Re: irredicuble polynomials
- From: Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Fri, 25 Nov 2005 01:05:48 GMT
In article
<6753587.1132828228028.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
eugene <jane1806@xxxxxxx> wrote:
> I am stuck with the proof of the following fact:
>
> Prove that for every prime power q, one has that for all n there exists and
> irreducible polynomial of degree n.
>
> At the begginning the proof suggest the following identity
>
> \product_{k=1^\infty} (1/(1-x^k)^I(k))=\sum_{n=1}^{\infty}q^n x^n
>
> where I(k)-the number of irredicuble polynomials of degree k(we are working
> in our field F_q).
> Could you please explain me this identity.
> Thanks in advance
I hate to harp on typos, but the word you are looking for
is "irreducible."
Anyway, the right side of the identity, the coefficient of x^n
is the number of monic polynomials of degree n, right? Now, each
such polynomial has a unique expression as a product of (monic)
irreducible polynomials, right?
On the left side,
1 / (1 - x^k)^I(k) = (1 + x^k + x^(2k) + ... )^I(k)
Now all you have to do is identify each term in the expansion of
product_{k = 1 to infinity} (1 + x^k + x^(2k) + ... )^I(k)
with a factorization into irreducibles. Can you see it?
--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.
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