure
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| (i.e. im(rho_3) not GL_2(Z/3)). In other words, an E' of the
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| type we need exists. (To make sure E' is semistable, choose
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| it 5-adically close to E. Then it is semistable at 5, and at
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| other primes because rho_{E',5} = rho_{E,5}.)
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| | |
|
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| Referencesm:
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| | |
| | |
| American Mathematical Monthly
| |
| January 1994.
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| | |
| Notices of the AMS, Februrary 1994.
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|
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| ###########################################################################
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