Why is this element irreducible?

Question

I am following a proof that every PID $R$ is also UFD

The proof starts this way:

Define $$S := \{a \in R \setminus U(R) \mid a \neq 0 \land a \mathrm{\ is \ not \ a \ product \ of \ irreducible \ elements}\}$$

It then says: Because $R$ is Noetherian, it follows that the set $\{(a) \mid a \in S\}$ has a maximal element $(a)$ with $a \in S$. Clearly this element is reducible.

Why is this element reducible?

For me, a product is a binary operation, so if you want to say it is reducible, there should be a factorisation $a = bc$ where both $b,c$ are non-units.

Answer 1

In this context, an irreducible element is considered a product of irreducible elements.

By assumption, $a$ is not a product of irreducible elements, so it is not irreducible (a one element product is allowed here).

Hence it is reducible.

Now the contradiction is easily reached, because $a=bc$ with neither $b$ nor $c$ invertible. Therefore $(a)\subsetneq(b)$ and $(a)\subsetneq(c)$. By maximality, both $b$ and $c$ are product of irreducible elements.