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Aa = a Implies R is Commutative

  1. Jun 24, 2008 #1
    The problem statement, all variables and given/known data
    Suppose that R is a ring and that aa = a for all a in R. Show that R is commutative.

    The attempt at a solution
    Let a and b be two elements from R. I'm having a hard time figuring this out without multiplicative inverses. All I've been able to derive is ab = aabb = abab so aabb - abab = a(ab - ba)b = 0.
     
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  3. Jun 24, 2008 #2

    tmc

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    You don't need multiplicative inverses. Start with (ab-ba), multiply by b in a few places, then reduce the 'bb' to 'bb=b', and you'll find ab-ba=0.
     
  4. Jun 24, 2008 #3

    D H

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    That means at least one of a, b, or ab-ba is zero. Look at the special cases where a or b is zero (is multiplication by zero commutative?) and then look at the general case where a and b are non-zero.
     
  5. Jun 24, 2008 #4
    Thanks for the tip tmc. I got it know.
     
  6. Jun 24, 2008 #5
    I can't assume that.
     
  7. Jun 24, 2008 #6
    Now that I've bothered with the algebra, I've realized that I wrote the above prematurely. I have to start with an equation, not just with ab - ba, so I started with ab - ba = -(ba) + ab but this lead nowhere.
     
  8. Jun 24, 2008 #7

    tmc

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    You're right, that doesn't work.

    Try by contradiction: suppose ba ~= ab, and multiplying by b from either side will not change the inequality (assuming b non-zero; if b is zero, then ab=ba is trivial). You should pretty quickly find a contradiction along the lines of 0 ~= 0
     
  9. Jun 24, 2008 #8

    D H

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    Stupid zero divisors!
    Doesn't the same problem with the stupid zero divisors arise here?
     
  10. Jun 24, 2008 #9

    matt grime

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    What is (a+b)^2? What is (-1)^2 (we may as well assume it is unital)?
     
  11. Jun 24, 2008 #10
    The problem with the contradiction is that the algebra with the not-equal sign is not well defined: How do you know a != b implies ab != bb? It is possible for ab = bb but a != b.
     
  12. Jun 24, 2008 #11
    That's a good tip. If I consider a - b = (a - b)(a - b), I get the answer.
     
  13. Jun 24, 2008 #12

    matt grime

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    Actually, it's quite easy to show that 1=-1 in this ring in many ways, so I'm worried that you're writing a-b in there. With what you wrote you get

    a-b=(a-b)^2=a^2+b^2 -ab-ba = a+b -ab-ba

    i.e. 0=2b-ab-ba

    which is quite a long way from what you need to end up with (though it is still correct, and will lead to a proof).
     
  14. Jun 24, 2008 #13
    I wouldn't say I'm a long way from what I need since I end up with ba = a(-b). If I can show that -b = b, I'm done. This unfortunately I don't know how to demonstrate. By the way, you can't assume R has a multiplicative identity.
     
  15. Jun 25, 2008 #14

    HallsofIvy

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    Then look at (b+b)2= b+ b.
     
  16. Jun 25, 2008 #15

    matt grime

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    Or embed it in its unitalization - the same property of x^2=x will still hold. And you don't end up with that from considering (a=b)^2: where did the 2b go? That was my point about the minus signs, not the ab=-ba fact which comes up if you consider (a+b)^2 instead.
     
  17. Jun 25, 2008 #16
    HallsofIvy: Good tip. Thanks.

    matt grime: I don't understand your point. Nevertheless, I now have a satisfactory answer. Thanks.
     
  18. Jun 25, 2008 #17

    matt grime

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    Which bit don't you understand? The important bit is that you should consider a+b and not a-b as it is simpler. (Although it is entirely equivalent and you prove the same thing: 2x=0 for all x). You claimed that by considering a-b you only needed to show that ab=-ba, but in doing so you must have assumed that 2b=0, i.e b=-b.
     
  19. Jun 25, 2008 #18
    Right. Using a + b rather than a - b is simpler. I didn't follow your argument about unity and that -1 = 1. I guess you were trying to suggest that -a = a for all a in R.
     
  20. Jun 25, 2008 #19

    matt grime

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    Any ring can be embedded in a unital ring. If R is a non-unital ring, then define R' as follows.

    R' as a set is RxZ (the integers), i.e. ordered pairs (r,n) where r is in R and n an integer. Addition is defined as

    (r,n)+(s,m)=(r+s,n+m)

    and multiplication

    (r,n)(s,m)=(rs+ns+mr,sm).

    You can verify that (0,1) is the unit element, and that R is a subring of R' via the elements (r,0). If we now declare that x^2=x for all x in R' (i.e. form a quotient ring), we still have a unital ring with R as a subring with the same property. So wlog you may assume that R has a unit.
     
  21. Jun 25, 2008 #20
    Interesting. Thank you for the clarification.
     
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