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If $R$ in an integral domain, $K$ its quotient field and $M$ an $R$ module, and $m \in M, k \in K$ can one show
$m\otimes k =0$ implies $m\otimes l =0$ for all $l\in K$ directly from the properties of a tensor product ?

Edit: So does $m\otimes 1=0 \rightarrow \exists r \in R(rm=0)$ ? How to prove this is now my main question. If we have this then, $m\otimes k=rm\otimes \frac{k}{r}=0$ and conversely, if $m\otimes k=0$ then multiply by the denominator to get $m\otimes s=0$ for $s \in R$ and thus $sm\otimes 1=0$. And then there is $r$ such that $rsm=0$ and as above this gives $m\otimes l=0$ all $l$.

This question arose from my failure to answer another question on this site, so what I am looking for is some insight into the Tensor product construction, thus an elementary answer showing how this arises from the tensor product definition is desired.

Also in general is there a general condition as to when $m \otimes n =0$ holds in a tensor product $M\otimes_R N$ ?

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You need $k\neq 0$. Then $m\otimes k$ is the equivalence class of the pair $(m,k)$, where $(m,k)\sim (m',k')$ if there exists $s\neq 0$ in $R$ s.t. $mks=m'k's$. Then $mks=0$ implies $ml(ks/l)=0$, i.e., $(m,l)=0$. – adrido Jun 6 '14 at 0:32
But $\frac{ks}{l}$ may not be in $R$. Also I dont see why the condition you give follows from the definition of tensor, which is either as a universal object or as the quotient of free module by relations. – Rene Schipperus Jun 6 '14 at 0:59
you're right, i shouldn't divide by $l$. but $mks=0\implies ml(ks)=0$ and $ks$ is in $R$. I was using the fact that tensoring with $K$ is the same thing as localization at the prime ideal $(0)$. – adrido Jun 6 '14 at 1:34
I just realized it's easier to go with the definition. See my answer below. – adrido Jun 6 '14 at 2:01
up vote 2 down vote accepted

Let $M_0$ be localization of $M$ at the prime $(0)\subset R$. Look at the bilinear map $\beta:M\times K\to M_0$ defined via $\beta(m,k)=mk$. This must factor through an $R$-linear map $\alpha:M\otimes K\to M_0$. Now if $m\otimes k=0$, then $\beta(m,k)=0$ by linearity of $\alpha$. Suppose $k=r/s$ and write $\beta(m,r/s)=0$ for $r,s\in K$. Then, using bilinearity of $\beta$ twice, you get $\beta(m,1)=0$. Since $\beta(m,1)=m$, you get $m=0$ and thus $m\otimes 1=0$. Now if $m\otimes 1=0$ in $M_0$, it means $(m,1)\sim (0,1)$, i.e., $\exists r\neq 0$ in $R$ s.t $mr=0$.

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But $M$ is only an $R$ module, multiplication by $K$ is not defined. I think what I am dont understand is why, $m\otimes 1=0\rightarrow \exists r \in R (rm=0)$. – Rene Schipperus Jun 6 '14 at 2:26
yes, we first need to make $M$ a $K$-module. So I need to first localize at $(0)$ (this is the same thing as tensoring with $K$ but I'm not using this). Now if $m\otimes 1=0$, then $(m,1)\sim (0,1)$ in the localization of $M$. this gives $r\in R$ s.t. $mr=0$. – adrido Jun 6 '14 at 3:36
Sorry, I still dont see this, $M\otimes K$ is the solution to a universal problem if $f:M\times K\rightarrow L$ is a bilinear map to an $R$-module $L$, then $f$ factors through, $M\otimes K$. The equivalence relation you give is more like localization. How does it come out of the universal problem ? – Rene Schipperus Jun 6 '14 at 3:41
yes, but you can pick $L$ to be localization of $M$ at $(0)$. Then the bilinear map I defined must still factor through $M\otimes K$. – adrido Jun 6 '14 at 3:43
Oh, OK that is it! I could not figure out what to use for $L$ ! Take $L$ to be the localisation then with your $f$, $f(m,1)=m \in M_0$, now if $m\otimes 1=0$ then $m=0$ in $M_0$ and this implies that there is $r \in R$ such that $rm=0$ in $R$. Thank-you so much. – Rene Schipperus Jun 6 '14 at 3:49

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