0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

"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 $a_n=bc$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.