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Let $R$ be a commutative domain with fraction field $K$. It is known that $K_R$ is injective.

Now, if $M_R$ is a torsion-free module and we localize at $S=R-0$ we get $M⊗_RK=S^{-1}M⊇M$. My question is:

Why $M⊗_RK$ is a $K$- vector space, and what is a basis thereof? Thanks!

karparvar
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  • It's a vector space purely because you have a $K$-module structure by $\alpha\cdot(m\otimes \beta)=m\otimes(\alpha\beta)$. I don't think there is a nice intrinsic way of phrasing $\dim(M\otimes K)$. – Alex Youcis Aug 20 '14 at 08:03

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It is actually more general. Whenever $M$ is an $R$ module, and $S\subset R$ a multiplicative subset, $S^{-1}M$ is an $S^{-1}R$ module. One defines the $S^{-1}R$ action in the only possible manner, $$\frac{r}{s}\cdot\frac{m}{t}=\frac{rm}{st}.$$ Now, if $\{m_\alpha\}$ generate $M$ over $R$, then their images in $S^{-1}M$ generate it over $S^{-1}R.$

Amitai Yuval
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