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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?

Thanks!

2 Answers 2

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(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).

2

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.

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    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! :-)2012-08-01