Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We define the rank of free module as the number of elements on the basis of free module. It may be infinity. How do we define the rank of projective module?

share|improve this question
1  
What properties would you expect such a definition to have? –  Thomas Andrews Apr 18 '12 at 5:52
    
For an arbitrary module $M$ over an integral domain $R$, we define the rank of $M$ to be the dimension over $k$ of $k\otimes_RM$, where $k$ is the quotient field of $R$. –  Bruno Joyal Apr 18 '12 at 6:12
    
@Bruno I would be curious (I haven't thought about it) if we could define the rank $M$ to be the rank of the smallest free module which has $M$ as a direct summand. It seems to work fine for well-behaved rings such as integral domains. Not sure in general. Any ideas? –  Alex Youcis Apr 18 '12 at 17:17
add comment

1 Answer 1

The rank of a projective module $M$ over $R$ is the function $\mathrm{rk} : \mathrm{Spec}(R) \to \mathrm{Card}$, $\mathfrak{p} \mapsto \mathrm{dim}_{\mathrm{Quot}(R/\mathfrak{p})}(M \otimes_R \mathrm{Quot}(R/\mathfrak{p}))$. This is the dimension of the fiber of $\tilde{M}$ at $\mathfrak{p}$. One can show that if $M$ is finitely generated, then this rank function is locally constant (without any finiteness condition this may fail). In fact, then $\tilde{M}$ is locally free of finite rank. In particular, if $\mathrm{Spec}(R)$ is connected ($\Leftrightarrow$ $R$ has only the trivial idempotents $0,1$), this function is constant. Then you have just one rank $\mathrm{rk}(M)\in \mathbb{N}$. In particular, when $R$ is an integral domain, we have $\mathrm{rk}(M) = \mathrm{dim}_K(M \otimes_R K)$, where $K=\mathrm{Quot}(R)$.

share|improve this answer
    
What is $\tilde{M}$? Presumably, $\mathrm{Quot}$ is the quotient field of an integral domain? Also, you are missing a parenthesis at the end of the definition of $\mathrm{rk}$. –  Thomas Andrews Apr 18 '12 at 13:23
    
@ThomasAndrews, I suppose $\tilde M$ is the coherent sheaf on $\text{Spec} R$ associated to $M$ –  Bruno Joyal Apr 18 '12 at 17:59
2  
@BRuno You mean quasi-coherent. There's no reason for the sheaf attached to $M$ to be coherent (if $M$ is finitely generated and $R$ is Noetherian, then it will be, but not necessarily otherwise). –  Keenan Kidwell Apr 18 '12 at 18:12
    
@KeenanKidwell Absolutely! –  Bruno Joyal Apr 18 '12 at 18:15
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.