4
$\begingroup$

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.

$\endgroup$

1 Answer 1

5
$\begingroup$

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 $l_B(B)=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.