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.


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.

  • $\begingroup$ Dear Georges: Could you give an example of non-isomorphic finitely presented projective modules of same rank? Thank you in advance. (+1 of course!) $\endgroup$ – Pierre-Yves Gaillard Jan 7 '12 at 14:37
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    $\begingroup$ Here is a direct link to the proposition Georges refers to. $\endgroup$ – Pierre-Yves Gaillard Jan 7 '12 at 14:43
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    $\begingroup$ @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 $I = (2, 1 + \sqrt{-5})$. $\endgroup$ – Andrea Jan 7 '12 at 14:53
  • $\begingroup$ Dear @Andrea: Thanks! By the way, I found your question very interesting (and upvoted it). $\endgroup$ – Pierre-Yves Gaillard Jan 7 '12 at 15:28
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    $\begingroup$ 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) $\endgroup$ – Georges Elencwajg Jan 7 '12 at 16:03

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