Approaches to FLT In the early approaches to

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Approaches to FLT • In the early approaches to FLT several methods were used

Approaches to FLT • In the early approaches to FLT several methods were used (See for instance the proof of Germain Theorem). 1. Infinite descent 2. Congruences 3. 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.

Ernst Eduard Kummer (1810 -1893) • 1828 Kummer entered the University of Halle with

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.

Complex numbers • Complex numbers are pairs of real numbers (a, b) with the

Complex numbers • Complex numbers are pairs of real numbers (a, b) with the following operations. – (a, b)+(c, d)=(a+b, c+d) – (a, b)(c, d)=(ac-bd, ad+bc) • Notice that there is also subtraction and division. • One can also formally introduce the symbol i with the relation i 2=-1. Then we can write complex numbers (a, b) as z=a+ib. • Since a complex number is a pair of real numbers, it also has polar coordinates (r, f). – We can also write z=reif. – To convert one uses eif=cos f + i sin f and r=|z|=(a 2 -b 2)=(a+ib)(a-ib). – Then if z=reif and w=seiy then zw=rsei(f+y). • 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-wn 2) …(z-wnn-1) – wn=e 2 pi/n and wnn=e 2 pin/n=cos(2 p)+isin(2 p)=1.

Number systems • Just like considering √-1, we can introduce √-5 and look at

Number systems • Just like considering √-1, we can introduce √-5 and look at the numbers a+b√-5, but now with a and b in Z. • What changes? Unique factorization need not hold! • 6=2· 3=(1+√-5)(1 - √-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.

Where Lamé went wrong • Lamé factored xn+yn=(x+y)(x+wny) (x+wn 2 y) …(x+wnn-1 y) and

Where Lamé went wrong • Lamé factored xn+yn=(x+y)(x+wny) (x+wn 2 y) …(x+wnn-1 y) 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 a 0+a 1 w 23+a 2 w 232+…+a 21 w 2321 (actually w 2322=-1 -w 232 -…-w 2321) • But (1+w 2+w 4+w 5+w 10+w 11)(1+w+w 5+w 6+w 7+w 9+w 11) is divisible by 2, but none of the factors are! So there is no unique factorization!

Kummer’s “ideal numbers” • Kummer thought of the failure of unique factorization as caused

Kummer’s “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 <2>={2 n: n in Z}. • Then we can also look at the sets of the form A=<2, 1+√-5>={2 m+n(1+√-5): n, m in Z} and think of this set as the common divisors of 2 and 1+√ -5. • We would get A·A =<4, 2(1+√-5), -4+2√-5>=<2>. • Setting B=<3, 1+√-5> and C=<3, 1+√-5> one obtains <6>=<2><3>=A 2 BC=ABAC=<1+√-5, 1 -√-5>.