2. The space of transformation matrices

Throughout this paper we assume R to be a finitely generated, local (or graded) k-algebra without zerodivisors and any (any graded) module M as given as cokernel of a map of free modules. The presentation matrix of this map is assumed to represent a minimal system of generators of its image and is denoted by A(M) or simply A.

The case of a graded module is handled analogously. Therefore, we restrict ourself to the local settings.

Let two modules M and M' be given by their representation matrices A and A'. Assume the modules to be isomorphic. There are two quadratic matrices U and V such that UAV=A'. By the above properties of the representations both matrices have to be invertible, $U\in Gl_m(R)$ and $V\in Gl_m(R)$.

Now, let's look from the other side: Starting with two representation matrices A and A' a necessary condition for an isomorphism of the represented modules M and M' is an identical size of both matrices. Denote this size by $mxn,\ m,n\in {\bf N}$. The sufficient condition for an isomorphism is again the existence of matrices $U\in Gl_m(R)$ and $V\in Gl_n(R)$ with UAV=A'. That means, we have to resolve the equation

$$XA-A'Y=0,\ X\in M_m(R),\ Y\in M_n(R)$$ (1)

and to look for a pair (X₀,Y₀) of regular matrices among all solutions of (1).

First, we determine the module of all possible transformations $(X,Y)\in M_m(R) x M_n(R)$. Let us consider the matrix A as map

$$\phi(A): Hom(R^m,R^m)\longrightarrow Hom(R^n,R^m)$$

just by multiplying with A from the right. Analogously, we set

$$\psi(A'): Hom(R^n,R^n)\longrightarrow Hom(R^n,R^m)$$

to be the map induced by the multiplication with A' from the left. The map

$$\Phi_{(A,A')}: Hom(R^m,R^m) x Hom(R^n,R^n) \longrightarrow Hom(R^n,R^m)$$

given by $\Phi_{(A,A')}(X,Y)=XA-A'Y$ is a well defined module homomorphism and its kernel is the module Tr_(A,A') of possible transformations. Tr_(A,A') can be computed in a single syzygy computation.