# Thread: Rings of Fractions - Dummit and Foote Section 7.5

1. ## Rings of Fractions - Dummit and Foote Section 7.5

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

2. ## Re: Rings of Fractions - Dummit and Foote Section 7.5

in a ring of fractions:

a/b = c/d iff ad = bc.

your expression for i(ab) is not quite correct, i represents the embedding of R in Q, and "b" isn't IN Q (only i(b) = bd/d).

often we can use b/1 to represent b in Q, but ONLY if 1 is an element of the multiplicative subset D (it may not be, and furthermore R might not even have a one).

so i(ab) = (abd)/d

while i(a)i(b) = (ad/d)(bd/d) = (abd2/d2)

now d(abd2) = abd3 = (abd)d2

(in other words, to check equality of fractions, we "cross-multiply"),

so i(ab) = (abd)/d = (abd2)/d2 = (ad/d)(bd/d) = i(a)i(b) (R must be a commutative ring for this to be true).

*****************

let me give an example of this kind of ring, that is not a field.

let R = 2Z, and let D = {2k: k > 0 in Z}. surely you can see D is multiplicatively closed (but lacks an identity).

what Q is, in this case, is all rational numbers that when put in "reduced form" have a denominator of a power of two

(this "reduced form" in the rationals may differ from the form in D, see below):

note we HAVE to write integers as 2k/2, simply because "1" isn't in D (or R).

note that for any d in D, we have:

(d2/d)(2/2d) = 2d2/(2d2) = 2/2, which is the identity of Q.

in other words, we've turned all of D into units.

note Q HAS a multiplicative identity, even though R DOESN'T.

Q is NOT a field, though, because 6 (or rather, the embedded image of 6, 12/2) has no inverse in Q.

perhaps even easier to see, the element 6/2 of Q likewise has no inverse:

(if it did, we would have: (6/2)(2k/d) = 2/2 so:

24k = 4d for some integer k, and some positive power of 2, d.

since this is an equation between two INTEGERS, we can cancel, leaving:

6k = d. but 3 divides 6k, and does not divide d (the only prime that divides d is 2)).

3. ## Re: Rings of Fractions - Dummit and Foote Section 7.5

Thanks for the clarification Deveno ... appreciate your help

Thanks also for the excellent example ... definitely increased my understanding of rings of fractions and D&F Section 7.5 Theorem 15 in particular

I wish the textbooks would give more examples like the one in your post, would be really helpful for those engaged in self-study, but ...

Peter