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Math Help - Quotient ring and field of 9 elements

  1. #1
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    Quotient ring and field of 9 elements

    Hi guys, I need some help with the following exercise

    Consider the Guassian intergers. Let J be an ideal of Z[i]. Prove that the quotient Z[i]/J is a field of 9 elements.


    J= $\J = \left\{ {a + bi \in \mathbb{Z}\left[ i \right]:\left. 3 \right|a \wedge \left. 3 \right|b} \right\}\

    I know the quotient is a field but I donīt know why is has 9 elements.
    Last edited by orbit; May 21st 2011 at 05:54 PM.
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  2. #2
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    Quote Originally Posted by orbit View Post
    Hi guys, I need some help with the following exercise

    Consider the Guassian intergers. Let J be an ideal of Z[i]. Prove that the quotient Z[i]/J is a field of 9 elements.


    J= $\J = \left\{ {a + bi \in \mathbb{Z}\left[ i \right]:\left. 3 \right|a \wedge \left. 3 \right|b} \right\}\

    I know the quotient is a field but I donīt know why is has 9 elements.


    Hmmm....how exactly do you know the quotient is a field?

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Hmmm....how exactly do you know the quotient is a field?

    Tonio
    Because J es a maximal ideal.
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    Quote Originally Posted by orbit View Post
    Hi guys, I need some help with the following exercise

    Consider the Guassian intergers. Let J be an ideal of Z[i]. Prove that the quotient Z[i]/J is a field of 9 elements.


    J= $\J = \left\{ {a + bi \in \mathbb{Z}\left[ i \right]:\left. 3 \right|a \wedge \left. 3 \right|b} \right\}\

    I know the quotient is a field but I donīt know why is has 9 elements.
    It's not that hard to literally exhibit the elements of the ring. I mean, if you look at the cosets aren't they just \left\{a+bi+J:a,b\in\{0,1,2\}\right\}?
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  5. #5
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    sorry, I dont get it.
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  6. #6
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    Quote Originally Posted by orbit View Post
    sorry, I dont get it.
    A quotient ring, if anything, is the quotient group of the abelian group underlying the ring, right? So, in particular by definition \left|\mathbb{Z}[i]/J\right|=\left(\mathbb{Z}[i]:J\right) (the index), but I think you'll find that \mathbb{Z}[i]/J=\left\{a+bi:a,b\in\{0,1,2\}\right\}.
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  7. #7
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    The part I dont understand is why a,b belongs to {0,1,2}....
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  8. #8
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    suppose a = k+3m, b = n+3r.

    then doesn't a+bi + J = k+ni + J?
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  9. #9
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    Quote Originally Posted by Deveno View Post
    suppose a = k+3m, b = n+3r.

    then doesn't a+bi + J = k+ni + J?
    Yes, are the same... But... what do I know from that fact?
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  10. #10
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    what you know is that the coset a+bi + J is the same as a(mod 3) + b(mod 3)i + J. that is a,b = 0,1 or 2 (mod 3). this means we get exactly 9 (distinct) elements:

    0+0i + J, 0+i + J, 1+0i + J, 1+i + J, 0+2i + J, 2+0i + J, 1+2i + J, 2+i + J, 2+2i + J.
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  11. #11
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    Quote Originally Posted by Deveno View Post
    what you know is that the coset a+bi + J is the same as a(mod 3) + b(mod 3)i + J. that is a,b = 0,1 or 2 (mod 3). this means we get exactly 9 (distinct) elements:

    0+0i + J, 0+i + J, 1+0i + J, 1+i + J, 0+2i + J, 2+0i + J, 1+2i + J, 2+i + J, 2+2i + J.
    Ok, thank u so much!
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  12. #12
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    Quote Originally Posted by orbit View Post
    Ok, thank u so much!

    I sitll canīt figure out how did you manage to prove that ideal is maximal without finding out a little more

    about how its elements look and, thus, how easy it is to deduce that we have a field with 9 elements...

    Tonio
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  13. #13
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    as Z[i] is an integral domain, to show J is maximal it suffices to show that J is:____?
    Last edited by Deveno; May 22nd 2011 at 07:58 AM.
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