Prev: Helmholtz equation and the Maximum Principle
Next: Axiom of Regularity.
From: Andrew Usher on 14 Jan 2010 07:22
On Jan 14, 4:45 am, "Larry Hammick" <larryhamm...(a)telus.net> wrote:
> "Leonid Lenov" <leonidle...(a)gmail.com> wrote in message
>
> news:1219fc00-5d64-48ef-911c-05d0a4a9ec2b(a)j24g2000yqa.googlegroups.com...
>
> > Hello,
> > Diophantine equation y^2=x^3-3 can have only finitely many solutions.
> > Unfortunately, Z[sqrt{-3}] is not UFD so one cannot use unique
> > factorization to obtain it's solutions.
>
> Let u be a zero of the polynomial 1 + u + uu.
> ( y - u + uu)(y + u - uu) = yy +3
> and use the fact that Z[u] is a UFD.

As I said twice now, this is an unnecessary and inelegant
complication. Z[sqrt(-3)] restricted to 'odd' numbers, that is, a + b
sqrt(-3) with a,b not both even, does have unique factorisation
because all non-principal ideals of the ring divide by {2, 1 + sqrt
(-3)}, and clearly every ideal of the restricted set corresponds to
one of the full ring, thus all are principal i.e. we have unique
factorisation. This proves what he wants because b = 1 in the problem.

Andrew Usher
From: Leonid Lenov on 14 Jan 2010 07:51
On Jan 13, 10:51 pm, "Achava Nakhash, the Loving Snake"
<ach...(a)hotmail.com> wrote:
> Also, your set appears to me to be closed under multiplication, but
> maybe not under addition.  After all (2 + sqrt(-2)) = alpha and (1 +
> 2*sqrt(-3)) = beta are both in your set, but alpha + beta = 3 + 3*sqrt
> (-3) is not in your set since both coefficients are odd.  Unless I
> have your definition wrong?

I think you did. The only case that cannot happen is when both a and b
are even.

Of course, your method is quite right it is just that I was wondering
what is the smallest ring that would do...
From: Leonid Lenov on 14 Jan 2010 07:57
On Jan 13, 10:39 pm, Andrew Usher <k_over_hb...(a)yahoo.com> wrote:
> On Jan 13, 9:37 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:
>
> > Can we just use the ring R={2^k (a+b*sqrt{-3})| a,b,k in Z and either
> > one of a or b is odd} with the norm N(2^k(a+b*sqrt{-3}))=a^2+3b^2? Now
> > units will be +/-1 and numbers of the form +/-2^k where k in Z.
> > If R is a UFD it follows that y-sqrt{-3}=+/-2^k(x+iy)^3 which has no
> > solutions.
>
> I don't see this ring as being any more a UFD than Z[sqrt(-3)] irself
> - the famous factorisation 4 = 2*2 = (1+sqrt(-3))(1-sqrt(-3)) still
> works, and shows that your norm is not multiplicative besides.

Unique factorization means that every non-unit can be factored
uniquely into the product of irreducible elements up to the unit. The
way I defined the norm, 4 is a unit, and of course, it can be written
as a product of other units in many ways.
From: Leonid Lenov on 14 Jan 2010 08:05
On Jan 14, 1:57 pm, Leonid Lenov <leonidle...(a)gmail.com> wrote:
> On Jan 13, 10:39 pm, Andrew Usher <k_over_hb...(a)yahoo.com> wrote:
>
> > On Jan 13, 9:37 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:
>
> > > Can we just use the ring R={2^k (a+b*sqrt{-3})| a,b,k in Z and either
> > > one of a or b is odd} with the norm N(2^k(a+b*sqrt{-3}))=a^2+3b^2? Now
> > > units will be +/-1 and numbers of the form +/-2^k where k in Z.
> > > If R is a UFD it follows that y-sqrt{-3}=+/-2^k(x+iy)^3 which has no
> > > solutions.
>
> > I don't see this ring as being any more a UFD than Z[sqrt(-3)] irself
> > - the famous factorisation 4 = 2*2 = (1+sqrt(-3))(1-sqrt(-3)) still
> > works, and shows that your norm is not multiplicative besides.
>
> Unique factorization means that every non-unit can be factored
> uniquely into the product of irreducible elements up to the unit. The
> way I defined the norm, 4 is a unit, and of course, it can be written
> as a product of other units in many ways.

I see now that what I wrote is not entirely correct so I will no
longer insist on my method. Tbe best ways to solve the problem is as
Andrew Usher and Achava Nakhash suggested.
From: Andrew Usher on 14 Jan 2010 08:21
On Jan 14, 6:57 am, Leonid Lenov <leonidle...(a)gmail.com> wrote:

> > > Can we just use the ring R={2^k (a+b*sqrt{-3})| a,b,k in Z and either
> > > one of a or b is odd} with the norm N(2^k(a+b*sqrt{-3}))=a^2+3b^2? Now
> > > units will be +/-1 and numbers of the form +/-2^k where k in Z.
> > > If R is a UFD it follows that y-sqrt{-3}=+/-2^k(x+iy)^3 which has no
> > > solutions.
>
> > I don't see this ring as being any more a UFD than Z[sqrt(-3)] irself
> > - the famous factorisation 4 = 2*2 = (1+sqrt(-3))(1-sqrt(-3)) still
> > works, and shows that your norm is not multiplicative besides.
>
> Unique factorization means that every non-unit can be factored
> uniquely into the product of irreducible elements up to the unit. The
> way I defined the norm, 4 is a unit, and of course, it can be written
> as a product of other units in many ways.

4 and 2 are units but 1+sqrt(-3) is not! You have two non-units
multiplying to give a unit, which violates that principle.

Andrew Usher
First  |  Prev  |  Next  |  Last
Pages: 1 2 3 4 5
Prev: Helmholtz equation and the Maximum Principle
Next: Axiom of Regularity.