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Presentation of modules

  1. Jul 13, 2011 #1
    1. The problem statement, all variables and given/known data
    Let k be a field and k[x] be the set of polynomials over that field. Given that M is a module with presentation
    [tex] \begin{pmatrix} 1+ 3x & 2x & 3x \\ 1 + 2x & 1+ 2x -x^2 & 2x \\ x & x^2 & x \end{pmatrix} [/tex]
    determine M.

    2. Relevant equations
    One can apply elementary row and column operations. In the event that one reduces a row or column to the point that there is only one non-zero element and that element is a unit, we can remove the row and column of that unit and the corresponding presentation is isomorphic to the original one.


    3. The attempt at a solution
    So what we have here originally is a exact sequence
    [tex] k[x]^3 \xrightarrow{\phi} k[x]^3 \to M \to 0 [/tex]
    so that we may take [itex] M \cong \text{coker}\phi [/itex].

    If I play around with the matrix a bit, I can reduce it to (x) modulo mistakes in my matrix manipulation. Thus we get a new homomorphism
    [tex] k[x] \xrightarrow{\varphi} k[x] \to \tilde M \to 0 [/tex]
    with [itex] \tilde M \cong M [/itex].

    So here's the part I'm not too sure about. We have [itex] \varphi: k[x] \to k[x] [/itex], so does this mean that [itex] 1 \mapsto x [/itex]?
     
  2. jcsd
  3. Jul 13, 2011 #2

    micromass

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    Yes. In fact, we have here that [tex]M\cong k[X]/\xi(k[X])[/tex], where [itex]\xi[/itex] is the map associated with the matrix. So, we have [tex]M\cong k[X]/(X)\cong k[/tex] here.
     
  4. Jul 13, 2011 #3
    Thanks micromass.

    Yeah, that's what I had originally. Though I started doubting myself as to whether the image of the homomorphism was really just the ideal generated by x.
     
  5. Jul 13, 2011 #4

    micromass

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    No need to doubt yourself :smile: The image of a k[X]-module is always a k[X]-module. Thus you can expect the image of k[X] to be a submodule of k[X], and submodules of k[X] are exactly the ideals of k[X].
     
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