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The defining section of ${\mathcal{O}}_X(D)$ is nowhere zero on $C$, so its restriction is trivial and has degree zero. Every translated line bundle also has degree zero on $C$, by the preceding part. For any $t\in X$, either $C\subseteq D-t$ or the defining section of $T_t^{\ast }{\mathcal{O}}_X(D)$ restricts to a nonzero section on $C$. In the second case its zero effective divisor has degree zero, so it has no zeros. Thus a translate of $D$ either contains all of $C$ or misses all of $C$.

Take $d\in D$ and $x_1,x_2\in C$. The translate $D-(d-x_1)$ contains $x_1$, hence all of $C$. In particular $d+(x_2-x_1)\in D$. This proves $D+(x_2-x_1)\subseteq D$; exchanging $x_1,x_2$ proves the reverse inclusion. Since the prime Weil divisor $D$ is reduced and translation is an automorphism, equality of these sets gives equality of Weil divisors. Remembering that pullback by $T_t$ translates the divisor by $-t$, the answer is

The line bundle translation stabilizer is the closed subgroupEach difference $x_1-x_2$ lies in $K(L)$, by the equality of Weil divisors just proved. Thus the subgroup they generate lies in $K(L)$, and closedness puts its Zariski topology closure there too. Therefore $\boxed{Y\subseteq K(L)}$. The closure of a subgroup is again a subgroup: multiplication and inversion preserve its closure by continuity. In this setting it is connected as well, since $C-C$ is connected and contains zero, and the increasing finite sums of $C-C$ are connected with a common point. Its reduced closure is an abelian subvariety.

For the curve generation criterion for an abelian variety, first suppose $Y=X$ and an irreducible Weil divisor $D$ avoids $C$. The preceding part gives $K({\mathcal{O}}_X(D))=X$, so ${\varphi }_{{\mathcal{O}}_X(D)}=0$ and ${\mathcal{O}}_X(D)\in {\mathrm{Pic}}^0(X)$. A nonzero effective divisor cannot have this property: if $H$ is an ample line bundle and $g=\dim X$, then the intersection product of Cartier divisors $D\cdot H^{g-1}$ is positive, whereas an algebraically trivial line bundle has zero intersection with every curve. The latter follows from constancy of degree along a connected family defining algebraic equivalence; the former is the positive projective degree of $D$ after replacing $H$ by a very ample power. This contradiction proves that $C$ meets every irreducible Weil divisor.

Conversely suppose $Y\ne X$. Fix $c_0\in C$, so $C\subseteq c_0+Y$. Use the permitted fiber theorem to choose a morphism of varieties $f:X\to Z$ with $f^{-1}(z_0)=Y$. This morphism is nonconstant because $Y$ is proper. Choose an affine neighborhood $U$ of $z_0$. The irreducible image $f(X)$ has positive dimension, so its intersection with $U$ does too; some regular function $h$ on $U$ is nonconstant on this intersection. Then $F=h\circ f$ is a nonconstant rational function on $X$, regular on $f^{-1}(U)$.

The pole divisor avoiding a fiber construction applies: its pole Weil divisor is nonzero. Indeed on the smooth, hence normal, projective variety $X$, a rational function with no codimension-one poles extends to a global regular function; every global regular function on a connected projective variety is constant. Every pole component $D_0$ lies outside $f^{-1}(U)$ and hence avoids $Y$. Therefore $D_0+c_0$ is an irreducible Weil divisor disjoint from $c_0+Y$, and in particular from $C$. This proves the converse using precisely the fiber fact allowed in the question, without requiring a projective target $Z$. The criterion is

The printed construction is not well defined: the next tangent must pass through $x_1$, rather than $x_0$. If $x_0\ne x_1$ and ${\ell }_1\ne {\ell }_0$, a second line through $x_0$ cannot also pass through $x_1$, because the unique line through both points is ${\ell }_0$. Here is a concrete counterexample satisfying all the conic hypotheses. In the affine chart takewith ${\ell }_0:x=1$, $x_0=(1,3\sqrt{3}/2)$ and $x_1=(1,-3\sqrt{3}/2)$. The second tangent through $x_0$ is $-23x+12\sqrt{3}y=31$. It does not contain $x_1$: its left side there is $-77$. The conics meet transversely at the four complex points with $x^2=32/5$, $y^2=-27/5$. Thus this is a defect in the original PDF, not just in its conversion.

For the corrected construction, let ${\iota }_{\ell }$ exchange the two points of $C_2$ on a fixed tangent line, and let ${\iota }_x$ exchange the two tangents to $C_1$ through a fixed point of $C_2$. Both projections from $E$ are degree-two morphisms to a smooth plane conic: for the line projection this follows from intersecting a line with $C_2$, which has no line component. Since $E$ is smooth and the ground field has characteristic zero, their quadratic function-field extensions define regular involutions on the whole curve. At a ramification point “the second” point is the same point, counted with multiplicity. Each switch is an involution of a degree-two map from a genus one curve. Hence the corrected step is the everywhere-defined automorphism $F={\iota }_x\circ {\iota }_{\ell }$.

Choose an origin $O$ on the genus one curve. The Abel-Jacobi map of a genus-one curve identifies $E$ with ${\mathrm{Pic}}^0(E)$ by $p\mapsto {\mathcal{O}}_E(p-O)$. Fibers of each degree-two projection are linearly equivalent Weil divisors, since they are pullbacks of points of ${\mathbb{P}}^1$. Thus their group sums are constant: for suitable $a,b\in E$,This also holds at the ramification points, where $2p=a$ or $2p=b$. Therefore the corrected step is a translation on an elliptic curve. If $F^n(p_0)=p_0$ for one point, then $n(b-a)=0$. It follows that $F^n(p)=p$ for every $p\in E$. The corrected construction has the Poncelet porism: one periodic orbit implies all orbits are periodic, with the same least period. The literal printed construction fails before this conclusion; the proof establishes the intended, explicitly corrected assertion.

For an effective Cartier divisor $F$, the divisor restriction exact sequence givesThe scheme $F$ has dimension at most $n-1$, so the permitted scheme version of part (i) bounds $h^0(F,{\mathcal{O}}_F(kD))$ by $C{k}^{n-1}$. For the infinitely many $k$ in the hypothesis,once $k$ is sufficiently large. Thus infinitely many such $k$ have a nonzero section of $kD-F$. This dimension-drop argument is the section subtraction lemma for big divisors.

If “effective divisor” is interpreted as an effective Weil divisor on a normal variety, use its coherent divisor ideal instead. The quotient by that ideal is supported in dimension at most $n-1$, so the polynomial bound for sections of a fixed divisor gives the same conclusion. For $F=0$ the assertion is immediate.

A standard sufficient hypothesis is that $X$ is both a Noetherian scheme and an integral scheme, and is regular in codimension one; in particular, a Noetherian normal scheme qualifies. A Weil divisor is then a finite sumover integral codimension-one closed subschemes $Z$. The local ring at the generic point of each $Z$ is a discrete valuation ring, so every nonzero rational function $g\in k(X{)}^{\times }$ has a principal divisorThe divisor class group is

A Weil divisor on $X$ is a finite formal sumover integral codimension-one closed subschemes $Z$. Since $X$ is regular in codimension one, the local ring at the generic point of each $Z$ is a discrete valuation ring. A nonzero rational function $f\in k(X{)}^{\times }$ therefore defines the principal divisor $\mathrm{div}(f)={\sum }_Z{v}_Z(f)[Z]$. The divisor class group is

The standard affine charts of ${\mathbb{P}}^3$ are copies of ${\mathbb{A}}^3$, and the product charts of ${\mathbb{P}}^1\times {\mathbb{P}}^1$ are copies of ${\mathbb{A}}^2$. Their local rings are localizations of polynomial rings over $k$ and hence are regular local rings. Both schemes are therefore regular.

On a regular integral scheme every Weil divisor is Cartier, so the divisor class group is naturally the Picard group. Pullback of line bundles along the Segre embedding therefore definesThe two groups areThe Segre coordinates are bihomogeneous of bidegree $(1,1)$, so ${\iota }^{\ast }{\mathcal{O}}_{{\mathbb{P}}^3}(1)=\mathcal{O}(1,1)$. In these bases the map is $m\mapsto (m,m)$ and

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.

For a Noetherian separated integral scheme regular in codimension one, a Weil divisor is a finite integer combination of integral codimension-one closed subschemes. Principal divisors are the valuation divisors of nonzero rational functions, and the divisor class group is

For an open immersion $U\subseteq X$, restriction induces a surjectionevery prime divisor of $U$ closes to a prime divisor of $X$, while divisors supported in $X\setminus U$ form the kernel. Therefore $\mathrm{Cl}(X)=0$ implies $\mathrm{Cl}(U)=0$.

Affine schemes need not have trivial class group. For example,is a normal affine surface with $\mathrm{Cl}(X)\cong \mathbb{Z}/2\mathbb{Z}$. The height-one prime $(x,z)$ represents the nonzero class: twice this divisor is the principal divisor of $x$, but the divisor itself is not principal.

For a smooth integral variety, every Weil divisor is a Cartier divisor, so the Picard group equals the divisor class group. If $U$ is obtained by removing irreducible Weil divisors $D_j$, the localization sequence for the divisor class group gives the displayed exact sequence. A rational trivialization of a line bundle on $U$ has divisor supported on the removed divisors, describing the kernel concretely.

Suppose a nonconstant morphism of varieties $f:X\to Z$ has fiber $Y$ over $z_0$, where $X$ is normal, connected and projective. Choose an affine neighborhood $U$ of $z_0$ and a regular function on $U$ nonconstant on $f(X)\cap U$. Its pullback is a nonconstant rational function on $X$, so its pole Weil divisor is nonzero. A pole-free rational function on a normal variety is regular, and global regular functions on a connected projective variety are constant. The pole divisor avoids $f^{-1}(U)$, hence avoids $Y$. No projectivity assumption on $Z$ is needed.

On a normal integral Noetherian scheme, a nonzero rational function defines the displayed Weil divisor, with the sum over codimension-one prime divisors. The local ring at each such divisor is a discrete valuation ring, so the order is well defined. The divisor class group is the group of Weil divisors modulo principal Weil divisors. On a smooth curve this agrees with a principal divisor on an algebraic curve.

A Weil divisor is Q-Cartier if some positive integer multiple is a Cartier divisor. Its Cartier index is the smallest such integer. It is ample in the Q-Cartier sense when such a multiple is an ample Cartier divisor. This need not make the original divisor itself Cartier.