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.
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 $a\neq b\in \mathbb P^1$