1. ## Unique Factorization

Hello All,

Can someone prove unique factorization for the set of polynomials in $x$ with integer coefficients?

I know that the analogous Euclidean algorithm functions for these polynomials using "degree" instead of "norm."

Any help is appreciated!

2. To me, this problem seems like it would be easy given the wording of the question and knowing that the Euclidean algorithm applies for these polynomials. Is this more complex ? Can anyone offer any help?

3. Originally Posted by Samson
To me, this problem seems like it would be easy given the wording of the question and knowing that the Euclidean algorithm applies for these polynomials. Is this more complex ? Can anyone offer any help?
Try this on for size

Polynomial ring - Wikipedia, the free encyclopedia (subheading: Factorization in K[X])

4. I looked at it but that didn't seem to get me too far. What was it that I was supposed to notice? I failed to see how it connected for the set of polynomials in x.

5. Originally Posted by Samson
I looked at it but that didn't seem to get me too far. What was it that I was supposed to notice? I failed to see how it connected for the set of polynomials in x.
Samson, did you ever get a response or figure this problem out? I am interested in this as well.

If not, could someone post an explanation for the OP's question? --Brim

6. Originally Posted by Brimley
Samson, did you ever get a response or figure this problem out? I am interested in this as well.

If not, could someone post an explanation for the OP's question? --Brim
Hi Brimley,

No one ever did respond to my last post and I never did get this figured out. I'm still hoping that someone will post a solution proving unique factorization for the set of polynomials in $x$ with integer coefficients.

7. Well it turns out the link I gave is not directly applicable since $\,\mathbb{Z}$ is not a field. But $\,\mathbb{Z}$ is a unique factorization domain (UFD), and it can be proven that the ring of polynomials over a UFD is also a UFD. Apologies for not noticing this.

I am looking into the best way to present a proof and/or give a reference. It should be noted that I am not an expert in this but am also learning, as time permits. I try to make sure everything I post is well-reasoned and accurate, but of course, it doesn't always work out quite like that.

I hope to post again before too many hours pass with something useful; I am busy at the moment.

8. Originally Posted by undefined
Well it turns out the link I gave is not directly applicable since $\,\mathbb{Z}$ is not a field. But $\,\mathbb{Z}$ is a unique factorization domain (UFD), and it can be proven that the ring of polynomials over a UFD is also a UFD. Apologies for not noticing this.

I am looking into the best way to present a proof and/or give a reference. It should be noted that I am not an expert in this but am also learning, as time permits. I try to make sure everything I post is well-reasoned and accurate, but of course, it doesn't always work out quite like that.

I hope to post again before too many hours pass with something useful; I am busy at the moment.
Thank you undefined. I will be looking forward to your post with great anticipation! --Brim

9. This is a two-step procedure.

1) Show that any polynomial $p(x)\in \mathbb{Z}[x]$ can be factored into irreducibles.
2) Show that the factorization is unique.

Try part one first. The trick is to use strong induction on the degree of $p$.

10. Originally Posted by roninpro
This is a two-step procedure.

1) Show that any polynomial $p(x)\in \mathbb{Z}[x]$ can be factored into irreducibles.
2) Show that the factorization is unique.

Try part one first. The trick is to use strong induction on the degree of $p$.
For part 1) The following text is from the Wikipedia article "Irreducible polynomial" -
Every polynomial p(x) in F[x] can be factorized into polynomials that are irreducible over F. This factorization is unique up to permutation of the factors and the multiplication of the factors by constants from F (because the ring of polynomials over a field is a unique factorization domain).

Any polynomial over F must share either no roots or all roots with any given irreducible polynomial; this is Abel's irreducibility theorem.
Is there a way to best explain this in general terms without citing all of Abel's irreducibility theorem? (Is there an easier way?) roninpro suggested using strong induction on the degree of p. I'm not entirely on top of how to do that.

For part 2) I found this link here: Unique Factorization of Polynomials
However, it shows this only for a specific example. Is there a more general version of this that is applicable to the OP's case?

11. Hey roninpro, I am not sure on how to use the strong induction that you had mentioned, although I know what it is. According to Wikipedia [link here], "Another generalization, called complete induction (or strong induction or course of values induction), says that in the second step we may assume not only that the statement holds for n = m but also that it is true for all n less than or equal to m.".

12. For part one, you can check a few base cases for degree 0 and 1. Suppose that every polynomial of degree $\leq n$ can be factored into irreducibles. Let $p$ be a polynomial of degree $n+1$. If it is irreducible, then we are done. On the other hand, suppose $p$ is reducible. Then, we can find polynomials $q, r$ of positive degree $ such that $p(x)=q(x)r(x)$. What can be said at this point?

13. Originally Posted by roninpro
For part one, you can check a few base cases for degree 0 and 1. Suppose that every polynomial of degree $\leq n$ can be factored into irreducibles. Let $p$ be a polynomial of degree $n+1$. If it is irreducible, then we are done. On the other hand, suppose $p$ is reducible. Then, we can find polynomials $q, r$ of positive degree $ such that $p(x)=q(x)r(x)$. What can be said at this point?
Can't $p(x)=q(x)r(x)$ be broken down even further then if it is reducible? I'm not sure how, but I'm just trying to follow what you're saying.

14. Yes. You use the induction hypothesis on $q$ and $r$.

15. Could you continue to show this? I'm following along but this is a weakpoint for me.

Here is what I dug up though:
if P(k) is true (called induction hypothesis), then P(k+1) is true
I don't know if thats applicable, but I'm not sure how I can break it down as you had mentioned using the induction hypothesis.

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