# Prime elements and maximal ideals in a PID

• Mar 19th 2009, 10:34 AM
nqramjets
Prime elements and maximal ideals in a PID
Here is the problem I have:

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

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...(Doh)
• Mar 19th 2009, 01:35 PM
clic-clac
Hi

One way is easy:

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

Now assume $(p)$ prime. Take $x,y\in R/(p)$ i.e. classes of elements which are not in $(p).$ Since $R/(p)$ is an integral domain, $xy=0\Rightarrow x=0\ \text{or}\ y=0.$
So imagine there is a $r\in R$ such that $(p) \subseteq (r)\subsetneq R.$
Is there a divisibility relation between $r$ and $p$? Then think of what that gives you in $R/(p).$ Using the fact that $R$ is a PID, conclude!
• Mar 19th 2009, 03:56 PM
nqramjets
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!