# Trouble understanding proof regarding elements of an integral domain which are not product of irreducible elements.

I'm having trouble understanding the proof of a proposition regarding elements which are not product of irreducible elements in integral domains. The proposition is the following:

Let $$A$$ be an integral domain and $$a\in A$$ different of 0 and not a unit. If $$a$$ is not product of irreducible elements then there exists a sequence $$\{a_n\}_{n\in N}$$ of elements of $$A$$ such that $$a_{n+1}$$ is a proper divisor of $$a_n$$ for every natural $$n$$.

The proof that has been given to me is this:

Let's build inductively a sequence with this property. Let $$a_0 = a$$ and suppose $$a_0,\dots,a_n$$, $$n\geq0$$ are already built. Since $$a_n$$ is not a product of irreducibles, it must be composite, thus $$a_n = bc$$ with $$b,c$$ proper divisors of $$a_n$$. Clearly at least one of the two factors must not be a product of irreducibles. We define $$a_{n+1}$$ as this factor.

The problem I have understanding this proof is that I don't get why not being product of irreducibles implies it is composite. Wouldn't such elements be irreducibles themselves? I mean if it is composite, couldn't we keep decomposing its factors until all of them are irreducible?

Thanks for your help.

## 1 Answer

"Composite" here just means "not irreducible (and not a unit)". Since $$a_n$$ is not a product of irreducibles, it in particular is not irreducible (since then it would be a product of $$1$$ irreducible, itself). So, by definition of "irreducible", this means there exist $$b$$ and $$c$$ which are not units such that $$a_n=bc$$.