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Abstract Algebra: Another Ring Proof

  1. Feb 6, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Let R be a ring and suppose r ∈R such that r^2 = 0. Show that (1+r) has a multiplicative inverse in R.

    2. Relevant equations
    A multiplicative inverse if (1+r)*x = 1 where x is some element in R.

    3. The attempt at a solution

    We know we have to use two facts.
    1. Multiplicative inverse (1+r)*x=1
    2. the fact that r^2 = 0.

    So I first tried, well, why don't we use (1+r)(1+r) = 0

    However, this only got me to 1+r^2+2r = 0 with r^2 =0 we have 1+2r = 0
    But that doesn't appear to be what we wanted to show.

    Althought, we could subtract -1 from both sides, and subtract r from both sides to get
    r = -1-r
    r = -(1+r)
    But this still isn't what we wanted to show.

    If I divide both sides by r, then I get 1 = -1/r (1+r) and this would work, however, I can't assume element r can be divided to get 1. (at least I dont think I can based off the few multiplicative axioms I have for rings/fields).

    I have to think that I am on the right path, but any hints to get me closer?
     
  2. jcsd
  3. Feb 6, 2016 #2

    Samy_A

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    If r²=0, then (-r)²=0.
    So if this exercise is correct, what other element of the ring besides 1+r must also have a multiplicative inverse?
     
  4. Feb 6, 2016 #3

    RJLiberator

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    -(1+r) must have an inverse then?

    (-(1+r))^2=0
    (-(1+r))*\s=1

    (-(1+r)) = (-1-r)

    (-1-r)*s=1
    (1+r)*x=1

    (-1-r)(-1-r) = 1+2r+r^2 = 1+2r = 0

    Hm

    What if we try:
    (1+r)(-1-r) = 0
    -1-2r-r^2=0
    No, that doesn't help there.
    Hm...
     
  5. Feb 6, 2016 #4

    Samy_A

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    No, I didn't mean -(1+r).

    The exercise says: if (+r)²=0, then 1+r has a multiplicative inverse.
    How would you translate this literrally for (-r)²? Set s=-r, write down what the exercise claims about s, and then formulate it again in terms of r.
     
  6. Feb 6, 2016 #5

    RJLiberator

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    Hm, I see what you are saying.
    I think what you mean is the following:

    If (+r)^2 = 0 then 1+r has a M.I.
    So, (-r)^2=0 then 1-r has a M.I.

    We want to show that:
    (1-r)*x=1
    (1+r)*c=1

    (1-r)(1-r) = 1-2r=0
    (1+r)(1+r) = 1+2r=0

    1-2r=1+2r
    so r=0
    1=1
    1 always has a multiplicative inverse in a ring.
    (1+r) = 1
    Er, is that the correct way of showing it? Somehow one thing just lead to another...

    since r = 0
    c = 1
    x = 1
     
  7. Feb 6, 2016 #6

    Samy_A

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    Yes, that's what I meant.

    I'm not sure how you got that.

    But you are making this way too complicated.
    You now have two ring elements (1+r, 1-r) in search of a multiplicative inverse. Maybe they could help each other?
     
  8. Feb 6, 2016 #7

    RJLiberator

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    I was thinking, since r^2 = 0, just multiply (1+r)^2 = 0 and (1-r)^2 = 0 and then when you factor it out, the r^2 = 0. so that's how I got it :p.

    (1+r)(1-r) = 1-r^2 but since r^2 = 0 by assumption,
    (1+r)(1-r) = 1
    and thus, they are the multiplicative inverses.

    Hm, that makes more sense.
     
  9. Feb 6, 2016 #8

    Samy_A

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    But why should (1+r)²=0 be correct?

    Correct!
     
  10. Feb 6, 2016 #9

    RJLiberator

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    Ahhh, that makes sense..
    I was thinking just set (1+r) = r. But that seems like an 'illegal' move. Could be out of bounds of r.

    I appreciate the help here. Simple proof, but hard to see at first (as you could probably tell by my attempts in this thread :p).
     
  11. Feb 6, 2016 #10

    fresh_42

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    This is equivalent to 1 = 0.

    Very illegal if you have more than one element to deal with.

    It has nothing to do with r.
     
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