# Polishcuk Ex. 1 pg. 107

Setup: homo. or abel. var. and its stein factorization; i.e.

- and
- p is finite.

Then B’ has group structure making it an abel. var. such that f’ is homo. and ker(f’) is abel. sub var. of A.

pf: Note 1. implies the set theoretic image of f’ is dense in B’. Its a standard result (Hartshorne ex. II.4.4) that the image of a complete variety is closed and complete, so and B’ is complete.

The hard part is getting the group structure, i.e a map . Now the stein factorization says in particular that . And the universal property of relative Spec says to give a map into B’ is the same as a map and a map of sheaves . We use the commutative diagram:

The diagram gives a map , so we just need a map of sheaves. Enough to define the map on an open affine cover: of B. So the problem reduces to finding a ring homomorphism

(*)

Now, . Via the maps I identify with . Similarly, identify can be identified with . Set and . From the previous commutative diagram we get:

Set and . From the morphism we get a sheaf map , in particular we get a ring homo.

(here we are using identification of inverse images of the multiplication map discussed above. I claim

Indeed, a section s of the first ring defines a morphism via the map out of k[x] mapping x to s. Since A is complete any fiber is a map from a complete scheme to an affine scheme, so it must have image a point. The Rigidity lemma then implies that the map factors are . The latter map is equivalent to a ring element of , in this way we get the desired identification. The second isomorphism follows because B’ is also complete. Putting the two bullets points together we get the desired ring homorphism (*).

Thus by the universal property of relative Spec, we have a morphism which is in particular, compatible with the morphism (because it is on the level of sheaves on an affine cover of B). It remains to show satisfies all the group axioms i.e. associativity, identity, inverse etc. But the commutative of the first large diagram, and the face that and satisfy the group axioms is enough to show does too. Actually proving it would involve large diagrams I’m not doing here.

To recap, we have that is a homomorphism of abel. var. So ker(f’) is a closed subgroup scheme. But any closed subscheme of a complete scheme is complete, so ker(f’) is also an abelian variety. QED

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- May 21, 2009 / 4:07 pm

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