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Let $A$ be an integral domain and write $S=A-\{0\}$. Then the total ring of fractions $S^{-1}A$ of $A$ is an abelian field. Note that $\varepsilon:A\rightarrow S^{-1}A,\,a\mapsto a/1$, is an injective ring homomorphism.

Let $E,F$ be two vector $(S^{-1}A)$-spaces. We can define $A$-module structures on $E$ and $F$ using the homomorphism $\varepsilon$; let $\varepsilon_*(E)$ and $\varepsilon_*(F)$ denote the sets $E$ and $F$ with these $A$-module structures, respectively. There exists a unique $\mathbf{Z}$-linear surjection $$\varphi:\varepsilon_*(E)\otimes_A\varepsilon_*(F)\rightarrow E\otimes_{S^{-1}A}F$$ such that $\varphi(x\otimes_A y)=x\otimes_{S^{-1}A}y$ for $x\in E$ and $y\in F$. I want to show that $\varphi$ is injective as well.

Attempt:

Let $z\in E\otimes_{A} F$ such that $\varphi(z)$= 0. There exists $\xi\in\mathbf{Z}^{(E\times F)}$ such that $z=\sum_{(x,y)\in E\times F}\xi_{xy}(x\otimes_Ay)$. Then $$0=\sum_{(x,y)\in E\times F}\xi_{xy}(x\otimes_{S^{-1}A}y).$$ But $(x\otimes_{S^{-1}A}y)_{(x,y)\in E\times F}$ is not a basis of $E\otimes_{S^{-1}A}F$ so $\xi$ is not necessarily zero. Any suggestions?

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Show that for $E$ as in your question, $E\otimes_A S^{-1}A=E$, under the natural map.

Next, $$E\otimes_A F=E\otimes_A(S^{-1}A\otimes_{S^{-1}A} F)=(E\otimes_A{S^{-1}A})\otimes_{S^{-1}A} F=E\otimes_{S^{-1}A} F.$$

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