# irreducible of a UFD is prime

Any irreducible element of a factorial ring $D$ is a prime element of $D$.

Proof.  Let $p$ be an arbitrary irreducible element of $D$.  Thus $p$ is a non-unit.  If  $ab\in(p)\smallsetminus\{0\}$,  then  $ab=cp$  with  $c\in D$.  We write $a,\,b,\,c$ as products of irreducibles:

 $a\;=\;p_{1}\cdots p_{l},\quad b\;=\;q_{1}\cdots q_{m},\quad c\;=\;r_{1}\cdots r% _{n}$

Here, one of those first two products may me empty, i.e. it may be a unit.  We have

 $\displaystyle p_{1}\cdots p_{l}\,q_{1}\cdots q_{m}\;=\;r_{1}\cdots r_{n}\,p.$ (1)

Due to the uniqueness of prime factorization, every factor $r_{k}$ is an associate of certain of the $l\!+\!m$ irreducibles on the left hand side of (1).  Accordingly, $p$ has to be an associate of one of the $p_{i}$’s or $q_{j}$’s.  It means that either  $a\in(p)$  or  $b\in(p)$.  Thus, $(p)$ is a prime ideal of $D$, and its generator must be a prime element.

Title irreducible of a UFD is prime IrreducibleOfAUFDIsPrime 2013-03-22 18:04:35 2013-03-22 18:04:35 pahio (2872) pahio (2872) 7 pahio (2872) Theorem msc 13G05 msc 13F15 PrimeElementIsIrreducibleInIntegralDomain