# Irreducible Polynomials f(x) = x^7 - x

• Jul 24th 2011, 02:53 PM
rpietrov
Irreducible Polynomials f(x) = x^7 - x
in Q[x], Z/2[x], and Z/7[x].

in Q[x] we think its x(x-1)(x+1)(x^2-x+1)(x^2+x+1)

in Z/2[x] we think its x(x-1)(x+1)

and we have no idea about Z/7[x] we've tried using Little Fermats Theorem.

THANKEEE YOU!!
• Jul 24th 2011, 09:22 PM
chisigma
Re: Irreducible Polynomials f(x) = x^7 - x
Quote:

Originally Posted by rpietrov
in Q[x], Z/2[x], and Z/7[x].

in Q[x] we think its x(x-1)(x+1)(x^2-x+1)(x^2+x+1)

in Z/2[x] we think its x(x-1)(x+1)

and we have no idea about Z/7[x] we've tried using Little Fermats Theorem.

THANKEEE YOU!!

The polynomial $P(x)= x^{7}-x$ contain in any case $x$. The other roots satisfy the equation...

$x^{6} \equiv 1\ , \text{mod}\ 7$ (1)

... or, equivalently, one of the equations...

$x^{3} \equiv 1\ , \text{mod}\ 7$ (2)

$x^{3} \equiv -1\ , \text{mod}\ 7$ (3)

The roots of (2) are $1$, $2$ and $4$, the roots of (3) are $-1$, $-2$ and $-4$ so that is...

$x^{7}-x= x\ (x+1)\ (x-1)\ (x+2)\ (x-2)\ (x+4)\ (x-4)$ (4)

Kind regards

$\chi$ $\sigma$
• Jul 24th 2011, 10:47 PM
FernandoRevilla
Re: Irreducible Polynomials f(x) = x^7 - x
Quote:

Originally Posted by rpietrov
in Q[x] we think its x(x-1)(x+1)(x^2-x+1)(x^2+x+1)

Right.

Quote:

in Z/2[x] we think its x(x-1)(x+1)
Try again, it should be $x^7-x=x(x+1)^2(x^2+x+1)^2$ .