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1) I want to learn about algebraic curves and i'm confused, please correct me if i'm wrong : when we say an Affine algebraic curve over the field $F$ :

here affine to distinguish it from projective and $F$ is the field of coefficients and zeros of the polynomials $P_i$ in the equations $P_i=0$ defining the algebraic curve. Being a curve, we have $n$ variables $x_1,...,x_n\in F$ and $n-1$ equations $P_i(x_1,...,x_n)=0$. When we have only one equation and 2 variables $x_1,x_2\in F$ we can add the word "plane curve" to the name of the curve. When $F=\mathbb R$ we can add the word "Real curve" instead of saying over the field $F$.

2) every curve is at the same time affine and projective: if we have an affine curve we can turn it into projective curve by a sort of change of variables and a multiplication to put the polynomials into homogeneous form.

3) In wikipedia http://en.wikipedia.org/wiki/Algebraic_curve we read " An algebraic curve likewise has topological dimension two; in other words, it is a surface. " what is a topological dimension here and what is a surface here? the line and the circle are algebraic curves but they are not surfaces!!!!

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Re: 3) the passage there adresses complex curves, (in the Hausdorff topology), which are of dimension $1$ as complex manifolds and of dimension $2$ as real manifolds. They look like handlebodies. As for topological dimension, it's the Lebesgue covering dimension of a topological space. –  t.b. Jun 22 '11 at 12:19
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I'll address point 2): It's not accurate to say that any curve is both affine and projective. You're right that you can "homogenize" the polynomials, and get a projective curve, but it's a new curve, since homogenizing introduces more solutions. You can think about the projective curve as consisting of the original affine curve, plus some extra points "at infinity".

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Regarding (3), while you need at least $n-1$ equations in $n$-variables to cut out a curve, you could need more. What happens is that the first $n-1$ equations could cut out a curve with multiple components, and then you will need additional equations to cut out the individual components.

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