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Proof of Fermat's last theorem - special case?

I've been told that Fermat's last theorem (x^n + y^n = z^n has no nontrivial integer solutions for n > 2) is easy to prove for those n with the property that the ring of integers in the algebraic number field obtained by adjoining a primitive nth root of unity is a PID. I recall reading this is Pierre Samuel's "Algebraic Theory of Numbers", for example. Does anybody know the proof or where it can be found?

Update:

Hi,

I want to ask a follow-up question before I give you the 10 points that are rightfully yours. Is the statement that p does not divide xyz equivalent to or related to the statement that the ring of integers in Q[w] is a PID, where w is a primitive pth root of unity? Thanks for your great answer!

3 Answers

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  • kb
    Lv 7
    1 decade ago
    Favorite Answer

    This is a fun topic!

    WLOG, one tries to prove this for primes p>2 instead of general n.

    The proof of Fermat's Last Theorem using algebraic numbers is broken down into Case 1 (p does not divide xyz) and Case 2 (p | xyz).

    The first case is proved in many books dealing with Algebraic Number Theory (like Stewart and Tall's "Algebraic Number Theory" or Ireland and Rosen's "A Classical Introduction to Modern Number Theory").

    J.S. Mine's also proves this case in his lecture notes to ALg. Number Theory. which can be found online.

    The considerably harder second case is harder to find. The one book that I know that covers this case is Emil Grosswald's "Topics From The Theory Of Numbers".

    Happy reading!

  • 1 decade ago

    The proof of this can be found in the first chapter

    of Washington's book Introduction to cyclotomic fields.

    More generally the result is proved for regular primes

    in the case where p doesn't divide xyz.

    A regular prime is one which doesn't divide the class number

    of Q(w).

    The statement that p doesn't divide xyz is called the

    first case of Fermat's last theorem. If p| xyz we have the

    much harder second case. This has nothing to do

    with the statement that the ring of integers in Q[w] is a PID.

    Incidentally, The largest p for which the ring of integers in Q[w]

    is a PID is p = 19. That's also proved in Washington's

    book, chapter 11.

  • pitt
    Lv 4
    5 years ago

    many of the solutions listed right here are good and maximum suitable yet for non-mathematicians right here is yet another: Fermat postulated that for x^n + y^n = z^n there substitute into no complete quantity answer for x, y and z the place n substitute into better than 2. So for 3^2 + 4^2 = 5^2 (3 squared plus 4 squared = 5 sq. it rather is a Pythagorean triplet) yet strengthen the flexibility to x cubed + y cubed = z cubed there are not any complete quantity suggestions and nor are there for any fee of n better than 2. it is his final theorum because of the fact while he wrote the thought down, he scribbled interior the margin of his pc some thing alongside the strains of "I surely have got here upon a outstanding evidence of this", regrettably he had run out of paper at this ingredient, much greater regrettably he died formerly he ought to write down the evidence. the undertaking itself has taxed the brains of mathematicians ever when you consider that, and alter into purely in the near previous solved by potential of a professor at Oxford (uk) college after 10 at as quickly as years engaged on the undertaking! And his evidence substitute into no longer outstanding in any respect, based on the 4 coloration map evidence it ran properly over a hundred pages. in simple terms approximately as quickly as he printed, yet another mathematician observed an blunders in his evidence (A eastern pupil - if I remember genuine) and that they the two labored for yet another three hundred and sixty 5 days or 2 formerly finally publishing the same old evidence. So did he resolve the undertaking? sure, actual. substitute into it Fermat's outstanding evidence? actual no longer. So the two Fermat substitute into incorrect, or on the instant 4 hundred years later we nevertheless have not proved ourselves to be as sensible as him. wish that enables.

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