Z^n-graded resolutions and Z^n-graded Betti numbers

[Martina]

Hi,

I would like to compute a Z^n-graded resolution and the Z^n-graded
Betti numbers of a homogeneous
monomial ideal. I read on the Macaulay webpage that this should be
possible but only found out how to compute a Z-graded resolution. Does
anyone know if
there an easy way to do the computations for the multi-graded case?
It would be great if someone could help me with this problem.

Best,

Martina

[David Cook II]

Martina,

This may not be the best way, but I've found that you can simulate
ZZ^n grading by using, for example:

R = QQ[a,b,c,d,e, Degrees => apply(5, i->rotate(-i,
{1,0,0,0,0}))];

Then the resolution and Betti tally of I are ZZ^n graded. (This is
example 1.25 from Miller & Sturmfels "Combinatorial Commutative
Algebra".)

C = res ideal(d*e, a*b*e, a*c*e, a*b*c*d);
B = betti C;

Notice, however, if you print B it's the ZZ-graded tally you will
see. You can use

peek B

to see the ZZ^n-graded Betti numbers. You can also use

poincare B

to see the Poincare polnomial of the ideal (example 1.14 from Miller &
Sturmfels).

Hope this helps,
David

[Martina]

Hi David,

thanks a lot.
This indeed helps and works.

Best,

Martina

[mike stillman]

I would suggest the following:

R = QQ[a..e, DegreeRank=>5]

This creates a ring with ZZ^5 grading, and the variables have the
desired degrees. Check the degrees of the variables:

degree a
degree d

or all of them:

(gens R)/degree

This really does more than "simulate" ZZ^5 grading, as all modules and
ideals over this ring will have a ZZ^5-grading. The rest of David's
post works verbatim, as his ring is the same as this one.

The betti function is also able to display other "slices" of the
degrees of the free resolution too. Try
viewHelp betti
or
viewHelp (betti, BettiTally)
for more on that.

-- Mike