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