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Suppose $R$ is a domain and let $Q$ be its field of fractions. Then $ 0 \to R \to Q $ is exact. Now suppose that $M$ is a torsion free $R$ module.

Is it necessary that $ 0 \to M \to Q \otimes M $ is exact ?

This is true when $R$ is a PID. But I am not sure whether assuming that it is a domain suffices. If it is not please give a counter example. Thanks

user90041
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    Is $M$ an $R$-module ? – WLOG May 06 '14 at 10:15
  • Anyway try to see this http://math.stackexchange.com/questions/27150/example-of-modules-that-are-projective-but-not-free-torsion-free-but-not-free/27210#27210 – WLOG May 06 '14 at 10:19
  • @WLOG Yes, $M$ is an $R$ module. Thanks for pointing it out, I did not indicate that earlier and have now edited the question. I looked at the link but sorry I am unable to see how that would answer my question. I am not asking whether torsion free implies flat or not for general domains which are not PIDs, just asking whether injection is preserved in this one particular case. – user90041 May 06 '14 at 10:33

2 Answers2

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Yes, $R$ being an integral domain suffices. See Corollary 4.27 of http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf.

KCd
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Maybe this works: It's easy to see that both $Q(R) \otimes_{R}(-)$ and $(R-0)^{-1}(-)$ are two functors that works as left-adjoint to the forgetful functor from $Q(R)$ modules to $R$ modules. This immediately tells you that the two functors are isomorphic. So this tells you that $Q(R) \otimes_{R}(M)$ and $(R-0)^{-1}(M)$ are naturally isomorphic. Now you have by construction that $\frac{m}{1}=0$ in $(R-0)^{-1}M$ iff there is $a \neq 0$ in R such that $am=0$. That is, iff m is a torsion element. So for torsion free elements, your map is injective. Hope it helps and hope it works

Water
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