Expressing Elements as Products of Irreducibles in Unique Factorization Domains

[EbolaPox]

I'm having a bit of trouble with some ring theory I've been reading about, specifically unique factorization domains. I'm not really clear on how one would go about showing that an element can be factored into irreducibles

Homework Statement

Let $$R$$ be an integral domain such that every prime ideal in R$ contains a principal prime ideal $$(a)$$ , where $$a \neq 0$$.

The Attempt at a Solution

Let $$b \in R$$. I need to show that I can express b as a product of irreducibles. I'm not too sure how my assumption can help me out here. I do know that if $$c$$ is irreducible, then $$(c)$$ is maximal in the set of proper principal ideals. So, maybe if I consider the principal ideal of b, I can find a $$c_1$$ such that $$(b) \subset (c_1)$$ and so $$c_1$$ divides b. Therefore, I have $$b = c_1 x$$ where $$c_1$$ is irreducible. I then thought I could apply the same idea to $$x$$ and get $$(x) \subset (c_2)$$, where $$c_2$$ is irreducible and hence $$x = x_2 c_2 , b = c_1 c_2 x_2$$. But, I'm not sure if this makes sense. I'm not sure if I can necessarily even find a maximal proper principal ideal containing b. I then thought I could "continue" this procedure, but I'm not sure how to know when this will end and I'm not using my assumptions at all. I need to somehow use the fact that a prime ideal contains a principal ideal. I have no idea how to use this idea.

I'm not looking for a full solution, I'd just like a helpful hint in the right direction. Thanks!EDIT: Never mind, I figured it out.

 

[lippyka]

You can use the fact that every prime ideal contains a principal ideal to construct a sequence of principal ideals whose intersection is the principal ideal generated by b. Then you can use the fact that a principal ideal is a product of irreducibles to factor b into irreducibles.