When fitting ideals determine the module? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:15:41Z http://mathoverflow.net/feeds/question/99738 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99738/when-fitting-ideals-determine-the-module When fitting ideals determine the module? Dmitry Kerner 2012-06-15T19:16:07Z 2013-01-04T13:16:21Z <p>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? </p> <p><a href="http://math.stackexchange.com/questions/97148/finitely-presented-modules" rel="nofollow">For example,</a> 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. </p> <p>What is known about other cases? (at least, for regular local rings, for Gorenstein rings of low dimensions, for Cohen-Macaulay modules...) </p> <p>Any strengthening of such "fitting type" invariants that determines the module?</p> <p>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)</p> <p>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</p> http://mathoverflow.net/questions/99738/when-fitting-ideals-determine-the-module/99741#99741 Answer by Will Sawin for When fitting ideals determine the module? Will Sawin 2012-06-15T19:56:05Z 2012-06-15T20:13:41Z <p>This should never really work in dimension at least two.</p> <p>Let $x$ and $y$ be linearly independent elements of $m/m^2$.</p> <p>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.</p>