Hallo can any one tell me what is the idea behind this?
the polynomial $p=x^{4}-x^{2}+2x+1 \in \mathbb Q[x]$ is irreducible, because in in $F_{2}[x]$ it can factored to $(x^2+x+1)^2$ and in $F_{3}[x]$ it can be factored to $(x-1)(x^3+x^2-1)$.
thank you in advance
Suppose that $p$ were reducible in $\mathbb{Q}[x]$; that is, suppose that $p=fg$, where $f,g\in\mathbb{Q}[x]$ are some monic, non-constant polynomials (in fact, $f$ and $g$ would have to be in $\mathbb{Z}[x]$ by Gauss's lemma).
For a monic polynomial $q\in\mathbb{Z}[x]$, let $\tilde{q}$ denote its reduction modulo $2$ to an element of $\mathbb{F}_2[x]$, and let $\overline{q}$ denote its reduction modulo $3$ to an element of $\mathbb{F}_3[x]$. Note that $\deg(q)=\deg(\tilde{q})=\deg(\overline{q})$.
Because the reduction maps $\mathbb{Z}[x]\to\mathbb{F}_2[x]$ and $\mathbb{Z}[x]\to\mathbb{F}_3[x]$ are ring homomorphisms, we have that $$\tilde{p}=\widetilde{fg}=\tilde{f}\tilde{g},\qquad \bar{p}=\overline{fg}=\bar{f}\bar{g}.$$ Because the factorization of $\tilde{p}$ into irreducibles in $\mathbb{F}_2[x]$ is $(x^2+x+1)^2$, where there are only two irreducible factors, the non-trivial factorization $\tilde{p}=\tilde{f}\tilde{g}$ must be this factorization, i.e. in $\mathbb{F}_2[x]$, $$\tilde{f}=\tilde{g}=x^2+x+1.$$ Thus $\deg(f)=\deg(g)=\deg(\tilde{f})=\deg(\tilde{g})=2$. However, we know that in $\mathbb{F}_3[x]$, the irreducible element $x^3+x^2-1$ divides $\bar{p}$, and since $\bar{p}=\bar{f}\bar{g}$, it must divide either $\bar{f}$ or $\bar{g}$. But this is impossible since $\deg(\bar{f})=\deg(\bar{g})=2$.