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Let $M$ be a module over a local ring $(R,m)$, everything is finitely generated/presented. The fitting ideals, $I_j(M)$ carry a lot of information about the module. When do they actually determine the module?

For example, there is no hope for a positive answer unless the ring is local or graded. And the fitting ideals do determine the module over PID.

What is known about other cases? (at least, for regular local rings, for Gorenstein rings of low dimensions, for Cohen-Macaulay modules...)

Any strengthening of such "fitting type" invariants that determines the module?

upd. In view of W.Sawin's example, probably one should assume the module to be Cohen-Macaulay. Or, at least, some "purity" conditions on the support of $M$. For me the most interesting case is a Cohen-Macaulay module over hypersurface singularity. (alternatively, a module over a local ring, whose minimal presentation is a square matrix)

upd2. Well, the simplest example is: given $M$, let $A$ be its presentation matrix, i.e. $M=coker(A)$. Then $coker(A)$ and $coker(A^T)$ have the same fitting ideals, though the modules are non-isomorphic in general

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Why do you say "no hope unless the ring is local or graded"? There are non-local non-graded PIDs, such as $\mathbb Z$. –  Will Sawin Jun 15 '12 at 19:43

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up vote 4 down vote accepted

This should never really work in dimension at least two.

Let $x$ and $y$ be linearly independent elements of $m/m^2$.

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 kernels of the map $M \otimes R/m^2 \to M \otimes R/m$ have different dimensions as vector spaces over $R/m$ because there is a different number of relations, $1$ in the first case and $2$ in the second.

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Thanks! See upd. –  Dmitry Kerner Jun 15 '12 at 21:08

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