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Do we have any classification of all singular curves of genus $0$ and $1$ in $\mathbb{P}_2$? For example if it is of degree $2$ (a conic) then it is $L.L'$ where both $L$ and $L'$ are isomorphic to $\mathbb{P}_1$.

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No. There is a lot of research focusing on singular rational plane curves and there are many open problems here. For example, it is not known how many cusps a rational irreducible plane curve can have (although it is conjectured that this number is four, see e.g. this paper by Piontkowski or Moe's thesis). I don't know what the situation is for genus 1 curves, but I would not suspect that this problem is easier.

EDIT: In the above, I am assuming that you are talking about the genus of the normalization and not the arithmetic genus $p_a=(d-1)(d-2)/2$ (in which case the classification problem is trivial).

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Dear Ottem, Thanks for your answer, but the paper you pointed is for any rational curve. I am only interested in genus 0 and 1. So that the degree of the curve can not go too high (?) unless it has a high number of singularities so that the genus comes down!! –  user13684 May 29 '11 at 8:47
    
I also read the paper. It seems the conjecture is only for cuspidal curves!! –  user13684 May 29 '11 at 10:23
    
Dear unknown. The point is that plane curves with geometric genus 0 (that is, rational curves) are quite complicated objects and we don't know their classification, even in the case where the curves are cuspidal. –  J.C. Ottem May 29 '11 at 10:50
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Dear unknown and Ottem. What an interesting discussion. The subject of curves, and cuspidal curves in particular is by no means completely classified. The conjecture in Piontkowski's article has not been proved, but there is an article by Tono where an upper bound for the number of cusps on a cuspidal curve is given for curves of any given genus:

http://onlinelibrary.wiley.com/doi/10.1002/mana.200310236/abstract

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Welcome to Mathoverflow, K Moe! :) –  J.C. Ottem May 29 '11 at 14:15
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