Irreducible polynomials over ring of integers

1. Nov 19, 2011

pedja

Is it true that polynomials of the form :

$f_n= x^n+x^{n-1}+\cdots+x^{k+1}+ax^k+ax^{k-1}+\cdots+a$

where $\gcd(n+1,k+1)=1$ , $a\in \mathbb{Z^{+}}$ , $a$ is odd number , $a>1$, and $a_1\neq 1$

are irreducible over the ring of integers $\mathbb{Z}$?

http://en.wikipedia.org/wiki/Eisenstein%27s_criterion" [Broken] cannot be applied to the polynomials of this form.

Example :

The polynomial $x^4+x^3+x^2+3x+3$ is irreducible over the integers but none of the criteria above can be applied on this polynomial.

Last edited by a moderator: May 5, 2017
2. Nov 26, 2011

COURAGE_WOLF

I hope this helps.

The polynomial x4+x3+x2+x +1 is the cyclotomic polynomial $\Phi$5(x). I believe the general polynomial that you described is also the monic cyclotomic polynomial $\Phi$n(x) in Z[x] with degree $\varphi$(n). These polynomials are irreducible. The proof is a little tedious and not exactly immediate. Check out this paper

http://www.math.umn.edu/~garrett/m/algebra/notes/08.pdf

I think it does a pretty good job. Then again, if that general polynomial turns out to not be cyclotomic, I'll have to start over and come up with something new

3. Nov 26, 2011

pedja

These polynomials are not cyclotomic polynomials.
$f_n$ can be rewritten into form :

$f_n=\displaystyle \sum_{i=0}^n x^{i}+(a-1)\cdot \displaystyle \sum_{i=0}^k x^{i}$ ,or

$f_n=\frac{x^{n+1}+(a-1)x^{k+1}-a}{x-1}$

4. Nov 26, 2011

COURAGE_WOLF

Maybe if you give a different example it would be easier to see. I don't really see much of a correlation between the general case and the example you gave. Anyways, you can always fall back on the general technique for finding irreduciblity.

Let I be a proper ideal in the integral domain R and let p(x) be the monic polynomial in R[x]. If the image of p(x) in (R/I)[x] cannot be factored in (R/I)[x] into two polynomials of smaller dgree, then p(x) is irreducible in R[x].

Start reducing mod some n and see where that gets you. If it's small degree then it's pretty obvious. A higher degree will probably take a little work. Unfortunately, this technique doesn't always work, but I can't think of any other irreducibility criteria.

5. Nov 26, 2011

COURAGE_WOLF

Wait I just thought of another one: Hilbert's Irreducibility Theorem.

6. Nov 26, 2011

COURAGE_WOLF

Also Kronecker's Method