Rank of a module - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:28:24Z http://mathoverflow.net/feeds/question/73903 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73903/rank-of-a-module Rank of a module javad 2011-08-28T17:16:59Z 2011-08-28T17:42:51Z <p>I have seen the definition of a module,not neccessary free, the alternatin sum of free modules in a free resolution of that module. it's clear that when the module is free our definition Coincide the rank of free module. but why this definition is Well-defined?</p> http://mathoverflow.net/questions/73903/rank-of-a-module/73906#73906 Answer by unknown (google) for Rank of a module unknown (google) 2011-08-28T17:41:13Z 2011-08-28T17:41:13Z <p>We have a ring $R$, a module $M$ over $R$, and a (finite length) free-resolution of $M$: $$\cdots\to R^{\oplus n_3}\to R^{\oplus n_2}\to R^{\oplus n_1}\to R^{\oplus n_0}\to M\to 0$$ At any prime ideal $\mathfrak p$ of $R$, you can talk about the rank of $M$ over $\mathfrak p$. Concretely, this is just the dimension of $\operatorname{Frac}(R/\mathfrak p)\otimes_RM$ as a vector space over $\operatorname{Frac}(R/\mathfrak p)$ (where $\operatorname{Frac}$ denotes fraction field).</p> <p>If we tensor the free resolution with $\operatorname{Frac}(R/\mathfrak p)$, then we get a complex: $$\cdots\to\operatorname{Frac}(R/\mathfrak p)^{\oplus n_2}\to\operatorname{Frac}(R/\mathfrak p)^{\oplus n_1}\to \operatorname{Frac}(R/\mathfrak p)^{\oplus n_0}\to\operatorname{Frac}(R/\mathfrak p)\otimes_RM\to 0$$ This might not be exact, since $R/\mathfrak p$ may not be flat over $R$. Thus we cannot necessarily say that <code>$\dim_{\operatorname{Frac}(R/\mathfrak p)}\operatorname{Frac}(R/\mathfrak p)\otimes_RM=\sum_{i=0}^\infty(-1)^in_i$</code>.</p> <p>There is however a case when $R/\mathfrak p$ is flat over $R$, namely when $\mathfrak p=(0)$ (supposing $R$ is a domain). Then we do have equality: $$\dim_{\operatorname{Frac}(R)}\operatorname{Frac}(R)\otimes_RM=\sum_{i=0}^\infty(-1)^in_i$$ Thus if $R$ is a domain, then your notion of rank is well-defined and does have a nice interpretation: it's just the dimension over the fraction field of $R$ of $M$ tensored with the fraction field. In the scheme-theoretic sense, this is the "generic" rank of $M$.</p>