For ease of notation, define the degree of a variety to be the sum of the degrees of its irreducible components. The generalized Bézout theorem (due to Fulton and Macpherson) states that, for $V_1$, $V_2$ varieties in $\mathbb{A}^n$ or $\mathbb{P}^n$, $$\deg(V_1\cap V_2) \leq \deg(V_1) \deg(V_2).$$ See section 8.4 in Fulton's Intersection Theory (or II.3.2.2 in Danilov-Shokurov, say).
Here $V_1\cap V_2$ just means the variety whose set of points over an algebraically closed field $K$ is $V_1(K)\cap V_2(K)$. There are ways of assigning multiplicities to the components of $V_1\cap V_2$ so that the inequality above is an equality, but let us not get into that.
I was recently having lunch and tea (involving donuts of the wrong genus) with two algebraic geometers visiting Uganda. I was talking to them about some things that directly follow from the above, and one of them that they could not possibly be right without additional conditions. When I showed them the statement above, they expressed surprise that such a thing could be true without qualifications.
It turned out that we were having a bit of a communication difficulty (not helped by my not being an algebraic geometer; they had to explain to me what "Cohen-Macaulay" means - if you assume that condition, everybody agrees). Here is an example they gave me. Bottom line: if by $V_1\cap V_2$ you mean what I said before (let us call this the "naïve" definition), then everything works out -- the statement of the generalized Bézout theorem is true; however, if you interpret $V_1\cap V_2$ as they do (apparently the "right" definition), the statement of the theorem is not true.
Example. Let us work in $\mathbb{A}^4$. Let $V_1$ be the union of the xy-plane through the origin and the zw-plane through the origin. Let $V_2$ be the hyperplane given by $x=z$. Then, using the naïve definition, we see that $V_1\cap V_2$ is the union of the line $x=z=w=0$ and the line $x=y=z=0$. So all is well: $\deg(V_1\cap V_2)=2$, $\deg(V_1)=2$, $\deg(V_2)=1$.
However, the "right" definition involves working with the ideals $I_1$ and $I_2$ defining $V_1$ and $V_2$ (namely, $I_1=(x z, x w, y z, y w)$ and $I_2=(x-z)$) and taking the sum $I_1+I_2$ (which here equals $(x z, x w, y z, y w, x-z)$). I can reply that I also do that: what I understand by $V_1\cap V_2$ is just the variety defined by $I_1+ I_2$, or, what is the same (or so I thought), by the radical of $I_1+ I_2$.
But no, they were not happy with that: taking the radical destroys information. It took me a bit of time to understand that they were telling me that the right thing to do was to take the primary decomposition of $I_1+ I_2$, and define $V_1\cap V_2$ to consist of one component for each ideal in the primary decomposition.
Even I can figure out that $I_1\cap I_2$ has primary decomposition $Q_1\cap Q_2\cap Q_3$, where $$Q_1=(x,y,z),\;\;\;\;\;\;\;\; Q_2 = (x,z,w),\;\;\;\;\;\;\;\; Q_3 = (x-z,y,z^2,w).$$ Here $Q_1$ and $Q_2$ just give us the lines $x=z=w=0$ and the line $x=y=z=0$, whereas $Q_3$ is a slightly fattened-up version of the point at the origin (its radical is $(x,y,z,w)$, which defines the point at the origin). So, the degree of $V_1\cap V_2$ would have to be $3$ (or greater than $3$, if you somehow defined a fat point to have degree greater than $1$; this is just a stray thought on my part) and so the generalized Bézout theorem would not hold.
End of example.
Now, I am not too bothered, since, for my applications, the naïve definition of intersection is completely appropriate; it seems I can just go on working as before. Still, I am a bit bothered: how come one gets an elegant, useful, very general inequality if one uses the "wrong" (naïve) definition, and one has to assume additional conditions (which are a pain to check, as they are not inherited by unions) if one wants the same inequality for the "right" definition of $V_1\cap V_2$? Also, why is the "right" definition the right one, and the naïve one naïve? Or is it just two definitions that are appropriate for different purposes?
PS. What is the better name for the "naïve" intersection: "set-theoretic intersection" or "geometric intersection"? The algebraic geometers in this (true) story called the "right" definition "ideal-theoretic"; as I said above, the naïve definition looks ideal-theoretic to me as well - one just needs to take the radical before decomposing, no?
