I am working through Theorem 15 (Rings of Fractions) of Dummit and Foote Chapter 7 (see attachment)

I am unable to completely follow the details of the argument for the uniqueness property of Q - see page 263 of D&F (see attachment) where D&F's argument is as follows:

"It remains to establish the uniqueness propoerty of Q.

Assume $\displaystyle \phi \ : \ R \rightarrow S $ is an injective ring homomorphism such that $\displaystyle \phi (d)$ is a unit in S for all $\displaystyle d \in D$.

Extend $\displaystyle \phi $ to a map $\displaystyle \Phi \ : \ R \rightarrow S $ by defining $\displaystyle \Phi (r d^{-1}) = \phi(r) (\phi(d))^{-1} $ for all $\displaystyle r \in R, d \in D $.

The map $\displaystyle \Phi $ is well defined since $\displaystyle r d^{-1} = s e^{-1} $ implies $\displaystyle re = sd$ , so $\displaystyle \phi (r) \phi (e) = \phi(s) \phi(d) $ and then

$\displaystyle \Phi (r d^{-1} ) = \phi(r) {\phi(d)}^{-1} \phi(s) {\phi(e)}^{-1} = \Phi (se^{-1} )$.

It is straightforward to check that $\displaystyle \Phi $ is a ring homomorphism - details left as an exercise.

Finally, $\displaystyle \Phi $ is injective because $\displaystyle r d^{-1} \in ker \ \Phi $ implies $\displaystyle r \in ker \ \Phi \cap R = ker \ \phi $

Since $\displaystyle \phi $ is injective ths forces r and hence $\displaystyle r d ^{-1} $ to be zero.

This completes the proof"

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My problems are as follows:

1. I cannot follow the argument that $\displaystyle r d^{-1} \in ker \ \Phi $ implies $\displaystyle r \in ker \ \Phi \cap R = ker \ \phi $

Can someone please give an explicit and formal set of steps that establishes this.

2. in the statement of Theorem 15 on page 261 (see attachment) D&F state that Q is unique and that the ring Q is the "smallest" ring containing R in which all elements of D become units

Can someone please explain how the above proof actually establishes these facts.

Peter