Re: Ring problem, integral elements

From: Alireza (alireza_abdollahi_at_yahoo.com)
Date: 12/20/04 

Date: 19 Dec 2004 19:55:05 -0800

Dear Todd,

The obtained polynomial does not work for the case x=y=-1 and a=1 (in
the ring of complex numbers) that is, in this case

f(1-1)=f(0)=/=0

also for x= 1/4+1/4*5^(1/2)-1/4*I*2^(1/2)*(5-5^(1/2))^(1/2)

and y=1/4+1/4*5^(1/2)+1/4*I*2^(1/2)*(5-5^(1/2))^(1/2)

(I=sqrt(-1))
and where a=1 (in fact x and y are two distinct fifth root of unity)
we have

f(x-y)=-2069375/2+910525/2*5^(1/2)=/=0

These show that, maybe somthing is wrong in your calculations, but
perhaps I can find out it ...........???

Anyway thank you very much for your information.

Best Regards

Alireza

Todd Trimble wrote:
> On 19 Dec 2004, Alireza wrote:
> >Dear All,
> >
> >Excuse-me in my question R is not assumed to be commutative. and
the
> >correct (complete) question is the following (I think!!):
> >
> >Suppose that R is a ring with identity such that the centre Z(R) of
R,
> >is a field. Assume that x and y are two commuting
> >elements of R (i.e. xy=yx) such that
> >
> >
> >x^5+a=0
> >y^5+a=0
> >
> >
> >for some a in Z(R). In fact x and y are integral over
> >Z(R).
> >
> >
> >Question: Construct a non-zero polynomial f with coeficceint in
Z(R)
> >such that f(x-y)=0.
> >
>
> We may restrict attention to the commutative ring Z(R)[x, y],
> and solve the problem universally by replacing Z(R) by Z[a],
> a an indeterminate (Z the integers), and Z(R)[x, y] by the
extension
>
> E = Z[x, y, a]/(x^5 = -a = y^5) over Z[a].
>
> If we find a monic polynomial for x - y with coefficients in
> Z[a], then we are done (since we can just apply some obvious
> rings homomorphisms Z[a] --> Z(R), E --> R to the equation
> f(x-y) = 0 in E).
>
> Obviously y/x is a fifth root of unity (denoted w) over the
> field Q(a), so
>
> (x - y)^5 = a(w - 1)^5
>
> and the problem reduces to finding a monic polynomial (over Z)
> for (w - 1)^5. This is routine if a bit tedious; one works
> over the complex numbers and calculates the coefficients of
>
> (u - r1)(u - r2)(u - r3)(u - r4)
>
> where ri = (w^i - 1)^5. There is a lot of cancellation due
> to the fact that the ri are pure imaginary and come in conjugate
> pairs; my back-of-envelope calculations give the last polynomial
> as
> u^4 + 3936u^2 + 3125
>
> Substituting u = (x - y)^5/a and clearing denominators, I get
>
> (x - y)^20 + 3936a^2(x - y)^10 + 3125a^4 = 0.
>
> Todd Trimble

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