Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 18 / 4

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.

Pull $L$ back by the morphism of varietiesThe fiber of this line bundle over $x$ is ${\phi }^{\ast }{T}_x^{\ast }L$. An abelian variety is connected, so the preceding constancy of line bundle degree in a family applies. Consequently $\boxed{\mathrm{deg}({\phi }^{\ast }{T}_x^{\ast }L)=\mathrm{deg}({\phi }^{\ast }L)}$ for every $x\in X$.

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