# About Fulton's Intersection theory Appendix Lemma A 4.1

The assumption for Lemma A.4.1 is $$A \to B$$ is flat. The second assumption is that $$A$$ and $$B$$ are Artinian rings. From this Lemma A.4.1 states that $$l_B(B) = l_A(A) \cdot l_B(B/mB)$$ where $$m$$ is the maximal ideal of $$A$$. The proof is given as follows :

Take a chain of ideals $$I_i$$ of $$A$$ such that $$A \subset I_1 \subset I_2 \cdots \subset I_r = 0$$ where $$I_{i-1}/I_i$$ is isomorphic to $$A/m$$. Then we have the chain $$B \subset I_1B \subset I_2B \subset \cdots \subset I_rB = 0$$ with $$I_{i-1}B/I_iB \approx (I_{i-1}/I_i) \otimes_A B \approx B/mB$$.

However I think we do not need flatness condition for the proof since taking tensor product is right exact and hence we have $$I_{i-1}B/I_iB \approx (I_{i-1}/I_i) \otimes_A B \approx B/mB$$ without any assumption. Am I right? If so, then why does Fulton use this condition for the construction of pull back of algebraic cycles? For the pull back of algebraic cycles I guess we only need surjectivity between the generic points of both sides of the map.

No. Whereas $$A/I\otimes_A B=B/IB$$ by right-exactness $$I\otimes_A B\to IB$$ is not an isomorphism in general. The exact sequence
$$0\to I_{i-1}\to I_i\to I_i/I_{i-1} \to 0$$
$$I_{i-1}\otimes_A B\to I_i\otimes_A B \to I_i/I_{i-1}\otimes_A B\to 0$$
We obtain a surjective map $$I_i/I_{i-1}\otimes_A B\to I_iB/I_{i-1}B$$ but it is not injective in general.
Here's a counter-example. Let $$A = k[e]/(e^2)$$ and $$B=A/(e)=k$$ and choose the filtration $$I_0=0$$ and $$I_1=(e)$$ and $$I_2=A$$. Then $$I_0B=I_1B=0$$ and $$I_2B=k$$. But $$(I_1/I_0)\otimes_A B=k\otimes_A k = k$$. We have that $$l_A(A)=2$$, $$l_B(B/mB)=1$$ but $$l_B(B)=1$$.