I have a Dedekind domain $R$ with field of fractions $K$ and a non-zero prime ideal of $P$ of $R$. Let $L$ be a torsion-free $R$-module.
How can I show that $R_P\otimes_R L$ is torsion-free as an $R_P$-module, where $R_P$ is the localization of $R$ at $P$?
This is what I've tried:
Take a non-zero element $\sum_{i=1}^n x_i\otimes_R l_i\in R_P\otimes_R L$ which is killed by an element $r=a/b\in R_P$. If I write $x_i=r_i/s_i$ for each $i$ then I get $\sum_i(ar_i/bs_i)\otimes l_i=0$ which rewrites as $\sum_i(bs_i)^{-1}\otimes ar_il_i=0$. It's not obvious from this how I can deduce that $a=0$. Am I missing something easy?
Many thanks!
The tensor product $R_p\otimes_R L$ is "just" the localization of the module $L$ with respect to $R\setminus p$. So suppose
$ \frac{r}{s}\cdot\frac{m}{t}=0 $
for some $r\in R$, $s,t\in R\setminus p$, $m\in L\setminus 0$. Then by the definition of localization we have
$ urm=0 $
for some $u\in R\setminus p$. Hence $r=0$ by assumption about $L$.
As one sees, the only requirement for the statement to be true is that $R$ is a domain.