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Math Help - Rings- a^2 = a then R must be commutative

  1. #1
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    Rings- a^2 = a then R must be commutative

    Hi, just need a hint with this one...

    Let R be a ring such that a^2 = a for every a in R. Show that R must be commutative

    Thanks in advance.

    Oh and also having some trouble with...

    Show that \bb{Z}_{n} is a field if and only if n is a prime.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by liedora View Post
    Let R be a ring such that a^2 = a for every a in R. Show that R must be commutative

    Using (x+x)^2=x+x prove that

    x=-x,\quad \forall x\in R

    Using (a+b)^2=a+b prove that

    ba=-ab,\quad \forall a,b \in R

    Conclude.


    Fernando Revilla
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by liedora View Post
    Show that \bb{Z}_{n} is a field if and only if n is a prime.
    Hint ;

    If n=pq then, [p][q]=[0]



    Fernando Revilla
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    Hint ;

    If n=pq then, [p][q]=[0]

    Fernando Revilla
    ...and for the other way, use the Euclidean algorithm...
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  5. #5
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    Ok managed to get the first question out, but still having difficulties with the second. Any more hints would be greatly appreciated!
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by liedora View Post
    Ok managed to get the first question out, but still having difficulties with the second. Any more hints would be greatly appreciated!
    What have you tried?. At any rate, if you know that a finite integral domain is a field then, this fact can help you.


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  7. #7
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    I don't know how much of a stickler your instructor is, but (presumably - I mean I can't imagine you not) you've been working with [tex]\mathbb{Z}_n[\math] quite a bit so you get a lot free in the definition of a field. I would say the only thing you need prove is that [tex]\mathbb{Z}_n[\math] contains mult. inverses (as far as going in the forward direction is concerned - going backwards ". . is prime. . .field" shouldn't be too hard).
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