Let $R$ an UFD and $a,b,c \in R$ be such that $a$ and $b$ are nonzero and noninvertible. Suppose $a = bc$. How can I show that the number of irreducible factors of $a$ is greater than or equal to the number of irreducible factors of $b$?
I could easily check that for the case where $c$ is not invertible, the above holds.
Let $n(x)$ be the number of irreducible factors of the element $x$.
Let us note that $n(c) \geq 1 \geq 0$, then $n(b)+n(c) \geq n(b)$ and since $a = bc$, $n(a) = n(b)+n(c) \geq n(b)$.
But for the case where $c$ is invertible, it is not so clear to me how the above fact is satisfied. I have only been able to think that $a$ will have another decomposition into product of irreducible factors in which it differs from an invertible factor (can I claim in this situation that $n(a) = n(bc)$?), which in this case is $c$.