Let $R$ be an integral domain with a field of fractions $F$. Show that if $M$ is a $R$-module, then:
$1)$ $M_F=F\otimes_R M$ is a divisible $R$-module
$2)$ $x\rightarrow 1\otimes x$ is an essential monomorphism of $M$ into $M_F$ if $M$ is torsion-free.
Attempts: I have read over several of the other posts to piece the above together, as well as other resources, but I am still having trouble for both parts of this particular question. I was focused on the following: Let $R$ be an integral domain and let $S=R\setminus 0$. Then $Tor(R)=0$ such that $S^{-1}R$ is the field of fractions of $R$. Any help is greatly appreciated.
By definition an $R$-module $A$ is divisible if for every $x\in A$ and every non-zero divisor $r\in R$ there exists $y\in A$ such that $x=ry$. As $F$ is the field of fractions of $R$ and $M_F=F\otimes_RM$, every $x\in M_F$ is of the form $$x=\sum_{i=1}^nf_i\otimes m_i=\sum_{i=1}^n\frac{r_i}{s_i}\otimes m_i,$$ where $m_i\in M$ and $f_i\in F$ and $r_i,s_i\in R$ with $s_i\neq0$, because $R$ is an integral domain. It follows that for every non-zero $r\in R$ we have $$y:=r\cdot\sum_{i=1}^n\frac{r_i}{rs_i}\otimes m_i,$$ satisfies $x=ry$, so $M_F$ is divisible.
It is clear that the map $$M\ \longrightarrow\ M_F:\ x\ \longmapsto\ 1\otimes x,$$ is a monomorphism, and its image (which I will also denote $M$) is an essential submodule of $M_F$ because if $N\subset M_F$ is a submodule of $M_F$ such that $M\cap N=0$, then as before, for every $n\in N$ we can write $$n=\sum_{i=1}^k\frac{r_i}{s_i}\otimes m_i,$$ for some $k\in\Bbb{N}$ and $r_i,s_i\in R$ with $s_i\neq0$ and $m_i\in M$. Clearing denominators and setting $t_i:=\frac{r_i}{s_i}\prod_{i=j}^ks_j\in R$ we find that $$(s_1\cdots s_k)n=\sum_{i=1}^kt_i\otimes m_i=\sum_{i=1}^k1\otimes t_im_1=1\otimes\sum_{i=1}^kt_im_i,$$ and so $(s_1\cdots s_k)n=0$. Because $M$ is torsion-free it follows that $n=0$, and hence that $N=0$. This means $M$ is an essential submodule of $M_F$.