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?
Rank of projective module
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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
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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0@KeenanKidwell Absolutely! – 2012-04-18