Well it works with a simple ruse: One just considers the induced sequence of tensor products with Q, then we have an exact sequence and the trace won't change.
Hi! Let a finitely generated free abelian group and a homomorphism, then one can define a trace as usually as sum of the diagonal elements (of the matrix form of ).
If is not free, then is free ( denotes the torsion group) and one has an induced homomorphism . Therefore we can define the trace of as .
Now my question:
Let finitely generated abelian groups and
a commuative diagramm with exact rows (there should be vertical arrows: , but i do not know how to produce this. I hope it is clear what i mean).
Why do we have in this case?
My problem is that the induced sequence between the quotients is not longer exact anymore, since we lose the property .
Does somebody have a clue how to prove this?