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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.

For the same pair, $\boxed{S=\mathbb{Z}[X]\subset \mathbb{Q}[X]}$. Since $\mathbb{Z}$ is a unique factorization domain, Gauss lemma for polynomials proves that $S$ is a unique factorization domain.

To prove that $S$ is not a principal ideal domain, consider the ideal $I=(2,X)$. A generator $g$ would divide 2 and $X$. Divisibility of 2 forces $g$ to be a nonzero constant integer; divisibility of $X$ then forces that integer to divide its coefficient 1, so $g=\pm 1$. This would make $I=S$. But evaluation at $X=0$ followed by reduction modulo 2 maps every element of $I$ to zero and maps 1 to one. Hence $1\notin I$, a contradiction. This is the integer polynomial ring is not a principal ideal domain example, with both clauses satisfied by a single inclusion.

First, every nonzero nonunit in a principal ideal domain factors into irreducible elements. Otherwise repeatedly choose a proper nonunit factor $a_{j+1}$ of $a_j$; then the principal idealsform a strictly increasing chain, contradicting the ascending chain condition, since every principal ideal domain is a Noetherian ring.

Next every irreducible $p$ is a prime element. If $p\mid ab$ but $p\nmid a$, then $\gcd (p,a)=1$. The Bezout identity gives $up+va=1$, so multiplying by $b$ shows $p\mid b$. Thus irreducibles are prime.

Existence of an irreducible factorization and primality give uniqueness: an irreducible on one side divides the product on the other, hence is associate to one factor; cancel and continue. Therefore every principal ideal domain is a unique factorization domain.

Suppose $X=f_1\cdots f_s$ were a product of irreducible elements of $R$. Since degree of a polynomial is additive under products, exactly one factor has positive degree and all the others are constant. By the preceding classification, the positive-degree factor has constant term $\pm 1$, while every constant irreducible is $\pm p$ for a prime $p$. Their product therefore has a nonzero constant term, contradicting $X(0)=0$.

Thus $X$ is not a product of irreducibles, so this integral domain is not an atomic domain and in particular is not a unique factorization domain.

The resolution principle for sheaf cohomology states that an exact sequencewith $H^q(X,{\mathcal{A}}^j)=0$ for every $q>0$ computes $H^i(X,\mathcal{F})$ as the cohomology of $\Gamma (X,{\mathcal{A}}^{\bullet })$. In particular one may use a flasque resolution.

An affine morphism $\varphi :X\to Y$ is one for which the inverse image of every affine open subset of $Y$ is affine. We use vanishing of quasi-coherent cohomology on an affine scheme: a quasi-coherent sheaf has zero higher sheaf cohomology on an affine variety. Take a flasque resolution $\mathcal{F}\to {\mathcal{I}}^{\bullet }$. Each direct image sheaf ${\varphi }_{\ast }{\mathcal{I}}^j$ is flasque, because restriction on $V\subseteq U\subseteq Y$ is restriction on ${\varphi }^{-1}V\subseteq {\varphi }^{-1}U$. On an affine open $U\subseteq Y$, the augmented complex of sections isIts positive-degree cohomology vanishes by the affine theorem, and its degree-zero kernel is $\Gamma ({\varphi }^{-1}U,\mathcal{F})$. Checking exactness on this affine basis shows that ${\varphi }_{\ast }{\mathcal{I}}^{\bullet }$ is a flasque resolution of ${\varphi }_{\ast }\mathcal{F}$. The global-section complexes are identical, proving cohomology under an affine morphism:

For projective space, put $S=k[x_0,\dots ,x_n]$ with its usual grading and $U_i=\{x_i\ne 0\}$. Construct the twisting sheaf on projective space by setting its sections on $U_i$ to be the degree-$m$ part $(S_{x_i}{)}_m$, viewed as a module over $(S_{x_i}{)}_0$. It is free of rank one with generator $x_i^m$, even when $m$ is negative. On overlaps the generators differ by the unit $(x_i/x_j{)}^m$, so these modules glue to a line bundle $\mathcal{O}(m)$.

Assume first $n\ge 1$. These local modules all embed in the graded fraction field, and agreement on overlaps identifies global sections with ${\bigcap }_i(S_{x_i}{)}_m$. Since $S$ is a unique factorization domain and at least two distinct variables occur, a reduced fraction belonging to every $S_{x_i}$ has no nonconstant denominator. Thus the intersection is $S_m$. Counting monomials by stars and bars givesHere $h_n^i(m)={\dim }_k{H}^i({\mathbf{P}}^n,\mathcal{O}(m))$. The printed formula needs the convention that a counting binomial coefficient is zero when its upper integer is smaller than its nonnegative lower integer; the generalized polynomial convention for negative upper integers would give incorrect dimensions. For $n=0$, every twist on the single point ${\mathbf{P}}^0$ is trivial, so $h_0^0(m)=1$ for every $m$ and its higher cohomology is zero. This case supplies the base of dimension induction; the displayed top-cohomology formula below is for $n\ge 1$.

The standard $n+1$ affine opens have affine finite intersections. The acyclic cover theorem and affine vanishing identify sheaf cohomology with a Čech cochain complex ending in degree $n$. In particular $H^i({\mathbf{P}}^n,\mathcal{O}(m))=0$ for $i>n$.

Let $H=\{x_n=0\}\cong {\mathbf{P}}^{n-1}$, with inclusion $i$. Multiplication by $x_n$ gives the hyperplane exact sequence for twisting sheavesExactness is checked in the local trivializations, where a hyperplane equation is a non-zero-divisor and the quotient is its restriction to $H$. Since a closed immersion is an affine morphism, the first part identifies the cohomology of $i_{\ast }{\mathcal{O}}_H(m)$ with that on $H$.

For $n=1$, the long exact sequence in sheaf cohomology and $H\cong {\mathbf{P}}^0$ giveStarting from the given vanishing for sufficiently large $m$ and descending gives $h_1^1(m)=0$ for $m\ge -1$ and $h_1^1(m)=-m-1$ for $m\le -2$.

Now let $n\ge 2$, and assume the formulas in dimension $n-1$. Restrictionis surjective for every $m$: for $m\ge 0$ it is polynomial restriction, and for $m<0$ both groups are zero since $\dim H\ge 1$. The long exact sequence in sheaf cohomology therefore makes $H^1({\mathbf{P}}^n,\mathcal{O}(m-1))\to H^1({\mathbf{P}}^n,\mathcal{O}(m))$ injective. For $2\le j<n$, dimension induction gives $H^{j-1}(H,{\mathcal{O}}_H(m))=0$, so the same sequence makesinjective. Composing these injections up to a sufficiently large twist, where the target is zero by the given hypothesis, proves $H^j({\mathbf{P}}^n,\mathcal{O}(m))=0$ for every $m$ and $0<j<n$.

The remaining part of the long exact sequence in sheaf cohomology is consequentlywhere the last zero uses the affine-cover dimension bound on $H$. Hence $h_n^n(m-1)=h_n^n(m)+h_{n-1}^{n-1}(m)$. Using dimension induction and eventual vanishing, downward summation yieldsby telescoping Pascal's identity; for $m\ge -n$ it is zero. Thus, without invoking Serre duality in the induction,Finally, the canonical bundle of projective space is ${\omega }_{{\mathbf{P}}^n}\cong \mathcal{O}(-n-1)$, obtained by taking determinants in the dual Euler sequence. Serre duality therefore predictsIt pairs the two computed extreme-degree dimensions and pairs zero intermediate groups with zero intermediate groups, exactly as required.

Every Euclidean domain is a principal ideal domain: in a nonzero ideal choose an element of least positive Euclidean value and divide every other element by it; minimality forces every remainder to vanish. Every principal ideal domain is a unique factorization domain: the ascending-chain condition supplies factorizations into irreducibles, and Euclid lemma follows because every irreducible generates a prime ideal. Therefore

Put $R=k[x,y,z]/(xy-z^3)$ and let $X=\mathrm{Spec}R$. The ideal $P=(x,z)$ is a height-one prime ideal, because $R/P\cong k[y]$ and $\dim R=2$. It therefore defines a prime Weil divisor $D=V(x,z)$.

Localizing at $x$ eliminates $y$, givinga unique factorization domain. The Nagata theorem for divisor class groups says that $\mathrm{Cl}(X)$ is generated by the height-one primes containing $x$. Sincethe only such prime is $P$, and hence $[D]$ generates. At the generic point of $D$, $y$ is a unit in a ring and $x=z^3/y$, so the order of vanishing is three:Thus $3[D]=0$.

This relation has exact order three. Indeed, if a principal divisor were supported on $D$, its defining rational function would be a unit on $D(x)$. The units of $R_x=k[z][x,x^{-1}]$ are precisely $c{x}^m$ with $c\in k^{\times }$ and $m\in \mathbb{Z}$, whose divisors are $3mD$. Consequentlygenerated by $[V(x,z)]$. This is the $n=3$ case of the Divisor class group of an A-type surface singularity.

First prove that $A$ is an integrally closed domain. If $u\in \mathrm{Frac}A$ is integral over $A$, then it is integral over $A_x$. Since the unique factorization domain $A_x$ is integrally closed, write $u=a/x^n$ with $a\in A$ and $n\ge 0$ minimal. If $n>0$, an integral equation for $u$ gives, after multiplication by a suitable power of $x$,Hence $a^m\in (x)$. The ideal $(x)$ is prime, so $a\in (x)$, contradicting minimality of $n$. Therefore $n=0$ and $u\in A$.

Now apply the Nagata theorem for divisor class groups. The class group of $A_x$ vanishes by the divisor-class criterion for unique factorization. Hence $\mathrm{Cl}(A)$ is generated by height-one primes that meet $\{1,x,x^2,\dots \}$. Such a prime contains $x$ and must equal $(x)$, because the Krull principal ideal theorem makes the nonzero prime $(x)$ itself height one. Its divisor class is principal, so $\mathrm{Cl}(A)=0$. A second application of the divisor-class criterion yieldsThis argument is the Nagata criterion for unique factorization domains.

The height of a prime ideal $P$ is

The Krull principal ideal theorem states that if $R$ is Noetherian and $P$ is minimal among the primes containing a proper principal ideal $(a)$, thenSuppose otherwise that $P_0⊊Q⊊P$. Quotient by $P_0$ and localize at $P$; it is enough to consider a Noetherian local domain whose maximal ideal $P$ is the only prime containing $(a)$ and which has $0⊊Q⊊P$.

For $n\ge 1$, putThis is a $Q$-primary ideal. Since $R/(a)$ is a zero-dimensional Noetherian ring, it is Artinian, and the descending chain $(Q^{(n)}+(a))/(a)$ eventually stabilizes. Thus, for all sufficiently large $n$, every $q_n\in Q^{(n)}$ can be writtenNow $ra\in Q^{(n)}$, while $a\notin Q$; $Q$-primaryness gives $r\in Q^{(n)}$. HenceThe Nakayama lemma applied to $Q^{(n)}/Q^{(n+1)}$ gives $Q^{(n)}=Q^{(n+1)}$. Localizing at $Q$ makes all sufficiently large powers of the nonzero maximal ideal $Q{R}_Q$ equal. A nonzero element of the stable power then belongs to ${\bigcap }_n{Q}^n{R}_Q$, contradicting the Krull intersection theorem. This proves the theorem.

Now let $R$ be a Noetherian integral domain. If $R$ is a unique factorization domain and $P$ has height one, choose $0\ne x\in P$ and an irreducible factor $p$ of $x$. In a UFD, $p$ is prime, soHeight one forces $P=(p)$.

Conversely, suppose every height-one prime is principal. Noetherianity makes $R$ an atomic domain. Given an irreducible element $p$, choose a prime $P$ minimal over $(p)$. The principal ideal theorem gives $\mathrm{ht}P=1$, so $P=(q)$ by hypothesis. Since $p=qr$ and $p$ is irreducible, $r$ is a unit; hence $(p)=P$ is prime. Thus every irreducible is a prime element, and an atomic domain with this property is a UFD. Therefore

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

Suppose first that every ideal of $R$ is finitely generated. For an ascending chainthe union $I={\bigcup }_n{I}_n$ is an ideal. Its finite generating set lies in one $I_N$, so $I_n=I_N$ for all $n\ge N$. Thus $R$ satisfies the ascending chain condition and is a Noetherian ring.

Conversely, if an ideal $I$ is not finitely generated, choose $a_1\in I$, and after choosing $a_1,\dots ,a_n$, chooseThis creates a strictly ascending chain of ideals, contradicting Noetherianity. Hence every ideal is finitely generated.

If $\phi :R\to S$ is surjective and $J⊲S$, then ${\phi }^{-1}(J)$ is an ideal of $R$. If it is generated by $a_1,\dots ,a_m$, then $J$ is generated by $\phi (a_1),\dots ,\phi (a_m)$. Thus every ideal of $S$ is finitely generated, so $S$ is Noetherian.

The Hilbert basis theorem states that if $R$ is a commutative Noetherian ring, then $R[X]$ is Noetherian. To prove it, let $I⊲R[X]$. The leading coefficients of polynomials in $I$ generate an ideal of $R$; choose generators that occur as leading coefficients of $f_1,\dots ,f_s\in I$, and let $N={\max }_i\mathrm{deg}{f}_i$. For each degree below $N$, the leading coefficients of members of $I$ of at most that degree likewise form a finitely generated ideal; choose finitely many corresponding polynomials. Any $f\in I$ of degree at least $N$ can have its leading term cancelled by a linear combination of monomial multiples of the $f_i$. Repeating lowers its degree below $N$, where the second finite list completes the reduction. These finitely many selected polynomials generate $I$, proving the theorem.

For every $\alpha \in \mathbb{C}$, the ring $\mathbb{Z}[\alpha ]$ is Noetherian. If $\alpha$ is transcendental, evaluation identifies it with $\mathbb{Z}[X]$. If $\alpha$ is algebraic, it is a quotient of $\mathbb{Z}[X]$. The integers are Noetherian, the Hilbert basis theorem handles $\mathbb{Z}[X]$, and quotients preserve Noetherianity.

For a unique factorization domain that is not Noetherian, takeEvery polynomial and every factorization uses only finitely many variables, so existence and uniqueness of factorization reduce to a finite-variable polynomial ring, which is a UFD. Butis a strictly ascending chain, so the ring is not Noetherian.

Finally, the ring $C(\mathbb{R},\mathbb{R})$ is not Noetherian. For $n\ge 1$, letThese are ideals and $I_n⊊I_{n+1}$: the distance function to $[-1/(n+1),1/(n+1)]$ belongs to $I_{n+1}$ but not to $I_n$. This strict ascending chain proves the claim.

PutSinceis an integral domain, $(x,z)$ is a prime ideal. Moreover $\dim R=3$ and $\dim R/(x,z)=2$, so its height of an ideal is one. Therefore $D=V(x,z)$ is a prime Weil divisor.

Localizing at $x$ eliminates $y$ and givesa unique factorization domain. The Nagata theorem for divisor class groups says that $\mathrm{Cl}(R)$ is generated by the height-one primes containing $x$. Sincethese are $D=V(x,z)$ and $E=V(x,w)$. Both occur with multiplicity one in the principal divisorThe units of $R_x\cong k[z,w][x,x^{-1}]$ are exactly $c{x}^n$ with $c\in k^{\ast }$ and $n\in \mathbb{Z}$. Consequently the only relation supplied by localization is $[D]+[E]=0$, andThis is the divisor class group of the three-dimensional affine quadric cone.

A polynomial over a unique factorization domain is primitive when its coefficients have no common nonunit factor. If a prime divided every coefficient of a product of two primitive polynomials, reduction modulo that prime would turn the product of two nonzero polynomials over an integral domain into zero, impossible. This proves Gauss lemma for polynomials.

If positive-degree $f,g\in R[X]$ became non-coprime in $F[X]$, they would share a positive-degree primitive factor after clearing content. Gauss's lemma then makes that factor divide both in $R[X]$, contrary to coprimality. Hence they remain coprime over the fraction field.

Write $A=k[x,y]$. If $\dim Z=1$, its prime ideal has height one and is generated by an irreducible polynomial $f$, because $A$ is a unique factorization domain. Then $U=D(f)$ is a principal open subscheme andIf $\dim Z=0$, its complement has codimension two. Since ${\mathbb{A}}_k^2$ is normal, regular functions extend across that codimension-two subset, givingThis includes the calculation for the punctured affine plane. If $Z={\mathbb{A}}_k^2$ has dimension two, then $U$ is empty and its ring of sections is the zero ring.

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 $F[X]$.

To deduce the final assertion, use the characterization that an integral domain is a unique factorization domain when it is atomic and every irreducible element is prime. Factoring the contents in $R$ and then the primitive parts by degree shows that $R[X]$ is atomic.

An irreducible constant of $R$ is prime in $R[X]$. Every irreducible polynomial of positive degree is, up to a constant unit, primitive. Part (i) makes it irreducible in $F[X]$; because $F[X]$ is a principal ideal domain, it is prime there, and part (ii) makes it prime in $R[X]$. Thus every irreducible of $R[X]$ is prime, andThis is the polynomial ring over a unique factorization domain theorem.

Choose a coprime presentationFor a polynomial $r$, write $r^{\ast }(X,Y)=r(Y,X)$. Symmetry of $f$ givesSince $\mathbb{C}[X,Y]$ is a unique factorization domain and $\gcd (p,q)=1$, one has $p\mid p^{\ast }$ and $q\mid q^{\ast }$. Exchanging the variables preserves total degree, sofor nonzero constants $\lambda ,\mu$. The displayed identity gives $\lambda =\mu$, and applying the exchange twice gives ${\lambda }^2=1$.

If $\lambda =-1$, then $p(X,X)=q(X,X)=0$, so $X-Y$ divides both polynomials, contradicting coprimality. Therefore $\lambda =1$, and both $p$ and $q$ are symmetric. Taking $g=p$ and $h=q$ proves the claim, which is the symmetric rational function in two variables result.

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.