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Ring and commutative ring

  1. Oct 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Let R be a ring. If any x in R satisfies xx=x, prove that R is a commutative ring.
    [1 is not necessarily in R in the definition of ring according to this particular book]


    2. Relevant equations



    3. The attempt at a solution
    I made some attempts but failed. I have ab=aabb=abab. However, a(ab-ba)b=0 does not implies ab-ba=0, since cancellation law does not necessarily hold...
    Any hint? I fail to draw any conclusion more than just ab=aabb=abab....
     
  2. jcsd
  3. Oct 17, 2008 #2

    cristo

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    I'm a bit rusty on this sort of thing, but haven't you shown that R is commutative? If you've got aabb=abab, then can't you just multiply on the left with a-1 and on the right with b-1, giving ab=ba?


    Edit: Probably ignore that: rings don't necessarily have multiplicative inverses, do they?
     
  4. Oct 17, 2008 #3

    morphism

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    Nope.

    Instead I'd first show that 2=0 in this ring. Then I'd think about how one would run into something that looks like ab-ba (=ab+ba) -- and especially something that involves squares.
     
  5. Oct 17, 2008 #4
    1, I make no progress...
    2, since 1 is not required to be in the ring according to the definition of ring in this book ...perhaps, 2=0 is not necessarily be true, as it is proved (perhaps?) by 2=1+1 and is therefore in R. and 2=2*1=2*(2-1)=2*2-2*1=2-2=0
    Okay, now I find something weird...is it proper to write xx+x=x(x+1) if 1 is not in R? I'm getting more confused:confused: I'll check the book later.
    3, perhaps it is ok for me to assume 1 is in R. We will see whether and how 1 is used in a proof that works.
    4, I make no progress.....
     
  6. Oct 20, 2008 #5
    Anyone?
     
  7. Oct 21, 2008 #6

    morphism

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    We don't need 1 to be in R.

    (1) By considering (x+y)^2, prove that xy+yx=0.
    (2) Then prove that xy+xy=0.

    Now here's a challenge for you: can you find a (nonzero) ring R without 1 such that x^2=x for all x in R?
     
  8. Oct 22, 2008 #7
    Here's what I thought:
    R, {3k|k is integers} mod 6, is such a commutative ring without '1' in it. And we have 3*3=3, 0*0=0.
    however, it seems that 3 really acts as 1 in this ring, since 3*3=3 and 3*0=0. So it is not a ring without 1...but I've no idea how to get a ring without a real 1 at the moment...

    BTW, I have one thing to clarify...If 1 is required in the definition of ring, should the subring inherit the same 1 from its original ring? I guess no. If yes, it would cause something strange...but I'm not quite sure about it right now..

    back to the topic..could you give me hint to find a ring without 1 such that xx=x?

    Thanks for all the help...
     
    Last edited: Oct 22, 2008
  9. Oct 27, 2008 #8

    morphism

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    Sorry for the late reply. I forgot about this thread!

    I guess we're discussing this in your other thread.

    Take the power set of some set S. Define A+B to be symmetric difference of A and B, and define AB to be the intersection of A and B. Verify that S is a ring and that AA=A for all A in S. S does have a '1' though - what is it?

    Can you use this example to come up with a ring without 1 such that xx=x?

    Or how about taking an infinite direct sum of copies of Z_2?
     
  10. Oct 28, 2008 #9
    I've been thinking about The Boolean ring defined on S (since it is one of the few rings with the property xx=x that I know). However, if S is well defined, it seems that S is always the 1...
    I'll be thinking on this carefully later, and will raise another thread if I encounter problems. Thanks very much:)
     
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