Existence of polynomial in R^2

[ihggin]

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

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

My attempt: say [tex]x(t)[/tex] is of degree [tex]a[/tex] and [tex]y(t)[/tex] is of degree [tex]b[/tex]. Then we can take [tex]n=ab[/tex], as we will then have two terms: [tex]c_{0a}y^{a}[/tex] and [tex]c_{b0}x^{b}[/tex] with the same highest degree of [tex]t[/tex]: [tex]ab[/tex]. I then tried to prove that the number of ordered pairs [tex](i,j)[/tex], such that [tex]ia+jb \leq ab[/tex], is greater than [tex]ab[/tex], so that we would have at least as many variables [tex]c_{ij}[/tex] as we have equations ([tex]ab[/tex]) to solve, so that we can always find a solution to [tex]f(x(t),y(t))=\sum_{i,j} c_{ij} x(t)^i y(t)^j[/tex] 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 [tex] 0\leq i,j\leq n[/tex] it seems more natural to pick the degree of f(x,y) as a polynomial, so find n for which [tex] 0\leq i+j\leq n[/tex]. 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