Thread: Prime elements and maximal ideals in a PID

1. Prime elements and maximal ideals in a PID

Here is the problem I have:

Let $\displaystyle R$ be a PID and $\displaystyle p \in R$. Prove that the ideal $\displaystyle \langle p \rangle$ is maximal in $\displaystyle R$ if and only if $\displaystyle p$ is a prime element in $\displaystyle R$. (Recall $\displaystyle p$ is a prime in $\displaystyle R$ if $\displaystyle p$ is not a unit and if $\displaystyle p|ab$ then $\displaystyle p|a$ or $\displaystyle p|b$)

I know that prime elements generate prime ideals (and I think I can prove that if I have to) and the only trick I had up my sleeve was that prime ideal -> ID and finite ID's are fields -> maximal, but it's not finite, so I am having difficulties constructing a proof...

2. Hi

One way is easy:

$\displaystyle (p)\ \text{maximal ideal}\Leftrightarrow R/(p)\ \text{field}\Rightarrow R/(p)\ \text{integral domain}\Leftrightarrow$ $\displaystyle (p)\ \text{prime ideal}$

Now assume $\displaystyle (p)$ prime. Take $\displaystyle x,y\in R/(p)$ i.e. classes of elements which are not in $\displaystyle (p).$ Since $\displaystyle R/(p)$ is an integral domain, $\displaystyle xy=0\Rightarrow x=0\ \text{or}\ y=0.$
So imagine there is a $\displaystyle r\in R$ such that $\displaystyle (p) \subseteq (r)\subsetneq R.$
Is there a divisibility relation between $\displaystyle r$ and $\displaystyle p$? Then think of what that gives you in $\displaystyle R/(p).$ Using the fact that $\displaystyle R$ is a PID, conclude!

3. This is very much like what I managed to come up with. I finally broke down and showed that prime elements iff prime ideals, and then went that route.

The only tricky part was getting the two different divisibility cases for prime->maximal. One was obvious containment, thus maximal, and the other showed that the multiple was a unit, and thus containment.

Thanks a bunch!