# Polynomials written as a product of irreducible polynomials

• January 6th 2010, 04:11 PM
pseudonym
Polynomials written as a product of irreducible polynomials

Prove by induction that every polynomial in R[x] can be written as a product of irreducible polynomials in R[x] (Note: R is the real numbers)

Thank you!
• January 6th 2010, 05:43 PM
Shanks
Hint: prove by induction on the degree of polynomial in R[x].
• January 7th 2010, 04:07 AM
HallsofIvy
Quote:

Originally Posted by Shanks
Hint: prove by induction on the degree of polynomial in R[x].
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• January 7th 2010, 07:46 PM
Roam
Quote:

Originally Posted by pseudonym

Prove by induction that every polynomial in R[x] can be written as a product of irreducible polynomials in R[x] (Note: R is the real numbers)

Thank you!

Here's a hint:

Base case: for $n=1$ every polynomial of degree 1 is irreducible so it can be written as itself.

Inductive step: We can assume the statement is true for polynomials with degree $1,...,n$. You need to show that it is true for polynomials with degree $n+1$.

Can you continue from here?
• January 7th 2010, 08:58 PM
tonio
Quote:

Originally Posted by Roam
Here's a hint:

Base case: for $n=1$ every polynomial of degree 1 is irreducible so it can be written as itself.

Inductive step: We can assume the statement is true for polynomials with degree $1,...,n$. You need to show that it is true for polynomials with degree $n+1$.

Can you continue from here?

1) Any real polynomial of odd degree has a real root (m.v.t. for continuous functions) , and:

2) If $z\in\mathbb{C}$ is a root of a real polynomial, then also $\overline{z}$ is , or in other words: complex non-real roots of a real polynomial come in conjugate pairs.

Tonio
• January 8th 2010, 12:46 AM
Swlabr
Quote:

Originally Posted by tonio

1) Any real polynomial of odd degree has a real root (m.v.t. for continuous functions) , and:

2) If $z\in\mathbb{C}$ is a root of a real polynomial, then also $\overline{z}$ is , or in other words: complex non-real roots of a real polynomial come in conjugate pairs.

Tonio

Are these hints necessary to the proof? I mean, you start the induction (strongly!) and end up with two cases. These two cases easily break down by your inductive argument. Nothing too fancy is needed.

Unless you are hinting towards a non-inductive proof?
• January 8th 2010, 01:52 AM
Roam
To prove the inductive step you have to note that all polynomials of degree $n+1$ are either irreducible (if so they can be written as a themselves), or written as a product of two polynomials with smaller degree. Then you can use the assumption and write down each of them as a product of irreducible polynomials.
• January 8th 2010, 03:13 AM
tonio
Quote:

Originally Posted by Swlabr
Are these hints necessary to the proof? I mean, you start the induction (strongly!) and end up with two cases. These two cases easily break down by your inductive argument. Nothing too fancy is needed.

Unless you are hinting towards a non-inductive proof?

As many other times in the past I misread, or better: didn't read completely. The induction proof works for polynomials over any integral domain, and since I read "real" I assumed (because, and I checked, I did NOT read it in the OP) that he meant : any real polynomial can be written as the product of irreducible polynomials....of degree 1 and/or 2 .

My hints, thus, as huge overkill.

Tonio