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

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.

• No. A priori your polynomial could be the product of two irreducible quadratics.
– lulu
Jul 28 '21 at 19:48
• Being reducible over $\mathbb{Q}$ does not mean the polynomial must have a rational root. Jul 28 '21 at 19:50
• @lulu But then those quadratics will be irreducible. I dont understand. Jul 28 '21 at 19:53
• What don't you understand? There are irreducible quadratics over $\mathbb Q$ and the product of two such is a (reducible) quartic.
– lulu
Jul 28 '21 at 19:55

(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 $$\mathbb Q$$.

• this is ingenious..Thank you Jul 29 '21 at 18:53

$$f$$ 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.