Results 1 to 2 of 2

Math Help - induced sequence of quotients

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    34

    induced sequence of quotients

    Hi! Let A a finitely generated free abelian group and 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 f).

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

    Now my question:
    Let A, \, B, \, C finitely generated abelian groups and
    <br />
0 \longrightarrow A \overset{\text{i}}{\longrightarrow} B \overset{\text{p}}{\longrightarrow} C \longrightarrow 0<br />

    <br />
0 \longrightarrow A \overset{\text{i}}{\longrightarrow} B \overset{\text{p}}{\longrightarrow} C \longrightarrow 0<br />
    a commuative diagramm with exact rows (there should be vertical arrows: 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 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 ker(\tilde p) \subset im(\tilde i).

    Does somebody have a clue how to prove this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2008
    Posts
    34
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. induced partition?
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: September 22nd 2010, 05:55 PM
  2. Induced Representations
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 25th 2010, 10:34 PM
  3. induced subgraphs
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: December 3rd 2009, 02:58 AM
  4. induced norm
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: November 23rd 2009, 06:58 PM
  5. back emf ,induced emf difference..
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: August 29th 2009, 09:47 AM

Search Tags


/mathhelpforum @mathhelpforum