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I have started working through the textbook Ideals, Varieties, and Algorithms by Cox, Little, and O’Shea and I am stuck on one part of an introductory question.

The question begins by getting one to show

(1) If $f(t)$ and $g(t)$ are polynomials of degree $\le n$ in $t$, then for $m$ large enough, the monomials

$$[f(t)]^e[g(t)]^f$$

with $e+f\le m$ form a linearly dependent set in $k[t]$.

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The next part of the question asks one to deduce

(2) If $C\,:\,x=f(t),\,y=g(t)$ is any polynomial parametic curve in $k^2$, then $C$ is contained in $\textbf{V}(F)$ for some $F\in k[x,y]$.

I have demonstrated (1), but I am at a loss as to how to use that fact to prove (2). It doesn’t seem like it is supposed to be a hard problem, but never the less I am having trouble wrapping my head around it. Additionally I assume that for (2) we are not interested in the trivial solution $F=0$.

*[Added for completeness]*

To do part (1), we recognize that if $\deg(f),\deg(g)\le n$, then $$\deg\left([f(t)]^e[g(t)]^f\right)\le ne+nf\le nm$$

Hence if we have more than $nm+1$ polynomials of the form $[f(t)]^e[g(t)]^f$ then we have a dependent set (since $d+1$ is the dimension of the vector space of polynomials of degree less than or equal to $d$). Now we simply count the number of polynomials in this form (this was an exercise before the problem) – it is $\binom{m+2}{2}=\frac{1}{2}(m+1)(m+2)$ from a standard application of stars and bars. Ensuring $m$ large enough (specifically $m>2n-3$) gives us the conclusion.

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To say that

$$[f(t)]^e[g(t)]^f$$

with $e+f\le m$ form a linearly dependent set in $k[t]$ is the same as saying that there are $A_{e, f} \in k$, not all zero, so that

$$\sum_{e + f \le m} A_{e, f}[f(t)]^e[g(t)]^f = 0$$

(I suppose you are saying that the set is dependent in $k[t]$, treating $k[t]$ as a $k$-vector space). Thus the curve $(f(t), g(t))$ lie in the zero set of

$$F(x, y) = \sum_{e+f \le m} A_{e, f} x^e y^f$$

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