Polynomials written as a product of irreducible polynomials

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January 6th 2010, 04:11 PM
pseudonym

Polynomials written as a product of irreducible polynomials

Can someone please help me with this question:

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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Absolutely! I will help him as soon as he has show he has helped himself!(Wink)
January 7th 2010, 07:46 PM
Roam

Quote:

Originally Posted by pseudonym

Can someone please help me with this question:

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?

Two hints more (well read and better thought, the following hints are about 90% of the answer to your question):

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

Two hints more (well read and better thought, the following hints are about 90% of the answer to your question):

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