# Unique factorization

• Oct 31st 2006, 06:24 PM
beta12
Unique factorization
Hi all,
If you have idea about the questions that I posted lastly , please do help.
I found the quadratic field chapter is very difficult. Even I read finished the book but still got stuck in the questions most of the time. :(

Question
Prove unique factorization for the set of polynomials in x, with integer coefficients.

Hint: The analogous Euclidean algorithm works for such polynomials,
using " degree" instead of " norm" . Prove this and then apply similar arguments to show unique factorization.
• Oct 31st 2006, 07:18 PM
topsquark
Quote:

Originally Posted by beta12
Hi all,
If you have idea about the questions that I posted lastly , please do help.
I found the quadratic field chapter is very difficult. Even I read finished the book but still got stuck in the questions most of the time. :(

Question
Prove unique factorization for the set of polynomials in x, with integer coefficients.

Hint: The analogous Euclidean algorithm works for such polynomials,
using " degree" instead of " norm" . Prove this and then apply similar arguments to show unique factorization.

This is an open invitation for someone to pick my proof apart and tell me where I got lazy. :)

Beta12, please let me know if and where I lose you in this.

Let Q[x] be your field of polynomials.

Given an element f(x) in Q[x] either f(x) is irreducible or it is not. If it is irreducible then we are done factorizing f(x). If not then f(x) = g(x)h(x) for some polynomials g(x), h(x) with deg(g(x)) < deg(f(x)) and deg(h(x)) < deg(f(x)). If both g(x) and h(x) are irreducible the f(x) = g(x)h(x) is our factorization of f(x) into irreducibles. If not then at least one of g(x), h(x) must be reducible. We may continue in this fashion. The process is guarenteed to stop at some point because the degree of the factor polynomials is less at each stage. In this fashion we may always factor the polynomial f(x) into irreducible polynomials in Q[x].

Now, say we have two factorizations of f(x) into irreducible polynomials:
$f(x) = p_1(x)p_2(x)...p_s(x) = q_1(x)q_2(x)...q_t(x)$

We may assume $s \leq t$ and we may assume t > 1. (If t = 1 then f(x) was irreducible and, of course, both factorizations will be the same, up to a constant.)

Now, $f(x) = q_1(x)q_2(x)...q_t(x)$ and we know that $p_1(x)$ divides f(x) since it is a factor in the first list. Thus we know that $p_1(x)$ divides $q_1(x)q_2(x)...q_t(x)$. But all the q(x)'s are irreducible and so is $p_1(x)$. Thus $p_1(x)$ and one of the q(x)'s must be the same up to a constant.

We may relabel the list of q(x)'s such that this particular q(x) is $q_1(x)$ for convenience. Define $p_1(x) = c_1q_1(x)$. Thus
$f(x) = p_1(x)p_2(x)...p_s(x) = c_1p_1(x)q_2(x)...q_t(x)$

Now consider $p_2(x)...p_s(x) = q_2(x)...q_t(x)$
We may use the same process as above to find a q(x) such that $p_2(x) = c_2q_2(x)$ shuffling the order of the q(x)'s as needed. Thus:
$f(x) = p_1(x)p_2(x)...p_s(x) = c_1c_2p_1(x)p_2(x)...q_t(x)$

We may continue this process for all the p(x)'s:
$f(x) = p_1(x)p_2(x)...p_s(x) = c_1c_2...c_sp_1(x)p_2(x)...p_s(x)q_{s+1}(x)...q_t( x)$

Now, comparing the list of factors we see that the polynomial $p_1(x)p_2(x)...p_s(x)$ must be the same as the polynomial $p_1(x)p_2(x)...p_s(x)q_{s+1}(x)...q_t(x)$ up to a constant. Thus by looking at the degrees of the p(x)'s and the remaining q(x)'s we see that $deg(q_{s+1}(x)...q_t(x)) = 0$. Thus all of the remaining q(x)'s can only have a degree of 0, ie they can only be constants.

In this manner we see that the two factorizations are the same, up to an overall constant.

-Dan
• Oct 31st 2006, 07:21 PM
ThePerfectHacker
Quote:

Originally Posted by beta12
Question
Prove unique factorization for the set of polynomials in x, with integer coefficients.

The most fundamental result with polynomials over a field is that if,
$p(x)$ is an irreducible polynomial over a field and it divides $f(x)g(x)$ then it divides either $f(x)$ or $g(x)$.

(I will assume you are familar with field theory).

Proof: Assume $p(x)$ divides $f(x)g(x)$ then $f(x)g(x)\in $ (principal ideal). But $$ is a maximal ideal because $p(x)$ is an irreducible polynomial. Thus, $$ is a prime ideal. Which means, $f(x)\in $ or $g(x)\in $, thus, $p(x)$ divides either $f(x)$ or $g(x)$.

It is not difficult to show each polynomial over a field factors into irreducible polynomials.

Nnow we can show uniquess in a field factorization.
If,
$p_1(x)p_2(x)...p_n(x)=q_1(x)q_2(x)...q_m(x)$
Then the right hand side is divisible by $p_1(x)$ thus since it is odd it divides some $q_i(x)$ but since of irreducibility $p_1(x)=kq_i(x)$. Continuing in this fashion we are left with,
$1=k_1k_2k_3...k_sq_1(x)q_2(x)...q_r(x)$
Thus, $n=m$ so that the equation would become,
$1=k_1k_2\cdot...\cdot k_n$
Where the constants cancel each other out.

I know, I know that the the messiest proof ever but I was trying to show the idea to why we have uniqueness in factorization.

Quote:

Originally Posted by topsquark
This is an open invitation for someone to pick my proof apart and tell me where I got lazy

You never metion the important propery about irreducible polynomials. Nor prove it.
• Nov 1st 2006, 09:49 AM
beta12
Hi Topsquark,

Very nice proof. :)

Also , Perfecthacker, I got what you mean too. Thanks.
• Nov 1st 2006, 05:33 PM
topsquark
Quote:

Originally Posted by beta12
Hi Topsquark,

Very nice proof. :)

Also , Perfecthacker, I got what you mean too. Thanks.

You can thank the Mr. B. Sethuraman who wrote "Rings, Fields, and Vector Spaces." That's where I got most of the proof from. (It's a very nice little book if you can get your hands on it. It doesn't go very deep, but gives a nice overview on geometric constructibility.)

-Dan
• Nov 1st 2006, 05:47 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
You can thank the Mr. B. Sethuraman who wrote "Rings, Fields, and Vector Spaces." That's where I got most of the proof from. (It's a very nice little book if you can get your hands on it. It doesn't go very deep, but gives a nice overview on geometric constructibility.)

-Dan

I try to memorize proofs. Because sometimes you can create your own theorem using a very clever trick that you seen used in another theorem.

The Algebra book I have is, Abstract Algebra 7/e. Awesome book, you can ignore the bad reviews because those are the lesser mortal people that were not able to understand it and thus need to complain.

By the way... topsquark... the question on Sylow theorems is developed just like in Hungerford (that is what the author says). Does it use class equations?
• Nov 1st 2006, 05:54 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
I try to memorize proofs. Because sometimes you can create your own theorem using a very clever trick that you seen used in another theorem.

The Algebra book I have is, Abstract Algebra 7/e. Awesome book, you can ignore the bad reviews because those are the lesser mortal people that were not able to understand it and thus need to complain.

By the way... topsquark... the question on Sylow theorems is developed just like in Hungerford (that is what the author says). Does it use class equations?

I'm not entirely sure as I don't know what a "class equation" is. (I looked it up on the "Wolfram Mathworld" site and couldn't make sense of it.) If they are in the book I didn't get that far into it, or the author didn't specifically give them a name when he used them.

-Dan
• Nov 1st 2006, 06:17 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
I'm not entirely sure as I don't know what a "class equation" is.

Given a Group G
And a set G (the group itself)
Define group action on G as the conjugation map (inner automorphism)
The orbits from this action are conjugate classes.
There is an equation involving these conjugate class.
Called the class equation.
• Nov 2nd 2006, 10:22 AM
beta12
Quote:

Originally Posted by topsquark
You can thank the Mr. B. Sethuraman who wrote "Rings, Fields, and Vector Spaces." That's where I got most of the proof from. (It's a very nice little book if you can get your hands on it. It doesn't go very deep, but gives a nice overview on geometric constructibility.)

-Dan

I see. :)