9
$\begingroup$

I heard that one can prove special cases of FLT by using unique factorization in $\mathbb{Z}[\xi]$ (whenever this is possible), where $\xi$ is a primitive $n$-th root of unity. How can one do this in detail, and is it known for which natural $n$ $\mathbb{Z}[\xi]$ is not a UFD? I mean assuming that $n$ is such that $\mathbb{Z}[\xi]$, how does one proceed?

A reference is good enough.

$\endgroup$
  • $\begingroup$ See en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/… - in particular, there are only finitely many $n$ for which the cyclotomic ring is a UFD. $\endgroup$ – Steven Stadnicki Apr 21 '15 at 22:03
  • $\begingroup$ @StevenStadnicki thanks, this answers the second question. $\endgroup$ – ulli Apr 21 '15 at 22:13
  • 2
    $\begingroup$ See the first chapter from Marcus' book "Number Fields". $\endgroup$ – Joe Apr 26 '15 at 8:17
  • 1
    $\begingroup$ It's probably done in Harold M. Edwards' book, Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory. $\endgroup$ – Gerry Myerson Apr 26 '15 at 8:51
  • 1
    $\begingroup$ There is a sketch of the proof you want at cs.uwaterloo.ca/~alopez-o/math-faq/mathtext/node9.html under the discussion of whether Fermat could have had a proof. $\endgroup$ – Gerry Myerson Apr 26 '15 at 8:58
11
$\begingroup$

I think you refer to a theorem of Kummer: Fermat's Last Theorem is true for an odd prime $p$ if and only if $p$ doe not divide the class number of the cyclotomic extension $\mathbf Q(\zeta_p)$, i. e. the order of the group of fractionary ideals of this field modulo principal ideals.

Such a prime number is called a regular prime.

Kummer criterion:

An odd prime $p$ is regular if and only if $p$ does not divide the numerators of the Bernoulli numbers: $\, B_2, B_4,\dots, B_{p-3}$.

All odd primes up to $31$ regular. It is not known if there is an infinity of regular primes.

In case $\mathbf Z(\zeta_p)$ is a UFD, which is equivalent to being a PID, the class number is equal to $1$, hence $p$ doen't divide it, so Fermat Last Theorem is true for these primes, of which the complete list is: $$\{3,5,7,11,13,17,19\}$$

$\endgroup$
  • $\begingroup$ I have seen this notion of a regular prime everywhere, but why can't one just say: ok we assume that $n$ is s.t. $\mathbb{Z}[\xi_n]$ is a UFD, and now we prove it? Just assuming that it is a UFD. Do you know how to do this? Apparently the proof shouldn't be too long. $\endgroup$ – ulli Apr 21 '15 at 22:39
  • $\begingroup$ Fermat's Last Theorem holds for any odd prime, including irregular ones. Kummer was only able to prove it for regular primes. $\endgroup$ – anomaly Apr 21 '15 at 22:47
  • $\begingroup$ There are only 30 values of $n$ such that $\mathbf Z[\zeta_n]$ is a UFD. It is not equivalent, for $ n$ an odd prime, to be regular, , although all odd primes among the 30 are regular.. $\endgroup$ – Bernard Apr 21 '15 at 22:50
  • $\begingroup$ Yes I know, this was pointed out above. I still think the proof should be very interesting, even though it is probably a special case of this theorem of Kummer. I also expect the proof to be much simpler $\endgroup$ – ulli Apr 21 '15 at 22:56
  • $\begingroup$ @ulli It's more or less given in the first link in my comment: we have factorizations in $\mathbb{Z}[\zeta_n]$ of $x^n+y^n$ and of $z^n$, and can show that they aren't related by multiplying by units (as they would need to be if unique factorization holds). $\endgroup$ – Steven Stadnicki Apr 21 '15 at 22:58
6
+500
$\begingroup$

Since you told that "a reference is good enough", since first chapter of Marcus' book "Number Field" is the very right place to go and since writing down a complete and detailed answer is a huge work -even if rewarded with +$500$ as already pointed out-, I'm writing this reference suggestion as an answer, simply because reasonably this IS a good answer.

$\endgroup$
  • $\begingroup$ I actually wanted to encourage you to post this as an answer, but I was too lazy. You will probably get the 500, unless someone gives a more detailed answer. If no one does this, I will perhaps edit your answer and include a (very rough) sketch of the proof, unless you disagree with this. $\endgroup$ – Mister Benjamin Dover Apr 27 '15 at 14:59
  • 1
    $\begingroup$ I studied on Marcus' book: any sufficiently detailed proof is really a huge work. The other possibility is, as you said, a rough sketch of the proof. To me there is no problem if you want to edit my answer provided that you write your name under the edit and it is (in its roughness), precise, well done. Obvious requests, do you agree? :-) $\endgroup$ – Joe Apr 27 '15 at 16:23
  • 1
    $\begingroup$ I agree, of course. I actually thought that the proof would be much shorter, but little did I know. The relationship between FLT and the factoriality of $\mathbf{Z}[\xi_n]$ is well-known, but the usual description merely consists of writing down the factorization of $x^n+y^n$, without giving more details. So I think the book of Marcus is rare in this aspect, and I am very grateful to you for providing this reference. I somehow never managed to find a reference (apart from references for the more general Kummer result). $\endgroup$ – Mister Benjamin Dover Apr 27 '15 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.