Prove $(F/R) \otimes_R (F/R)=0$ for $R$ an integral domain and $F$ its field of fractions

Question

Suppose $R$ is an integral domain and $F=\mathrm{Frac}(R)$ is considered as an $R$-module. I want to prove that
$$(F/R) \otimes_R (F/R)=0.$$

My attempt:

I first considered the exact sequence $$R \rightarrow F \rightarrow F/R \rightarrow 0$$ and from this I obtained the exact sequence $$F/R \otimes_R R \overset{f}{\rightarrow} F/R \otimes_R F \overset{g}{\rightarrow} F/R \otimes_R F/R \rightarrow 0$$ If I can show that Im$f= F/R\otimes_R F$ then since Im$f=\ker g$ $$(F/R) \otimes_R (F/R)=0$$ but $F/R\otimes_R R \cong F/R$. This implies I need to show that $$F/R \cong F/R\otimes_R F$$ I'm not quite sure if they are really isomorphic to each other.

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Complete the proof: Consider the map $\left(\frac{b}{a}+R \right) \otimes \frac{b'}{a'} \mapsto \frac{b}{a}+R$ for $a, a', b, b' \in R$. It is clearly surjective and $R$-linear. Now for $a'', b'' \in R$, we have $$\left(\frac{b}{a}+R \right) \otimes \frac{b'}{a'}=a\left(\frac{b}{a}+R \right) \otimes \frac{b'}{aa'}=0\otimes \frac{b'}{aa'}=0=0\otimes \frac{b''}{aa''}=a\left(\frac{b}{a}+R \right) \otimes \frac{b''}{aa''}=\left(\frac{b}{a}+R \right) \otimes \frac{b''}{a'}$$ Hence the map is injective.

Answer 1

This is covered in Cartan and Eilenberg, Homological Algebra, Chapter VII, Proposition 1.8. The proof goes like this.

Let $a\otimes b$ be an element of your tensor product. $F/R$ is a torsion $R$-module so there is a nonzero $r \in R$ for which $rb=0.$ But $F/R$ is also a divisible $R$-module, that is, its elements are divisible by nonzero elements of $R$. Hence, there is an element $a'$ of $F/R$ such that $a=ra'.$ So, $a\otimes b = (ra')\otimes b = a' \otimes rb = a' \otimes 0 = 0.$ Hence, your tensor product is $0$.

Answer 2

Let $x,y\in F $; then there exists $a,b\in R $ and $s,t\in R\setminus\{0\} $ such that $x=\frac as$ and $y=\frac bt $. Then \begin{align} (x+R)\otimes (y+R)&=\left(\frac as+R\right)\otimes\left(\frac bt+R\right)\\ &=\left(\frac{at}{st}+R\right)\otimes\left(\frac bt+R\right)\\ &=\left(\frac{a}{st}+R\right)\otimes\left(\frac{bt}t+R\right)\\ &=\left(\frac{a}{st}+R\right)\otimes(b+R)\\ &=0 \end{align}