Polishcuk Ex. 1 pg. 107

Setup: \ A \xrightarrow{f} B homo. or abel. var. and \ A \xrightarrow{f'} {B'} \xrightarrow{p} B its stein factorization; i.e.

  1. \ f'_*\mathcal{O}_A \cong \mathcal{O}_{B'} and 
  2. 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 \ f'(A) = B' and B’ is complete.  

The hard part is getting the group structure, i.e a map \ B' \times B' \to B'.  Now the stein factorization says in particular that \ B' = \mbox{Spec}_B f_*\mathcal{O}_A.  And the universal property of relative Spec says to give a map into B’ is the same as a map \ B'\times B' \xrightarrow{g} B and a map of sheaves \ f_*\mathcal{O}_A \to g_*\mathcal{O}_{B' \times B'}.  We use the commutative diagram:

\ \begin{array}{ccc} A \times A & \xrightarrow{m_A} & A \\ \downarrow & \mbox{ } & \downarrow \\ B' \times B' & -?- & B' \\ \downarrow & \mbox{ } & \downarrow \\ B\times B & \xrightarrow{m_B} & B \end{array}

The diagram gives a map \ B' \times B' \to B, so we just need a map of sheaves.  Enough to define the map on an open affine cover: \ \{U_i = \mbox{Spec}R_i \} of B.  So the problem reduces to finding a ring homomorphism 

\ \mathcal{O}_A(f^{-1}U_i) \to \mathcal{O}_{B' \times B'}((p,p)^{-1}m_B^{-1}U_i)   (*)

Now, \ m_B^{-1}U_i = \cup_{b \in U_i} \{ (a + b, -a)| a \in B \}.  Via the maps \ U_i \times B \xrightarrow{(m_B, [-1] \circ p_2)} B\times B I identify \ m_B^{-1}U_i with \ U_i \times B.  Similarly, identify \ m_A^{-1}f^{-1}U_i can be identified with \ f^{-1}U_i \times A. Set \ AR_i := \mathcal{O}_A(f^{-1}U_i) and \ AU_i = \mbox{Spec} AR_i. From the previous commutative diagram we get:

\ \begin{array}{ccc} f^{-1}U_i \times A & \xrightarrow{\ \ m_A \ \ } & f^{-1}U_i \\ \downarrow & \mbox{ } & \downarrow \\ AU_i \times B' & \mbox{ } & AU_i \\ \downarrow & \mbox{ } & \downarrow \\ U_i \times B & \xrightarrow{\ \ m_B \ \ } & U_i \end{array}

Set \ AR_i := \mathcal{O}_A(f^{-1}U_i) and \ AU_i = \mbox{Spec} AR_i.  From the morphism \ m_A we get a sheaf map \ \mathcal{O}_A \to m_*\mathcal{O}_{A\times A}, in particular we get a ring homo.

  • \ AR_i \to \mathcal{O}_{A\times A}(f^{-1}U_i \times A) 

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

  • \ \mathcal{O}_{A\times A}(f^{-1}U_i \times A) \cong AR_i \cong \mathcal{O}_{B'\times B'}((p,p)^{-1}m_B^{-1}U_i)

Indeed, a section s of the first ring defines a morphism \ f^{-1}U_i \times A \to \mathbb{A}^1 via the map out of k[x] mapping x to s.  Since A is complete  any fiber \ A \times \{pt\} \to \mathbb{A}^1 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 \ f^{-1}U_i \times A \to f^{-1}U_i \to \mathbb{A}^1.  The latter map is equivalent to a ring element of \ AR_i, 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 \ B' \times B' \to B' which is in particular, compatible with the morphism \ A\times A \to A (because it is on the level of sheaves on an affine cover of B).  It remains to show \ B' \times B' \to B' satisfies all the group axioms i.e. associativity, identity, inverse etc.  But the commutative of the first large diagram, and the face that \ A\times A \to A and \ B\times B \to B satisfy the group axioms is enough to show \ B'\times B' \to B' does too.  Actually proving it would involve large diagrams I’m not doing here.  

To recap, we have that \ A \xrightarrow{f'} B' 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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