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Let $C$ be a local complete intersection curve in $\mathbb{P}^3$ (not irreducible or smooth) of degree $e$. Suppose $f_1, f_2$ (and $e_i=\deg(f_i)$) are two of the lowest degree generators of $I(C)$. Then:

1) Is it true that the minimal number of generators of $I(C)$ is less than or equal to $2+(e_1e_2-e)$? (Intuitively I would think this to be true by using the degree formula from intersection theory which says that $e_1e_2=\sum_im_i\deg(C_i)$ where $C_i$ ranges over the irreducible components of the curve defined by $f_1$ and $f_2$ and $m_i$ are their multiplicities)

2) Is there any known result/approach to compute the upper bound on the minimal number of generators of $I(C)$ in terms of its degree?

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"every locally complete intersection curve in P3 can be defined by four equations."

look at http://www.math.binghamton.edu/somnath/Notes/curves.pdf and http://www.math.tifr.res.in/~publ/ln/tifr62.pdf

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  • $\begingroup$ @Kalinin: I am aware of this result. I was asking for a bit stronger result which is to say the number of generators of the ideal of a curves i.e., the radical of this ideal generated by these $4$ equations. Is there a bound on the number of generators of the radical of an ideal based on the number of generators of the original ideal? Thanks anyways for your answer. $\endgroup$ Nov 15 '12 at 11:19

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