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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.)

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    $\begingroup$ Of course it is $[Y]\smile [Z]$! This is clear in your example. Try again! $\endgroup$
    – abx
    Commented Jan 27, 2014 at 14:17
  • $\begingroup$ Why do you think so? $\endgroup$
    – Alex Gavrilov
    Commented Jan 27, 2014 at 16:11
  • $\begingroup$ This is well known. A 4-dimensional smooth quadric $X$ has $H^2(X,\Bbb{Z})=\Bbb{Z}.h$ and $H^4(X,\Bbb{Z})=\Bbb{Z}.p+\Bbb{Z}.q$, with $p^2=q^2=1$, $p.q=0$ and $h^2=p+q$. With your notation, $[Y]=p$ and $[Z]=q$. $\endgroup$
    – abx
    Commented Jan 27, 2014 at 17:04
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    $\begingroup$ A more general statement is, if $X$ smooth complete contains $Y,Z$ where $Y$ is a hypersurface containing no component of $Z$, then $[Y\cap Z] = [Y]\smile[Z]$. $\endgroup$
    – Allen Knutson
    Commented Jan 27, 2014 at 18:06
  • $\begingroup$ abx, what do you mean? With my notation, $[Y]=[Z]=h$, because these are hyperplane sections. $\endgroup$
    – Alex Gavrilov
    Commented Jan 28, 2014 at 8:02

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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 $[P]\smile[Q]=0$.

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