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!

  • $\begingroup$ The first thing one might try is to suppose you have a torsion element in the tensor module, and then see if it can be chased back to the original L. Did you take a look at that? $\endgroup$ – rschwieb Jul 2 '14 at 12:48
  • $\begingroup$ I can't tell either way without taking some time later to think about it :) In the meantime, could you add that to the original post above? It makes the question much more tempting to answer when the poster has come as far as you have in thinking about it. $\endgroup$ – rschwieb Jul 2 '14 at 15:08

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.


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