Unique Factorization: Proving for Polynomials in x

In summary, Omar provided a summary of the content. He said that Z[x] is a ufd, due to gauss, and that this proof should appear in every abstract algebra book. He also said that Q[x] is a ufd, and that you can prove this using the idea you mentioned.
  • #1
stoolie77
7
0
Hello everybody. I had been reading up on Unique Factorization again and I came across an interesting question.

Can someone prove unique factorization for the set of polynomials in [itex]x[/itex], with integer coefficients?

From what I understand, the analogous Euclidean algorithm works for such polynomials using "degree" instead of "norm".

If someone could answer this, that would be great as it might clear up some of my other thoughts I have with Unique Factorization.

Gracias - Omar
 
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  • #2
Can anyone post a solution to this? I would like to see this as well. --Brim
 
  • #3
stoolie77 said:
Hello everybody. I had been reading up on Unique Factorization again and I came across an interesting question.

Can someone prove unique factorization for the set of polynomials in [itex]x[/itex], with integer coefficients?

From what I understand, the analogous Euclidean algorithm works for such polynomials using "degree" instead of "norm".

If someone could answer this, that would be great as it might clear up some of my other thoughts I have with Unique Factorization.

Gracias - Omar

see the book by Ireland and Rosen "A Classical Introduction to Modern Number Theory" 2nd edition pp 6-8
 
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  • #4
I don't know exactly where to find page 7 and 8, but here is page 6. I still don't know how to relate it directly in terms that aren't advanced as this.
 

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  • #5
the fact that Z[X] is a ufd is due to gauss. this proof should appear in every abstract algebra book, e.g. in most of the free ones on my web page.

it is NOT a euclidean domain however. the crucial concept is tod efine the "content" o=f a polynomial as the gcd of the coefficients, then prove the content of a product is the product of the contents. then one can reduce the proof to the case of Q[X].
 
  • #6
mathwonk said:
the fact that Z[X] is a ufd is due to gauss. this proof should appear in every abstract algebra book, e.g. in most of the free ones on my web page.

it is NOT a euclidean domain however. the crucial concept is tod efine the "content" o=f a polynomial as the gcd of the coefficients, then prove the content of a product is the product of the contents. then one can reduce the proof to the case of Q[X].

Hello Roy, I reviewed your webpage although I didn't find any free sources that gave a lot of insight on this, and I didn't entirely make sense of your response. I know that you were using Z[x] as a UFD, however I'm trying to prove unique factorization for the set of polynomials in [itex]x[/itex] with integer coefficients. You talk about defining the "content" of a polynomial as the GCD of its coefficients, then proving the content of a product is the product of its contents. I'm totally lost with what you're saying.
 
  • #7
Z[x] is my notation for the ring of polynomials in one variable x with integer coefficients. The proof this is a ufd is what I thought you wanted. this is due to gauss, and is reduced down to the case of assuming that Q[x] is a ufd, where Q is the rational numbers.

let me give you specific refernces to my web page.

math 844.1, section 4, page 18. that Z[x] is a ufd is specifically proved in detail.

math 80006b,pages 19-24, more terse than above.

the last two sections on my math 4000 notes also discuss the ideas. In particular they discuss the fact that Q[x] is a principal ideal domain, hence also a unique factorization domain, using the concept of degree as euclidean norm. Moreover they prove that any element of Z[x] that factors in Q[x] already factors in Z[x], the main step to proving that Z[x] is a ufd, "Gauss lemma."so indeed this topic is discussed in most of the notes there.
 
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  • #8
If you use the idea you mentioned, you can prove that Q[x] is a ufd. Then the attached notes (extracted from my web page) allow you to deduce it for Z[x], as gauss did.
 

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Related to Unique Factorization: Proving for Polynomials in x

1. What is unique factorization for polynomials in x?

Unique factorization for polynomials in x is the process of breaking down a polynomial with coefficients in a given field into a product of irreducible polynomials. This means that the polynomial cannot be factored further into smaller polynomials with coefficients in the same field.

2. Why is unique factorization important for polynomials in x?

Unique factorization is important for polynomials in x because it allows us to simplify and solve polynomial equations. It also helps us to determine the roots or zeros of a polynomial, which can be useful in various applications including engineering, physics, and economics.

3. How do you prove unique factorization for polynomials in x?

The proof of unique factorization for polynomials in x involves showing that every polynomial can be factored into a product of irreducible polynomials and that this factorization is unique up to order and multiplication by a constant. This can be done using mathematical induction and the fundamental theorem of algebra.

4. What is an irreducible polynomial?

An irreducible polynomial is a polynomial that cannot be factored into smaller polynomials with coefficients in the same field. In other words, it has no factors other than itself and the constant 1. Examples of irreducible polynomials include x^2 + 1 and x^3 + 2x + 1.

5. Can unique factorization be applied to all polynomials?

No, unique factorization can only be applied to polynomials with coefficients in a given field. For example, unique factorization cannot be applied to polynomials with coefficients in the set of integers, as seen in the case of x^2 + 1 which cannot be factored into linear factors in the set of integers.

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