Prve that $x^4+x^2+x+1$ is irreducible over $\mathbb{Q}$.

Question

I try Eisenstein's Criterion, replacing $x$ as $x+1$, $x+2$ or $x-1$. But they are the forms we need.

it has no roots since it is always strictly positive on $\mathbb R$. This means it has two couples of complex roots, so you have to check it doesn't split into the product of two quadratic polynomial with rational coefficients — Exodd, Dec 15 '17 at 23:41

score 1 · Accepted Answer · answered Dec 15 '17 at 23:46

Let $f(x)=x^4+x^2+x+1$. Suppose that $f$ is not irreducible. $f$ has no rational root by the rational root theorem. Hence $f$ does not factor as $(x-\alpha)(x^3+\ldots)$ where $\alpha \in \mathbb Q$.

Thus $f$ is the product of two polynomials of degree $2$. Since the leading coefficient of $f$ is $1$, and the constant coefficient is $1$ as well, either $\forall x, f(x)=(x^2+ax+1)(x^2+bx+1)$ or $\forall x, f(x)=(x^2+ax-1)(x^2+bx-1)$.

In the first case, equating coefficients of degree $3$ and $1$ yields $0=a+b=1$. In the second case, the same reasoning yields $0=a+b=-1$.

Prve that $x^4+x^2+x+1$ is irreducible over $\mathbb{Q}$.

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