# Vector spaces over an integral domain and the canonical isomorphism between the tensor products

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?

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