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.


"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 $a_n=bc$.


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