# Projections of real algebraic curves

Suppose that the real algebraic curve $\gamma$ in $\mathbb{R}^3$ is the intersection of the zero sets of the polynomials $p_1,...,p_k \in \mathbb{R}[x,y,z]$. Is the projection of $\gamma$ on a generic plane (isomorphic to $\mathbb{R}^2$) a real algebraic curve? I.e., is it the zero set of one polynomial $p \in \mathbb{R}[x,y]$?

It is obvious that it is not true that the projection of a real algebraic curve from $\mathbb{R}^2$ to $\mathbb{R}$ is again a real algebraic curve (if the initial curve is closed, then its projection is a line segment, which is not the zero set of a polynomial).

I would also like to ask:

Is the projection of a complex algebraic curve from $\mathbb{C}^3$ to $\mathbb{C}^2$ a complex algebraic curve?

I hope my questions are not trivial; I have searched in books, but I was unable to find an answer.

Thank you very much!

• Well, if the curve is given by $x=y=0$ and you are projecting onto the $(x,y)$-plane, then you certainly don't get a curve. If however, $\gamma$ is not a line, you can consider the elimination ideal $(p_1,\ldots,p_k)\cap \mathbb{k}[x,y]$ which defines the image of the projection Jan 19 '12 at 14:52
• Thank you! In the case of the line, the projection on a generic plane is still a line (there is only one plane the projection on which is not a line). Jan 19 '12 at 15:36
• You can only ensure that the projection is semi-algebraic. See en.wikipedia.org/wiki/Tarski%E2%80%93Seidenberg_theorem Jan 19 '12 at 17:41

• semialgebraic sets in the case of real closed fields (e.g. $\mathbb{R}$),
• constructible sets (boolean combinations of zero sets) in the case of algebraically closed fields (e.g. $\mathbb{C}$).
So, finally, I understand that the answer to my question is no in general. In particular, the projection of a variety from $\mathbb{K}^3$ to $\mathbb{K}^2$ may not even be a variety. Also, if $\mathbb{K}=\mathbb{C}$ (algebraically closed), then the variety defined by the elimination ideal $(p_1,...,p_k) \cap \mathbb{C}[x,y]$ is the smallest variety containing the projection of the initial curve on the $(x,y)$-plane, and only under certain assumptions (for example, closure theorem) it is equal to the projection.
If, however, $\mathbb{K}=\mathbb{R}$, then the variety defined by the elimination ideal may not even be the smallest one containing the projection, which, as mentioned, might not even be a variety.