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Eisenstein's criterion. Let $f(X)=a_n{X}^n+\cdots +a_0\in \mathbb{Z}[X]$ be a nonconstant primitive polynomial. If a prime number $p$ satisfies $p\nmid a_n$, $p\mid a_j$ for $0\le j<n$, and $p^2\nmid a_0$, then $f$ is an irreducible polynomial over $\mathbb{Q}$.

To prove the Eisenstein criterion, suppose a nontrivial factorization exists over $\mathbb{Q}$. Gauss lemma for polynomials gives $f=gh$ with $g,h\in \mathbb{Z}[X]$ of positive degree. Their leading coefficients are not divisible by $p$, because their product is $a_n$. Reduction modulo $p$ therefore preserves both degrees, andSince the only irreducible factor of the right side is $X$, both reduced factors are nonzero monomials of positive degree. Their constant coefficients are consequently divisible by $p$. This would make $p^2\mid g(0)h(0)=a_0$, a contradiction.

For the requested polynomial, shift the variable by one. The binomial theorem givesIt is monic, all coefficients below the leading one are divisible by $p$, and the constant coefficient is $p$, not divisible by $p^2$. The Eisenstein criterion proves irreducibility of this shifted polynomial. The substitution $X\mapsto X+1$ is an automorphism of $\mathbb{Q}[X]$, with inverse $X\mapsto X-1$, so it preserves nontrivial factorizations. HenceThis is the cyclotomic polynomial for a prime index.

A unique factorization domain is an integral domain in which every nonzero nonunit is a finite product of irreducible elements, uniquely up to reordering and multiplication of factors by a unit in a ring. A principal ideal domain is an integral domain in which every ideal is generated by one element.

Gauss lemma for polynomials says that over a unique factorization domain $D$, the product of two primitive polynomials is primitive, and the polynomial content satisfies $c(fg)=c(f)c(g)$ up to a unit. In particular a primitive polynomial of positive degree is irreducible over $D$ exactly when it is irreducible over the fraction field of $D$. Consequently a polynomial ring over a unique factorization domain is itself a unique factorization domain.

The Eisenstein criterion says that if $f(X)=a_n{X}^n+\cdots +a_0\in D[X]$ has $n\ge 1$ and there is a prime element $p$ with $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$, then $f$ is irreducible over the fraction field. If $f$ is primitive, it is also irreducible in $D[X]$. Over $\mathbb{Z}$ one may take $p$ to be an ordinary prime number. The primitivity qualification prevents a nonunit constant content from being overlooked.

Gauss lemma for polynomials says that the product of two primitive polynomials in $\mathbb{Z}[X]$ is primitive. Equivalently, a primitive polynomial in $\mathbb{Z}[X]$ is reducible over $\mathbb{Q}$ exactly when it is reducible over $\mathbb{Z}$.

Gauss lemma for polynomials states that the product of two primitive polynomials over a unique factorization domain is primitive. In particular, a primitive polynomial in $\mathbb{Z}[X]$ is reducible over $\mathbb{Q}$ exactly when it is reducible over $\mathbb{Z}$.

Eisenstein criterion says that a primitive polynomialis irreducible over $\mathbb{Q}$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. To prove it, suppose $F=GH$ with both factors of positive degree. After removing contents, Gauss's lemma lets us take $G,H\in \mathbb{Z}[X]$ primitive. Reduction modulo $p$ givesso both reduced factors are monomials of positive degree. Their constant terms are therefore divisible by $p$, making $p^2$ divide $a_0=G(0)H(0)$, a contradiction.

An algebraic integer is a complex number satisfying a monic polynomial in $\mathbb{Z}[X]$. Let $\alpha$ be one and let $m_{\alpha }\in \mathbb{Q}[X]$ be its monic minimal polynomial. The usual proof using polynomial content shows that $m_{\alpha }\in \mathbb{Z}[X]$. The setis the kernel of the polynomial-evaluation ring homomorphism, hence an ideal. Every $f\in I_{\alpha }$ is divisible by $m_{\alpha }$ in $\mathbb{Q}[X]$. Because $m_{\alpha }$ is monic, polynomial division of $f$ by $m_{\alpha }$ takes place in $\mathbb{Z}[X]$, and the remainder must vanish by minimality. Thereforewhere $m_{\alpha }$ is monic and irreducible.

Fortranslation givesThis is Eisenstein at $p=3$, so $f(X+1)$ and therefore $f(X)$ are irreducible over $\mathbb{Q}$. HenceIrreducibility and Gauss's lemma also show that $(f)$ is a prime ideal of $\mathbb{Z}[X]$, soIt is not a field. The class of $2$ is nonzero but not a unit: reducing further modulo $2$ gives a nonzero quotientin which the image of $2$ is zero, whereas a unit must remain a unit under every unital quotient map. Thus

A polynomial over a unique factorization domain is called a primitive polynomial when the greatest common divisor of its coefficients is a unit.

Assume that the primitive polynomial $f$ factors in $F[X]$ as $f=gh$, with both factors of positive degree. Clearing denominators and removing contents writeswhere $a,b\in F^{\times }$ and $g_0,h_0\in R[X]$ are primitive. ThenBy Gauss lemma for polynomials, $g_0{h}_0$ is primitive. Since $f$ is also primitive, comparison of contents forces $ab$ to be a unit of $R$. Absorbing that unit into one factor gives a nontrivial factorization of $f$ in $R[X]$, contrary to its assumed irreducibility. Hence $f$ is irreducible in $F[X]$.

Letbe a primitive polynomial. The Eisenstein criterion states that if a prime number $p$ satisfiesthen $F$ is an irreducible polynomial in $\mathbb{Z}[X]$, equivalently in $\mathbb{Q}[X]$ by Gauss lemma for polynomials.

To prove it, suppose that $F=GH$ with $G,H\in \mathbb{Z}[X]$ of positive degree. Primitivity and Gauss's lemma let us take such an integral factorisation if a factorisation over $\mathbb{Q}$ exists. Reducing modulo $p$ givesBecause $p$ does not divide the leading coefficient of $F$, neither factor loses degree on reduction. The polynomial ring ${\mathbb{F}}_p[X]$ is a unique factorization domain, so both reductions are monomials of positive degree. In particular, $p$ divides both constant terms $G(0)$ and $H(0)$. It follows that $p^2$ dividesa contradiction. This proves the criterion.

For a prime $p$, translate the geometric sum by one:Its leading coefficient is one, every other coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. It is therefore Eisenstein at $p$. Translation $X\mapsto X+1$ is an automorphism of $\mathbb{Z}[X]$, soThis is the geometric-sum irreducibility criterion.

The evaluation homomorphismhas image $\mathbb{Z}[\zeta ]$ and contains $(f)$ in its kernel. Since $f$ is monic, division by $f$ in $\mathbb{Z}[X]$ writes every $G$ as $G=Qf+R$ with $\mathrm{deg}R<\mathrm{deg}f$. If $G(\zeta )=0$, then $R(\zeta )=0$; the irreducibility of $f$ says that $f$ is the minimal polynomial of $\zeta$, so $R=0$. Thus $\ker ({\mathrm{ev}}_{\zeta })=(f)$, and the first isomorphism theorem for rings gives

Now take $p=3$. Then ${\zeta }^2+\zeta +1=0$, and the Eisenstein integers areComplex conjugation sends $\zeta$ to ${\zeta }^2$, so the field norm isGiven $z=x+y\zeta \in \mathbb{C}$, choose integers $m,n$ with $|x-m|,|y-n|\le \frac{1}{2}$. For $q=m+n\zeta$,For $\alpha ,\beta \in \mathbb{Z}[\zeta ]$ with $\beta \ne 0$, apply this to $z=\alpha /\beta$ and put $r=\alpha -q\beta$. ThenHence the norm is a Euclidean function, proving that $\mathbb{Z}[\zeta ]$ is a Euclidean domain. This is the Euclidean norm on the Eisenstein integers.

Finally suppose $A\in {\mathrm{GL}}_n(\mathbb{Z})$ satisfies $A^2+A+I=0$. Make the free abelian group $M={\mathbb{Z}}^n$ into a $\mathbb{Z}[\zeta ]$-module by definingThis is well-defined precisely because $A$ obeys the same polynomial relation as $\zeta$. The module is finitely generated. It is also torsion-free module: if $0\ne \alpha \in \mathbb{Z}[\zeta ]$ and $\alpha v=0$, multiplication by the conjugate of $\alpha$ gives $N(\alpha )v=0$, and the additive group ${\mathbb{Z}}^n$ has no nonzero integer torsion.

A Euclidean domain is a principal ideal domain, and the structure theorem for finitely generated modules over a principal ideal domain says that a finitely generated torsion-free module over one is free. Thusfor some $m$. Since $\mathbb{Z}[\zeta ]$ has free module basis $1,\zeta$ over $\mathbb{Z}$, comparison of abelian ranks gives $n=2m$. There can consequently be no such matrix when $n$ is odd.

If $n=2m$, choose a $\mathbb{Z}[\zeta ]$-basis $v_1,\dots ,v_m$. Thenis a $\mathbb{Z}$-basis, and $A^2{v}_i=-v_i-A{v}_i$. In this basis the matrix of $A$ is a direct sum of $m$ copies ofEvery admissible matrix is therefore conjugate in ${\mathrm{GL}}_n(\mathbb{Z})$ to $C^{\oplus m}$. Hence there is exactly one conjugacy class for even $n$, and none for odd $n$. This is the classification of integral matrices satisfying the third cyclotomic polynomial.