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Math Help - Rings problem

  1. #1
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    Unhappy Rings problem

    Hi guys, not sure if this is the correct forum for Rings but I'll try it here anyways

    I'm working on exam questions and I am very lost. Any help or pointers on the following would be really appreciated as I'm running out of revision time!

    Q1
    Let R be the subring of C defined by R={a+2bi e Z[i]:a,b e Z}.
    Show that the set I={2m+2ni e R: m,n e Z} is an ideal of R.

    Show that I is not a principal ideal of R

    Q2
    Let f(x)=x^3+x+2 e Z_5[x] and let I=<f(x)>. How many distinct cosets of I are there?

    In q1 I think you can look at the elements 2 and 2i in I to show that it is not a principal ideal?

    Thanks!
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  2. #2
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    First part of Q1

    Ok, I think I've proved that I is an ideal of R as follows:

    You need to show that p+q e I, 0 e I, -p e I and rp e I where r e R and p e I

    The first three are easy enough, the fourth I did the following(I think its right!)
    Let r=m+2ni e R
    Then rp= (m+2ni)(2a+2bi)=2ma+2mbi+4nai-4nb
    =2(ma-2nb)+2(mb+2na)i which is clearly an element of I

    I still don't know how to show that I is not a principle Ideal, probably because I don't know how to get the Principal Ideal!
    Anyone any ideas?
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  3. #3
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    Quote Originally Posted by musicmental85 View Post

    Q1
    Let R be the subring of C defined by R={a+2bi e Z[i]:a,b e Z}.
    Show that the set I={2m+2ni e R: m,n e Z} is an ideal of R.

    Show that I is not a principal ideal of R
    What does principal mean? It means that I = \left < a \right> for some a\in I (this is the set \{ ra|r\in R\}).
    Try to argue by assumption that there is a 2a+2bi such that \left< 2a+2bi \right> = I and reach a contradiction.
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  4. #4
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    I know the dfinition for a principal Ideal. I think the problem is that I don't really understand what <a>=I actually means so Assuming <2a+2bi>=I isn't ringing any bells as to where I can go from there?
    Do I need to look at each element as part of I? like 2=2(1)+2(0)i which is an element of I
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  5. #5
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    Quote Originally Posted by musicmental85 View Post
    Q2
    Let f(x)=x^3+x+2 e Z_5[x] and let I=<f(x)>. How many distinct cosets of I are there?
    The division algorithm says that if p(x) is any polynomial in Z_5[x] then we can divide p(x) by f(x) and get a remainder which must be of degree at most 2. So p(x) = q(x)f(x) + r(x) \in I+r(x). This says that p(x) is in the coset I+r(x), and there is one coset for each polynomial r(x) of degree at most 2. All you have to do is to count how many such polynomials there are.
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  6. #6
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    For your second question maybe this will help.
    Last edited by ThePerfectHacker; December 3rd 2007 at 06:12 PM.
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  7. #7
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    That link doesn't open but I think that I has precisely 9 distinct cosets in Z_i? I'm rushing so I can't state the proof yet but this would imply that I has 9 distinct cosets in Z_5 also?
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  8. #8
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    Quote Originally Posted by musicmental85 View Post
    That link doesn't open but I think that I has precisely 9 distinct cosets in Z_i? I'm rushing so I can't state the proof yet but this would imply that I has 9 distinct cosets in Z_5 also?
    Try it now.
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  9. #9
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    Thanks! That did help
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