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!!

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- Jul 24th 2011, 02:53 PMrpietrovIrreducible 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 PMchisigmaRe: Irreducible Polynomials f(x) = x^7 - x
The polynomial $\displaystyle P(x)= x^{7}-x$ contain in any case $\displaystyle x$. The other roots satisfy the equation...

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jul 24th 2011, 10:47 PMFernandoRevillaRe: Irreducible Polynomials f(x) = x^7 - x