Suppose that we have a parametrization via polynomials as follows:


$$t\longrightarrow (f\_1(t),\ldots,f\_n(t)),$$


where $t$ is a vector in $\mathbb{C}^r$ and $f\_i$ are polynomials of arbitrary degree.


Can we always find equations such that the image is an affine algebraic variety?


The question is motivated by Exercise 1.11 in Hartshorne:



> 
> Let $Y\subseteq A^3$ be the curve given parametrically by $x = t^3, y= t^4, z = t^5$. Show
>  that $I(Y)$ is a prime ideal of height 2 in $k[x,y,z]$ which cannot be generated by
>  2 elements.
> 
> 
> 


I am not interested in the exercise in particular. Finding the variety is easy sometimes, for instance $t\rightarrow (t^2,t^3)$ is given by $I(x^3-y^2)$.


I am looking for a result which says that the image is always an affine algebraic variety AND a procedure to find the ideal.



Are all parametrizations via polynomials algebraic varieties?
I can't comment (b/c I'm not a registered user) but let me add: in case the dimension of the domain is 1 (as in your motivating example) the image is in fact an affine variety. To see this, note that the map can always be extended to a map from the projective line to projective space by homogenizing things (compare with Dan's example---if you tried homogenizing his map, you'd get $[x:y:z] \mapsto [xy:yz:z^2]$ which is not defined if $x=z=0$ or $y=z=0$), and use the fact that the image of a projective variety by a regular map is closed. Finally, observe that the image of the affine line you started with is precisely the intersection of the image of the homogenized map with the affine coordinate patch determined by your homogenization. Therefore the image of the map you started with *is* an affine variety. So Hartshorne's example is not an accident.


