Let me list some definitions to avoid confusion. A ring will always be assumed to be associative.
Let $R$ be a commutative ring with identity $1_R \neq 0_R$.
- $u \in R$ is called a unit if there exists $v \in R$ such that $uv = 1_R$.
- $a \in R$ is called an associate of $b \in R$ if there exists a unit $u \in R$ such that $a=bu$.
- $a \in R$ is called a divisor of $b \in R$ if there exists $c \in R$ such that $b=ac$.
- Nonzero element $a \in R$ is called irreducible if it is not a unit and its only divisors are units and its associates.
My question is the following.
Let $a$ be an element in $R$ which is not a unit. Consider the following two conditions on $a$.
- $a$ is irreducible.
- Whenever $a=bc$ for some $b, c \in R$, either $b$ or $c$ is a unit.
Are these conditions equivalent?
The answer is true when $R$ is an integral domain. Let me write down a proof for convenience.
(1=>2) Suppose $a=bc$. If $b$ is not a unit, then there exists a unit $u \in R$ such that $a=bu$. Now we have $bc=a=bu$. Since $R$ is an integral domain and $b$ is nonzero, we obtain $c=u$ and $c$ is a unit.
(2=>1) Let $b \in R$ be a divisor of $a$. Then $a=bc$ for some $c \in R$. If $b$ is not a unit, then $c$ is a unit. Since $cv=1_R$ for some $v \in R$, we have $b=bcv=av$. Now $b$ is an associate of $a$ because $v$ is a unit.
The assumption that $R$ is an integral domain is used only in the direction (1=>2). I tried to prove it without such assumption but failed. Is it still true or is there a counterexample?
