Is it true that if $R$ is a domain with quotient field $K$ and $M$ is a finitely generated torsion-free $R$-module then $M\otimes_R K$ is isomorphic with $K^n$ for some $n$?
I know that the first is the localization of $M$ at $R-\{0\}$.
Thanks for any help!
Yes, this is true. In fact, you don't even need $M$ to be torsionfree or $K$ to be the field of fractions of $R$; all you need is for $K$ to be some field equipped with a ring-homomorphism $R\to K$. Given such a homomorphism, $K$ can be considered as an $R$-module, so we can form the tensor product $M\otimes_R K$. Furthermore, this tensor product is a $K$-module, defining $a\cdot(m\otimes b)=m\otimes ab$ for $a,b\in K$, $m\in M$. If $\{m_1,\dots,m_N\}$ is a set of generators for $M$ as an $R$-module, then $\{m_1\otimes 1,\dots,m_N\otimes 1\}$ will generate $M\otimes_R K$ as a $K$-module. Since $K$ is a field, this means $M\otimes_R K$ is a finite-dimensional $K$-vector space, and is isomorphic to $K^n$ for some $n\leq N$.