TITLE: Irreducible elements in a PID are prime
QUESTION [7 upvotes]: How can I see that all irreducible elements in a principal ideal domain are prime? 
$u$ is irreducible when $u_1 u_2 = u \implies u_1 $ or $u_2$ is a unit.
$u$ is prime  when $u | ab \implies u|a$ or $u|b$.

REPLY [12 votes]: Suppose $u$ is irreducible in a principal ideal domain $R$. We will show that the principal ideal $(u)$ is maximal. Assume, to the contrary, that $(u)\subsetneq I\subsetneq R$ for some ideal $I$. Since $R$ is a PID, $I=(a)$ for some $a\in R$. Now, $u\in I$ gives $u=ar$ for some $r\in R$. Since $u$ is irreducible, and $a$ is not a unit (since $I=(a)\neq R$), it follows that $r$ is a unit. But then, $a=u r^{-1}$ and so $(u)=I$ which is a contradiction.
Conclusion: $(u)$ is a maximal ideal in $R$, and so in particular it is a prime ideal. Thus, $u$ is prime.