April 4

Plan:

• Questions
• Cyclotomy (homework 3)
• Lagrange's theory of equations

Discussion:
The discussion page can be found here, for observations and remarks.

In class we discussed the solution to problem 7 of Homework 4 (4.38 in the text)

$a = (1 + \omega^{2} + \omega^{4} + \omega^{5} + \omega^{6} + \omega^{11})$
$b = (1 + \omega + \omega^{5} + \omega^{6} + \omega^{7} + \omega^{9} + \omega^{11})$

The product $ab$ is

$c = 1 + \omega + \omega^{2} + \omega^{3} + \omega^{4} + 3 \omega^{5} + 3 \omega^{6} + 3 \omega^{7} + \omega^{8} + 3 \omega^{9} + 3 \omega^{10} + 7 \omega^{11} + 3 \omega^{12} + 3 \omega^{13} + \omega^{14} + 3 \omega^{15} \\ + 3 \omega^{16} + 3 \omega^{17} + \omega^{18} + \omega^{19} + \omega^{20} + \omega^{21} + \omega^{22}$

Although neither a nor b is divisible by 2, the product c is. This demonstrates that the FTA fails for $\omega^{23}$. Consider the roots of unity, write 3 as 2+1 and 7 as 6+1. c can be written as

$2( \omega^{5} + \omega^{6} + \omega^{7} + \omega^{9} + \omega^{10} + 3 \omega^{11} + \omega^{12} + \omega^{13} + \omega^{15} + \omega^{16} + \omega^{17}) + d$

where d is the collection of terms that are not divisible by 2:
$( \omega^{22} + \omega^{21} + ... + \omega + 1)$

$\omega$ is a solution to the equation $x^{23} -1 =0$ which can be factored into $(x-1)(x^{22} + x^{21} + ... + x + 1)$ . $\omega$ is also a solution to the second factor, this means that $d = 0$ and the product is divisible by 2.

Solving this problem led to a review of what we had learned in number theory about the proof that $x^{p} + y^{p} = z^{p}$ has no solutions for $p>2$. The homework problem is an example of when the FTA does not work. Kummer used $\omega^{23}$ as a counter example to show that the proofs for the FLT have assumed that $Z[\omega_{p}]$ has unique factorization.