In the paper “Commutative non-Koszul Algebras having a linear resolution of arbitrarily high order”, Roos gave an example of a family of algebras $S_\alpha$, one for each integer $\alpha\geq 2$, in which the resolution of $k$ (the coefficient ring of characteristic zero) over $S_\alpha$ is linear for for $\alpha$-many steps and non-linear in homological degree $\alpha+1$ (he also gave examples in positive characteristic). The rings are
$$S_\alpha=k[x,y,z,u,v,w]/(x^2,xy,yz,z^2,zu,uv,vw,w^2, xz+azw-uw,zw+xu+(a-2)uw)$$
I find it fascinating that it’s possible to create a family with this homological behavior in which the coefficients of the polynomials depend on the parameter in such a simple way. The relationship between homological behavior and coefficients of the polynomials is usually difficult to so precisely determine.
At first glance, the relations look pretty random and it seems impossible that anyone one could come up with examples like these, but their construction is surprisingly natural, coming from taking the ring $k[x,y]/(x^2,xy)$ (the smallest quadratic non-ci).
I was curious if a smaller example could be constructed by trying the “obvious”: drop each of the relations, one at a time, to see if the quotient by the ten remaining polynomials still exhibited the desired behavior. Here’s the Macaulay2 code (note that I use $\mathbb{Z}/101$ instead of a field of characteristic zero, but this will only affect the resolution of the residue field in higher homological degrees than those we’re interested in):
k=ZZ/101;
alpha = 2;
Q = k[x,y,z,u,v,w];
relSet = {x^2,x*y, y*z, z^2,z*u, u^2, u*v, v*w,w^2, x*z + alpha*z*w - u*w,z*w+ x*u + (alpha - 2)*u*w};
subs = subsets(relSet, 10);
for relset in subs list {
R = Q/(ideal relset);
F = res (coker vars R,LengthLimit=>5);
betti F}
For brevity, I won’t print all the Betti tables here, but the key observation to make is that, as we change $\alpha$ from $2$ to $3$, we still see non-linearity in degree $3$ in every single Betti table. This shows that every relation is essential for this example to work. For example, if we drop $x^2$, we get the following table for $\alpha=2$:
0 1 2 3 4 5 6 7
total: 1 6 25 96 364 1385 5287 20207
0: 1 6 25 92 320 1084 3625 12046
1: . . . 4 44 301 1646 7905
2: . . . . . . 16 256
And the following for $\alpha=3$
0 1 2 3 4 5 6 7
total: 1 6 25 96 360 1341 4986 18529
0: 1 6 25 95 349 1266 4570 16465
1: . . . 1 11 75 415 2048
2: . . . . . . 1 16