
Re: Can a small degree polynomial exist in this Ideal?
Posted:
May 14, 2013 9:31 AM


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 onevariable 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 inversion"
etc. etc. etc,

