In a Principal Ideal Domain (PID), a Prime Ideal is a Maximal Ideal

Problems and solutions of ring theory in abstract algebra

Problem 175

Let $R$ be a principal ideal domain (PID) and let $P$ be a nonzero prime ideal in $R$.
Show that $P$ is a maximal ideal in $R$.
 
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Definition

A commutative ring $R$ is a principal ideal domain (PID) if $R$ is a domain and any ideal $I$ is generated by a single element $a\in I$, that is $I=(a)$.

Proof.

Since $R$ is a PID, we can write $P=(a)$, an ideal generated by an element $a\in R$.
Since $P$ is a nonzero ideal, the element $a\neq 0$.

Now suppose that we have
\[P \subset I \subset R\] for some ideal $I$ of $R$.

We can write $I=(b)$ for some $b \in R$ since $R$ is a PID.
The element $a\in (a) \subset (b)$ and so there is an element $c \in R$ such that $a=bc$.

Since $a=bc$ is in the prime ideal $P$, we have either $b \in P$ or $c \in P$.
If $b\in P$, then it follows that $I=(b)\subset P$, and hence $P=I$.
If $c \in P=(a)$, then we have $d\in R$ such that $c=ad$.

Then we have
\begin{align*}
a=bc=bad
\end{align*}
and since $R$ is a domain and $a\neq 0$, we have
\[1=bd.\] This yields that $b$ is a unit and hence $I=(b)=R$.

In summary, we observe that whenever we have $P \subset I \subset R$, we have either $I=P$ or $I=R$. Thus $P$ is a maximal ideal.


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