Suppose $R$ has two elements
Let $x$ and $y$ that satisfy no relations beyond the obvious and are non-zero-divisors. Two generators of $m/m^2$ in any ring be linearly independent elements of dimension at least $2$ should do.m/m^2$. Consider the modules$R^2/(ya-xb)$and$R\oplus R/(x,y)$. They both have the same fitting ideals$I_0(M)=0$,$I_1(M)=(x,y)$,$I_2(M)=1$. These modules are nonisomorphic because the first module can be embedded into kernels of the map$R$($a\to x$, M \otimes R/m^2 \to M \otimes R/m$ have different dimensions as vector spaces over $b\to y$) and so R/m$because there is no elements are$x$-torsion or a different number of relations,$y$-torsion while 1$ in the other module has $x$-torsion first case and $y$-torsion.2$in the second. 1 This should never really work in dimension at least two. Suppose$R$has two elements$x$and$y$that satisfy no relations beyond the obvious and are non-zero-divisors. Two generators of$m/m^2$in any ring of dimension at least$2$should do. Consider the modules$R^2/(ya-xb)$and$R\oplus R/(x,y)$. They both have the same fitting ideals$I_0(M)=0$,$I_1(M)=(x,y)$,$I_2(M)=1$. These modules are nonisomorphic because the first module can be embedded into$R$($a\to x$,$b\to y$) and so is no elements are$x$-torsion or$y$-torsion while the other module has$x$-torsion and$y\$-torsion.