Using Macaulay2 to Write the Canonical Module as a Quotient of a Free Module

Question

Let $S$ be a polynomial ring over a field. Let $I$ be a homogeneous parameter ideal of $S.$ Observe that $S/I$ is an Artinian local ring, so it is Cohen-Macaulay, and it is finitely generated as an $S$-module. Even more, the projection of $S$ onto $S/I$ maps the homogeneous maximal ideal of $S$ onto the maximal ideal of $S/I,$ hence there exists a canonical module of $S/I.$ Explicitly, it is $\operatorname{Ext}_S^{\dim S}(S/I, S).$ One can find this canonical module by writing a (finite) minimal free resolution of $S/I$ as an $S$-module; applying $\operatorname{Hom}_S(-, S)$; and taking cohomology.

Let $S = \Bbb Q[x_1, \dots, x_5]$ and $I = (x_1 x_3, x_1 x_4, x_2 x_4, x_2 x_5, x_3 x_5) + (x_i^2 \mid 1 \leq i \leq 5).$ (One might recognize this as the sum of the Stanley-Reisner ideal of the five-cycle and the ideal generated by the squares of all of the variables.) I am attempting to use Macaulay2 to express the canonical module of $S/I$ as a quotient of a free $S$-module. Unfortunately, when I use the command Ext^5(S^1/I, S^1) in Macaulay2, it produces the following output that I am unable to interpret.

cokernel {-7} | x_5 x_4 x_2 0 x_3 0 0 0 0 0 0 0 0 0 0 x_1 0 0 0 0 |

{-7} | 0 0 0 x_5 -x_4 x_3 x_2 0 0 x_1 0 0 0 0 0 0 0 0 0 0 |

{-7} | 0 0 0 0 0 0 0 x_5 x_3 -x_2 x_1 x_4 0 0 0 0 0 0 0 0 |

{-7} | 0 0 0 0 0 0 0 0 0 0 0 x_5 x_4 x_3 x_1 0 0 x_2 0 0 |

{-7} | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x_5 x_4 -x_3 x_2 x_1 |

If I am correct, {-7} refers to the degree of the generators as a graded homomorphism; however, I'm not sure how this matrix $A$ is acting. Considering that this is a $5 \times 20$ matrix with entries in $S,$ I believe that this is acting by left-multiplication on the column vectors of $S^{\oplus 20},$ but I am not sure what a vector there looks like.

I have searched extensively for an interpretation of this output in the Macaulay2 documentation, but I have not found anything to settle my question. Ultimately, if I can decipher the output $\operatorname{cokernel}(A),$ then I would like to use the fact that $\omega_{S/I} = \operatorname{Ext}_S^5(S/I, S) = S^{\oplus 5} / \operatorname{coker}(A)$ is the canonical module of $S/I$ to determine the idealization of $\omega_{S/I}$ over $S.$ I would appreciate any insight or suggestions.

Answer 1

I am not sure if this will clear your confusion, but let me just go through some of what you say in your post.

First, notice that if $\Delta$ is some abstract simplicial complex over a set $X$, then the Stanley--Reisner ideal $I_\Delta$ is generated by the square-free monomials $x_{i_1}\cdots x_{i_k}$ where $\{i_1,\ldots,i_k\}$ is not in $\Delta$. If you take the $5$-cycle to be $12,23,34,45,51$, then what you wrote is somewhat the opposite of this, though if you remove the squares you will get the SR-ideal for the cycle $13524$,

For the resolutions. You can ask M2 to tell you what the resolution of $I$ as an $S$ module looks like, and then ask for the transpose of the $5$th differential:

o53 = R

o53 : PolynomialRing

i54 : I = ideal(x1*x2,x2*x3,x3*x4,x4*x5,x5*x1,x1^2,x2^2,x3^2,x4^2,x5^2)

                                                  2    2    2    2    2
o54 = ideal (x1*x2, x2*x3, x3*x4, x4*x5, x1*x5, x1 , x2 , x3 , x4 , x5 )

o54 : Ideal of R

i55 : M = module R/I

o55 = cokernel | x1x2 x2x3 x3x4 x4x5 x1x5 x1^2 x2^2 x3^2 x4^2 x5^2 |

                             1
o55 : R-module, quotient of R

i56 : C = res M 

       1      10      25      31      20      5
o56 = R  <-- R   <-- R   <-- R   <-- R   <-- R  <-- 0
                                                     
      0      1       2       3       4       5      6

o56 : ChainComplex

i57 : (Hom(C,module R)).dd_(-4)

o57 = {-7} | x5    0   -x4 x3  0  -x2 x1 0  0   0   0   0  0  0  0  0   0  0   0  | 
      {-7} | 0  -x5 0   0   -x4 0  0   0  x3 0   -x2 x1  0  0  0  0  0   0  0   0  |  
      {-7} | 0  0   -x5 0   0   x4 0   0  0  -x3 0   -x2 x1 0  0  0  0   0  0   0  |  
      {-7} | 0  0   0   0   0   x5 0   0  0  0   0   0   0  x4 0  x3 -x2 x1 0   0  |   
      {-7} | 0  0   0   0   0   0  0   x5 0  0   0   0   0  0  x4 0  x3  0  -x2 x1 |
      --------------------------------------------------------------------------
              5       20
o57 : Matrix R  <--- R

Thus, you can see that the matrix that computes $\mathrm{Ext}^5_S(S/I,S)$ is the above, I assume the one that M2 gave you is obtained from this one by some elementary operations. At any rate, you compute this module as the quotient of $S^5$ by the image of this matrix.