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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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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