EDIT: this is a stupid question (see the comments and the answer). 


Let $X$ be a projective complex manifold. Consider two different irreducible hypersurfaces $Y,Z\subset X$, with cohomology classes $[Y],[Z]\in H^2(X)$. 
Normally, one would expect that $[Y\cap Z]=[Y]\smile[Z]\in H^4(X)$. 
But, as I understand it, this is not always the case.
A simple example I know is $X=V(xy+zw+uv)\subset P^5$, $Y=X\cap V(u)$, $Z=X\cap V(z)$. 
In this case the intersection $Y\cap Z$ has two irreducible components and, 
no matter how you look at it, its cohomology class is not $[Y]\smile[Z]$.


*Question* What is a correct formulation of the rule $[Y\cap Z]=[Y]\smile[Z]$ 
for hypersurfaces?


(I tried to find this in Fulton, but failed. Admittedly, this does not tell very much, as my
native subject is differential geometry.)



Cohomology class of the intersection of two hypersurfaces
Yes, the formula is true in this case, too. As abx says, I've made a mistake in my computation (what a shame!). We have $Y\cap Z=P\cup Q$, and the irreducible 
components $P$ and $Q$ have different cohomology classes $[P]=p,[Q]=q$, such that 
$p+q=h^2=[Y]\smile[Z]$ (see comments). I thought that $[P]=[Q]$, which was quite stupid, because it is not so difficult to check out that $[P]\smile[Q]=0$. 


