From: rusin@math.niu.edu (Dave Rusin)
Subject: Seeking counterexamples to FLT in other rings
Date: 6 Sep 00 21:48:54 GMT
Newsgroups: sci.math.numberthy
Summary: x^n+y^n=z^n has nontrivial solutions in some rings.
I often find myself trying to explain to students and amateurs why their
putative proofs of Fermat's Last Theorem are incorrect. (On a good day
I'll give the person a nice response, and watch to see whether they
seem like a learner or a crank.) One technique which is useful is to
produce counterexamples to FLT in other rings; this can point to places
where they have made an algebra error or a leap in logic.
Given my druthers I would love to have three elements x,y,z of a ring R
for which x^p + y^p = z^p where the ring R is very similar to the
ring of integers. Some useful features would be:
R is associative, commutative, has no zero divisors
R has unique factorization
R has no units other than +- 1 (or maybe other roots of unity?)
p (is not zero and) generates a prime ideal in R
R/pR has exactly p elements
xyz is relatively prime to p in R
p > 3
(More suggestions also welcome.)
Just what features ought to be present in an example depends on what
gaps in a "proof" should be illustrated, so it could be useful to
collect examples of several types.
I have looked in particular at the case R is the ring of integers of a
quadratic extension of the rationals. There are many examples of the form
(a + b sqrt(d) )^3 + (a - b sqrt(d) )^3 = c^3
with a, b, and c rational integers. Such solutions exist precisely
when the elliptic curve
Y^2 = X^3 - 432 d^3
has rational points. Current technology makes it easy to decide whether
or not there are such solutions for any particular d.
However, these examples may not always suffice to locate quickly the
errors in a particular proposed "proof"; these examples might not
satisfy sufficiently many of the conditions I set above. Unfortunately,
these are the only counterexamples to FLT I could generate for
quadratic extension rings of Z. So I would like to ask the following:
(1) Are there any known counterexamples to FLT in quadratic extensions
of Z with exponent p >= 5 ?
(2) Are there any solutions in these rings with p=3 besides the ones
of the form above? (Of course I can hide the form in simple ways
e.g. by multiplying through by a non-rational cube.
Surely the answer to my query is yes; the question asks
whether the group of Q(sqrt(d))-rational points of the
curve Y^2 = X^3 - 432 d^3 has larger rank than the group of
rational points. I don't have much experience computing ranks
over other fields than Q.)
Of particular interest to me would be the cases in which the ring R
has unique factorization and only trivial units, that is, the rings of
integers in Q(sqrt(d)) where d = -1, -2, -3, -7, -11, -19, -43, -67, -163.
Can someone tell me the ranks of the Fermat curve a^3 + b^3 = 1
(that is, y^2 = x^3 - 432) over these nine fields?
(Only when d=-2 and d=-11 are there points with a and b conjugate
over Q as discussed above.)
Learned discourse welcome as well as simple examples, of course :-)
dave
rusin@math.niu.edu
==============================================================================
From: Franz Lemmermeyer
Subject: Re: Seeking counterexamples to FLT in other rings
Date: Thu, 7 Sep 2000 03:10:17 +0200 (METDST)
To: Dave Rusin
Hi Dave,
> (2) Are there any solutions in these rings with p=3 besides the ones
> of the form above?
This question was studied by people like Fueter and Aigner in the
1930s and 1950s; I assume you have access to MathSciNet, so here goes
Aigner: 14,452a; 14,621d; 17,464b; 17,945a; 19,120e
I can't seem to find Fueter's papers, but I think Aigner may
have given some references to preceding work.
franz
==============================================================================
From: Dave Rusin
Subject: Re: Seeking counterexamples to FLT in other rings
Date: Wed, 6 Sep 2000 21:11:47 -0500 (CDT)
To: hb3@ix.urz.uni-heidelberg.de
Great! These are just the pointers I need. (There are some good links to
Fueter's work by searching for Anywhere="Fueter and Fermat" in
MathSciNet, but since his work goes back much further it is more
productive to check the online versions of Zentralblatt and the
Jahrbuch uber die Fortschritte der Mathematik.) Guess it's time to
head to the library...
Thanks again for responding.
dave
==============================================================================
From: John Cremona
Subject: Re: Seeking counterexamples to FLT in other rings
Date: Thu, 07 Sep 2000 08:24:26 +0100
To: Dave Rusin
Hi Dave,
I passed your nmbrthry contribution on to Frazer Jarvis at Sheffield
(which is where Neil Dummigan has a permanent position as of around now)
as I know that he is working on this sort of question. I have some
texed notes he sent me about it, but am not sure of their status. He is
a student of R Taylor and is trying to use Wiles-type methods to extend
FLT to some other fields -- but he has also been investigating some
down-to-earth methods of finding nontrivial solutions over certain small
degree number fields.
[deletia --djr]
John
--
Prof. J. E. Cremona |
University of Nottingham | Tel.: +44-115-9514920
School of Mathematical Sciences | Fax: +44-115-9514951
University Park | Email: John.Cremona@nottingham.ac.uk
Nottingham NG7 2RD, UK |
==============================================================================
From: John Cremona
Subject: Re: Seeking counterexamples to FLT in other rings
Date: Thu, 07 Sep 2000 08:27:05 +0100
To: Dave Rusin
One more point on your posting:
The Q(sqrt(d)) rank of Y^2=X^3-432 is equal to the rank of the curve
over Q + the rank of its d-twist. Sine the curve has rank 0 over Q (by
Fermat!) the rank over Q(sqrt(d)) is equal to the rank over Q of the
twisted curve Y^2 = X^3 - 432 d^3. So all quadratic points on the p=3
Fermat curve arise this way.
John
--
Prof. J. E. Cremona |
University of Nottingham | Tel.: +44-115-9514920
School of Mathematical Sciences | Fax: +44-115-9514951
University Park | Email: John.Cremona@nottingham.ac.uk
Nottingham NG7 2RD, UK |
==============================================================================
From: "A.F.Jarvis"
Subject: Solutions to FLT in other rings
Date: Thu, 7 Sep 2000 12:11:25 +0100
To: rusin@math.niu.edu
Dear Dave,
John Cremona kindly forwarded your message regarding solutions
to FLT in other rings, as I am not a subscriber to the appropriate
newsgroup!
I was a student of Richard Taylor a few years ago, so am
reasonably well acquainted with the proof of FLT, and as I've been
working on Hilbert modular forms, it seemed natural to suggest to
my PhD student that he try to prove FLT over certain totally real
number fields, following the work of Ribet and Wiles, proving that
the Frey curve is both not modular and modular. This will establish
FLT over these fields for "sufficiently large exponent". We've been
looking particularly at real quadratic fields Q(sqrt{p}) with p prime,
although this is purely for convenience so far; other real quadratic
fields can be treated in the same way. While my student, Paul
Meekin, has been doing this, I have, in my spare moments, been
thinking about solutions when the exponent is not "sufficiently
large". I have a few results of relevance to your question, and some
references from others. We intend to publish these in due course,
probably combining his results on FLT for sufficiently large
exponent (he has several examples of fields for which the Frey
curve is not modular, and now we want to think about generalising
Wiles's stuff, following Fujiwara's work) with my results on lower
exponent.
As you point out, the Fermat cubic has points in Q(sqrt{d})
precisely when Y^2 = X^3 - 432 d^3 has rational points. Perhaps
surprisingly, there is an exact correspondence between rational
points on this elliptic curve and points on the Fermat cubic with
coordinates in Q(sqrt{d}). This is not completely obvious; it's a
corollary of FLT for exponent 3 over Q! This means that one can
compute the rank of the Fermat cubic over Q(sqrt{d}) by knowing
the rank of these Mordell curves over Q. In particular, if d=2, one
finds that the curve has rank 1, generated by (X,Y)=(28,136),
leading to the solution:
(378 + 357sqrt{2})^3 + 127^3 = (451 + 306sqrt{2})^3.
If d=3, the Mordell curve has rank 0, so there are no points on the
Fermat cubic over Q(sqrt{3}). For d=5, there is the point
(9 + sqrt{5})^3 + (9 - sqrt{5})^3 = 12^3,
which is the "easiest" solution I know of.
Here's a way of constructing points on the Fermat cubic. Take a
line with rational slope which passes through one of the three (in
projective space) rational points on the cubic. Then this line
necessarily intersects the Fermat cubic in two further points (by
Bezout's theorem, if we work in P^2(C)); it is easy to see that
these lie in some quadratic field and are conjugate. Some easy
Galois theory leads to the conclusion that all quadratic points arise
in this way. It is precisely the knowledge of FLT for exponent 3 over
Q that means that we know all rational points on the Fermat cubic,
and, by considering a general line of rational slope through these
rational points, we can deduce the theorem I stated above. (The
same method means that it is easy to compute the rank over any
quadratic field of an elliptic curve with rank 0 over Q; X_0(15) would
be interesting, as Wiles uses it in his work on FLT...)
To construct points on the Fermat quartic, take a line with rational
slope passing through a rational point; then this line will meet the
quartic at three further points, all lying in some cubic field. If the
line goes through two of the rational points (there are four rational
points, so six such lines), we get points on the quartic lying in
quadratic fields. If I remember, four of these are solutions like
0^4 + i^4 = 1,
but the other eight lie in Q(sqrt{-7}). Faddeev (1960) proved that
these were the only points on the Fermat quartic in quadratic
fields, and further that all points in cubic fields arose in the manner
above.
For exponent 5, one can find two points in Q(sqrt{-3}) by
considering the line going through all three rational points; here it is:
(1 + sqrt{-3})^5 + (1 - sqrt{-3})^5 = 2^5;
this is basically equivalent to observing that the sum of the two
primitive 6th roots of unity is 1. So it is also a solution for all
exponents congruent to plus or minus 1 mod 6. There is a theorem
of Tzermias that these are the only points on the Fermat quintic
lying in quadratic or cubic fields, if I remember correctly. But you
should be able to construct points lying in quartic extensions using
this intersection method.
I believe that Tzermias has results also for exponent 7 and possibly
11 as well.
You might also like to read a paper of Debarre and Klassen in
which more applications of this intersection method are presented,
and a general conjecture formulated.
I've been writing up a few of these more naive thoughts on FLT for
small exponent over small fields in some informal notes, not
intended for publication; I can send you a copy if you are interested.
Best wishes,
Frazer Jarvis
Dr A.F.Jarvis,
Department of Pure Mathematics,
Hicks Building,
University of Sheffield,
Sheffield S3 7RH
(0114)2223845
http://www.shef.ac.uk/~pm1afj
==============================================================================
From: Bill Daly
Subject: Re: Seeking counterexamples to FLT in other rings
Date: Wed, 13 Sep 2000 10:37:26 -0400
To: Dave Rusin
This may not fit your requirements perfectly, but have you noticed that
FLT has a (fairly trivial) nonzero solution in Z(w), for w a primitive
6th root of 1, for all odd primes p > 3, namely
(1)^p + (w^2)^p = (w)^p
If also p = 2 mod 3, then p is prime in Z(w).
Regards,
Bill
==============================================================================
From: Dave Rusin
Subject: Re: Seeking counterexamples to FLT in other rings
Date: Wed, 13 Sep 2000 10:23:14 -0500 (CDT)
To: bill.daly@tradition-ny.com
Yes, thanks. This observation was made by others too:
(1 + sqrt{-3})^5 + (1 - sqrt{-3})^5 = 2^5;
and likewise for other exponents. I'm embarrassed I didn't spot this myself.
Thanks again for responding. When I get a few minute's peace I'll
summarize the other responses I received to the list.
dave
==============================================================================
Note [djr]: the solutions implied by Jarvis for n=4 are essentially
(1+sqrt(-7))^4 + (1 - sqrt(-7))^4 = 2^4
allowing of course for sign changes absorbed by the fourth powers,
multiplying through by constants in the ring, and so on.
In cubic extensions we have solutions (x,y) = (m*t, 1+t) where
t is a root of 4+6*t+4*t^2+(m^4+1)*t^3 . Here m could be any rational
number (except +- 1, in which case the cubic has a rational root).
Roughly speaking we understand that there are few solutions to the
Fermat equations which are algebraic of low degree. More precise results
are available for exponents 5 and 7, and are conjectured in general.
See for example these articles and their reviews in Math Reviews:
Tzermias, Pavlos
Algebraic points of low degree on the Fermat curve of degree seven.
Manuscripta Math. 97 (1998), no. 4, 483--488.
99j:11075 11G30
Sall, Oumar
Points algébriques de petit degré sur les courbes de Fermat.
C. R. Acad. Sci. Paris Sér. I Math. 330 (2000), no. 2, 67--70.
2001a:11104 11G30 (14G05)
So the hope of the original poster -- to find nearly-rational solutions
to a Fermat equation -- is pretty much limited to unimpressive cases!
(Not nearly as impressive as Elkies' solution
3472073^7 + 4627011^7 = 4710868^7
which can be verified with any 20-digit calculator!...)
==============================================================================
From: "Dik T. Winter"
Subject: Re: FLT for Gaussian integers
Date: Wed, 31 Jan 2001 01:24:49 GMT
Newsgroups: sci.math
In article <94th88$p0s$1@agate.berkeley.edu> chernoff@math.berkeley.edu (Paul R. Chernoff) writes:
> There has been a lot of discussion on this newsgroup of "elementary"
> proofs of FLT for the usual integers. I want to change the subject a
> bit. Is FLT true (or false) for the *Gaussian integers* Z[i], i.e.
> the set of all complex numbers of the form a = m + ni, where m and n
> are ordinary integers.
I do not think much work has been done specifically for the Gaussian
integers. And I think it is still much more difficult and intractable
than FLT for the integers themselves. Research has been done for
particular exponents in arbitrary quadratic number fields: 2, 3, 4, 6
and 9. I am not aware of other exponents being researched.
I understand results are complete for 2, 4, 6 and 9, and possibly
complete for 3. And as far as I know in most cases only those
numbers in Q(sqrt(m)) (with m square free, != 0 and 1) are considered
of the form a + b.sqrt(m) with a and b integer, so not always all
algebraic integers. Considering the results I can only say that
they are looking, eh, complex.
See: n = 2 provides additional non-trivial solutions only when
m is not divisible by a prime factor of the form +-3 mod 8.
n = 4 provides solutions when m = -7 (but only one essential
primitive solution). n = 6 and n = 9 have never any solution.
n = 3 is bizarre. Aigner has done much work on that one, but
class numbers of the field come in here again.
On the other hand Gauss' proof of FLT3 for the Eisenstein integers is
closely related to finding rational points on elliptic curves. So
there is hope.
Another generalisation is x^k + y^k = n.z^k; which also has not yet
been solved in the integers. The example I gave uses k = 5 and
n = 68101. It is known that 68101 is the smallest n with k = 5 such
that there are solutions (and only one primitive solution). More
such n are known, but it is not known whether there are infinitely
many or not. But I think it has been shown (Faltings?) that if
there are solutions there are only finitely many for any particular n.
So in this direction Wiles' proof might offer opportunities.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
==============================================================================
From: Jan Kristian Haugland
Subject: Re: FLT for Gaussian integers
Date: Mon, 29 Jan 2001 00:10:01 +0100
Newsgroups: sci.math
Jamas Enright wrote:
[deletia -- djr]
> Another related question might be: what conditions do you need for FLT to
> be false?
>
> Numerous finite rings fail FLT, pick a number n and a power p > 2, then in
> Z mod n^p, you get n^p + n^p = n^p (=0). :)
>
> Also p-adics fall over, such as Plutonium's much loved 10-adic solutions.
>
> In many ways it looks like that if the field of interest is too large or
> too small then FLT fails, so it's needs to be the right size. (For an as
> yet to be defined value of 'right size' of course :).
Size isn't everything: With w = (-1 + i sqrt(3))/2,
x^n + y^n = z^n have non-zero solutions in Z[w] for
all integers n not divisible by 2 or 3, for instance
w^p + (-1-w)^p = (-1)^p.
--
Jan Kristian Haugland
http://home.hia.no/~jkhaug00
==============================================================================
From: Bill Daly
Subject: FLT in algebraic rings
Date: Tue, 23 Oct 2001 08:52:12 -0700 (PDT)
To: Dave Rusin
I was reminded of your earlier question concerning
solutions of FLT in algebraic rings by a reference
from mathpuzzle.com. I have a few simple comments.
If there is a solution of x^n + y^n = z^n with x,y,z
in an algebraic ring R, then w = norm_R(z)/z is also
in R, thus if x' = xw, y' = yw, z' = zw = norm_R(z),
we obtain a solution x'^n + y'^n = z'^n with x,y in R
and z in Z. Thus, it is sufficient to look for
solutions with z in Z.
This means that all solutions in quadratic rings may
be obtained from solutions of (u + sqrt(v))^n + (u -
sqrt(v))^n = w^n.
For n = 3, this expands to u^3 + 3uv = w^3 / 2. This
is linear in v, thus for arbitrary u and w, there is a
solution with v = (w^3 / 2 - u^3) / (3u).
For n = 4, we get u^4 + 6u^2 v + v^2 = w^4 / 2. As a
quadratic in v, the discriminant is 32u^4 + 2w^4,
which must be a square if there is to be a solution
with v in Z. This is equivalent to solving a^4 + b^4 =
2c^2, corresponding to [u,w] = [a,2b]. An obvious
solution is [a,b] = [1,1], giving [u,w] = [1,2], from
which we obtain the solutions v = 1, leading to 2^4 +
0^4 = 2^4, and v = -7, leading to (1+sqrt(-7))^4 +
(1-sqrt(-7))^4 = 2^4. I assume that there are no other
solutions, but I haven't checked it.
We should also consider solutions of x^4 + y^4 + z^4 =
0, which has no nontrivial solutions in real quadratic
rings, but which may be solvable in imaginary
quadratic rings. This leads to the solvability of a^4
- b^4 = 2c^2, with again the obvious solution [a,b] =
[1,1], giving [u,w] = [1,2]. This leads to a single
solution v = -3, corresponding to (1+sqrt(-3))^4 +
(1-sqrt(-3))^4 + 2^4 = 0. I'm fairly sure that this is
the only solution.
For n = 5, we get u^5 + 10u^3 v + 5uv^2 = w^5 / 2,
again a quadratic in v whose discriminant 80u^6 +
10uw^5 must be a square. The obvious [u,w] = [1,2]
gives solutions with v = 1 and v = -3, corresponding
to the trivial 2^5 + 0^5 = 2^5 and (1+sqrt(-3))^5 +
(1-sqrt(-3))^5 = 2^5. Presumably, there are no other
solutions.
Regards,
Bill
[sig deleted --djr]
==============================================================================
From: rusin@math.niu.edu
Subject: Re: FLT in algebraic rings
Date: Sat, 27 Oct 2001 22:50:00 -0500 (CDT)
To: billdaly99@yahoo.com
You wrote me recently about solving FLT, especially in quadratic
extension fields of Q.
>If there is a solution of x^n + y^n = z^n ...
>Thus, it is sufficient to look for solutions with z in Z.
OK
>This means that all solutions in quadratic rings may
>be obtained from solutions of (u + sqrt(v))^n + (u - sqrt(v))^n = w^n.
Not OK. It's probably "almost true" somehow but at the very least you
have to allow for the possibility that y/(x_bar) is some other root
of unity besides 1. (I'm mostly thinking of n=3 here, but even in
the general case I'm not 100% sure I see what to do.)
>For n = 3, this expands to u^3 + 3uv = w^3 / 2
OK: this constructs solutions to FLT3 in quadratic fields. Modulo your
claim in the previous paragraph, these are (close to) the only solutions.
Usually people ask me about specific quadratic fields, and it's less
clear from your construction whether it will give elements in any
pre-assigned field, but it has been proved that the case n=3 for
quadratic number fields reduces to finding rational points on a particular
elliptic curve, so (for d not too terribly large) this is a solvable problem.
Your comments about n=4 and n=5 are also OK, although it looks like
you missed the solutions for n=4 which happen to have x y z = 0.
More generally, it seems people have made progress classifying solutions
to FLT with exponent n in number fields of degree less than n,
for n up through about 11. You can see some of what I learned about
this at http://www.math-atlas.org/newstuff/flt_rings
Unless you have an objection I will probably tack on your message to that file.
OK?
dave
==============================================================================