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Thread: Ideals

  1. #1
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    Ideals

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

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

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

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

    We also know $\displaystyle a = b\cdot p^n $ for some $\displaystyle 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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    $\displaystyle R = \mathbb{Z}[i]/(p) = \{a+(p)|a\in\mathbb{Z}[i]\} $.

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

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

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

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

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

    Finally we get $\displaystyle \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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