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How to prove that any finite set in the affine plane is realizable as intersection of TWO plane curves?

Thanks.

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  • $\begingroup$ Why the downvote? Does anyone know a quick down-to-earth proof of this? In link.springer.com/article/10.1007%2FBF01418923 Eisenbud and Evans prove more generally, that any algebraic set in affine $n$-space is the intersection of $n$ hypersurfaces. Not sure if $n=2$ is so much easier. $\endgroup$
    – Peter Mueller
    Commented Oct 18, 2013 at 13:11

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What you want to show is that a radical ideal $I$ in $k[X,Y]$ must be equal to the radical of a two-generated ideal.

The image of $I$ in $k(X)[Y]$ is certainly principal (because $k(X)[Y]$ is a PID). So we can write $s(X)I\subset (f)$ for some $f\in k[X,Y]$.

Let $A=k[X]/(s)$ and let $B$ be $A$ mod nilpotents. Let $J$ be the radical of the image of $I$ in $A[Y]$ and let $\overline{J}$ be the image of $J$ in $B[Y]$. Then $\overline{J}$ is principal because $B$ is a product of fields. It follows that $J$ is prinicipal (because it's a radical ideal and it's principal mod nilpotents). Lift a generator to $g\in I$.

It's straightforward to check that $I$ is equal to the radical of $(f,g)$.

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