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

• Mar 12th 2012, 11:01 AM
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.
• Mar 12th 2012, 03:55 PM
NonCommAlg
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.