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Math Help - Rings of Fractions - Dummit and Foote Section 7.5

  1. #1
    Super Member Bernhard's Avatar
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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

     i : R \rightarrow Q  by    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  \in Q since

    i(a+b) =  \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

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

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

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

    but why (exactly!!) is this the case

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

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

    Can anyone clarify this for me?

    Peter
    Last edited by Bernhard; November 10th 2012 at 01:45 AM.
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    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)).
    Thanks from Bernhard
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    Super Member Bernhard's Avatar
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    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
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