How can we show that $P=x^4-5x^3+3x-2$ is irreducible over the rationals. I tried to use Eisenstein criterion and many shifts $P(x+a)$ but did not work. I tried to show it is irreducible modulo $2$ or $3$ but it did not work.
I tried to use the rational root test to prove that $P$ does not have any rational root. Also $P(-1)=1$, $P(0)=-2$, $P(1)=-3$ and $P(5)=13$.
So we know that there are two irrational roots $\alpha\in (-1,0)$ and $\beta \in (1,5)$ so $\beta \not = -\alpha$, the other two roots are conjugate complex but I don't know how to prove it, is it enough to say that $P$ has two irrational non opposite roots and two conjugate complex roots to say it is irreducible over $\mathbb Q$?
Best Answer
By Gauss' lemma, irreducibility over $\mathbb{Q}$ is equivalent to irreducibility over $\mathbb{Z}$ in this case. Since there are no rational roots, the only way $f$ can be reducible is being a product of two monic polynomials $g, h\in\mathbb{Z}[x]$ of degree $2$. But now reduce the equation $f=gh$ modulo $2$. You get that the polynomial $\overline{f}=x^4+x^3+x\in\mathbb{F_2}[x]$ is a product of two monic polynomials of degree $2$ in $\mathbb{F_2}[x]$. But that is a contradiction, because:
$\overline{f}=x(x^3+x^2+1)$
is the factorization of $\overline{f}$ into irreducibles in $\mathbb{F_2}[x]$, and such factorization is unique. (if it was a product of two polynomials of degree $2$, it wouldn't have an irreducible factor of degree $3$)