Consider a polynomial ring S=\mathbb{Z}[x_0,\dots,x_r], and a quotient Q=S^p/N where N is a homogeneous submodule generated in degrees at most d. Suppose that Q_d is free of rank n. We then have a short exact sequence 0\to N_d\to S_d^p\to Q_d\to 0 where also N_d is free. Thus, N_d\otimes S_1\to S^p_{d+1}\to Q_{d+1}\to 0 is a free presentation of Q_{d+1}. If the basis of Q_d can be chosen as a so called Gotzmann set, then Q_{d+1} is free of rank n if and only if the (n1)'th Fitting ideal of Q_{d+1} is zero.
As an example, we consider a quotient Q of S^2=\mathbb{Z}[x_0,x_1]^2 such that Q_1 is free of rank 1. As S^2_1=\mathbb{Z}<xe_1,ye_1,xe_2,ye_2> is a free \mathbb{Z}^4module we have that Q_1 is the cokernel of a (1\times 4)matrix.



Suppose that Q_1 has a basis given by xe_1, corresponding to the 0'th column. We check that this is a Gotzmann set, and calculate a free presentation of Q_2.




N_2 is the matrix corresponding to the map N_1\otimes S_1\to S^2_2, so Q_2 is the cokernel. We want to calculate the (n1)'th Fitting ideal of N_2 with n=1.

Thus, the obstruction for Q_2 to be free of rank 1 is the equation a_2a_3a_4.
A result is that the Quot scheme of rank n quotients of \mathcal{O}^p is cut out by a single (n1)'th Fitting ideal in the Grassmannian of locally free rank n quotients of a push forward of \mathcal{O}(d)^p for d\ge n.
In the case above, we have the Grassmannian Gr(1,4)=\mathbb{P}^3, and the Quot scheme is given by a (n1)'th Fitting ideal. All of the above calculations can also be done directly by:

We can calculate much bigger examples with these function than with the ordinary Fitting ideal function. As an example, we consider the following with rank 2 quotients of S^2.








Note that our method is a bit faster for this small example, and for rank 2 quotients of S^3=\mathbb{Z}[x,y]^3 the time difference is massive.