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Let $X$ be a nonsingular, integral projective variety of dimension at least 2 over $k$ algebraically closed. Let $Y$ and $Z$ be two codimension 1 subschemes (effective Weil divisors) of $X$. Must they intersect?

Moreover, letting $Y$ and $Z$ be arbitrary subschemes, does one have the usual dimension estimate: $dim(Y\cap Z) \geq dim Y +dim Z - dim X $?

I tried modifying the proof for $X = \mathbb{P}^n$ (or $\mathbb{A}^n$) but stumbled over the fact that a coordinate ring for $X$ may not be a UFD (i.e. may have non principal height 1 primes). But on a stalk it will be a UFD, so perhaps there's a workaround?

References, rationale, or counterexamples if applicable, please. Thanks.

Cass
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  • Have you looked at exercise 11.3.F of Vakil, by first embedding $X$ into some $\mathbb{P}^n$, and then ignoring $X$ all together? At the end of the day, you can still apply Krull's height theorem to get the result you want, unless I am misunderstanding something. – Alex Youcis Jul 30 '14 at 08:51
  • We would have to know that $n$ is small. If $n \geq 2*dimX$, then $Y$ and $Z$ have large codimension in $\mathbb{P}^n$ and would not be guaranteed by dimension reasons to intersect (if we forget $X$). – Cass Jul 30 '14 at 09:02

1 Answers1

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The answer is no:
Let $X\subset \mathbb P^3$ be a smooth quadric and $Y,Z\subset X$ be two distinct lines of the same ruling.
These lines have codimension $1$ in $X$ but don't intersect.

Remark:
Up to isomorphism we have $X=\mathbb P^1\times \mathbb P^1, Y=\{a\}\times \mathbb P^1,Z=\{b\}\times \mathbb P^1$ for $a\neq b\in \mathbb P^1$

Georges Elencwajg
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