Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
Because of Schanuel's 'long' lemma. –  Mariano Suárez-Alvarez Aug 28 '11 at 17:27
    
javad: you should probably edit the question to include more relevant tags---'writing' is rather unconnected to your question. Also, there are other sites where this kind of questions will be more at home, like math.stackexchange.com and the others listed in the FAQ, mathoverflow.net/faq –  Mariano Suárez-Alvarez Aug 28 '11 at 17:39
add comment

1 Answer 1

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).

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 $\dim_{\operatorname{Frac}(R/\mathfrak p)}\operatorname{Frac}(R/\mathfrak p)\otimes_RM=\sum_{i=0}^\infty(-1)^in_i$.

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 $M$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.