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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!

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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 –  J.C. Ottem 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). –  Marina Iliopoulou Jan 19 '12 at 15:36
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You can only ensure that the projection is semi-algebraic. See en.wikipedia.org/wiki/Tarski%E2%80%93Seidenberg_theorem –  Lucas Kaufmann Jan 19 '12 at 17:41
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2 Answers

up vote 2 down vote accepted

As already mentioned, the answer is negative for varieties.

However, if you are interested in a more general setting when the answer is positive (i.e. a more general class of sets that is closed under projections), you can look at

  • 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. $\mathbb{C}$).

This is related to quantifier elimination. The theory of real closed fields (RCF) and the theory of algebraically closed fields (ACF) admit quantifier elimination (see e.g. David Marker, Model Theory, §3.2, 3.3.), which means that if you take a definable set in RCF or ACF, then its projection is also a definable set. And the definable sets are semialgebraic sets and constructible sets respectively.

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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.

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