# A question on irreducible elements in principal ideal domains.

In my book it says that if $d$ is an element in a principal ideal domain $D$ that cannot be written as a product of irreducible elements then it follows that $d$ is not irreducible. I dont understand this, im thinking that IF it is irreducible THEN it cannot be written as a product of irreducible elements...Am I missing something?

Claim: Any principal ideal domain is a UFD

Proof: "Let $D$ be a principal ideal domain and let $d$ be a nonzero element of $D$ that is not a unit. Suppose that $d$ cannot be written as a product of irreducible element. Then $d$ is not irreducible, and so $d = a_1b_1$, where neither $a_1$ nor $b_1$ is a unit and either $a_1$ or $b_1$ cannot be written as a product of irreducible elements. Assume that $a_1$ cannot be written as a product of irreducible element. Now $b_1$ is not a unit, and so we have $dD \subset a_1D$. We can continue this argument to obtain a factor $a_2$ of $a_1$ that cannot be written as a product of irreducible elements such that $a_1D \subset a_2D$. Thus the assumption that $d$ cannot be written as a product of irreducible elements allows us to construct a strictly ascending chain of ideals $dD \subset a_1D \subset a_2D \subset a_3D \subset.....$" (according to the last lemma, this chain contradicts the fact that $D$ is a principal ideal domain)

This is half the proof, namely they proved the first condition for an integral domain to be a UFD, however there are several parts i dont understand, like the one i pointed out in the beginning of the post. Can anyone explain this proof to me?

• Strange. In a PID any element $a$ can be written as $a=up_{1}\cdots p_{k}$ where $u$ is a unit and the $p_{i}$ are irreducible elements. Special cases are $k=0$ (then $a$ is a unit) and $k=1$ (then $a$ is irreducible). – drhab Jun 11 '14 at 8:30
• Its from the proof that every principal ideal domain is a UFD, from the book abstract algebra by Beachy and Blair, page 413. Should I write down the whole proof? Maybe im missunderstanding something. @drhab – user117449 Jun 11 '14 at 8:32
• Maybe they mean to say: any element that cannot be written as a (non-empty) product of irreducibles must be a unit, hence is not irreducible. – drhab Jun 11 '14 at 8:38
• I will edit my post with the proof, one second @drhab – user117449 Jun 11 '14 at 8:44
• Their point: If $d$ is irreducible then it can be written as a product of irreducibles (a product with one factor). So if it cannot be written as a product of irreducibles then it is not irreducible. It was excluded allready that $d$ is a unit so they come to $d=a_1b_1$ et cetera. I will delete my answer. It is not applicable. Cheers now (time lacks). – drhab Jun 11 '14 at 9:06

They are using the convention that any element may be considered as a product (of one factor). Thus if $\,d\,$ is irreducible then it is a (one-factor) product of irreducibles, so any counterexample is reducible. Here is another way to view the proof.
Suppose a counterexample exists. By ACCP, the set of counterexample principal ideals contains a maximal element $(r).\,$ By above, $r$ is reducible $\,r = st.\,$ But by maximality, $\,s,t\,$ have irreducible factorizations hence, appending them, so too does $\,r = st,\,$ contradiction.
Remark $\$ Interpreted positively, the proof shows if $S\,$ is the set of principal ideals disjoint from a monoid $M$ then $(r)$ is maximal in $\,S \iff r\,$ is irreducible. This is a generalization of the well-known case where $\,M = \{1\},\,$ i.e. $\ (r)$ is maximal among principal ideals $\iff r$ is rreducible.
This is a specal case of a wide class of results where elements satisfying such maximality properties are irreducible or prime, e.g. see this answer. One beautiful example along these lines is a famous theorem of Kaplansky that a domain is a UFD $\iff$ every prime ideal $\ne 0$ contains a prime $\ne 0$ (or, equivalently, every prime ideal is generated by primes).