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

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

    It seems to me that an infinity of ideals would necessarily imply an infinite chain (intuitively), since any ideal of $\displaystyle $R/A$$ is going to be of the form $\displaystyle $C/A$$, where $\displaystyle $C$$ is contained in $\displaystyle $A$$, 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.

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