math guru's help...purely abstract stuff..

semidevil

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Joined
Jun 23, 2002
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190
I'm trying to write a paper on Fermat's Last theorem, and browsing through the web, I found these 2 theorems:

1. If there is a solution(x, y, z, n) to Fermat's theorem, then the elliptic curve defined by the equation
Y^2 = X(x - x^n)(X + Y^n) is semistable but not modular

2. All Semistable elliptic curves w/ rational cofficients are modular.

My resources from the net gives proofs to the 2 theorems, thereby proving the Fermat problem. but I am having trouble relating the theorems to the Fermat problem. what does it mean for an elliptic curve to be semistable and modualar? I just don't see how it relates.
 
Joined
Nov 24, 2002
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Equations of the form x^n + y^n = z^n, called Diophantine equations when n is an integer, can be translated to describe a certain set of elliptic curves. These curves represent the surface of a torus, an object shaped like a smooth doughnut.

Taniyama suggested that for that certain set of elliptic curves in Euclidean geometry (where parallel lines never meet, even if infinitely extended) there are corresponding structures in the hyperbolic (non-Euclidian) plane (where parallel lines can both converge and diverge).

Frey suggested a connection between that certain set of elliptic curves and Fermat's last theorem, namely that if there were solutions in violation of the theorem, they would generate a subset of "semistable" elliptic curves, curves that could not be represented in the hyperbolic plane.

Wiles accepted Ribet's proof of Frey and reasoned that if he (Wiles) could prove Taniyama, at least for the "Fermat subset" of semistable elliptic curves (if not for that larger certain set of elliptic curves), no solutions to Fermat's last theorem could exist, thereby implying a proof of FLT.

Wiles then developed an unconventional method of counting both the Euclidian semistable elliptic curves and their hyperbolic (non-Euclidian) counterparts in such a way as to demonstrate a one-to-one correspondence between the two groups. In this way, he claims to have proved Taniyama for the "Fermat subset" of semistable elliptic curves.
A function f is said to be an entire modular form of weight k if it satisfies

1. f is analytic in the upper half-plane H,
Code:
2. f [U](at+b)[/U] = (ct+d)^k * f(t)
     (ct+d)
whenever [a b] ( <-matrix here) is a member of the modular group gamma, 
         [c d]
- (t should read Tau, Albert)

3. The Fourier series of f has the form
 f(t) = Sigma (n=0 to Infinity) c*(n)*e^(2*Pi*i*n*t)
- (Sigma is of course the Greek capital letter for summation and
 once again t should read Tau)
Hate to admit defeat but it is easier to just look at the top of this page for the def of modular:
http://mathworld.wolfram.com/ModularForm.html

Very good description of the whole shebang here:

http://mathworld.wolfram.com/Taniyama-ShimuraConjecture.html

My brain hurts now! :( Too many miles on the clock to remember this stuff, but sadly it comes back too too easily. Frankly I'd rather be Hemmeling! ;)
 

semidevil

Thread Starter
Joined
Jun 23, 2002
Messages
190
albert,

thank you so much....for both problems..... I got a 94% on my exam...hehe
 
Joined
Nov 24, 2002
Messages
2,438
If that is for what you did yourself under exam conditions then accept my heartiest congratulations Semidevil! A 94%!!!!! Being able to share ideas and get the odd pointer from others is what advancement in any subject is about. Glad to have helped a little. :)

So have you unified those fields yet then? ;)
 

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