My line of proof is as follows:
- $\pm1$ are the only candidates for being rational roots (Rational Root Theorem)
- Since $f(1)=5$ and $f(-1)=1$, none is a root
And hence the given polynomial is irreducible over the rationals $\mathbb{Q}$.
Is this proof correct? I have doubt because the solution given in the book uses Eisenstein's criterion by modifying the polynomial to $f(x+1)=\frac{(x+1)^5-1}{x}$ and so on… which seems complicated to me.
Best Answer
(Alternative proof, not using Eisenstein's criterion.)
The polynomial is reciprocal and can be easily factored over the reals:
$$ \begin{align} x^4+x^3+x^2+x+1 &= x^2 \left(\left(x^2+\frac{1}{x^2}\right)+\left(x+\frac{1}{x}\right)+1\right) \\ &= x^2 \left(\left(x+\frac{1}{x}\right)^2+\left(x+\frac{1}{x}\right)-1\right) \\ &= x^2\left(x+\frac{1}{x}-\frac{-1+\sqrt{5}}{2}\right)\left(x+\frac{1}{x}-\frac{-1-\sqrt{5}}{2}\right) \\ &= \left(x^2+\frac{1-\sqrt{5}}{2}x + 1\right)\left(x^2+\frac{1+\sqrt{5}}{2}x + 1\right) \end{align} $$
Since neither quadratic has real roots this is the unique irreducible factorization over $\mathbb R$, and since the coefficients are not rational the polynomial is irreducible over $\mathbb Q$.