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Math Help - Ring theory, some questions...

  1. #1
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    Ring theory, some questions...

    Hey everybody, thank you for reading.
    I have an exam soon, and I have accumulated some questions which I haven't found a solution for... I'd appreciate it if I could get a push...

    1)Given a ring in which every x in R satisfies x^3=x, I need to show it's commutative, and fail to see why. (note: this question was given a little after teaching the basic definitions and notions of "ideals".)
    2) Given a set which is almost a ring, besides the fact that we're not given that a+b=b+a (abelian group under addition), that has a unit element, I need to show that the condition a+b=b+a must be satisfied anyhow.
    'note: there's a hint here: expand (a+b)(1+1) in 2 ways. Haven't mannaged to do anything with that, however.

    That's it for now :-)
    Thanks!!!
    Tomer.
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  2. #2
    Newbie Halmos Rules's Avatar
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    What do you mean by "has a unit element"?

    Maybe this helps.....
    (a+b)(1+1)=(a+b)1+(a+b)1
    and
    (a+b)(1+1)=a(1+1)+b(1+1)
    Last edited by mr fantastic; February 20th 2009 at 04:04 PM. Reason: Merged posts
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  3. #3
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    By "has a unit element" I mean there exists an element 1 in the set that satisfies 1*x=x*1=x for all x... That's not one of a Ring's axioms, and apperently it's existence implies that a+b=b+a (that's the Ring axiom I'm not given).

    And, thanks for the tip, but I obviously tried this and other combinations and couldn't see where it leads me....
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  4. #4
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    Consider (1+1)(a+b). We can distribute from the left:

    (1+1)a + (1+1)b = a + a + b + b

    We can also distribute it from the right:

    1(a+b) + 1(a+b) = a + b + a + b.

    Setting these two equal to each other and canceling the a on the left and the b on the right yields a + b = b + a.
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  5. #5
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    Thanks
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