Zion Wheeler

2022-06-26

To prove $f={x}^{4}+{x}^{3}+{x}^{2}+{x}^{1}+1$ is irreducible over the $\mathbb{Q}$

benedictazk

Expert

The polynomial is reciprocal and can be easily factored over the reals:
$\begin{array}{rl}{x}^{4}+{x}^{3}+{x}^{2}+x+1& ={x}^{2}\left(\left({x}^{2}+\frac{1}{{x}^{2}}\right)+\left(x+\frac{1}{x}\right)+1\right)\\ & ={x}^{2}\left({\left(x+\frac{1}{x}\right)}^{2}+\left(x+\frac{1}{x}\right)-1\right)\\ & ={x}^{2}\left(x+\frac{1}{x}-\frac{-1+\sqrt{5}}{2}\right)\left(x+\frac{1}{x}-\frac{-1-\sqrt{5}}{2}\right)\\ & =\left({x}^{2}+\frac{1-\sqrt{5}}{2}x+1\right)\left({x}^{2}+\frac{1+\sqrt{5}}{2}x+1\right)\end{array}$
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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