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An atomic domain is an integral domain in which every nonzero nonunit is a finite product of irreducible elements. Every unique factorization domain is atomic, but atomicity alone does not imply uniqueness.

Two nonzero elements with no common nonunit irreducible factor. In a unique factorization domain this is equivalent to their greatest common divisor being a unit. Coprimality of numerator and denominator makes a fraction reduced, so simultaneous restrictions on its denominator can force it to be a regular function.

A Noetherian ring that is an integrally closed domain is a unique factorization domain exactly when its divisor class group vanishes. Indeed, vanishing says every prime Weil divisor, equivalently every height-one prime ideal, is principal.

If a finite group has no nontrivial homomorphism to ${\mathbb{C}}^{\times }$, its polynomial invariant ring is a unique factorization domain. Factor an invariant in the ambient polynomial ring and collect its irreducible factors into orbit products. Each orbit product transforms by a linear character, hence is invariant. It is prime in the invariant ring, and the invariant factorization is a product of these primes.

For a module over a commutative ring presented by a matrix with n generators, its zeroth Fitting ideal is generated by all n-by-n minors of the relation matrix. It is independent of the chosen finite presentation. Over a unique factorization domain, the greatest common divisor of these minors defines the module order up to a unit. This does not require a Smith normal form over the ring. The square Seifert-matrix presentation of the Alexander module makes the order its determinant.

// Target: knot-theory.bigb

The polynomial ring $\mathbb{Z}[X]$ is a unique factorization domain by Gauss lemma for polynomials, but its ideal $(2,X)$ is not principal. Any generator would divide both 2 and $X$, hence be a unit: a divisor of 2 has degree zero, and a constant dividing $X$ is $\pm 1$. Yet evaluation at zero followed by reduction modulo 2 annihilates the ideal and not 1. It embeds in the principal ideal domain $\mathbb{Q}[X]$, illustrating that the property need not pass to a subring.

A local defining function for a complex analytic hypersurface $Y$ near $x$ is a holomorphic function $f$ on a neighbourhood $U$ such that $Y\cap U=\{f=0\}$. It may be chosen reduced, with each local irreducible factor occurring once. The local ring ${\mathcal{O}}_{X,x}$ of a complex manifold is a regular local ring and hence a unique factorization domain; each height-one prime of the hypersurface germ is principal, and the product of generators for its finitely many local branches gives $f$.

Let $A$ be an integral domain that is a Noetherian ring, and let $x\in A$ generate a prime ideal. If the localization of a ring $A_x$ is a unique factorization domain, then $A$ is a unique factorization domain. Indeed, integrality of $a/x^n$ over $A$, with $n$ minimal, forces $x\mid a$ and hence proves that $A$ is an integrally closed domain. The Nagata theorem for divisor class groups then says that $\mathrm{Cl}(A)$ is generated by the height-one primes containing $x$; the only one is the principal prime $(x)$, so $\mathrm{Cl}(A)=0$.

At a generic smooth point of an irreducible complex analytic hypersurface $Y$, a nonzero meromorphic function has the form $f=t^{m}u$, where $t$ is a reduced transverse local equation and $u$ is a holomorphic unit. The integer $m$ is its order along $Y$: positive for a zero and negative for a pole. The unique factorization domain property of the local ring of a complex manifold makes this independent of the equation and propagates it along $Y$. These orders are the coefficients of the principal divisor on a complex manifold of $f$.

Put $A=R/(p^k)$. A principal ideal domain is a unique factorization domain, so every nonzero element of this quotient ring can be writtenIndeed, remove the largest power of the prime element $p$ from a representative that is not divisible by $p^k$. Since $s$ and $p^k$ have greatest common divisor $1$ up to a unit, Bézout's identity in the principal ideal domain gives $xs+y{p}^k=1$. Thus $\overline{s}$ is a unit in $A$.

Consequently every nonzero principal ideal in $A$ is generated by one of $1,\overline{p},\dots ,{\overline{p}}^{k-1}$. Include the zero ideal by assigning it exponent $e=k$. These ideals are distinct: their inverse images in $R$ are $(1),(p),\dots ,(p^k)$, and equality of two different powers would, by cancellation in the integral domain, make the nonunit $p$ a unit.

Now suppose $(\overline{a})=(\overline{b})$. If this is the zero ideal, both generators are zero and are associate elements. Otherwise their exponents agree, so writewith $\overline{s},\overline{t}$ both units. Thenand the multiplier is a unit. ThusThe exponent detects the principal ideal, while its remaining factor is invertible; this is why zero divisors cause no obstruction here.

First extract the polynomial content from $f$:The primitive quartic is Eisenstein at the prime $2$: every nonleading coefficient is divisible by $2$, while the constant term is not divisible by $4$. Eisenstein's criterion makes it irreducible over $\mathbb{Q}$, and Gauss's lemma makes it irreducible in $\mathbb{Z}[X]$. The constant factor $2$ is itself irreducible in $\mathbb{Z}[X]$.

The polynomial $g$ is already irreducible over $\mathbb{Q}$: it is Eisenstein at $3$. Its nonleading coefficients, including the missing zero coefficients, are divisible by $3$, and $-21$ is not divisible by $9$. Thus $g$ is its own sole nonconstant irreducible factor over $\mathbb{Q}$.

Over the field ${\mathbb{Z}}_3$, grouping terms givesThe first two factors are linear and therefore irreducible. The quadratic takes the values $1,2,2$ at $0,1,2$ respectively, so it has no root in ${\mathbb{Z}}_3$. A degree-two polynomial over a field is reducible exactly when it has a root, proving its irreducibility.

All three factorizations are unique up to order and multiplication by units. The ring $\mathbb{Z}$ is a unique factorization domain, hence so is $\mathbb{Z}[X]$ by Gauss's lemma. A polynomial ring over a field is a Euclidean domain, so both $\mathbb{Q}[X]$ and ${\mathbb{Z}}_3[X]$ have unique factorization. The units are $\pm 1$ in the first ring, nonzero rational constants in the second, and $1,2$ in the third.

The usual field norm $N_{L/K}$ is defined for a finite field extension. The printed hypothesis “algebraic” alone allows an infinite extension, for which this determinant norm is not defined. We therefore assume $[L:K]<\infty$. For an arbitrary algebraic extension the meaningful elementwise statement is instead $N_{K(b)/K}(b)\in A$, and we prove this as well.

First a unique factorization domain is an integrally closed domain. Indeed, write an integral fraction as $u/v$ with relatively prime numerator and denominator. A monic integral equation, after multiplication by $v^r$, shows that $v$ divides $u^r$. Unique factorization forces $v$ to be a unit.

Let $m_b(T)=T^r+a_{r-1}{T}^{r-1}+\cdots +a_0\in K[T]$ be the monic minimal polynomial of $b$. Because $b$ is an integral element over $A$, it satisfies a monic polynomial $f\in A[T]$, and $m_b$ divides $f$ in $K[T]$. Every root of $m_b$ in an algebraic closure is therefore integral over $A$. This assertion includes repeated roots when the extension is inseparable. The coefficients of $m_b$ are elementary symmetric expressions in these roots. The integral elements form a ring: finitely many integral elements generate a finite $A$-module algebra, and the determinant trick shows every element of that algebra is integral. Thus all $a_i$ are integral over $A$. Since $a_i\in K$ and $A$ is integrally closed, $a_i\in A$.

Multiplication by $b$ on $K(b)$, in the basis $1,b,\dots ,b^{r-1}$, has the companion matrix of $m_b$, whose determinant is $(-1{)}^r{a}_0$. If $e=[L:K(b)]$, take a $K(b)$-basis of $L$. Multiplication by $b$ preserves each summand in the resulting direct sum of $e$ copies of $K(b)$, so its determinant over $K$ isIn particular $N_{K(b)/K}(b)=(-1{)}^r{a}_0\in A$. The argument requires no separability; in fact it holds for every integrally closed domain, not just a unique factorization domain.

For the connection with dimension, let $B$ be the coordinate ring of an irreducible affine variety. By Noether normalization there is a finite integral inclusion $A=k[t_1,\dots ,t_d]\subset B$, with $d=\dim B$. Their fraction fields form a finite extension. For any $0\ne b\in B$, its minimal-polynomial constant term $a_0$ is nonzero and lies in $A$. Its equation givesThus a nonzero function on the normalized variety produces a nonzero function on the base that vanishes wherever $b$ vanishes. Up to sign, the field norm is a power of this $a_0$, so it too belongs to $bB\cap A$.

Under the finite morphism to ${\mathbb{A}}^d$, the image of $V_B(b)$ is therefore contained in the proper Zariski-closed set $V_A(a_0)$ of dimension at most $d-1$. Finite morphisms preserve the dimension of a closed subvariety and its image. Applying this to a nonzero element of the ideal of any proper closed subvariety proves proper closed subvarieties have strictly smaller dimension. This is the useful role of the norm argument: nonzero equations upstairs give nonzero equations downstairs under a finite normalization.

Let $S=k[X_0,\dots ,X_n]$ with its usual grading, and put $S(m{)}_d=S_{m+d}$. Define the twisting sheaf on projective space byMore explicitly, on $U_i=D_{+}(X_i)$ its sections are the homogeneous degree-$m$ part of $S_{X_i}$. They are a free rank-one module over $(S_{X_i}{)}_0=k[X_j/X_i:j\ne i]$, with generator $e_i=X_i^m$. These generators make sense for every integer $m$, since $X_i$ is invertible on this chart. Their transition rule isThe ratios are regular units on overlaps and satisfy the cocycle identity, so they glue to an invertible sheaf. In particular $\mathcal{O}(m)\otimes \mathcal{O}(r)\cong \mathcal{O}(m+r)$.

For $n\ge 1$, a global section is a compatible homogeneous degree-$m$ rational expression lying in every $S_{X_i}$. Inside the fraction field of $S$,To justify this, write a fraction in lowest terms in the unique factorization domain $S$. Membership in $S_{X_i}$ forces every irreducible divisor of its denominator to be associated with $X_i$. Since there are at least two nonassociated variables, the denominator must be a unit. ThereforeThe degree-$m$ monomials are indexed by nonnegative integers $a_0,\dots ,a_n$ with sum $m$. The stars and bars count givesHere and below, an all-integer binomial shorthand must use the counting convention $\left(\genfrac{}{}{0ex}{}{a}{n}\right)=0$ when $a<n$, including negative $a$. The polynomial extension of binomial coefficients to negative upper arguments does not give the stated dimensions. In dimension zero, ${\mathbb{P}}^0$ is one point and every ${\mathcal{O}}_{{\mathbb{P}}^0}(m)$ is trivial, so its space of global sections is $k$ for every $m$; this case is treated separately.

For a closed inclusion $i:H\hookrightarrow {\mathbb{P}}^n$, direct image sheaf formation is exact: its stalk is the original stalk at a point of $H$ and zero outside $H$. It preserves flabby sheaves, since restrictions on the target are restrictions between the corresponding open subsets of $H$. Push forward a flasque resolution of $\mathcal{F}$; exactness gives a flasque resolution of $i_{\ast }\mathcal{F}$, and its complex of global sections is the original one becauseTaking cohomology of these identical complexes proves sheaf cohomology under a closed inclusion:This argument applies even without quasi-coherence.

We derive the remaining sheaf cohomology dimensions using the given eventual vanishing, the long exact sequence in sheaf cohomology, and induction on dimension. Multiplication by the linear equation of a hyperplane $H\cong {\mathbb{P}}^{n-1}$ gives the hyperplane exact sequence for twisting sheavesExactness can be checked in each chart by quotienting its polynomial ring by the equation of $H$; multiplication is injective because that equation is a non-zero-divisor. We use the closed-inclusion cohomology identification to replace the cohomology of the final sheaf by that on $H$.

The dimension-zero base is a point, whose global-section functor is exact. Thus all its higher sheaf cohomology is zero. For $n=1$, the relevant long exact sequence isWhen $m\ge 0$, restriction of homogeneous polynomials to the hyperplane point is surjective: after choosing that point as $[1:0]$, the section $X_0^m$ restricts to a generator. Therefore the consecutive $H^1$ groups are isomorphic for $m\ge 0$. The given vanishing for large $m$ propagates down to $m=-1$. For $m<0$, both displayed $H^0$ groups are zero, and the sequence becomesStarting with $H^1(\mathcal{O}(-1))=0$, induction downward gives $h^1(\mathcal{O}(m))=-m-1$ for $m\le -2$. For degrees $i>1$, the point contributes no cohomology to the exact sequence, so consecutive $H^i$ groups are isomorphic for every $m$; the given eventual vanishing makes them all zero. Hence the complete positive-degree formula holds for ${\mathbb{P}}^1$.

Now let $n\ge 2$ and assume the formulas for $H\cong {\mathbb{P}}^{n-1}$. For every $m$, the restrictionis surjective: for $m\ge 0$ extend a polynomial in the hyperplane coordinates to the same polynomial on ${\mathbb{P}}^n$, and for $m<0$ both groups vanish because $H$ has positive dimension. Consequently the exact sequence gives an injectionFor $2\le i\le n-1$, the induction hypothesis says $H^{i-1}(H,{\mathcal{O}}_H(m))=0$, and gives the same injection in degree $i$. For each fixed $m$, composing finitely many injections reaches a sufficiently large twist, whose higher cohomology is zero by the given hypothesis. ThereforeFor $i>n$, the two neighbouring hyperplane groups $H^{i-1}(H,{\mathcal{O}}_H(m))$ and $H^i(H,{\mathcal{O}}_H(m))$ are zero by induction. Consecutive $H^i$ groups on ${\mathbb{P}}^n$ are then isomorphic, and eventual vanishing again makes them zero. This derives vanishing above the dimension too, without needing a special projective-space cohomology theorem.

For top cohomology the exact sequence now reduces toWrite $b_n(m)=h^n({\mathcal{O}}_{{\mathbb{P}}^n}(m))$. The induction hypothesis and eventual vanishing give the recurrenceConsecutive top groups agree for $m\ge -n+1$, so descending from a large twist gives $b_n(m)=0$ for $m\ge -n$. Below this range, iterating the recurrence gives a finite sum:The last equality follows by summing Pascal's identity. These short exact sequences also prove that every group whose dimension is being counted is finite-dimensional. Thus, for $n\ge 1$,With the stated counting convention, the latter is the requested single binomial formula. The base ${\mathbb{P}}^0$ has $h^0=1$ for all twists and no higher cohomology; it is not governed by the negative-twist $H^0$ rule for positive-dimensional projective space.

Finally calculate the canonical line bundle of a smooth variety from its transition functions. On $U_0$ write $t_j=X_j/X_0$ for $1\le j\le n$. On $U_i$, $i>0$, take the coordinates $u_j=X_j/X_i$ in increasing order of $j\ne i$. ThusIn their wedge product, $d{u}_0=-t_i^{-2}d{t}_i$ and
$d{u}_j=t_i^{-1}d{t}_j-t_j{t}_i^{-2}d{t}_i$. Every term involving a second $d{t}_i$ vanishes, so reordering the remaining differentials givesThis computation is valid in every characteristic. Multiply the ordered local generator on $U_i$ by $(-1{)}^i$. The resulting generators ${\eta }_i$ satisfy ${\eta }_i=(X_i/X_0{)}^{-n-1}{\eta }_0$ on $U_i\cap U_0$, and hence the corresponding transition rule on every overlap. They are exactly the transitions of $\mathcal{O}(-n-1)$. Therefore the canonical bundle of projective space is

For a line bundle, Serre duality statesThe intermediate groups on both sides vanish. In the two outer degrees, the top group at twist $m$ has the same dimension as the global-section group at twist $-m-n-1$:The polynomial degree on the right is nonnegative precisely when $m\le -n-1$, and its dimension is $\left(\genfrac{}{}{0ex}{}{-m-1}{n}\right)$. Interchanging $m$ and $-m-n-1$ checks the other outer degree. On ${\mathbb{P}}^0$ both sides are one-dimensional. Thus all the formulas are consistent with Serre duality, which was used only for this final consistency check, not to prove the cohomology formulas.

Take the integral domain $R=\mathbb{Z}[\sqrt{-5}]$, embedded in $\mathbb{C}$. It exhibits failure of unique factorization in the ring of integers generated by square root of minus five:Here is why these are genuinely distinct factorizations into irreducible elements. The multiplicative algebraic norm $N(a+b\sqrt{-5})=a^2+5b^2$ is a nonnegative integer and equals one only for $\pm 1$, which are therefore the only units. Neither two nor three is a value of $N$: if $b\ne 0$ the value is at least five, and if $b=0$ it is a square. A factorization of $2$ into nonunits would require norms two and two; one of $3$ would require norms three and three; one of $1\pm \sqrt{-5}$ would require norms two and three. All are impossible. Thus all four factors are irreducible. The factors of norms four and nine cannot be associate elements of those of norm six, and $1+\sqrt{-5}$ and $1-\sqrt{-5}$ are not related by a sign. Therefore $R$ is an integral domain which is not a unique factorization domain.

A nonzero polynomial of degree $n$ over a field has at most $n$ distinct roots in that field. For a proof, the factor theorem gives $(t-a)\mid f(t)$ whenever $f(a)=0$. Distinct linear factors are coprime in $K[t]$, a unique factorization domain, so for distinct roots $a_1,\dots ,a_k$ their product divides $f$. Its degree is $k$, giving $k\le n$. One may equivalently remove one linear factor at a time and argue by induction.

If the field has at least $n$ elements, this bound is attained by a product of $n$ distinct linear factors. For a fixed finite field with fewer than $n$ elements, the largest possible number of distinct roots is $\min (n,|K|)$ instead; multiplicities do not create more distinct roots. The zero polynomial is excluded since it has no finite degree.

The dense algebraic torus has coordinate ring $k[M]$, a Laurent polynomial ring and hence a unique factorization domain. Its divisor class group is therefore zero. Restrict an arbitrary Weil divisor $D$ to the algebraic torus and choose a rational function on an algebraic variety $f$ whose divisor there is $D{|}_T$. Then $D-\mathrm{div}(f)$ is supported on the algebraic torus boundary. The prime divisors in that boundary are precisely the $F_j$ corresponding to the toric rays, by the torus orbit-cone correspondence. ConsequentlySince the variety is smooth, every Weil divisor is Cartier. This gives invariant representatives for all divisor classes, without assuming that the original divisor was invariant.

Now use such a representative. The associated vector space isThe principal divisor on a toric variety formula gives $\mathrm{div}({\chi }^m)={\sum }_j⟨m,e_j⟩F_j$. Define the lattice polytope of a toric divisorFor a maximal toric cone $\sigma$, its basic toric ray vectors are a basis of $N$, so there is a unique integral $m_{\sigma }$ with $⟨m_{\sigma },e_j⟩=-a_j$ on its toric rays. The algebraic torus character ${\chi }^{m_{\sigma }}$ is a local frame for $\mathcal{O}(D)$ on $U_{\sigma }$. Every global section restricts on the algebraic torus to a finite Laurent polynomial ${\sum }_m{c}_m{\chi }^m$. Relative to this frame, regularity on $U_{\sigma }$ means ${\sum }_m{c}_m{\chi }^{m-m_{\sigma }}\in k[{\sigma }^{\vee }\cap M]$. Since distinct algebraic torus characters are independent, this holds exactly when each exponent with nonzero coefficient satisfies the inequalities for the toric rays of $\sigma$. Taking all charts provesCompleteness makes $P_D$ bounded when nonempty: a nonzero recession vector would be nonnegative on every toric ray of a complete toric fan, and therefore on all of $N_{\mathbb{R}}$, which is impossible. The displayed space is consequently finite-dimensional.

At the torus-fixed point $p_{\sigma }$ in the smooth chart $U_{\sigma }\cong {\mathbb{A}}^n$, a section ${\chi }^m$ has local monomial ${\chi }^{m-m_{\sigma }}$. Its value is nonzero exactly when this monomial is constant, namely when $m=m_{\sigma }$. Thus if the complete linear system of a divisor has no base point, some section is nonzero at $p_{\sigma }$, forcing $m_{\sigma }\in P_D$. Conversely, if $m_{\sigma }\in P_D$ for every maximal toric cone, its algebraic torus character is a global section whose local expression on $U_{\sigma }$ is one. It therefore vanishes nowhere on that whole chart, and these charts cover $X$. This proves the toric basepoint-free criterion:The analogous toric ampleness criterion, stated without proof, replaces the inequalities off $\sigma$ by strict ones:Equivalently, the normal fan of a polytope $P_D$ is exactly $\Sigma$.

For the final fan subdivision, take $n\ge 2$ so that $\tilde{e}$ is a new toric ray, as the construction requires. The vectors $e_1,\dots ,e_n$ are a lattice basis, so their sum is primitive. Replacing any one basis vector by this sum gives another basis. Thus the star subdivision gives a smooth toric blowup at a torus-fixed point $\pi :X_{{\Sigma }^{\prime }}\to X_{\Sigma }$, with algebraic exceptional divisor $E=\tilde{F}$. On the old chart, a local equation of $D$ is ${\chi }^{-m_{\sigma }}$, whose order on the new toric ray is $-⟨m_{\sigma },\tilde{e}⟩={\sum }_{i=1}^n{a}_i=a$. All old toric ray coefficients are unchanged. ThereforeWe prove ampleness of $D_c=c{D}^{\prime }-E$ directly by inequalities, giving the ample divisor after blowing up a torus-fixed point result.

Let $u_1,\dots ,u_n$ be the basis of $M$ dual to $e_1,\dots ,e_n$. On the new maximal toric cone ${\sigma }_i$, which omits $e_i$, the required local algebraic torus character isIndeed its pairings on $e_j$, $j\ne i$, are $-c{a}_j$, while its pairing on $\tilde{e}$ is $-ca+1$, exactly the negative of the new coefficient $ca-1$. Its inequality on the omitted old toric ray has strict gap one:For any toric ray $e_j$ outside the original toric cone, put ${\delta }_j=⟨m_{\sigma },e_j⟩+a_j$. The original ampleness gives positive integers ${\delta }_j$, and the new strict gaps arefor all sufficiently large $c$.

For any unchanged maximal toric cone $\eta \ne \sigma$, the local algebraic torus character is $c{m}_{\eta }$. All strict inequalities on old toric rays follow from those for $D$. Its inequality on the new toric ray has gapEach summand is nonnegative, and at least one is positive because $\eta \ne \sigma$ cannot contain all its toric rays. Hence ${ϵ}_{\eta }$ is a positive integer, making this gap positive for $c\ge 2$. There are only finitely many remaining inequalities. For example, any integer satisfyingworks, with the inner maximum omitted when its index set is empty. The toric ampleness criterion now givesThe new-ray assumption is essential to the printed construction. In dimension one, $\tilde{e}=e_1$ and the fan subdivision is unchanged, so the two purported distinct divisors coincide. If the displayed definition of $D^{\prime }$ is applied literally on ${\mathbb{P}}^1$ with coefficients $a_1=-2$, $a_2=3$, then $D$ has degree one but $D^{\prime }$ has degree $-1$, contradicting the conclusion. The argument above applies to the intended blowup in dimension at least two.

An irreducible analytic subvariety of a complex manifold is a closed subset locally defined by finitely many holomorphic functions which cannot be expressed as the union of two proper closed analytic subsets. In a complex $n$-manifold, codimension $k$ means pure complex dimension $n-k$. A divisor on a complex manifold is a locally finite formal integer combinationof irreducible analytic hypersurfaces. Compactness makes the set of components in this sum finite. Negative coefficients are permitted; a local equation for such a divisor on a complex manifold is generally meromorphic rather than holomorphic.

We use these precise local facts. In holomorphic coordinates at $x$, the local ring is ${\mathcal{O}}_{X,x}\cong \mathbb{C}\{z_1,\dots ,z_n\}$, a Noetherian regular local ring and a unique factorization domain. Each analytic hypersurface germ has finitely many irreducible local branches; their prime ideals have height one and are generated by irreducible holomorphic germs. A holomorphic germ is a unit exactly when its value at $x$ is nonzero. Consequently a nonzero meromorphic germ with order zero on every local hypersurface branch is a holomorphic unit, by cancelling its irreducible factors in numerator and denominator. These are the auxiliary local-ring properties required below.

After shrinking a neighbourhood $U_i$ around each point, choose local branch equations $h_{i\nu a}$, and putIts zero and pole orders give exactly the local divisor on a complex manifold $D{|}_{U_i}$. A globally irreducible hypersurface may have several branches locally; all of them must be included with the indicated multiplicity. On an overlap, $f_i/f_j$ has zero order along every branch, so the local-ring facts show that it extends to a nowhere-zero holomorphic function. These germwise extensions patch uniquely to a holomorphic unit on the overlap. This describes a local defining function of $D$ and proves why multiplicities, not just zero sets, matter.

Set $g_{ij}=f_i/f_j$ and glue line-bundle frames by $e_j=g_{ij}{e}_i$. The identities $g_{ij}{g}_{jk}=g_{ik}$ give a holomorphic line bundle associated to a divisor, denoted $[D]$. Equivalently, the frame $e_i$ can be represented by $1/f_i$ in the sheaf of meromorphic functions, so $[D]={\mathcal{O}}_X(D)$. If alternative equations are $f_i^{\prime }=u_i{f}_i$, where $u_i$ are holomorphic units, the corresponding frames are identified by $e_i^{\prime }\mapsto u_i^{-1}{e}_i$. This respects the changed transitions. Refining the cover gives the same identifications, proving that $[D]$ is well-defined up to holomorphic bundle isomorphism. The expressions $f_i{e}_i$ glue to its canonical meromorphic section, whose divisor is $D$.

For two divisors on a complex manifold use a common cover with equations $f_i$ and $h_i$. Their sum has equation $f_i{h}_i$, and the tensor product of vector bundles has frame transitionsSending the sum-divisor frame to the tensor product of the two frames givesThis also yields $[-D]\cong [D{]}^{\ast }$.

For a nonsingular complex analytic hypersurface $Y$, choose reduced local equations $f_i$ with $d{f}_i$ nonzero along $Y$. The holomorphic conormal bundle iswhose fibre is the annihilator of $T^{1,0}Y$. It has local frame $d{f}_i{|}_Y$. Here this restriction is as a covector in the ambient cotangent bundle along $Y$, not its pullback to ${\Omega }_Y^1$, where it would vanish. The line bundle $[-Y]$ has local frame $a_i=f_i$, and if $f_j=u_{ji}{f}_i$, thensince $f_{i}d{u}_{ji}$ vanishes along $Y$. The frame identifications $a_i{|}_Y\mapsto d{f}_i{|}_Y$ therefore patch and are fibrewise nonzero. This proves the conormal line of a smooth analytic hypersurface formulaEquivalently it identifies ${\mathcal{O}}_X(-Y){|}_Y={\mathcal{I}}_Y/{\mathcal{I}}_Y^2$ with the analytic conormal sheaf.

Since $10\equiv 2(\mathrm{mod}4)$, the ring of integers is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{10}]$, with field norm $N(x+y\sqrt{10})=x^2-10y^2$. Norms $\pm 2$ and $\pm 3$ are impossible: modulo five they would require a square congruent to 2 or 3. An algebraic integer is a unit exactly when its norm is $\pm 1$.

If 2 factored into two nonunits, their absolute norms would both be 2, since $N(2)=4$; this is impossible. The same argument for $N(3)=9$ excludes a factorization of 3. Finally $ϵ+1=4+\sqrt{10}$ has norm 6, so a nonunit factorization would require absolute norms 2 and 3, also impossible. Hence 2, 3 and $ϵ+1$ are irreducible. The conjugate $\overline{ϵ}+1$ is irreducible by the same argument.

The equality $6=2\cdot 3=(4+\sqrt{10})(4-\sqrt{10})$ gives inequivalent factorizations into irreducibles: no norm-six factor is associated to either 2 or 3, whose norms are 4 and 9. A principal ideal domain is a unique factorization domain. Consequently ${\mathcal{O}}_K$ is not a principal ideal domain.

Suppose the map were nonconstant. Since projective space is proper, its image $Z$ is a closed projective subvariety of ${\mathbb{P}}^n$, of positive dimension $r\le n$. Fix $y\in Z$. Regard the map as surjective onto $Z$; the fiber dimension theorem just proved givesChoose a target hyperplane $H$ not containing $y$. It meets $Z$: the affine cone over $Z$ has dimension $r+1\ge 2$, and the hyperplane equation vanishes at its vertex but not identically. By the principal hypersurface dimension lemma its zero set on the cone has dimension $r\ge 1$, so it contains a nonzero cone point and hence a point of $Z\cap H$.

The inverse image $D={\varphi }^{-1}(H)$ is therefore nonempty and proper. Locally the hyperplane equation is a single regular equation, after trivializing the pulled-back line bundle. On the irreducible smooth source it is not identically zero, so the same hypersurface dimension lemma shows that each component of $D$ has dimension $m-1$. Choose one component $D_0$. Its cone is a codimension-one irreducible closed subset of affine $(m+1)$-space, whose prime ideal is principal because the polynomial ring is a unique factorization domain. Thus $D_0$ is a projective hypersurface defined by a homogeneous polynomial $P$ of positive degree.

Choose a positive-dimensional irreducible component $F$ of ${\varphi }^{-1}(y)$. Since $y\notin H$, $F$ is disjoint from $D_0$. On the cone over $F$, the equation $P$ is consequently nonzero everywhere except at the vertex. This contradicts the hypersurface dimension lemma: the cone has dimension at least two, so its nonempty proper zero set for $P$ has dimension at least one and cannot consist just of the vertex. Therefore $\boxed{\varphi \text{ is constant}}$. The argument proves morphisms from projective space to smaller projective space are constant by combining the fibre bound with the forced intersection of a positive-dimensional projective variety and a hypersurface.

On a complex manifold, a divisor on a complex manifold is a locally finite formal sum $D={\sum }_Y{m}_{Y}Y$, where $Y$ ranges over irreducible complex analytic hypersurfaces and $m_Y\in \mathbb{Z}$. The local algebra needed here is that the local ring ${\mathcal{O}}_{X,x}$, identified in coordinates with convergent power-series germs, is a regular local ring and a unique factorization domain. In particular, each height-one prime is generated by one irreducible germ, and a nonzero meromorphic germ is a quotient of holomorphic germs. Reduced local equations of the same hypersurface differ by a holomorphic unit. These facts include local branches when a globally irreducible hypersurface has several branches at a singular point.

The order of a meromorphic function along a hypersurface is the exponent of its reduced local equation in the quotient's factorization. Equivalently, at a generic smooth point of $Y$, write $f=t^{m}u$, with $t$ a transverse local equation of $Y$ and $u$ a holomorphic unit; then ${\mathrm{ord}}_Y(f)=m$. Changing a reduced equation multiplies it by a unit and leaves $m$ unchanged. These local orders agree on overlaps and are constant along the generic smooth locus of an irreducible hypersurface. A nonzero global meromorphic function therefore defines its principal divisor on a complex manifoldPositive coefficients record zeros and negative coefficients record poles.

A local defining function $f_i$ for $D$ on $U_i$ is a nonzero meromorphic function whose orders there equal the coefficients of $D$. It can be constructed as the product of local reduced hypersurface equations to the prescribed integer powers. On an overlap, $f_i/f_j$ has no zero or pole along any hypersurface. The unique factorization domain property then makes it a nowhere-zero holomorphic function, after choosing representatives of the germs. The holomorphic line bundle associated to a divisor is obtained by gluing holomorphic frames usingThese ratios satisfy the cocycle condition, and the expressions $f_i{e}_i$ agree on overlaps. They form the canonical meromorphic section of a holomorphic line bundle with divisor $D$.

If $D=\mathrm{div}(f)$ globally, $f_i/f$ is a holomorphic unit on $U_i$, and $(f_i/f)e_i$ is a global nowhere-zero holomorphic section. Hence $[D]$ is holomorphically trivial. Conversely, if $[D]$ has a nowhere-zero global holomorphic frame $e$, write $e_i=a_{i}e$ with $a_i$ holomorphic units. The coefficients $f_i{a}_i$ of the canonical meromorphic section of a holomorphic line bundle agree on overlaps and give a nonzero global meromorphic function $f$. Since multiplication by a unit does not change orders, $\mathrm{div}(f)=D$. Therefore

For a smooth hypersurface $Y\subset X$, the adjunction formula isIndeed, the holomorphic conormal sequence has first term $N_{Y/X}^{\ast }\cong [-Y]{|}_Y$; taking its determinant yields $K_X{|}_Y\cong N_{Y/X}^{\ast }\otimes K_Y$, which gives the formula. Here the canonical bundle is the top exterior power of the holomorphic cotangent bundle.

Apply this first to $W\subset {\mathbb{P}}^4$. Its degree-$d_1$ equation is a section of $\mathcal{O}(d_1)$, so $[W]\cong \mathcal{O}(d_1)$. Since $K_{{\mathbb{P}}^4}\cong \mathcal{O}(-5)$, we obtain $K_W\cong {\mathcal{O}}_W(d_1-5)$. The restricted second equation cuts out $S$ in $W$ and has associated bundle ${\mathcal{O}}_W(d_2)$. A second application of adjunction formula gives the canonical bundle of a regular projective complete intersection

It remains to prove that a nonzero exponent cannot yield a trivial holomorphic line bundle on $S$. Choose a connected component $S_0$ and a hyperplane meeting but not containing it. Such a hyperplane exists: a positive-dimensional compact projective submanifold cannot lie entirely in an affine chart, since all its affine coordinate functions would be global holomorphic functions and hence constant. A hyperplane through one point and not another consequently meets $S_0$ properly. Its equation gives a holomorphic section of ${\mathcal{O}}_{S_0}(1)$ that has both a zero and a nonzero value. For $m>0$, the $m$th power has the same two properties. Were ${\mathcal{O}}_{S_0}(m)$ trivial, this section would be represented by a global holomorphic function, contradicting its constancy on compact connected $S_0$. For $m<0$, triviality would imply triviality of its positive dual power. This proves nontrivial hyperplane powers on a compact projective submanifold and henceThe degrees are positive because the two equations define a hypersurface and a surface of the specified dimensions.

The calculation uses “non-singular” in the usual smooth complete-intersection sense: the equations define reduced smooth divisors, with independent differentials along $S$. If only their reduced zero sets are required to be smooth, polynomial multiplicities cannot determine the canonical bundle. For example, $Q_1=Z_0^m$ and $Q_2=Z_1^4+Z_2^4+Z_3^4+Z_4^4$ have the same smooth quartic surface in the hyperplane $Z_0=0$ for every $m\ge 1$. That surface has trivial canonical bundle, while the displayed polynomial degree sum is $m+4$. Thus unreduced equations must not be substituted into the adjunction formula for the reduced surface.