Let $X$ be a smooth projective complex algebraic variety. Let $V_i$, for $i=1,\dots, n$, be a collection of (smooth) connected hypersurfaces such that, for all $I\subseteq [n]$, the intersection $\cap_{i \in I} V_i$ is smooth.

Is the intersection $\cap_{i=1,\dots, n} V_i$ equidimensional?

Added later: If we take $V_i$ to be smooth effective (possible non ample) divisors, is the intersection $\cap_{i=1,\dots, n} V_i$ equidimensional?

  • 1
    $\begingroup$ If $V_i$ are allowed to be effective divisors (as opposed to ample divisors; cf. the comments below Sándor Kovács's answer), it seems likely that the following can happen: $V_1$, $V_2$, and $V_3$ are smooth irreducible divisors, $V_{12}$, $V_{23}$, and $V_{31}$ each consist of two smooth components, one of which is in common between all three, and the others intersect in a lower-dimensional variety. Then $V_{123}$ has components of different dimensions. I've tried to construct such an example, but so far have been unsuccessful. $\endgroup$ – R. van Dobben de Bruyn Jul 28 '18 at 22:27

If the intersection $\cap_{i=1}^nV_i$ is irreducible, then it is equidimensional. Otherwise let $r<n$ be such that $\cap_{i=1}^rV_i$ is irreducible, but $A:=\cap_{i=1}^{r+1}V_i$ is not. If $\dim\cap_{i=1}^rV_i\geq 2$, then $A$, being an effective ample divisor, is connected, so if it is smooth, then it is irreducible, contradicting the choice of $r$, so $\dim\cap_{i=1}^rV_i\leq 1$. Then $A$ is either empty or a finite set and hence equidimensional. (You can also remark, that it is not needed that all intersections are smooth, just that there is a sequence of getting to $\dim\cap_{i=1}^nV_i$ through smooth intersections).

On the other hand, if you only assumed that the ultimate intersection is smooth, then this is not true: Let $V_1,V_2\subseteq \mathbb P^3$ be two quadric surfaces that share a tangent plane. Then $V_1\cap V_2$ is the union of two intersecting lines, say $\ell_1, \ell_2$. Now let $V_3$ be a third quadric that contains $\ell_1$ but does not contain $\ell_2$. Then $V_3$ intersects $\ell_2$ in two distinct points, one of which is on $\ell_1$. Let $P$ be the intersection point of $V_3$ and $\ell_2$ which is not on $\ell_1$. Then $V_1\cap V_2\cap V_3=\ell_1\cup \{P\}$. Which is smooth but not equidimensional. Of course, this can only be done if the intersection is the union of a positive dimensional irreducible component and a set of points.

  • 2
    $\begingroup$ Why is $A$ ample? Effective, yes. Are the $V_i$ supposed to be ample? $\endgroup$ – Zach Teitler Jul 27 '18 at 20:21
  • 4
    $\begingroup$ I guess the OP didn't make this entirely clear. The $V_i$ are hypersurfaces, no? That means ample. I suppose you are saying that they are simply Cartier divisors? I don't think "hypersurface" makes sense other than being defined by a single equation everywhere in which case it is ample. $\endgroup$ – Sándor Kovács Jul 27 '18 at 20:54
  • $\begingroup$ Okay, that makes sense. $\endgroup$ – Zach Teitler Jul 27 '18 at 21:02
  • $\begingroup$ Zach, I think it would make sense to ask what you are suggesting, so this is a good point, I just didn't think of it. I'd have to think about this more general question (or someone else might know)... $\endgroup$ – Sándor Kovács Jul 27 '18 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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