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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?

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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

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

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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
@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

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