# 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: $\displaystyle a \;=\; p_1\cdots p_l, \quad b \;=\; q_1\cdots q_m, \quad c \;=\; r_1\cdots r_n$ 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.

it may be too simple but; why $a \in (p)$ instead of $p_1 \in (p)$? is it because : since $p$ has to be an associate of one of the $p_i$'s or $q_j$'s. let s say $p_2$ is an associate of $p$. so, $p_2$=pw, $w\in R$. since a=$p_1p_2...p_l$ then a=$p_1 pwp_3...p_l$ and a=$p(p_1p_3...p_lw)$, $p_1p_3...p_lw \in R$ so a is generated by $p$. hence $a\in (p)$ ?

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Your reasoning looks correct to me. –  Zach L. Dec 13 '12 at 15:12
Yes, more simply $\rm\: p\mid p_i\mid a,\:$ i.e. it follows by transitivity of "divides" (or "contains", if expressed using ideals) –  Bill Dubuque Dec 13 '12 at 15:32