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While studying, I found the following problem:

Let $f, g \in F[t]$. Prove that $\exists p \in F[x, y], p \neq 0 : p(f(t), g(t)) = 0$

I'd thank any hints that point me in the right direction.

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    $\begingroup$ Try starting with $n=\operatorname{lcm}(\operatorname{deg}f,\operatorname{deg}g)$, $p(x,y)=x^{n/\operatorname{deg}f}-x^{n/\operatorname{deg}g}$. $\endgroup$ – Alexander Gruber Dec 16 '14 at 4:59
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I think this question is more related to elimination theory than to linear algebra: this is about finding a polynomial that vanishes on the parametric curve $t\mapsto(f(t),g(t))$, thus "eliminating" the parameter $t$.

However I think the following linear-algebra inspired proof can be given. Consider the usual subspaces $F[t]_{<d}$ of polynomials in$~t$ of degree${}<d$ and similarly $F[x,y]_{<d}$ of polynomials in$~x,y$ of total degree${}<d$. Then $\dim(F[t]_{<d})=d$ and $\dim(F[x,y]_{<d})=\binom{d+1}2$. For the given $f,g\in F[t]$ let $L_{f,g}:F[x,y]\to F[t]$ be the substitution map $p\mapsto p[x:=f,y:=g]$, which is clearly $F$-linear; any nonzero element $p\in\ker(L_{f,g})$ will answer our question. Now with $m=\max(\deg(f),\deg(g))$ and any $d$, the restriction of$~f$ to $F[x,y]_{<d}$ has its image contained in $F[t]_{<md}$. Since $\binom{d+1}2$ grows quadratically with$~d$, one will have $\binom{d+1}2>md$ for sufficiently large$~d$; indeed this happens as soon as $d\geq2m$. For such $d$ the restriction of $L_{f,g}$ to $F[x,y]_{<d}$ cannot be injective for dimension reasons, completing the proof.

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