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Suppose first that $A$ is a unique factorization domain, and let $\mathfrak{p}$ be a minimal nonzero prime ideal. Choose $0\ne a\in \mathfrak{p}$ and factor it into irreducibles. In a UFD each irreducible is a prime element, so one factor $\pi$ belongs to $\mathfrak{p}$. The nonzero prime ideal $(\pi )\subseteq \mathfrak{p}$ must equal $\mathfrak{p}$ by minimality.

Conversely, the ascending chain condition in the Noetherian domain implies that every nonzero nonunit factors into irreducibles. Let $\pi$ be irreducible and choose a prime $\mathfrak{p}$ minimal over $(\pi )$. The Krull principal ideal theorem gives $\mathrm{ht}\mathfrak{p}\le 1$. Since $A$ is a domain and $\mathfrak{p}\ne (0)$, it is a minimal nonzero prime and hence is principal, say $\mathfrak{p}=(q)$. The divisibility $q\mid \pi$ and irreducibility of $\pi$ force $q$ to be associate to $\pi$, so $(\pi )=\mathfrak{p}$ is prime. Thus every irreducible is prime, proving that $A$ is a UFD. This is the Minimal-prime criterion for a Noetherian unique factorization domain.

Let $A$ be a principal ideal domain that is not a field. A PID is Noetherian and is a unique factorization domain. Every nonzero prime ideal is generated by a prime element. Ifthen $a\mid p$, so primality of $p$ makes $a$ a unit or an associate of $p$. The proper alternative is $(a)=(p)$, and every nonzero prime is therefore maximal. Since $A$ has a nonzero prime ideal, its Krull dimension is exactly one.

A Cartier divisor on an integral scheme is an open cover $(U_i)$ together with nonzero rational functions $f_i$ such that every ratio $f_i/f_j$ is a regular unit on $U_i\cap U_j$.

Every prime Weil divisor on ${\mathbb{P}}_k^n$ is cut out by an irreducible homogeneous polynomial because the polynomial ring $k[x_0,\dots ,x_n]$ is a unique factorization domain. Hence any Weil divisor can be represented by a homogeneous rational expressionof some total degree $d$. On the standard chart $U_i=D_{+}(x_i)$, put $f_i=F/x_i^d$. This is a degree-zero rational function, and on $U_i\cap U_j$,is a regular unit. These local equations form a Cartier divisor whose associated Weil divisor is the original one.

Now put $P=X\times {\mathbb{P}}^n$ and let $H=X\times {\mathbb{P}}^{n-1}$ be the hyperplane divisor at infinity. Its complement is $X\times {\mathbb{A}}^n$. Iterating the given invariance under multiplication by ${\mathbb{A}}^1$ givesThe localization sequence for the divisor class group shows that every class on $P$ is a pullback of a class on $X$ plus an integer multiple of $[H]$. Restriction to the generic fiber ${\mathbb{P}}_{K(X)}^n$ kills pullbacks from $X$ and sends $[H]$ to the generator of $\mathrm{Cl}({\mathbb{P}}_{K(X)}^n)\cong \mathbb{Z}$. Therefore the sum is direct, provingThis is the divisor class group of a projective-space bundle with trivial vector bundle.

An irreducible complex analytic hypersurface in a complex manifold $X$ is a closed irreducible analytic subset of pure complex codimension one. A local defining function of a complex analytic hypersurface $Y$ at $x$ is a holomorphic function $f$ on a neighbourhood $U$ such thatThe necessary local algebra is that the stalk ${\mathcal{O}}_{X,x}$ is a regular local ring, hence a unique factorization domain, and that the local branches of a hypersurface germ determine finitely many height-one prime ideals. Each is principal; the product of their generators gives $f$, and removing repeated factors makes it reduced. This also covers a globally irreducible hypersurface that has several local branches at a singular point.

A divisor on a complex manifold is a locally finite formal sum $D={\sum }_Y{n}_{Y}Y$ of irreducible analytic hypersurfaces with integer coefficients. On a sufficiently small $U_{\alpha }$, local defining functions give a meromorphic equation $f_{\alpha }$ for $D$. The quotients $f_{\alpha }/f_{\beta }$ are nowhere-zero holomorphic functions. Gluing frames byproduces the holomorphic line bundle associated to a divisor $[D]$, and $f_{\alpha }{e}_{\alpha }$ gives its canonical meromorphic section with divisor $D$.

The Euler sequence on complex projective spaceimplies $\det T{\mathbb{CP}}^n\cong \mathcal{O}(n+1)$. Taking the dual determinant yields the canonical bundle of complex projective spacewhere $H$ is a hyperplane divisor.

The hypotheses on the homogeneous polynomial $p$ say thatis a smooth projective hypersurface of degree $k$, so its divisor line bundle is $[V]\cong \mathcal{O}(kH)$. The Adjunction formula givesThis is the canonical bundle of a smooth projective hypersurface.

Now fix an isomorphism $\Phi :[P]\to [Q]$ and regard $s_P$ and ${\Phi }^{-1}{s}_Q$ as holomorphic sections of the same holomorphic line bundle. They have no common zero because $P\ne Q$. Their homogeneous coordinates therefore define a well-defined holomorphic mapIn a local frame, the quotientis a meromorphic function with divisor $(f)=P-Q$. Thus $F^{-1}(0)=P$ and $F^{-1}(\infty )=Q$, both with multiplicity one. The degree of a holomorphic map $F$ is therefore one. A nonconstant degree-one holomorphic map between compact connected Riemann surfaces is a biholomorphism, so the displayed map is biholomorphic.

False. LetThe element $1/y$ does not belong to $A$: assigning degrees $\mathrm{deg}x=\mathrm{deg}y=1$ and $\mathrm{deg}(x/y)=0$ shows that every element of $A$ has nonnegative total degree, whereas $1/y$ has degree $-1$.

If $A=T^{-1}R$ for some multiplicative subset $T$, the membership $x/y\in T^{-1}R$ would give $x/y=a/t$ for some $a\in R$ and $t\in T$. Since $R$ is a unique factorization domain and $x,y$ are coprime, the equation $xt=ay$ implies $y\mid t$. Write $t=yu$. As $t$ is invertible in $T^{-1}R$, so iscontrary to $1/y\notin A$. Thus a subring of a localization of a ring need not itself be a localization of the original ring.

The polynomial ring $\mathbb{C}[X,Y]$ is a two-dimensional unique factorization domain. Let $\mathfrak{p}$ be a prime ideal containing $(f,g)$. Since $f\ne 0$, the prime $\mathfrak{p}$ is nonzero. If it were not maximal, it would have height one, so the height-one prime in a unique factorization domain would give $\mathfrak{p}=(h)$ for an irreducible polynomial $h$. Then $h$ would divide both $f$ and $g$, contrary to the hypothesis. Every prime ofis therefore maximal.

The Hilbert basis theorem makes $R$ Noetherian, and it has Krull dimension zero by the preceding paragraph. The Noetherian dimension-zero criterion for an Artinian ring now shows that

At the generic point of $L_0$, the functions $x_1,x_2$ are units and the equation gives $x_0=x_3^3/(x_1{x}_2)$. Taking $x_3$ as a uniformizer yieldsAt $L_1$, the functions $x_0,x_2$ are units and $x_1=x_3^3/(x_0{x}_2)$, so ${\nu }_{L_1}(x_0/x_3)=-1$. Symmetrically, ${\nu }_{L_2}(x_0/x_3)=-1$.

The complement of the three lines is $U=D_{+}(x_3)$, withThis unique factorization domain has trivial divisor class group, so part b says that $[L_0],[L_1],[L_2]$ generate $\mathrm{Cl}(X)$. The units on $U$ are scalar multiples of $u_0^a{u}_1^b$, and their boundary valuation vectors are generated byThe resulting integer matrix has Smith normal form $\mathrm{diag}(1,3,0)$. Consequently

Write $A=k[x,y]$ and $\mathfrak{m}=(x,y)$. Since $f(0,0)\ne 0$, the principal open subscheme $D(f)=\mathrm{Spec}{A}_f$ contains $\mathfrak{m}$, andCover $V$ by the two affine opens $D(xf)$ and $D(yf)$, whose intersection is $D(xyf)$. The degree-zero part of the resulting Čech cohomology complex givesinside the fraction field of $A$. The last equality follows because $A_f$ is a unique factorization domain and a rational function regular after localizing at both $x$ and $y$ has no possible prime factor left in its denominator.

The same affine cover is acyclic, so its degree-one Čech group computes sheaf cohomology and givesBefore localizing at $f$, the quotienthas the $k$-basisWriting $f=c+h$ with $c=f(0,0)\in k^{\times }$ and $h\in (x,y)$, multiplication by $h$ is locally nilpotent on $Q$: for each negative monomial, a sufficiently high power of $(x,y)$ moves every term into $A_x+A_y$. Hence $c+h$ acts invertibly on $Q$ by a finite geometric series on each element. Localizing at $f$ therefore leaves $Q$ unchanged, andThe displayed infinite basis proves that this vector space is infinite-dimensional.

For a group acting linearly on $W$, act on the coordinate ring by $(g\cdot f)(w)=f(g^{-1}w)$. The polynomial invariant ring is the fixed subalgebra $\mathbb{C}[W{]}^G$. It inherits the degree grading. Even over the complex numbers it need not be a unique factorization domain, as the quadratic cone invariant ring demonstrates.

If $R$ is a unique factorization domain with fraction field $F$, then $R[X]$ is a unique factorization domain. A primitive polynomial irreducible in $R[X]$ remains irreducible, hence prime, in the principal ideal domain $F[X]$; Gauss lemma for polynomials brings divisibility back to $R[X]$.

A Noetherian local ring $(R,\mathfrak{m})$ is regular when the minimal number of generators of its maximal ideal equals its Krull dimension. Every regular local ring is an integral domain and a unique factorization domain.

Every nonzero nonunit in a unique factorization domain is a finite product of irreducible elements. Such a factorization is unique up to permutation and multiplication of the factors by units. Irreducibles are prime elements, so divisibility can be checked by comparing their multiplicities. This applies to a polynomial ring over a field and is useful for intersections of localizations inside its fraction field.