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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

by "

D&F then state that i is a ring homomorphism

I am OK with i(a + b) = i(a) + i(b) for a, b Q since

i(a+b) =

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

{is this actually correct?}

and

So it looks as if we need to demostrate that

but why (exactly!!) is this the case

Now we could write

It looks as if ... but why exactly??

Can anyone clarify this for me?

Peter

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) = (abd^{2}/d^{2})

now d(abd^{2}) = abd^{3} = (abd)d^{2}

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

so i(ab) = (abd)/d = (abd^{2})/d^{2} = (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 = {2^{k}: 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:

(d^{2}/d)(2/2d) = 2d^{2}/(2d^{2}) = 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)).

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