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For a line bundle $M$ on $C\times T$, with $C$ a smooth projective curve, this function is locally constant. Twist both $M$ and $M^{-1}$ by a large power of an ample line bundle on $C$. Near a chosen fiber their first sheaf cohomology groups vanish. The Riemann-Roch theorem and upper semicontinuity of their zeroth sheaf cohomology groups then bound the degree from above and below by the degree on that fiber. Connectedness of $T$ is necessary for a single constant across all fibers.

A smooth projective curve $C\subset A$ generates $A$ by its point differences if and only if it meets every prime Weil divisor. If a prime Weil divisor avoids $C$, its restriction to $C$ has degree zero; constancy of line bundle degree in a family shows that every translate either contains $C$ or misses it. This forces invariance under $C-C$ and puts the generated subgroup inside the line bundle translation stabilizer. If this subgroup were all of $A$, the divisor would be algebraically trivial, contradicting its positive intersection with an ample line bundle. Conversely a proper closed generated subgroup admits a disjoint divisor by the pole divisor avoiding a fiber construction, after translating the fiber to contain $C$.

For a finite morphism $f:X\to Y$, the pullback of an ample line bundle is ample. If $f$ is also surjective, ampleness is equivalent in both directions. These results hold for proper schemes over a field, including nonreduced schemes. Cohomological ampleness and the exact finite pushforward prove the forward direction; coherent-sheaf devissage gives finite-surjective descent. In particular, a finite map to projective space makes the pulled-back hyperplane bundle ample.

If $A$ is an ample line bundle on a projective variety and $L$ is any line bundle, then $A^{\otimes m}\otimes L$ is very ample for all sufficiently large integers $m$. Taking $L$ to be trivial also makes $A^{\otimes m}$ very ample.

We regard $N$ as positive and invertible in $k$; $N=1$ gives the trivial group and the same arguments apply. Put $A={\mathrm{Pic}}^0(C)$, an abelian variety of dimension $g$.

First compute the prime-to-characteristic torsion of an abelian variety. The differential of $[N]:A\to A$ at the origin is $N\mathrm{id}$, and translation gives the same assertion everywhere. Consequently $[N]$ is an étale morphism, its fibres have dimension zero, and properness makes it finite. Its image is closed of dimension $g$, hence all of $A$. Choose an ample line bundle $L_0$ on $A$ and put $L=L_0\otimes [-1{]}^{\ast }{L}_0$, which is symmetric and ample. The Theorem of the Cube gives the recurrenceAfter fixing a trivialization at the origin, induction from $n=0,1$ gives $[n{]}^{\ast }L\cong L^{\otimes n^2}$. Comparing top intersection products under the finite map $[N]$ givesThe intersection number is positive because $L$ is ample, so $\mathrm{deg}[N]=N^{2g}$. Since $[N]$ is étale and $k$ is algebraically closed field, its kernel has precisely $N^{2g}$ points.

For a prime power ${\ell }^r\mid N$, write the finite abelian group $A[{\ell }^r]$ as a product of cyclic groups of orders ${\ell }^{a_j}$, with $1\le a_j\le r$. Its subgroup killed by $\ell$ is $A[\ell ]$, so the number of factors is $2g$. Its total order is ${\ell }^{2gr}$, so $\sum a_j=2gr$ forces every $a_j=r$. Combining the prime-power factors by the Chinese remainder theorem provesThe same reasoning works for the dual abelian variety $\hat{A}$, also of dimension $g$.

To define the Weil pairing for an abelian variety on two copies of $A[N]$, we first justify the required theta identification with $\hat{A}$. Identify $X={\mathrm{Pic}}^{g-1}(C)$ with an $A$-torsor and use ordinary translation $T_q(x)=x+q$. The Theorem of the square, a consequence of the Theorem of the Cube, makesa group homomorphism $A\to \hat{A}$. Fix the negative Abel map of an algebraic curve $j:C\to A$, $P\mapsto [P_0-P]$, and let $\rho :\hat{A}\to {\mathrm{Pic}}^0(C)=A$ be pullback by $j$. Pullback by a translate of $j$ gives the same map: algebraically trivial line bundles on an abelian variety are translation-invariant as Picard classes, by the Theorem of the square.

For any $q\in A$, choose a degree-$g$ class $D$ such that both $D$ and $D+q$ have one-dimensional section spaces. This is possible because the nonspecial degree-$g$ classes form a nonempty open subset of the irreducible Picard scheme of a curve, and its translate has a nonempty intersection with it. Since $T_q{\varphi }_D={\varphi }_{D+q}$, the restriction of the theta divisor to an Abel curve proved above givesEach ${\varphi }_D$ differs from $j$ by a translation. Thus $\rho \lambda ={\mathrm{id}}_A$. This equality on closed points is equality of morphisms, since $A$ is reduced and the target is separated. In particular $\lambda$ is a closed immersion, being a section of the separated morphism $\rho$. Its image has dimension $g=\dim \hat{A}$, so it is all of the reduced irreducible abelian variety $\hat{A}$. Therefore $\lambda$ is an isomorphism of schemes. This proves the needed theta autoduality of a Jacobian, including possible infinitesimal issues in positive characteristic.

Here is a definition of the Weil pairing for an abelian variety that also proves nondegeneracy. For $M\in \hat{A}[N]$, the Theorem of the Cube gives $[N{]}^{\ast }M\cong M^{\otimes N}\cong {\mathcal{O}}_A$. Choose a trivialization of $[N{]}^{\ast }M$. The finite étale cover $[N]:A\to A$ has deck group $G=A[N]$. Its natural descent action on $[N{]}^{\ast }M$ becomes multiplication by scalars under this trivialization: these scalars are constant because every invertible regular function on the proper connected abelian variety $A$ is constant. They form a characterwhose values lie in ${\mu }_N$, since $NP=0$ for every $P\in G$. Rescaling the trivialization changes none of these scalars. Translation gives multiplicativity in $P$, and tensor product gives ${\chi }_{M\otimes M^{\prime }}={\chi }_M{\chi }_{M^{\prime }}$. DefineThis fixes a convention for the Weil pairing for an abelian variety; reversing the convention for the descent action inverts all values and changes no nondegeneracy assertion.

If ${\chi }_M=1$, the trivialization is equivariant for the deck group. Descent then makes $M$ the trivial line bundle on the target. Thusis injective. The source and target both have $N^{2g}$ elements, since $k$ contains all $N$ distinct roots of unity and $A[N]\cong (\mathbb{Z}/N{)}^{2g}$. Hence it is an isomorphism. Characters of this group separate its points, so the Weil pairing for an abelian variety is nondegenerate in both variables. Composing with the isomorphism $\lambda$ proves the pairing on $A[N]$ is perfect.

For a line bundle $M$ on $C\times T$, put $d(t)=\mathrm{deg}{M}_t$, and choose an ample line bundle $A$ on $C$. Fix $t_0$. For sufficiently large $q$, both $H^1(C,M_{t_0}\otimes A^q)$ and $H^1(C,M_{t_0}^{-1}\otimes A^q)$ vanish, by Serre duality and negativity of the degrees of their dual canonical twists. The semicontinuity theorem for coherent cohomology makes both groups vanish on a neighborhood of $t_0$. The family is flat over $T$: $C\times T\to T$ is flat and the sheaves in question are locally free.

On that neighborhood the Riemann-Roch theorem givesUpper semicontinuity of both zeroth sheaf cohomology dimensions, after shrinking again, gives $d(t)\le d(t_0)$ and $-d(t)\le -d(t_0)$. Hence $d(t)=d(t_0)$. The degree is locally constant, and thus constant on connected $T$. With the usual convention that a variety is irreducible, this proves the assertion as printed. If disconnected varieties are allowed, the necessary qualification is constancy on each connected component: take $T$ to be two points and the bundles ${\mathcal{O}}_{{\mathbb{P}}^1}$ and ${\mathcal{O}}_{{\mathbb{P}}^1}(1)$ to see that a global constant degree need not exist.

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

Put $L={\mathcal{O}}_X(D)$ and let $s_D$ be the canonical global section cutting out the effective Cartier divisor $D$. The divisor restriction exact sequence givesThe hypothesis says $L{|}_D$ is an ample line bundle. By Serre vanishing, $H^1(D,L^m{|}_D)=0$ for all sufficiently large $m$, so the long exact sequence in sheaf cohomology makessurjective for all such $m$. These are finite-dimensional vector spaces. Their dimensions form a nonincreasing sequence of nonnegative integers, so the maps are isomorphisms from some point on. Exactness then implies that the restrictionis surjective for all sufficiently large $m$.

Choose such an $m$ for which $L^m{|}_D$ is also a globally generated line bundle. Lift a generating collection of its global sections to $X$. At each point of $D$, one lift has nonzero image in the one-dimensional residue-field fibre, hence generates the stalk of $L^m$ by Nakayama lemma. Outside $D$, $s_D^m$ is nowhere zero and generates $L^m$. Together these sections generate it everywhere. ThereforeThis proof works on an arbitrary projective scheme because the defining section of an effective Cartier divisor is a non-zero-divisor. If $D$ is empty, $L\cong {\mathcal{O}}_X$ and the conclusion is immediate. Semiampleness is the conclusion: the pullback of a line avoiding the centre of a blowup of a smooth algebraic surface of ${\mathbb{P}}^2$ satisfies the hypothesis on its support but has zero intersection with the exceptional curve, and therefore is not ample.

If $N$ is invertible in the base field, multiplication by $N$ on an abelian variety is finite étale: its differential is $N$ times the identity, and properness turns its zero-dimensional fibres into finite fibres. The Theorem of the Cube gives $[N{]}^{\ast }L\cong L^{N^2}$ for a symmetric ample line bundle. Top intersection products give degree $N^{2\dim A}$. For each prime power ${\ell }^r\mid N$, the kernels killed by ${\ell }^s$ have orders ${\ell }^{2s\dim A}$; the classification of finite abelian groups forces $2\dim A$ cyclic factors of length $r$. Combining the prime powers gives the displayed group.

For an ample line bundle $A$ on a projective scheme and a coherent sheaf $\mathcal{F}$, $H^i(X,\mathcal{F}\otimes A^m)=0$ for every $i>0$ and all sufficiently large $m$.

A line bundle is very ample when its global sections define a closed embedding into projective space.

Over a field, a line bundle is very ample when it is the pullback of $\mathcal{O}(1)$ by a closed immersion into projective space. It is an ample line bundle, but an ample bundle need not itself be very ample.