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I know that one can compute Fitting ideals of a finitely presented module (over a commutative ring with identity). However, are they the only invariants of such a module?

In other words, my question is: if two finitely presented modules have the same Fitting ideals, then are they isomorphic? Obviously the answer is yes if the base ring is a PID. If the answer is no, how is a strategy to prove that two f.p. modules with the same Fitting ideals are not isomorphic?

Unfortunately, base change is not useful in this problem.

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up vote 3 down vote accepted

The answer to your question is no:
The Fitting ideals of a finitely presented module $M$ over a ring $A$ do no not determine the module.

Indeed suppose the ring $A$ is connected (no idempotents $\neq0,1$).
Then $all$ finitely presented projective modules $M$ of rank $r$ have the same sequence of Fitting modules, namely
$$F^0(M)=F^1(M)=... =F^{r-1} (M)=0\subsetneq F^r(M)=F^{r+1}(M)=...=A $$

You can find the proof in Eisenbud's Commutative Algebra, Proposition 20.8, which Google books is kind enough to show you in its entirety, proof included.

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Dear Georges: Could you give an example of non-isomorphic finitely presented projective modules of same rank? Thank you in advance. (+1 of course!) – Pierre-Yves Gaillard Jan 7 '12 at 14:37
Here is a direct link to the proposition Georges refers to. – Pierre-Yves Gaillard Jan 7 '12 at 14:43
@Georges, thank you a lot! My question was quite stupid, sorry. To Pierre, take a non principal ideal in a Dedekind domain: for example, $A = \mathbb{Z}[\sqrt{-5}]$ and $I = (2, 1 + \sqrt{-5})$. – Andrea Jan 7 '12 at 14:53
Dear @Andrea: Thanks! By the way, I found your question very interesting (and upvoted it). – Pierre-Yves Gaillard Jan 7 '12 at 15:28
Dear @Andrea, your question was not stupid at all. On the contrary: although it is quite natural, I have never seen it mentioned in a book. Moreover it is definitely not trivial: I know no simpler way to answer it than to invoke the beautiful but somewhat unexpected theorem mentioned in the answer. ( Needless to say, I have upvoted it) – Georges Elencwajg Jan 7 '12 at 16:03

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