Approaches to FLT - PowerPoint PPT Presentation

About This Presentation
Title:

Approaches to FLT

Description:

Then we can write complex numbers ... but none is a prime. So there is no way to unique write 6 as a the product ... Now this would be true if there were a unique ... – PowerPoint PPT presentation

Number of Views:23
Avg rating:3.0/5.0
Slides: 7
Provided by: RalphKa9
Category:
Tags: flt | approaches | none | then | there | were

less

Transcript and Presenter's Notes

Title: Approaches to FLT


1
Approaches to FLT
  • In the early approaches to FLT several methods
    were used (See for instance the proof of Germain
    Theorem).
  • Infinite descent
  • Congruences
  • Unique factorization into prime factors
  • In 1847 Lamé put forth a proof of Fermat using
    a combination of the above techniques which he
    thought to generalize.
  • A few weeks later Kummer showed that unique
    factorization does not generalize in the way that
    Lamé expected.

2
Ernst Eduard Kummer(1810-1893)
  • 1828 Kummer entered the University of Halle with
    the intention of studying Protestant theology.
  • 1831 Kummer was awarded a prize for a
    mathematical essay on a topic set by Scherk. For
    this, he was also awarded a doctorate.
  • 1831 became a teacher (for 10 years).
    Corresponded with Jacobi and Dirichlet.
  • 1836 elected into the Berlin Academy of Science.
  • 1842 became a full professor in Breslau.
  • 1855 became a professor in Berlin.
  • 1883 retired because of fading memory.
  • 1843 Kummer, realized that unique factorisation
    of integers did not extend to other rings of
    complex numbers and hence attempts to prove
    Fermat's Last Theorem broke down. To compensate
    he introduced ideal numbers.
  • 1857 Grand Prize of the Paris Academy of Sciences
    for his work toward FLT, although he never
    entered the contest.

3
Complex numbers
  • Complex numbers are pairs of real numbers (a,b)
    with the following operations.
  • (a,b)(c,d)(ab,cd)
  • (a,b)(c,d)(ac-bd,adbc)
  • Notice that there is also subtraction and
    division.
  • One can also formally introduce the symbol i
    ?with the relation i2-1. Then we can write
    complex numbers (a,b) as zaib.
  • Since a complex number is a pair of real numbers,
    it also has polar coordinates (r,f).
  • We can also write zreif.
  • To convert one uses eifcos f i sin f and
    rz(a2-b2)(aib)(a-ib).
  • Then if zreif and wseiy then zwrsei(fy).
  • Fundamental Theorem of Algebra Every polynomial
    equation of degree n with complex coefficients
    has n roots in the complex numbers.
  • Example zn-1(z-1)(z-wn) (z-wn2) (z-wnn-1)
  • wne2pi/n and wnne2pin/ncos(2p)isin(2p)1.

4
Number systems
  • Just like considering v-1, we can introduce v-5
    and look at the numbers abv-5, but now with a
    and b in Z.
  • What changes? Unique factorization need not hold!
  • 623(1v-5)(1- v-5)
  • Also we need a new concept of prime numbers.
  • We call m irreducible if m cannot be written as a
    product not containing 1.
  • We call p prime if p divides mn implies that
    either p divides m or p divides n.
  • Problem In the above factorization all the
    factors are irreducible, but none is a prime. So
    there is no way to unique write 6 as a the
    product of primes.

5
Where Lamé went wrong
  • Lamé factored xnyn(xy)(xwny) (xwn2y)
    (xwnn-1y) and then wanted to deduce that each
    linear factor has to be an n-th power.
  • Now this would be true if there were a unique
    factorization.
  • For n?23 however Kummer showed that this is not
    generally the case. For this consider numbers of
    the form a0a1w23a2w232a21w2321
  • (actually w2322-1-w23-w232--w2321)
  • But (1w2w4w5w10w11)(1ww5w6w7w9w11) is
    divisible by 2, but none of the factors are! So
    there is no unique factorization!

6
Kummers ideal numbers
  • Kummer thought of the failure of unique
    factorization as caused by hidden divisors. He
    found a way to treat the hidden divisors as
    ideal numbers.
  • For this a number is generalized to the subset it
    generates 2 ? lt2gt2nn in Z.
  • Then we can also look at the sets of the form
  • Alt2,1v-5gt2mn(1v-5)n,m in Z
  • and think of this set as the common divisors of
    2 and 1v-5.
  • We would get AA lt4,2(1v-5),-42v-5gtlt2gt.
  • Setting Blt3, 1v-5gt and Clt3,1v-5gt one obtains
  • lt6gtlt2gtlt3gtA2BCABAClt1v-5, 1-v-5gt.
Write a Comment
User Comments (0)
About PowerShow.com