7
$\begingroup$

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?

  • 1
    What properties would you expect such a definition to have?2012-04-18
  • 0
    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 $R$.2012-04-18
  • 1
    @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?2012-04-18

1 Answers 1

15

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 $K=\mathrm{Quot}(R)$.

  • 0
    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 $\mathrm{rk}$.2012-04-18
  • 2
    @ThomasAndrews, I suppose $\tilde M$ is the coherent sheaf on $\text{Spec} R$ associated to $M$2012-04-18
  • 3
    @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).2012-04-18
  • 0
    @KeenanKidwell Absolutely!2012-04-18