Factoring in Integral Domains: Can Every Element be Factored into Irreducibles?

[EbolaPox]

Homework Statement

Consider the integral domain a + b \sqrt{10}. Show that every element can be factored into a product of irreducibles, but this factorization need not be unique.

Homework Equations

The Attempt at a Solution

I know that this is not a unique factorization domain because 2,3, and 4 + \sqrt{10} are irreducible, but not prime. I also have that 2 * 3 = 6 = (4 + \sqrt{10}) (4- \sqrt{10}), so 6 has two different irreducible decompositions. So I know that I can't have unique factorizations. So, I now need to argue that every element in the integral domain can be factored into a product of irreducibles. I'm not sure how to show that. Could anyone give me a hint or a suggestion on how one would go about showing a factorization exists? I'm not looking for a full solution, just a hint in the right direction. Thanks!

 

[micromass]

Well, you could show that your ring is a Noetherian domain. This would suffices, since every Noetherian domain satisfies that every element can be written as a product of irreducibles (perhaps not uniquely).

 

[EbolaPox]

Thanks for the suggestion! Unfortunately I'm not yet familiar with Noetherian domains, so I'm not too sure how to apply that to the problem.

 

[micromass]

Hmm, are you familier with principal ideal domains? And have you seen the proof that PID implies UFD??

 

[EbolaPox]

Yes. I am familiar with that result. That's in fact the only thing I know about UFD's besides the definition of a UFD. However, I know that my integral domain in question is not a UFD and hence cannot be a PID. Can I adjust the proof they use to show a principal ideal domain is a UFD to factorize my integral domain?

How does this sound?
Let a \in R. Then (a) \subset (c) where c is irreducible. Then, c divides a. So, a = cx where c is irreducible. If x is not a unit, then I can repeat this on x to decompose a = c c_2 x_2. However, I need to show that this will eventually 'stop' and I only have finitely many irreducibles in the decomposition. I'll have to think about that

 

[micromass]

Ah yes. This is very good.

But in the proof of PID => UFD. You have shown that any element can be written as the product of irreducibles. But this proof (probably) never uses that our ring was a PID (at least not in a fundamental way). So the proof can be carried over to this context. So try to adapt the proof of PID=>UFD to this problem...

 

[micromass]

How does this sound?
Let a \in R. Then (a) \subset (c) where c is irreducible. Then, c divides a. So, a = cx where c is irreducible. If x is not a unit, then I can repeat this on x to decompose a = c c_2 x_2. However, I need to show that this will eventually 'stop' and I only have finitely many irreducibles in the decomposition. I'll have to think about that

This proof is very good. But you indeed have to show that this process stops. This is exactly what Noetherian means. In a Noetherian ring, such things will stop.

But first I'm puzzled about this. How did you find a c such that (a)\subseteq (c)?

 

[EbolaPox]

Ah, that is actually my problem. I can't claim there exists a (a) \subset (c). I have that [tex ] (a) [/tex] is contained in a maximal ideal. If my ring were a PID, I could claim that it is in a principal ideal generated by c, and I have a theorem that says that tells us c would be irreducible. I can't claim that since I don't have my assumption that R is a PID.