Two polynomial problem

Question

Suppose that there is a path $g(t) = (p(t), q(t))$ in a plane where $p$ and $q$ are polynomials. Is there always a polynomial $f(x,y)$ different from zero polynomial such that the image of $g$ is contained in the set $\{(x,y) | f(x,y) = 0\}$?

Answer 1

Let $f(x,y)$ be the resultant of $p(t) - x$ and $q(t) - y$ with respect to $t$.

Answer 2

I do not resist to give an elementary solution. Let $n,m$ the degrees of the polynomials $p,q$. Let $K$ an integer. There is a priori $(K+1)^2$ polynomials of the form $p^k q^l$ with $0\leq k\leq K$ and $0\leq l\leq K$. But the degree of such polynomial is $kn+lm\leq K(n+m)$. So they belongs to a vector space of dimension $K(n+m)+1$. For large $K$, we have $(K+1)^2>K(n+m)+1$, and the monomials $p^kq^l$ must be dependant over the base field, so there exists $f(x,y)$ not $0$ such that $f(p(t),q(t))=0$.

Answer 3

I strongly suggest that you take a look at Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms: they treat this exact problem in Chapter 3 Elimination, particularly in $\S3$ Implicitization. The main result is the following theorem.

Theorem 1 (Polynomial Implicitization) If $k$ is an infinite field, let $F: k^m \to k^n$ be the function determined by the polynomial parametrization \begin{align*} x_1 &= p_1(t_1, \ldots, t_m)\\ &\ \, \vdots\\ x_n &= p_n(t_1, \ldots, t_m) \, . \end{align*} Let $I = \langle x_1 - p_1, \ldots, x_n - p_n \rangle$ and let $I_m = I \cap k[x_1, \ldots, x_n]$ be the $m^\text{th}$ elimination ideal. Then $\mathbb{V}(I_m)$ is the smallest variety in $k^n$ containing $F(k^m)$.

One can compute a Gröbner basis $G$ for the elimination ideal, and any polynomial $f \in G$ not involving the $t_i$ will satisfy your condition.

So in your case, if you compute a Gröbner basis for the ideal $\langle x - p(t), y - q(t) \rangle \trianglelefteq k[x,y,t]$ with respect to the lexicographic monomial ordering $t > x > y$, you will find your desired $f$.

Answer 4

You can extend $g$ to a rational map $G:\mathbb P^1\dashrightarrow \mathbb P^2 $, which is actually a morphism (cf. Hartshorne Ch. I, Prop.6.8).
The image of $G$ is thus an algebraic curve (supposing $p,q$ are nor both constants), which has a polynomial equation $F(x,y,z)=0$ and the image of $g$ thus has equation $f(x,y)=F(x,y,1)=0$.
Note carefully that the image of a morphism $\mathbb C^2\to \mathbb C^2: z\mapsto (p(z),q(z))$ is not a closed subvariety of $\mathbb C^2$ in general.

Moral
Algebraic geometry is the art of not calculating, just as Galois theory is the art of not solving polynomial equations :-)