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Abstract Algebra: Unit question I am stumped!

  1. Feb 21, 2007 #1
    1. The problem statement, all variables and given/known data

    From An Introduction to Abstract Algebra by T. Hungerford
    Section 3.2 #29

    Let R be a ring with identity and no zero divisors.
    If ab is a unit in R prove that a and b are units.



    2. Relevant equations


    c is a unit in R if and only if there exists an element x in R s.t. cx=xc=1
    where 1 is the identity element of R.



    c is a zero divisor in R if and only if 1)c is not equal to 0 and 2)there exists
    and element d in R s.t. either cd=0 or dc=0.



    3. The attempt at a solution


    Any help please? Thank you.
     
  2. jcsd
  3. Feb 21, 2007 #2

    Dick

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    In a ring with unit and no zero divisors first prove that you have both left and right cancellation. Then since a*b is a unit you have a d such that (ab)d=d(ab)=1. So a(bd)=1. Now all you need to show is (bd)a=1. Time to multiply both sides by things and use your cancellation properties.
     
  4. Feb 21, 2007 #3
    for the left cancellation case, if a is nonzero and ab = ac then ab - ac = 0
    hence a(b-c) = 0, since there are no zero divisors, then b-c = 0 which
    implies b = c, hence ab = ac -> b = c -> left cancellation

    is this correct?


    I imagine that after I have proven that I have left and right cancellation
    that it goes along these lines:

    Starting off: (ab)d = 1

    a(bd) = 1
    a(bd)a = 1*a "right" multiplying both sides by a
    a(bd)a = a*1 commute the identity.
    (bd)a = 1 left cancellation.

    which goes to show that a is a unit.


    and similarly d(ab) = 1

    (da)b = 1
    b(da)b = b*1 "left" multiplication both sides by b
    b(da)b = 1*b commute the identity is always okay
    b(da) = 1 by right cancellation

    which shows that b is a unit

    is this the correct way?
     
  5. Feb 22, 2007 #4

    Dick

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    Very nice. But the real test is do you believe the proof?
     
  6. Feb 22, 2007 #5
    Thank you Dick for your help. I could not develop on my own the relevance that having no zero divisors had. WHen I worked the question before I posted for help, I encountered a need to use cancellation, yet since I hadn't proven I could use cancellation, I felt I was going in the wrong direction. I didn't know that I should have kept going in that direction, and proven cancellation.

    THe lesson I learned here, is the intuitive connection between a lack of zero divisors implies allowable use of cancellation.
     
  7. Feb 22, 2007 #6

    Dick

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    Exactly. You're quite welcome.
     
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