Computing intersection multiplicity using Tor - explicit example

(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).

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.