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Let $R$ be an integral domain.

Let $r \in R - R^*$ be an element, such that $r$ cannot be written as a product of irreducible elements $a,b \in R$.

Prove that $r$ is irreducible and that $r = bd, b, d \notin R^*$ where $R^*$ denote the set of units of $R$.

My idea: $r$ cannot be written as a product of irreducible elements imply that $r = bd$, where $b$ or $d$ is not irreducible. Suppose $b$ is not irreducible, then $r=jkd$, $j, k$ are non-units. Since $r=j(kd)$ and $kd\notin R^*$ this means $r$ is not irreducible and can be written as $j(kd)$ - a unit is irreducible imply $j$ is not a unit.

What if $r$ cannot be factored except $r=1r$, then $r$ is irreducible ? So we are looking at $r \in R$ that has non-trivial factorization ?

Thanks

Andrés E. Caicedo
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Shuzheng
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    If $r=bd$ is a product of two non-units, it is reducible (and not irreducible). I am not sure what you want to prove. – Dietrich Burde Oct 13 '13 at 20:13
  • This is something from a proof that in a pricipal ideal domain every $r \in R $ has a factorization as a product of irreducible elements.

    Suppose $r$ cannot be factored as a product of irreducible elements. Prove that $r$ is irreducible and that $r=bd, b, d \notin R^*$

     – Shuzheng Oct 13 '13 at 21:08
  • updated: $r \in R - R^*$ – Shuzheng Oct 13 '13 at 21:15
  • Yes, in a PID its true, because a PID is Noetherian. In general, it is false. See also http://math.stackexchange.com/questions/297271/does-any-integral-domain-contain-an-irreducible-element. – Dietrich Burde Oct 15 '13 at 09:58

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