Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

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

 Messages: [ Previous | Next ]
 Timothy Murphy Posts: 657 Registered: 12/18/07
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
> non-zero 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 non-zero 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
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College Dublin

Date Subject Author
5/14/13 eric.fennessey@gmail.com
5/14/13 Robin Chapman
5/14/13 eric.fennessey@gmail.com
5/14/13 Robin Chapman
5/14/13 eric.fennessey@gmail.com
5/14/13 Robin Chapman
5/15/13 Timothy Murphy