Existence of polynomial in R^2

[ihggin]

Here is a potentially neat problem. Let x(t),y(t) (for all t\in \mathbb{R}) be polynomials in t. Prove that for any x(t),y(t) there exists a non-zero polynomial f(x,y) in 2 variables such that f(x(t),y(t))=0 for all t. The strategy is to show that for n sufficiently large, the polynomials x(t)^{i}y(t)^{j} with 0\leq i,j \leq n are linearly dependent.

For example, suppose we are given x(t)=t and y(t)=t^2 + 1. Then the polynomial f(x,y)=1-y+x^2 would be a non-zero polynomial such that f(x(t),y(t)) = 1 - (t^2 +1) + t^2 = 0 for all t \in \mathbb{R}.

My attempt: say x(t) is of degree a and y(t) is of degree b. Then we can take n=ab, as we will then have two terms: c_{0a}y^{a} and c_{b0}x^{b} with the same highest degree of t: ab. I then tried to prove that the number of ordered pairs (i,j), such that ia+jb \leq ab, is greater than ab, so that we would have at least as many variables c_{ij} as we have equations (ab) to solve, so that we can always find a solution to f(x(t),y(t))=\sum_{i,j} c_{ij} x(t)^i y(t)^j being zero. However, I played around with trying to prove this inequality and I don't think it's true.

Does anyone have any ideas on how to solve this problem?

 

[Office_Shredder]

Rather than when looking at a choice of n for which 0\leq i,j\leq n it seems more natural to pick the degree of f(x,y) as a polynomial, so find n for which 0\leq i+j\leq n. If f(x,y) is of degree n, you have n+1 choices of i and j when i+j=n, n choices when i+j=n-1, all the way down to one choice of i and j when i+j=0. So the number of monomial terms in f(x,y) is (n+1)(n+2)/2 and this is the number of coefficients you are going to get to choose (one coefficient for each monomial in x and y in f(x,y))

On the other hand, the highest degree term of t is going to be n*deg(x) or n*deg(y) (whichever is larger). Heuristically (and you can try to prove this, I don't think it will be too hard) if we have fewer terms t^k than we have coefficients to pick, we'll be able to find a choice of coefficients that makes the whole thing zero. Since the former grows linearly and the latter quadratically, we know for large enough n we can find a solution, and it's easy to calculate exactly when this will occur