Let $f\colon X\longrightarrow Y$ be a surjective morphism of projective varieties. Consider a non-singular irreducible projective curve $T$ and surjective morphism $p_Y\colon Y\longrightarrow T$. Then we get the following commutative diagram $\require{AMScd}$ \begin{CD} X_t @>f_t>> Y_t\\ @V V V @VV V\\ X @>f>> Y\\ @Vp_X V V @Vp_YV V\\ T @>>\mathsf{id}_T> T \end{CD} so in particular an induced map between fibers $f_t\colon X_t\longrightarrow Y_t$. I would to check the following result:

Result: Let $V\subseteq X$ be an irreducible projective variety such that $\dim V=\dim W$, where $W:=f(V)$. Denote by $[V_t]$ the cycle associated to the projective variety $p_X^{-1}(t)\cap V$ and by $f_\ast$ the push-forward induced by $f$. Then $$(\#) \: \: \: \:(f_t)_\ast([V_t])=\big[f_\ast[V]\big]_t=\deg(f_{|V})[W_t].$$ This is a well know formula (eg. Prop. 10.1 of [Ful]), but in the reference I cited the Author proves it in a different way. Indeed, he consider intersection fibers instead of cycle associated to scheme theoretic fibers as I am doing.

My Attempt: We can assume that $V$ and $Y$ are affine and $f$ is dominant. Denote the coordinate rings of the fibers $V_t,W_t$ by $A:=k[V_t]$ and $R=k[W_t]$. Then we have $$(f_t)_\ast[V_t]=\sum_S\ell(A_{\mathfrak q_S})\deg({f_t}_{|S})f_t(S),$$ where $S$ runs over the irreducible components of $V_t$ such that $\dim f(S)=dim W-1$. On the other hand, we have $$\deg(f_{|V})\cdot[W_t]=[k(V):k(W)]\cdot\sum_Z\ell(R_{\mathfrak p_Z})Z,$$ where $Z$ runs over the irreducible components of $W_t$. Now fix an irreducible component $Z$ (corresponding to the minimal prime $\mathfrak p\subset R$) and consider the irreducible components $S_1,\ldots,S_n$ of $V_t$ (corresponding to the minimal primes $\mathfrak q_1,\ldots \mathfrak q_n\subset A$) whose image via $f_t$ is just $Z$. Proving $(\#)$ is then equivalent to provving the following formula $$[k(V):k(W)]\cdot\ell(R_\mathfrak p)=\sum_{i=1}^n[k(\mathfrak q_i):k(\mathfrak p)]\cdot\ell(A_{\mathfrak q_i}).$$

A similar aproach can be found here. More precisely, this lemma should be the key-point of the proof.

Add: The proof of this result would be important because one could check that the push-forward preserves algebraic equivalence of cycles by considering the following commutative diagram:

\begin{CD} \lbrace t\rbrace\times X @>f>> \lbrace t \rbrace\times Y\\ @V V V @VV V\\ T\times X @>\mathsf{id}\times f>> T\times Y\\ @Vp_X V V @Vp_YV V\\ T @>>\mathsf{id}_T> T \end{CD}

Thanks in advance for any help.

[Ful] Fulton, W. (1988) Intersection Theory. New York: Springer-Verlag.


1 Answer 1


Let me answer to my own question. At least, if someone will struggle with my same proplem in the future he will can find a solution.

The formula I was tempting to prove is actually true, with some modification. More precisely, in the Handbook of K-Theory (ed. Grayson and Friedlander), Thomas Geisser, in his paper Motivic Cohomology, K-Theory and Topological Cyclic Homology proves the following result.

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By mean of standard geometric arguments, He deduces that it is enough to check the claim in the affine case and consider $V$ in place of $X$ and $f(V)$ in place of $Y$. Thus, the proposition is proved by showing the formula

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