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Suppose I have a variety $Z$ in $\mathbb{C}^4$ which is a (set theoretic) complete intersection of the hypersurfaces $\mathbf{Z}(P)$ and $\mathbf{Z}(Q)$. How can I compute a polynomial $R$ such that $\mathbf{Z}(R) \cap Z$ is a 1--dimensional variety that contains the singular locus of $Z$? If $(P,Q)$ was a radical ideal, then I could look at when the $2\times2$ minors of $\binom{\nabla P}{\nabla Q}$ vanish, but it seems that if $(P,Q)$ is not radical, then all of the $2\times2$ minors of $\binom{\nabla P}{\nabla Q}$ might vanish identically on $Z$.

As a final complication, my hope is that the degree of $R$ is $O(\textrm{deg} P + \textrm{deg} Q)$. Thanks!

[edit: In response to Sándor Kovács's answer, I have updated the problem. Previously I asked that $\mathbf{Z}(R) \cap Z$ be the singular locus of $Z$, instead of merely requiring that $\mathbf{Z}(R) \cap Z$ contain the singular locus]

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The scheme-theoretic intersection of P and Q might be highly non-reduced (possibly generically singular), so I don't understand how you could say anything about the singular locus of $Z$ from these two equations.. –  J.C. Ottem Oct 26 '11 at 21:12

1 Answer 1

What makes you think that the singular locus is cut out by a hypersurface?

Let $P=x^2+y^2+z^2+t^2$ and $Q=t$. Then $Z$ is a quadric cone in $\mathbb C^3$ with a single isolated singular point. $Z(R)\cap Z$ for any $R\neq 0$ would be a curve, so there is no way you can find such an $R$.


Addition (after the question was changed):

If you only want an $R$ such that $Z(R)\cap Z$ contains $\mathrm{Sing} Z$ then you can take $R=P$, $R=Q$, or any polynomial combination of $P$ and $Q$, in particular $R=P\\ Q$, so you would have $\deg R=\deg P + \deg Q$.

I suppose you would want an $R$ such that $\mathrm{Sing}Z\subseteq Z(R)\cap Z\subsetneq Z$...

As JC commented, it is not clear how to get that from $P$ and $Q$. What you describe about $(P,Q)$ being or not being a radical ideal is exactly what he is saying about the scheme theoretic intersection being non-reduced everywhere.

The main issue is how you determine the singular locus of the corresponding reduced scheme based on $P$ and $Q$. My guess would be that you might be able to give some estimate on the minimal degree of such an $R$, but it seems doubtful that one could give you an explicit $R$.

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Oops; what I meant to say was that $Z\cap\mathbf{Z}(R)$ must contain the singular locus of $Z$. I have updated the original question to reflect this. –  Josh Zahl Oct 26 '11 at 21:28
    
Indeed, I want an $R$ such that $Z(R)\cap Z$ is 1-dimensional, and thus is properly contained in $Z$. An estimate on the minimal degree of such an $R$ would be enough. Do you know of any references that discuss this problem? I had no luck finding any. Thanks again! –  Josh Zahl Oct 28 '11 at 0:12

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