Exercise 1933Algebra › PolynomialsOlympiadAdded 4 Nov 2026

Irreducibility with Eisenstein's criterion

#Eisenstein#irreducibility#cyclotomic polynomial#shift trick

Statement

Eisenstein's criterion. Let $f(x)=a_nx^n+\dots+a_0$ have integer coefficients and let $p$ be a prime with $p\nmid a_n$ , $p\mid a_i$ for $i<n$ , and $p^2\nmid a_0$ . Then $f$ cannot be written as a product of two non-constant polynomials with integer coefficients.

Use it to prove:
1. $x^4+1$ is irreducible over $\mathbb{Q}$ .
2. For a prime $p$ , $\Phi_p(x)=x^{p-1}+x^{p-2}+\dots+x+1$ is irreducible over $\mathbb{Q}$ .

Hint

Eisenstein does not apply directly. Replace $x$ by $x+1$ : if $f(x+1)$ is irreducible, so is $f(x)$ .

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Solution

Shift trick. A factorisation $f(x)=g(x)h(x)$ gives $f(x+1)=g(x+1)h(x+1)$ with the same degrees, and conversely.

1. $(x+1)^4+1=x^4+4x^3+6x^2+4x+2$ . With $p=2$ : the leading coefficient $1$ is odd; $4,6,4,2$ are all even; and the constant term $2$ is not divisible by $4$ . Eisenstein applies: irreducible.

2. Since $\Phi_p(x)=\dfrac{x^p-1}{x-1}$ ,
$\Phi_p(x+1)=\frac{(x+1)^p-1}{x}=\sum_{k=1}^{p}\binom pkx^{k-1}=x^{p-1}+\binom p{p-1}x^{p-2}+\dots+\binom p2x+p.$
The leading coefficient is $1$ . For $1\le k\le p-1$ , $p\mid\binom pk$ . The constant term is $\binom p1=p$ , not divisible by $p^2$ . By Eisenstein, $\Phi_p(x+1)$ is irreducible, hence so is $\Phi_p(x)$ .

Remark. Thus $\Phi_p$ is the minimal polynomial of $e^{2\pi i/p}$ over $\mathbb{Q}$ .

Lesson. If a polynomial looks irreducible but Eisenstein fails, try the shifts $x\mapsto x+1$ or $x\mapsto x-1$ .

Irreducibility with Eisenstein's criterion

Statement

Solution

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