
Originally Posted by
peteryellow
Can somebody please check if the follwoing argument is correct:
Show that $\displaystyle x^4-4x-1$ is irrdeucible over rationals.
By the rational root theorem it has no rational roots.
Assume that
$\displaystyle x^4-4x-1 = (x^2+ax+b)(x^2+cx+d)$ then
$\displaystyle a+c=0,\ d+ac+b=0\ ad+bc=-4\ bd=-1.$
$\displaystyle a=-c$ implies that $\displaystyle d+b-c^2=0$,
$\displaystyle c(b-d)=-4$, $\displaystyle bd=-1$ Correct as far as here!
c=-4 or b-d=-4. NO!!!
Because bd=-1 it must hold that b=1,d=-1 or b=-1, d=1.
c=-4 shows that d+b=16 which is a contradiction.
if b-d=-4 we also get a contradiction. So the polynomial is irreducible.
Is it correct?