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Math Help - Units of integers mod m

  1. #1
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    Units of integers mod m

    Show that U(I_m) = {[k] in I_m : (k,m) = 1}.


    I_m is integers mod m.

    I know that if m is prime then the result is obvious. But do i need to break it into cases.???
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  2. #2
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    Re: Units of integers mod m

    How is the case (m prime) obvious?

    Why not use set inclusion, and that (m, k) = 1 iff am + bk = 1 for some a, b...
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  3. #3
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    Re: Units of integers mod m

    Should I assume (m,k)= 1 to try to show that [k] is in I_m?
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  4. #4
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    Re: Units of integers mod m

    I think I got it. Say (k,m)= 1 then there exist a, b that ak+bm =1 I can put bm on the other side to get

    ak-1=bm. Now we know that m| ak -1 and so ak is congruent to 1mod m. That proves [k] is in I_m. further more ak is congruent to 1 mod m therefore there exist a c such that kc =1 and that is the definition of a unit.

    Did I miss something?
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  5. #5
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    Re: Units of integers mod m

    a couple of things:

    [k] is, by definition, in I_m. what you are trying to prove is that [k] is in U(I_m). so you don't want a c such that kc = 1 (unless k =1, no such integer exists), you want a [c] such that [c][k] = [1].

    now you have shown that [ak] = [1], so you're almost there. the [c] that you're looking for is clearly [a]. what you haven't done, is show if 0 < c < m such that [c] = [a], that (c,m) = 1.
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  6. #6
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    Re: Units of integers mod m

    I have this piace of a proof from Rotman advanced algebra book. I dont understand how or what it means by solve of [x].

    If (a,m) = 1, then [a][x] = [1] can be solved for
    [x] in Im
    . Now(x,m) = 1, for rx + sm = 1 for some integer s, and so Proposition 1.13
    on page 5 gives (x,m) = 1; therefore, [x] ∈ U(Im),
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  7. #7
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    Re: Units of integers mod m

    we are given that (k,m) = 1 by the definition of U(I_m).

    so (by the defintion of gcd) we have a,b in Z such that ak + bm = 1.

    if two integers are equal, then they remain equal mod m, so

    [ak + bm] = [1]
    [a][k] + [b][m] = [1]
    [a][k] + [b][0] = [1]
    [a][k] = [1].

    now we just take the integer c, where 0 < c < m, so that [c] = [a] (in other words, reduce the integer a used in the gcd formula mod m).

    to give a concrete example, suppose m = 6 and k = 5. (5,6) = 1, so [5] is in U(6). note that 5(-7) + 6(6) = 1.

    so we can take a = -7. this means that [5][-7] = [1] (mod 6). but -1 isn't in the right range, so we add multiples of 6 until it is:

    -7 + 2(6) = 5. so [-1] = [5], so [5] is its own inverse in U(6). and indeed, [5][5] = [25] = [1], since 25 = 1 mod 6.

    well, here, we obviously have (5,6) = 1. ok, back to the general case:

    we know that [a] ( = [c]) comes from ak + bm = 1, so it's trivial that (a,m) = 1 (just switch the role of a and k).

    what's not so trivial, and what your proof from Rotman doesn't state explicitly, is that , in fact, (c,m) = 1.

    but this isn't that much extra difficulty: [c] = [a], so c = a + tm, for some integer t.

    therefore ak + bm = 1 --> (c - tm)k + bm = 1, and re-arranging slightly:

    kc + (b - tk)m = 1, so since there exists integers k, b - tk such that kc + (b - tk)m = 1, (c,m) = 1.
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