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# 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.

## Mathematics Subject Classification

13G05*no label found*13F15

*no label found*

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