# Is a smooth intersection of hypersurfaces equidimensional?

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?

• 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. – R. van Dobben de Bruyn Jul 28 '18 at 22:27

## 1 Answer

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.

• Why is $A$ ample? Effective, yes. Are the $V_i$ supposed to be ample? – Zach Teitler Jul 27 '18 at 20:21
• 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. – Sándor Kovács Jul 27 '18 at 20:54
• Okay, that makes sense. – Zach Teitler Jul 27 '18 at 21:02
• 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)... – Sándor Kovács Jul 27 '18 at 21:04