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Aug 18, 2013 at 1:53 comment added Dan Petersen Aah, absolutely, apologies. I'm too used to only thinking about ppav's.
Aug 17, 2013 at 22:07 comment added Damian Rössler Just one more thing (this is what threw me off): you have to consider products of abelian varieties together with the polarisation in the definition of decomposability (ie consider the product polarisations).
Aug 15, 2013 at 6:04 comment added Dan Petersen Being indecomposable is also equivalent to not being in the image of any of the finitely many "gluing" maps $A_h \times A_{g-h} \to A_g$.
Aug 14, 2013 at 21:56 comment added Damian Rössler Why is the property of being indecomposable an open condition ? This property is equivalent to the existence of an idempotent in the endomorphism algebra of the abelian variety and I don't see why this should be a constructible (let alone open) property (sorry if I missed an obvious point).
Aug 14, 2013 at 8:50 comment added Dan Petersen I mean not a product of lower-dimensional abelian varieties. They are allowed to be isogenous to such a product.
Aug 14, 2013 at 8:26 comment added Damian Rössler What do you mean by "indecomposable" ?
Aug 13, 2013 at 18:46 history asked Dan Petersen CC BY-SA 3.0