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Math Help - Quotient Ring

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    Quotient Ring

    Let r, s integers such that r/ ks+1. For some k belogs to [1,r)

    Prove that the subset {r,2r,...,(s-1)r} of Z/rsZ es a ring with identity ks+1.

    I tried to do it but I couldīt. Please help
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by orbit View Post
    Let r, s integers such that r/ ks+1. For some k belogs to [1,r)

    Prove that the subset {r,2r,...,(s-1)r} of Z/rsZ es a ring with identity ks+1.

    I tried to do it but I couldīt. Please help
    What precisely have you tried? Let us help you by leading you, not giving you the answer.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    What precisely have you tried? Let us help you by leading you, not giving you the answer.
    Ok, I thought that {r,2r,...} is a sibset of a ring, so if I want to see if is a ring or not, is only necessary to prove is a subring of Z/rsZ. So,

    I consider 2 different elements in {r,2r,...}, lets say $s_1r$ and $s_2r$. Then:

    $s_1r - s_2r$= $(s_1-s_2)r$ But, is this element is my subset?

    $(s_1r)(s_2r)$ = ??

    And I really dont know what to do with the identity element given....
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  4. #4
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    well if s2 < s1 < s, then certainly (s1-s2)r is in {r,2r,3r,.....,(s-1)r}. but hopefully you just forgot to include 0, because otherwise s1 = s2 leads to certain problems, eh?

    as for your second question, (s1r)(s2r) = (s1s2r)r, but remember, we are working mod rs, so we wind up with (s1s2r (mod rs))r.
    Last edited by Deveno; May 14th 2011 at 11:46 PM.
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  5. #5
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    But why does (s1r)(s2r) = (s1s2)r??
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  6. #6
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    my bad i had a typo. it should have been (s1s2r)r.

    after all s1,s2 and r are integers, correct? so (s1r)(s2r) is still SOME multiple of r (which may be some other multiple when s1s2r is reduced mod rs, but will still be some multiple of r).

    proving the set {0,r,2r,...,(s-1)r} is a subring of Z/(rs)Z isn't the interesting part.

    showing it has a multiplicative identity is.
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  7. #7
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    Ok, for the multiplicative identity I have done the following:

    First we know that r divides ks+1 then : ar= ks+1 , some interger a.

    Now , to prove ks+1 is the identity I have to check that if mr belongs tu my subset then mr (ks+1) = mr.
    So: mr (ks+1) = mr (ar) = (mra)r = (amr)r, but arm is congruent with 0(mod mr ) son arm = mr, so mr (ks+1) = mr (ar) = (mra)r = (amr)r = (mr)r Now I dont know what to do....
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  8. #8
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    well, look at it this way:

    ks+1 = ar, so ks+1 is in our set of multiples of r (mod rs) (it might look different when reduced).

    (ks+1)(mr) = (ks)(mr) + (mr) = (km)(rs) + (mr), and because we are working mod (rs)......?

    sometimes it helps to look at a specific example.

    suppose r = 4, and s = 13. then r divides 1 + 3(13), so we have the following set:

    {0,4,8,12,16,20,24,28,32,36,40,44,48} in Z/52Z.

    according to our theorem, 1+ks = 40 should be a multiplicative identity.

    for example (28)(40) = 1,120 = (52)(21) + 28, so (28)(40) = 28 (mod 52).

    and in fact, 40 = (3)(13) + 1, so

    (4m)(40) = (4m)((3)(13) + 1) = (3m)(52) + 4m = 4m (mod 52). see how that works?
    Last edited by Deveno; May 15th 2011 at 02:41 PM.
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