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

## 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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