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Math Help - Ideals

  1. #1
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    Ideals

    Suppose  p is irreducible in  \mathbb{Z}[i] .

    1. Let  A_n = (p^n) (the ideal generated by  p^n ). Prove  \mathbb{Z}[i]/(p) \cong A_n/A_{n+1} as additive abelian groups.

    2. Show  |\mathbb{Z}[i]/(p^n)| = |\mathbb{Z}[i]/(p)|^n .
    Last edited by Jim63; February 14th 2010 at 04:41 PM.
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  2. #2
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    So far I've got

    1.  A_n/A_{n+1} = \{a+(p^{n+1})|a\in (p^n)\}
    and  a=q\cdot p^{n+1} + r where  r=0 or  N(r) < N(p^{n+1}) .

    We also know  a = b\cdot p^n for some  b\in \mathbb{Z}[i] .

    I'm lost after this...
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  3. #3
    MHF Contributor chiph588@'s Avatar
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     R = \mathbb{Z}[i]/(p) = \{a+(p)|a\in\mathbb{Z}[i]\} .

     S = (p^n)/(p^{n+1}) = \{a+(p^{n+1})|a\in (p^n)\} = \{a\cdot p^n+(p^{n+1})|a\in\mathbb{Z}[i]\} .

    Define  \varphi: R \rightarrow S such that  a \mapsto p^n\cdot a .

     \varphi(0)=0 and  \varphi(a+b) = p^n(a+b) = p^n\cdot a + p^n\cdot b = \varphi(a)+\varphi(b) , hence  \varphi is a homomorphism.

    Suppose  \varphi(a) = \varphi(b) , then  p^n\cdot a = p^n\cdot b which forces  a = b since we're working in an integral domain. Thus  \varphi is injective.

    given  b\in S \Longrightarrow b = p^n\cdot a for some  a\in \mathbb{Z}[i] . So choose said  a to get  \varphi(a) = b . Hence  \varphi is surjective.

    Therefore  \varphi is an isomorphism with for these additive groups.

    Finally we get  \mathbb{Z}[i]/(p) \cong (p^n)/(p^{n+1}) as additive abelian groups.

    **For some reason I'm not too confident in this solution, so someone else might want to look it over.**
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Hint for #2:

    Use induction and use #1 in your yielding case.
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