TITLE: How prove this for sufficiently large $n$,the functions $x(t)^{i}y(t)^{j},0\le i,j\le n$ are linearly dependent
QUESTION [6 upvotes]: An algebraic curve in $R^2$ is the locus of zero of a polynomial $f(x,y)$ in two variables By a polynomial path in $R^2$,we mean a parametrized path $x=x(t),y=y(t)$,where $x(t),y(t)$ are polynomials in $t$.
show tha t every polynomial path lies on a real algebraic curve by showing that, for sufficiently large $n$,the functions $x(t)^{i}y(t)^{j},0\le i,j\le n$ are linearly dependent.

Thank you someone can help me   solution.
and this problem is my frend ask me, I consider somehours, and I can't solution,and he say this problem from a book(Specific the name of the book didn't tell me)

REPLY [5 votes]: show that every polynomial path lies on a real algebraic curve 

This is called implicitization, which is a special case of Zariski closure.
Implemented algorithms use Groebner bases, but for polynomial plane curves an 'explicit' answer is to write the curve as Resultant($x - x(t), y - y(t))=0$ with both terms regarded as polynomials in $t$.  An algorithmic answer is to eliminate $t$ from the pair of equations in variables $x,y,t$, to get one equation in $x,y$.

for sufficiently large $n$,the functions $x(t)^i y(t)^j,0≤i,j≤n$ are linearly dependent.

The number of $(i,j)$ with $\deg (x^i y^j) \leq n$ is approximately $Cn^2$, it is the number of lattice points in a triangle. For large $n$ this number exceeds $n+1$, the maximum number of linearly independent polynomials in $t$ of degree $\leq n$, and there must be a nonzero linear relation.
The problem appears to be from (Michael) Artin's textbook Algebra.