# Every Prime Ideal in a PID is Maximal / A Quotient of a PID by a Prime Ideal is a PID

## Problem 535

**(a)** Prove that every prime ideal of a Principal Ideal Domain (PID) is a maximal ideal.

**(b)** Prove that a quotient ring of a PID by a prime ideal is a PID.

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Contents

## Proof.

### (a) Prove that every PID is a maximal ideal.

Let $R$ be a Principal Ideal Domain (PID) and let $P$ be a nonzero prime ideal of $R$.

Since $R$ is a PID, every ideal of $R$ is principal.

Hence there exists $p\in R$ such that $P=(p)$.

Because $P$ is a nonzero ideal, we see that $p\neq 0$.

Let $I=(a)$ be an ideal of $R$ such that $P \subset I\subset R$.

To show that $P$ is a maximal ideal, we must show that $I=P$ or $I=R$.

Since $p\in (p)\subset (a)$, we have $p=ra$ for some $r\in R$.

As $p=ra$ is in the prime ideal $(p)$, we have either $a\in (p)$ or $r\in (p)$.

If $a\in (p)$, then it follows that $(a)\subset (p)$, and hence $(a)=(p)$.

So, in this case, we have $I=P$.

If $r\in (p)$, then we have $r=sp$ for some $s\in R$.

It yields that

\begin{align*}

p=ra=spa \quad \Leftrightarrow \quad p(1-sa)=0.

\end{align*}

Since $R$ is an integral domain and $p\neq 0$, this gives $sa=1$.

It follows that $1\in (a)$ and thus $I=(a)=R$.

We have shown that if $P\subset I \subset R$ for some ideal $I$, then we have either $I=P$ or $I=R$.

Hence we conclude that $P$ is a maximal ideal of $R$.

### (b) Prove that a quotient ring of a PID by a prime ideal is a PID.

Let $P$ be a prime ideal of a PID $R$.

It follows from part (a) that the ideal $P$ is maximal.

Thus the quotient $R/P$ is a field.

The only ideals of the field $R/P$ are the zero ideal $(0)$ and $R/P=(1)$ itself, which are principal.

Hence $R/P$ is a PID.

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