# Irreducible polynomials over ring of integers ?

• Nov 26th 2011, 10:38 PM
princeps
Irreducible polynomials over ring of integers ?
Is it true that polynomials of the form :

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

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

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

Note that general form of$\displaystyle f_n$ is :$\displaystyle f_n=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ , so condition $\displaystyle a_1 \neq 1$ is equivalent to the condition$\displaystyle k \geq 1$ . Also polynomial can be rewritten into form :

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

Eisenstein's criterion , Cohn's criterion , and Perron's criterion cannot be applied to the polynomials of this form.

Example :

The polynomial $\displaystyle x^4+x^3+x^2+3x+3$ is irreducible over the integers but none of the criteria above can be applied on this polynomial.
• Nov 27th 2011, 07:00 PM
NonCommAlg
Re: Irreducible polynomials over ring of integers ?
Quote:

Originally Posted by princeps
Is it true that polynomials of the form :

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

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

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

Note that general form of$\displaystyle f_n$ is :$\displaystyle f_n=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ , so condition $\displaystyle a_1 \neq 1$ is equivalent to the condition$\displaystyle k \geq 1$ . Also polynomial can be rewritten into form :

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

Eisenstein's criterion , Cohn's criterion , and Perron's criterion cannot be applied to the polynomials of this form.

Example :

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

i don't know the answer to the general case but it's easy to prove that if $\displaystyle n = 2^m - 1, \ m \geq 2, \ 2 \mid k$ and $\displaystyle a \equiv 3 \mod 4,$ then $\displaystyle f$ is irreducible.
to see this, note that the numbers $\displaystyle \binom{2^m}{i}, \ 0 < i < 2^m,$ and $\displaystyle a - 1$ are even and hence

$\displaystyle f(x+1) = \frac{(x+1)^{n+1}+(a-1)(x+1)^{k+1}-a}{x}=x^{n} + a_{n-1}x^{n-1}+ \ldots +$ $\displaystyle a_1x + a_0,$

where all $\displaystyle a_j$ are even and $\displaystyle a_0 = 2^m + (k+1)(a-1) \not \equiv 0 \mod 4.$ thus $\displaystyle f(x+1),$ and so $\displaystyle f(x),$ is irreducible by the Eisenstein's criterion.
• Nov 28th 2011, 01:54 AM
princeps
Re: Irreducible polynomials over ring of integers ?
Quote:

Originally Posted by NonCommAlg
$\displaystyle f(x+1) = \frac{(x+1)^{n+1}+(a-1)(x+1)^{k+1}-a}{x}=x^{n} + a_{n-1}x^{n-1}+ \ldots +$ $\displaystyle a_1x + a_0,$

where all $\displaystyle a_j$ are even and $\displaystyle a_0 = 2^m + (k+1)(a-1) \not \equiv 0 \mod 4.$

There is condition in the text of the question $\displaystyle a_k=a_{k-1}=\ldots =a_1=a_0=a$ , where $\displaystyle a$ is an odd number so $\displaystyle a_j$ cannot be even number.
• Nov 28th 2011, 03:13 AM
NonCommAlg
Re: Irreducible polynomials over ring of integers ?
Quote:

Originally Posted by princeps
There is condition in the text of the question $\displaystyle a_k=a_{k-1}=\ldots =a_1=a_0=a$ , where $\displaystyle a$ is an odd number so $\displaystyle a_j$ cannot be even number.

in my solution, $\displaystyle a_j$ are the coefficients of $\displaystyle f(x+1)$ not $\displaystyle f(x)$.
• Nov 28th 2011, 04:04 AM
princeps
Re: Irreducible polynomials over ring of integers ?
Can you give me an example of f(x+1) ?