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]

  • $\begingroup$ 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.. $\endgroup$
    – J.C. Ottem
    Commented Oct 26, 2011 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 $R$.

  • $\begingroup$ 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. $\endgroup$
    – Josh Zahl
    Commented Oct 26, 2011 at 21:28
  • $\begingroup$ 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! $\endgroup$
    – Josh Zahl
    Commented Oct 28, 2011 at 0:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.