Suppose we have a non-zero non unit $r$ in $R$ that cannot be written as a product of irreducibles ...

Question

Problem: Let $R$ be an integral domain. Suppose we have a non-zero non-unit $r$ in $R$ that cannot be written as a product of irreducibles, then my textbook states without proof that $r$ is "certainly not irreducible".

I'm not sure why this is true. If $r=ab$ then consider two cases:

1) $a,b$ are both reducible, in this case I have to show that $r=ab$ implies that both $a,b$ are non-units (I think). Not sure how to proceed

2) The case where $a$ is reducible, and $b$ is not (or vice versa).

I am very confused right now and I think that I am missing something basic. Insights appreciated.

My definition of irreducible is : An element $p \in R$ is irreducible if $p \neq 0$, $p$ is not a unit, and if $p=ab$ then either $a$ or $b$ is a unit. If an element does not satisfy this definition it is reducible.

This comes from the chapter "5.1: Irreducibles and Unique Factorization" from Nicholson's introduction to Abstract Algebra 4th edition page 255.

Answer 1

Note that "product of irreducibles" means "product of any number of irreducibles", not necessarily a binary product. So an irreducible element $r$ can always be written as a product of irreducibles, namely a unary product consisting just of $r$.

Answer 2

I suspect the issue is that the author is using a different definition of product than you are.

A fairly standard definition is something like the following:

Let $n$ be a natural number, regarded as an ordinal, so $n=\{0,\ldots,n-1\}$. Let $s:n\to A$ be a map to a ring (or more generally a monoid).

Then we define $\prod_{i\in n} s_i$, usually written as $\prod_{i=0}^{n-1} s_n$ recursively by $$\prod_{i\in 0} s_i =1,$$ and $$\prod_{i\in k+1} s_i = \left(\prod_{i\in k} s_i\right)\cdot s_k.$$

Note that this definition gives meaning to the notion of the empty and unary products.