About Fulton's Intersection theory Appendix Lemma A 4.1

Question

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.

Answer 1

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

$$0\to I_{i-1}\to I_i\to I_i/I_{i-1} \to 0$$

only gives rise to the right exact sequence

$$I_{i-1}\otimes_A B\to I_i\otimes_A B \to I_i/I_{i-1}\otimes_A B\to 0$$

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 $l_B(B)=1$.