[summary: While there are two distinct but closely related notions of "prime" and "irreducible", the two ideas are actually the same in a certain class of ring. This result is basically what enables the Fundamental Theorem of Arithmetic.]

Let $R$ be a ring which is a PID, and let $r \not = 0$ 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. %%note: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.%%

1. $r$ is irreducible.
2. $r$ is prime.
3. The generated [ideal_ring_theory ideal] $\langle r \rangle$ is [maximal_ideal maximal] in $R$.

[todo: 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 $\langle r \rangle$ is maximal, then it is prime. (Indeed, $r$ is prime if and only if its generated ideal is [prime_ideal prime].)

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

An ideal $I$ is [prime_ideal prime] if and only if the quotient $R/I$ is an Integral domain. ([ideal_prime_iff_quotient_is_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 $\langle r \rangle$ isn't simply the whole ring.

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

Suppose $\langle r \rangle$ is contained in the larger ideal $J$, then. Because we are in a principal ideal domain, $J = \langle a \rangle$, say, for some $a$, and so $r = a c$ for some $c$. It will be enough to show that $a$ is invertible, because then $\langle a \rangle$ 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 $\langle r \rangle$: that there is $j \in J$ which is not in $\langle r \rangle$. Since $j \in J = \langle a \rangle$, we can find $d$ (say) such that $j = a d$; so $j = r c^{-1} d$ and hence $j \in \langle r \rangle$, which is a contradiction.