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Topic: Can a small degree polynomial exist in this Ideal?
Replies: 6   Last Post: May 15, 2013 5:54 AM

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Robin Chapman

Posts: 280
Registered: 5/29/08
Re: Can a small degree polynomial exist in this Ideal?
Posted: May 14, 2013 9:31 AM
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On 14/05/2013 13:20, eric.fennessey@gmail.com wrote:
> On Tuesday, May 14, 2013 11:37:54 AM UTC+1, Robin Chapman wrote:
> <snip>... Yes</snip>
> Any chance of a pointer to somewhere I can look and see how the proof goes?

There are so many different approaches:

(i) show that X^n - 1 and Y^m -1 form a Groebner basis of I.

(ii) Prove that if sum a_{i,j} X^i y^j is in I
then for each r and s the sum of the a_{i,j} over all i and j
with i = r (mod n) and j = s (mod n) is zero.

(iii) Let M be the tensor product over Z of the modules
Z[X]/(X^n - 1) and Z[Y]/(Y^m - 1). Prove that wrt the obvious
action of Z[X,Y] on M that each element of I annihilates M
but no nonzero element of S does.

(iv) Show that if f is in I then f(u,v) = 0 whenever
u^n = v^m = 1. But show that the only element of S with this
property is zero, by noting that a one-variable
polynomial of degree k has <= k zeros.

(v) similar to (iv) but get a formula for the coefficients
of f, for f in S, in terms of the f(u,v) via "Fourier

etc. etc. etc,

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