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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.

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    $\begingroup$ No. A priori your polynomial could be the product of two irreducible quadratics. $\endgroup$
    – lulu
    Jul 28, 2021 at 19:48
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    $\begingroup$ Being reducible over $\mathbb{Q}$ does not mean the polynomial must have a rational root. $\endgroup$ Jul 28, 2021 at 19:50
  • $\begingroup$ @lulu But then those quadratics will be irreducible. I dont understand. $\endgroup$ Jul 28, 2021 at 19:53
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    $\begingroup$ What don't you understand? There are irreducible quadratics over $\mathbb Q$ and the product of two such is a (reducible) quartic. $\endgroup$
    – lulu
    Jul 28, 2021 at 19:55

2 Answers 2

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(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 $\mathbb Q$.

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$f$ having no real roots doesn't mean that $f$ is irreducible over $\mathbb{Q}$, take $(X^2+1)^2$ for instance. Notice that $f=\Phi_5$ is a cyclotomic polynomial therefore it is irreducible. A way to prove it is to use Eisenstein' criterion (https://en.wikipedia.org/wiki/Eisenstein%27s_criterion) : $$f(X+1)=\frac{(X+1)^5-1}{X}=X^4+5X^3+10X^2+10X+5$$ $5$ is a prime number that divides all the coefficients of $f(X+1)$ except the leading one, and $5^2$ doesn't divide the constant coefficient squared, therefore $f(X+1)$ is irreducible over $\mathbb{Q}$, so $f$ too.

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