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By "algebraic curve" here (both in $\mathbb{R}^3$ and in $\mathbb{C}^3$), I do not only mean that the dimension of the set as an algebraic variety is 1, but also that its dimension is 1 in a topological sense (i.e., that, roughly, it is locally a continuous parametric curve).

In this context, for example, the zero set of the polynomial $x^2 +y^2=0$ in variables $x,y,z$ is a line in $\mathbb{R}^3$ if the polynomial is seen in $\mathbb{R}[x,y,z]$, but a complex variety of dimension 2 in $\mathbb{C}^3$ if the polynomial is seen in $\mathbb{C}[x,y,z]$. However, if we consider the initial line in $\mathbb{R}^3$ as the zero set of the polynomials $x=0$ and $y=0$ (which is a more natural choice of polynomials), and we see these polynomials as elements of $\mathbb{C}[x,y,z]$, then their zero set in $\mathbb{C}^3$ is a line, i.e. a complex algebraic curve. Can we apply such a procedure for any real algebraic curve?

I would also like to ask something else: Is the projection of a real algebraic curve from $\mathbb{R}^3$ to a generic plane again a real algebraic curve? In other words, is the projection the zero set of one polynomial in $\mathbb{R}[x,y]$?

I do not have an algebraic geometry background, so I would really appreciate it if you could also tell me in which books I could search to understand these things better. Most of the literature deals with complex and not real varieties.

Thank you very much!

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Generally, the concept you're looking for is the algebraic closure. You just consider the set of all polynomials that vanish on the real algebraic curve in question, obviously, this will give you the smallest complex variety, since it has the highest possible amount of restrictions on points. I can't think of a proof that the algebraic closure of a real curve is a complex curve. I don't know much real algebraic geometry either. – Will Sawin Jan 16 '12 at 20:42
up vote 3 down vote accepted

The "right" way to look at these is through algebra. In the case of the curve defined by $x^2+y^2=0$, what you want to extend to $\mathbb C$ is its ideal, not any particular equation. If you take $I=\{f\in\mathbb R[x,y,z]] | f(P)=0\text{ for every }P=(x,y,z), \text{ for which } x^2+y^2=0\}$, then you get exactly the ideal $(x,y)$ and extending it over $\mathbb C$ gives you what you want. This is what you should do in general.

Let me also make a comment on your mentioning parametric curves. You have to be very careful with this. In algebraic geometry parametrization is very rare. Of course you can always have analytic parametrizations locally, but you will rarely have parametrizations in the Zariski topology. For example, any smooth curve defined by a degree $d$ equation on the plane for which $d>2$ does not have an algebraic parametrization. The "right" way to count dimension is by taking the associated ideal (see above), take the quotient of the polynomial ring by this ideal (this the ring of polynomial functions on your object) and count the dimension of this as a ring. For instance, if you work over a field and in a finite dimensional space, then this will be the transcendence degree of this ring over the base field.

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Thank you! But what I don't see is exactly why this procedure of extending this particular ideal of the real curve over $\mathbb{C}$ will always give a variety that is a curve, i.e. of dimension 1. Isn't it possible to get a complex variety of higher dimension? – Marina Iliopoulou Jan 20 '12 at 3:02
If you use the notion of dimension I described above, then you can see that this cannot happen. If you start with a curve, then the ring $\mathbb R[x,y,z]/I$ will have transcendence degree $1$ over $\mathbb R$. Extending over $\mathbb C$ on the level of rings is just tensoring with $\mathbb C$ over $\mathbb R$, but that does not change the transcendence degree over the base field. – Sándor Kovács Jan 20 '12 at 3:42

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