
Re: Can a small degree polynomial exist in this Ideal?
Posted:
May 15, 2013 5:54 AM


eric.fennessey@gmail.com wrote:
> I am feeling dim and my brain is failing me on this one. > > Let n,m >=2 and consider the subset, I, of polynomials in Z[X,Y] given by, > > I = {P.(X^n  1) + Q.(Y^m  1): P,Q in Z[X,Y]}, > > I.e. the ideal generated by X^n 1, Y^m 1 in Z[X,Y]. > > Let S be the subset of Z[X,Y] consisting of polynomials where each > nonzero term of the form n.X^i.Y^j has i<n and j<m. > > My question is: Is I intersected with S the zero polynomial? Or, can some > strange cancellation occur which wipes out all terms containing X^n's and > Y^m's and higher, but leave something nonzero behind?
It seems to me that there is an almost trivial proof of this result. Also, the result holds for the more general case of a field k in place of Z.
Suppose f(X,Y) = P(X,Y)(X^n  1) + Q(X,Y)(Y^m  1)
where f(X,Y) is of degree < n in X and < m in Y. Suppose P contains a term aX^rY^(m+s) of degree >= m in Y. One can in effect reduce this term to aX^rY^s by transferring part to Q, since
aX^rY^(m+s)(X^n  1)  aX^rY^s(X^n  1) = aX^r(X^n  1)(Y^m  1)
So one can assume that P is of degree < m in Y, and then it follows that Q must vanish, since otherwise there would be a term of degree >= m in Y.
 Timothy Murphy email: gayleard /at/ eircom.net tel: +353862336090, +35312842366 smail: School of Mathematics, Trinity College Dublin

