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Can ramification geometry of algebraic function be computed algebraically?

  1. Oct 15, 2012 #1
    Suppose I'm given a random function:

    [tex](-8+5 z+4 z^2)\text{}+(7 z+6 z^4-7 z^5)w+(3 z^2-z^3)w^2+(-8 z-2 z^4-2 z^5)w^3+(3-4 z+4 z^2+7 z^3+6 z^4-8 z^5)w^4+(-6 z+4 z^4)w^5=0[/tex]

    Is there no way to determine it's ramification geometry at each singular (critical) point algebraically? I'm pretty sure the multiplicity of the zeros at the critical points do not determine this geometry.

    Just seem there should be an algebraic way to do this or is there some theorem which deems this impossible?

    Also, I'm not sure if there are several definitions of "singular" point of algebraic function. The one I'm using is that for the function:

    [tex]f(z,w)=a_0(z)+a_1(z)w+a_2(z)w^2+\cdots+a_n(z)w^n=0[/tex]

    the point [itex]z_s[/itex] is a singular point when either a_n(z_s)=0 or the discriminant of f at [itex]z_s[/itex] is zero.

    Thanks,
    Jack
     
    Last edited: Oct 15, 2012
  2. jcsd
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