Unique factorization fails for cyclotomic integers starting with λ = 23 where λ represent the primitive root of unity that a given cyclotomic integer is based on. For details on cyclotomic integers start here. This blog is part of the general proof that shows that Fermat's Last Theorem is true for regular primes.
The content is today's blog is based on Harold M. Edwards' Fermat's Last Theorem: A Genetic Introduction for Algebraic Number Theory. In today's blog, I take one result from Edwards. Edwards goes over a much more complete list of calculations taken from Ernst Kummer's original article.
Theorem 1: Not all cyclotomic integers are characterized by unique factorization
Proof:
(1) The method of this proof is to show that unique factorization fails for the number 47 when λ = 23.
NOTE: We only need to show one case of failure to show that unique factorization fails.
(2) Norm of (1 - α + α21) = 47*139.
NOTE: These calculations are taken directly from Edwards. If you see any mistake, please compare it to Edwards' book for the correction. Edwards takes his calculations from Kummer's original paper.
(a) Nf(α) = f(α)*f(α2)*...*f(αλ-1) [Nf(α) = Norm of f(α) See Definition 2, here]
(b) Let G(α) = f(α)f(α4)f(α-7)f(α-5)f(α3)f(α-11)f(α2)f(α8)f(α9)f(α-10)f(α6)
where α-x = α23-x
(c) Then Nf(α) = G(α)G(α-2)
since G(α-2) = f(α-2)f(α-8)f(α14)f(α10)f(α-6)f(α22)f(α-4)f(α-16)f(α-18)f(α20)f(α-12)
To be clear:
G(α) = f(α)f(α4)f(α16)f(α18)f(α3)f(α12)f(α2)f(α8)f(α9)f(α13)f(α6)
G(α-2) = f(α21)f(α15)f(α14)f(α10)f(α17)f(α22)f(α19)f(α7)f(α5)f(α20)f(α11)
(d) Let:
g(α) = f(α)f(α4)
h(α) = g(α)f(α-7)
(e) Then:
G(α) = h(α)h(α-5)h(α2)g(α-10)
since this equals:
[f(α)f(α4)f(α16)][f(α18)f(α3)f(α12)][f(α2)f(α8)f(α9)][f(α13)f(α6)]
(f) Now, we can do the following computations:
g(α) = f(α)f(α4) = (1 - α + α21)(1 - α4 + α15) =
= α21 - α16 + α15 + α13 + α5 - α4 - α2 - α + 1.
h(α) = g(α)f(α-7) =
= (α21 - α16 + α15 + α13 + α5 - α4 - α2 - α + 1)(1 - α16 + α10) =
= α20 + α19 + α17 -3α16 + α13 + α12 + α9 - α8 - α7 + α5 - α2 - α + 1.
h(α-5) = α15 + α20 + α7 - 3α12 + α4 + α9 + α - α6 - α11 + α21 - α13 - α18 + 1.
h(α2) = α17 + α15 + α11 - 3α9 + α3 + α + α18 - α16 - α14 + α10 - α4 - α2 + 1
g(α-10) = α20 - α + α11 + α8 + α19 - α6 - α3 - α13 + 1
(g) So, G(α) = -44 + 15θ0 - 13θ1
where θ0 = α + α4 + α-7 + ... + α6 and θ1 = -1 - θ0.
(h) Now, we can check it to find:
θ0θ1 = 11 + 5θ0 + 5θ1 = 6.
(i) Now, we are ready to put the calculation all together so that:
N(1 - α + α21) = G(α)G(α-2) =
= (-31 + 28θ0)(-31 + 28θ1) = 312 - 31*28(θ0 + θ1) + 6*282 = 961 + 868 + 4704 = 6533 = 47*139.
(3) Norm(47) = 47*47*....*47 = 4722.
(4) Norm(139) = 139*139*...*139 = 13922
(5) But, then (1 - α + α21) doesn't divide 47 and (1 - α + α21) doesn't divide 139 since (from Lemma 6, here), a cyclotomic only divides another cyclotomic integer if the norm of the first cyclotomic integer divides the norm of the second cyclotomic integer.
NOTE: In other words, 47*139 does not divide 13922 and likewise, 47*139 does not divide 4722.
(6) And this violates Euclid's Lemma since (1 - α + α21) does divide 47*139 by the equation in step #2 but it doesn't divide 47 and it doesn't divide 139.
NOTE: The equation proved in step #2 is:
N(1 - α + α21) = (1 - α + α21)*(1 - α2 + α21)*...*(1 - α21 + α21) = 47*139.
(7) And this means that unique factorization does not exist in this situation. [See here for details on the Fundamental Theorem of Arithmetic and its dependence on Euclid's Lemma]
QED
Kummer then developed his theory of ideal numbers in order to save unique factorization for cyclotomic integers.
Edwards did not use 1 - x + x^22 as a counterexample. It was 1 - x + x^21. See: http://books.google.com. I did not check your calculations, so I do not know whose mistake it was.
ReplyDeleteAnother thing I wonder about is: we know that a certain cyclotomic number f(x) has norm equal to 47*139. We know that it cannot divide neither 47 nor 139. But do we know that f(x) is irreducible? I cannot find it in your reasoning. Edwards mentions that for λ = 23, there is no cyclotomic integer such that its norm is equal to 47... That argument would help.
Sorry for my English, and above all - thanks for these materials, they are really usesful.
Hi Arkadiusz,
ReplyDeleteThanks very much for noticing my typo. I've updated my proof. It looks like it is my mistake.
The reasoning for the proof is based on the nature of the norm.
The argument is that a norm provides a mapping between cyclotomic integers and the rational integers.
So, that we can use a norm to prove that f(x) is irreducible.
The key point is Lemma 6 here
If a cyclotomic integer x divides 1-α + α^22, then N(x) divides N(1 - α + α^22).
Now, since N(1 - α + α^22) = 47*139, we prove that it is irreducible if we prove that there are no norms that divide 47*139. Since 47,139 are primes, it follows that we only need to prove that there is no cyclotomic integer with a norm(47) and no cyclotomic integer with a norm (139).
I hope that helps.
-Larry