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Thread: induced sequence of quotients

  1. #1
    Junior Member
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    induced sequence of quotients

    Hi! Let $\displaystyle A$ a finitely generated free abelian group and $\displaystyle f: A \rightarrow A$ a homomorphism, then one can define a trace as usually as sum of the diagonal elements (of the matrix form of $\displaystyle f$).

    If $\displaystyle A$ is not free, then $\displaystyle \frac{A}{T(A)}$ is free ($\displaystyle T(A)$ denotes the torsion group) and one has an induced homomorphism $\displaystyle \tilde f: \frac{A}{T(A)} \rightarrow \frac{A}{T(A)}$. Therefore we can define the trace of $\displaystyle f$ as $\displaystyle tr(f):=tr(\tilde f: \frac{A}{T(A)} \rightarrow \frac{A}{T(A)})$.

    Now my question:
    Let $\displaystyle A, \, B, \, C$ finitely generated abelian groups and
    $\displaystyle
    0 \longrightarrow A \overset{\text{i}}{\longrightarrow} B \overset{\text{p}}{\longrightarrow} C \longrightarrow 0
    $

    $\displaystyle
    0 \longrightarrow A \overset{\text{i}}{\longrightarrow} B \overset{\text{p}}{\longrightarrow} C \longrightarrow 0
    $
    a commuative diagramm with exact rows (there should be vertical arrows: $\displaystyle f: A \rightarrow A, \, g: B \rightarrow B, \, h: C \rightarrow C$, but i do not know how to produce this. I hope it is clear what i mean).

    Why do we have $\displaystyle tr(g)=tr(f)+tr(h)$ in this case?
    My problem is that the induced sequence between the quotients is not longer exact anymore, since we lose the property $\displaystyle ker(\tilde p) \subset im(\tilde i)$.

    Does somebody have a clue how to prove this?
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  2. #2
    Junior Member
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    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.
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