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Suppose that $X$ is a curve in $\mathbb{A}^3$ (in the AG sense, let's say over an algebraically closed field $k$) that contains no lines perpendicular to the $xy$ plane, and that there exist two polynomials $f,g\in k[x,y,z]$ such that $\{f=0\}\cap\{g=0\}=X\cup l_1\cup\cdots\cup l_n$, where $l_i$ are lines perpendicular to the $xy$ plane (and can possibly intersect $X$). Is it possible to find a third polynomial $h$ such that the intersection $\{f=0\}\cap\{g=0\}\cap\{h=0\}=X$?

Since $X$ is algebraic, of course given a point that does not lie on $X$, there is a polynomial that is zero on $X$ and not zero on that point. I want to see if I can "cut out" $X$ with one other equation.

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Yes. I believe this is from a Shafarevich problem?

For instance, suppose $\ell_1$ intersects the $x-y$ plane at $(x,y) = (a,b)$. Consider the homomorphism $k[x,y,z] \rightarrow k[z]$ sending $x\mapsto a$ and $y \mapsto b$. The image of $I(X)$ is some prime ideal of $k[z]$, which is principal. Now look at the pullback of a generator.

To treat the general case, use this idea along with the Chinese remainder theorem.

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thanks! Yes, it's from Shafarevich's book. – rfauffar Feb 6 '12 at 3:04

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