Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When trying to compute the (Serre-generalized) intersection number of two varieties at a closed point, I came to a need to compute the following $\operatorname{Tor}$:

Let $k$ be an algebrically closed field, $A=k[x_1,x_2,x_3,x_4]$ and $\mathfrak m=(x_1,x_2,x_3,x_4)$. Let $M = k[x_1,x_2,x_3,x_4]/(x_1x_3,x_1x_4,x_2x_3,x_2x_4)$, $N=k[x_1,x_2,x_3,x_4]/(x_1-x_3,x_2-x_4)$. I want to compute $\operatorname{Tor}^i_{A_{\mathfrak m}}(M_{\mathfrak m},N_{\mathfrak m})$.

Any ideas how to do this?

I first noted that $N$ is gotten from $A$ by quotient by a a regular sequence, so that the Koszul complex of $(x_1-x_3,x_2-x_4)$ is a free resolution of $N$ over $A$. However, tensoring with $M$ computations became too hard, and I was not able to find the cohomology of the resulting complex.

Any ideas?


share|cite|improve this question

(I will omit the localization in this answer).

As $N$ has projective dimension $2$, you only need to worry about $i=0,1,2$. When $i=0$ the Tor is just tensor product, and tensoring with $N$ means setting $x_1=x_3$, $x_2=x_4$. Thus $M\otimes N = k[x_1,x_2]/(x_1^2,x_1x_2,x_2^2)$. The length is $3$.

For the rest, here is a key observation

(*) $(x_1-x_3)$ is a regular element on $M$.

We will use the short exact sequence $$0 \to L \to L \to N\to 0 $$ Here $L = A/(x_1-x_3)$ and the first map is multiplication by $(x_2-x_4)$. Because of (*), $\operatorname{Tor}_i(L,M)=0$ for $i>0$. So looking at the long exact sequence by tensoring with $M$ we immediately get $\operatorname{Tor}_2(M,N)=0$ and $\operatorname{Tor}_1(M,N)$ is the kernel of the map $$L\otimes M \to L\otimes M = M/(x_1-x_3)M=k[x_1,x_2,x_4]/(x_1^2,x_1x_2,x_1x_4,x_2x_4) $$ given by multiplication with $x_2-x_4$. This kernel is a vector space generated by the residue of $x_1$, so the length is $1$.

This is an example (simplest in some sense) where the naive count for multiplicity fails. Note that $M$ is not Cohen-Macaulay (this is not a coincidence).

share|cite|improve this answer

You can do this computation without thinking much, really.

The two generators of the ideal which defines $N$ form a regular sequence, so you can use a Koszul complex to projectively resolve $N$. Tensor it with $M$ and just compute the homology of the resulting complex.

share|cite|improve this answer
Mariano, the OP claimed he/she already tried it! That's why I had to find a simpler way (-: – curious Aug 1 '12 at 18:15
I am not the one who downvoted, by the way. – curious Aug 1 '12 at 18:40
I don't mind the downvote, really :-) My point is that at some point one needs to exercise the muscle which computes such homologies. If this particular example seems too hard it is just an indication that one's hardness-meter needs a readjustmente, and that is done by computing! :-) – Mariano Suárez-Alvarez Aug 1 '12 at 19:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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