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).

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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)).