I have another question about intersection theory "a la Fulton". This is about the projection formula, that states that if you have $$\require{AMScd} \begin{CD} {X'} @>{\pi'}>> {Y'}\\ @V{t'}VV @VV{t}V \\ X @>{\pi}>> Y; \end{CD}$$ a cartesian diagram with flat (with relative dimension) vertical arrows, and proper horizontal arrows. Then $$t^*\pi_*={\pi'}_*{t'}^*$$.
He says that we can assume that X and Y are varieties and $\pi$ is surjective which seems pretty clear to me.
But then he says "This is a local calculation involving local ring of irreducible components so we may assume that X and Y are spectra of fields, Y' is a local artinian ring, and X' is the tensor product that you imagine.
I've been trying out to work out the details of that assumption for almost a day now, and i cant seem to reduce the problem to what Fulton says it is reduced.
It's even worse because he says that in this case the reults follow form a lemma (fairly straightfoward) that is if $A\to B$ is a local moprhism of local rings, that induces an extension of degree d of residual fields, then the length of a B module seen as an A module is d times the length of the B module seen as a B module.
And i'm having the hardest time to see how this lemma implies the statement! To me there should be a sum of length of localisation at minimal prime ideals somewhere to compute the cycle [X'].
This leads me to a more general wondering. I'm getting a little bit hopeless after a couple days trying to understand that book. And i thought i had a decent formation in scheme theory. I often have a problem "reducing a geometric claim to its algebraic content"; like for instance when people say, "after base change we are left to prove that for a local domain etc....", this part always seems obscure to me, and i have to work out tediously the details to be convinced of the "reduction procedures". Is that supposed to be so? Because Futlon seems so casual about it, like it should be obvious, and if you dont see it, well, there's something that you have failed to understand.
In the maths i've done up to now, sometimes i said that something was "clearly" so, and i knew that if i wanted to i could work out the details, but i feel, in algebraic geometry, sometimes you cannot chop down an argument into tiny bits, you have to see it or not. So is it obvious to everyone why certains reductions work (like the one mentionned in that proof)? Or is it just me (and i'm probably bad at alg. geom. ?
Sorry for this long question. And thanks!
My solution is a little tricky. First of all, let's take $id\in Hom(\pi_*,\pi_*)$. Since there is a map $id\rightarrow t'_*t'^*$, therefore there is a distinguished element in $Hom(\pi_*,\pi_*t'_*t'^*)$. Since the diagram is Cartesian, $\pi_*t'_*=t_*\pi'_*$, so you got an element of $Hom(\pi_*,t_*\pi'_*t'^*)=Hom(t^*\pi_*,\pi'_*t'^*)$ by adjunction. I.e, there is a natural morphism $t^*\pi_*\rightarrow \pi'_*t'^*$. You want to show that it is an isomorphism. Ok, this only needs to be checked locally. You can assume that everything is (af)fine.
Now, you use that the vertical guys are flat. Let $Y=Spec S$, $X'=Spec R$, $X=Spec R'$ (just for confusion). Then
$t^*\pi_*F=R\otimes_S Hom_S(S,F)=Hom_R(R\otimes_S S,R\otimes_S F)=Hom_R(R,F\otimes_{R'} (R\otimes_S R'))=\pi_*'t'^*F$.
Since I did not use the properness condition, thus it is not needed.