Let $S$ be a domain. I want to determine whether or not, every element of $\text{Frac(S)}\otimes_S M$ is a simple tensor, where $M$ is any $S$-module.
I couldn't produce a tensor that is not pure in specific cases. I feel that every tensor is indeed pure, but I'm unable to see why. Any help is appreciated.
On
There is an important general fact which makes this result quite clear. Let $S \subseteq R$ be a multiplicative submonoid of a commutative ring. Let $M$ be an $R$ module. Then $R[S^{-1}] \otimes_R M \cong M[S^{-1}]$. Moreover, this isomorphism is natural in $M$. The map $M[S^{-1}] \longrightarrow R[S^{-1}] \otimes_R M$ is $\frac{m}{s} \mapsto \frac{1}{s} \otimes m$. The reverse map is $\frac{r}{s} \otimes m \mapsto \frac{rm}{s}$. These maps are well define as they are defined by the universal property of localization and the tensor product respectively. One can easily check that they are inverses and that this is natural (although this latter fact is irrelevant for the purposes of this result). Then in particular, the map $\frac{m}{s} \mapsto \frac{1}{s} \otimes m$ is onto, so every element here is a simple tensor.
Indeed every tensor is pure; this can be shown by expressing an arbitrary tensor as a finite sum of pure tensors, and then writing the coefficients from $\operatorname{Frac}S$ in such a way that they all have the same denominator.
Let $Q=\operatorname{Frac}S$. Every element $x\in Q\otimes M$ is of the form $$x=\sum_{i=1}^nq_i\otimes m_i,$$ for some $q_i\in Q$ and $m_i\in M$. Every $q_i\in Q$ is of the form $q_i=\tfrac{a_i}{b_i}$ for some $a_i,b_i\in S$ and hence $$q_i\otimes m_i=\tfrac{a_i}{b_i}\otimes m_i=\tfrac{1}{b_i}\otimes a_im_i.$$ Let $p=\prod_{i=1}^n\tfrac{1}{b_i}\in Q$ and for each $i$ let $c_i:=\prod_{j=1, j\neq i}^nb_j\in S$. Then $$q_i\otimes m_i=\tfrac{1}{b_i}\otimes a_im_i=pc_i\otimes a_im_i=p\otimes a_ic_im_i.$$ It follows that for $m:=\sum_{i=1}^na_ic_im_i$ we indeed have $$x=\sum_{i=1}^nq_i\otimes m_i=\sum_{i=1}^np\otimes a_ic_im_i=p\otimes\sum_{i=1}^na_ic_im_i=p\otimes m.$$