Let be a ring which is a PID, and let $r \neq 0$ be an element of $R$ . Then $r$ is irreducible if and only if $r$ is prime.

In fact, it is easier to prove a stronger statement: that the following are equivalent. ^[1]

$r$ is irreducible.

$r$ is prime.

The generated ideal $⟨ r ⟩$ is maximal in $R$ .

motivate the third bullet point

Proof

$2 \Rightarrow 1$

A proof that "prime implies irreducible" appears on the page for irreducibility.

$3 \Rightarrow 2$

We wish to show that if $⟨ r ⟩$ is maximal, then it is prime. (Indeed, $r$ is prime if and only if its generated ideal is prime.)

An ideal $I$ is maximal if and only if the quotient $R / I$ is a field. (Proof.)

An ideal $I$ is prime if and only if the quotient $R / I$ is an integral domain. (Proof.)

All fields are integral domains. (A proof of this appears on the page on integral domains.)

Hence maximal ideals are prime.

$1 \Rightarrow 3$

Let $r$ be irreducible; then in particular it is not invertible, so $⟨ r ⟩$ isn't simply the whole ring.

To show that $⟨ r ⟩$ is maximal, we need to show that if it is contained in any larger ideal then that ideal is the whole ring.

Suppose $⟨ r ⟩$ is contained in the larger ideal $J$ , then. Because we are in a principal ideal domain, $J = ⟨ a ⟩$ , say, for some $a$ , and so $r = a c$ for some $c$ . It will be enough to show that $a$ is invertible, because then $⟨ a ⟩$ would be the entire ring.

But $r$ is irreducible, so one of $a$ and $c$ is invertible; if $a$ is invertible then we are done, so suppose $c$ is invertible.

Then $a = r c^{- 1}$ . We have supposed that $J$ is indeed larger than $⟨ r ⟩$ : that there is $j \in J$ which is not in $⟨ r ⟩$ . Since $j \in J = ⟨ a ⟩$ , we can find $d$ (say) such that $j = a d$ ; so $j = r c^{- 1} d$ and hence $j \in ⟨ r ⟩$ , which is a contradiction.

^{^︎}
Every proof known to the author is of this shape either implicitly or explicitly, but when it's explicit, it should be clearer what is going on.