Proving that if coprime $\alpha_{i}\in R$ divide b, then $\alpha_{1}...\alpha_{n}$ divide b.

Question

Let $R$ be a principal ideal domain and let $\alpha_{1},...,\alpha_{n}\in R$ be such that $(\gcd(\alpha_{i},\alpha_{j}))=(1)$.

Let $b\in R$ such that each $\alpha_{i}|b$.

I want to show that $\alpha_{1}\alpha_{2}...\alpha_{n}|b$.

I have managed this for $\alpha_{1}$ and $\alpha_{2}$, as follows;

$\alpha_{1}|b$ so there exists $s_{1}\in R$ such that $b=\alpha_{1}s_{1}.$

By Bezout's lemma for principal ideal domains, there exists $r,s\in R$ such that $r\alpha_{1}+s\alpha_{2}=\gcd(\alpha_{1},\alpha_{2})=1$

Then $s_{1}=rb+s\alpha_{2}s_{1}$. Clearly $\alpha_{2}|s\alpha_{2}s_{1}$ and we know $\alpha_{2}|b$, so $\alpha_{2}|s_{1}$

So there exists $q\in R$ such that $s_{1}=\alpha_{2}q$

So $b=\alpha_{1}\alpha_{2}q$, and thus $\alpha_{1}\alpha_{2}|b$

I feel like this could be generalized to some kind of inductive argument, but I can't even seem to manage to show that $\alpha_{1}\alpha_{2}\alpha_{3}|b$.

Any help would be much appreciated, thanks!

Answer 1

Each principal ideal domain is factorical, so you can simply use factorization into prime elements. Then all you need to show is that being coprime means not having any prime factors in common (up to multiplication with units, etc.).