I am reading Dummit and Foote Section 7.5 Rings of Fractions - and in particular Theorem 15 - see attached.

I am at ease with the proof of Theorem 15 (see attachment) down to where D&F state near the bottom of page 262:

"Next we embed R into Q by defining

$\displaystyle i : R \rightarrow Q $ by $\displaystyle i : r \rightarrow \frac{rd}{d} $"

D&F then state that i is a ring homomorphism

I am OK with i(a + b) = i(a) + i(b) for a, b $\displaystyle \in $ Q since

i(a+b) = $\displaystyle \frac {(a + b)d}{d} = \frac{ad + bd}{d} = \frac{ad}{d} + \frac{bd}{d} = i(a) + i(b)$

However, my problem is that I am slightly uncomfortable with demostrating that i(ab) = i(a)i(b) since we have

$\displaystyle i(ab) = \frac{(ab)d}{d} = \frac{ad}{d} (b) $ {is this actually correct?}

and $\displaystyle i(a)i(b) = (\frac{ad}{d}) (\frac{bd}{d})$

So it looks as if we need to demostrate that $\displaystyle b = \frac{bd}{d}$

but why (exactly!!) is this the case

Now we could write $\displaystyle b = b.1 = b(\frac{d}{d}) $

It looks as if $\displaystyle b(\frac{d}{d}) = \frac{bd}{d}$ ... but why exactly??

Can anyone clarify this for me?

Peter