Three Equivalent Conditions for an Ideal is Prime in a PID

Problems and solutions of ring theory in abstract algebra

Problem 724

Let $R$ be a principal ideal domain. Let $a\in R$ be a nonzero, non-unit element. Show that the following are equivalent.

(1) The ideal $(a)$ generated by $a$ is maximal.
(2) The ideal $(a)$ is prime.
(3) The element $a$ is irreducible.

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Proof.

(1) $\implies$ (2)

Note that the ideal $(a)$ is maximal if and only if $R/(a)$ is a field. In particular $R/(a)$ is a domain and hence $(a)$ is a prime ideal.

(Note that this is true without assuming $R$ is a PID.)

(2) $\implies$ (3)

Now suppose that the ideal $(a)$ is prime.

Let $a=bc$ for some elements $b, c \in R$. Then the element $a=bc$ is in the prime ideal $(a)$, and thus we have either $b$ or $c$ is in $(a)$. Without loss of generality, we assume that $b\in (a)$.

Then we have $b=ad$ for some $d\in R$. It follows that we have
\[a=bc=adc\] and since $R$ is a domain, we have
\[1=dc\] and hence $c$ is a unit. Therefore the element $a$ is irreducible.

(3) $\implies$ (1)

Suppose that $a$ is an irreducible element.
Let $I$ be an ideal of $R$ such that
\[(a) \subset I \subset R.\]

Since $R$ is a PID, there exists $b\in R$ such that $I=(b)$.
Then since $(a)\subset (b)$, we have $a=bc$ for some $c\in R$.

The irreducibility of $a$ implies that either $b$ or $c$ is a unit.

If $b$ is a unit, then we have $I=R$. If $c$ is a unit, then we have $(a)=I$.
Therefore the ideal $(a)$ is maximal.


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