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Math Help - PID's factor rings have finite amount of principal ideals.

  1. #1
    Junior Member RaisinBread's Avatar
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    PID's factor rings have finite amount of principal ideals.

    Hey, I have a quick question about Principal Ideal Domains and Ideals.

    Let $R$ be a ring that is a Principal Ideal Domain (Integral domain such that all ideals are principals.), and $A\neq 0$ be an ideal of $R$.

    I must show that there is a finite number of ideals of $R/A$, and that they are all principal.

    Showing they are principal is quite easy. However I'm having problems with showing there is a finite number of them.

    What I've done so far;

    Suppose the number of distinct ideals of $R/A$ is infinite. If there exists an infinite chain $B_1\subset B_2\subset ...$ of ideals of $R/A$, I can easily show this leads to a contradiction from the fact that $R$ is a PID.

    It seems to me that an infinity of ideals would necessarily imply an infinite chain (intuitively), since any ideal of $R/A$ is going to be of the form $C/A$, where $C$ is contained in $A$, but I can't find a way to either prove of disprove this claim.
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  2. #2
    MHF Contributor

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    Re: PID's factor rings have finite amount of principal ideals.

    Ra \subseteq Rb if and only if a=rb, for some r \in R. so b is a factor of a and obviously the number of factors of a nonzero element in a PID is finite.
    Thanks from RaisinBread
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