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Lie theory - sum of nilpotent ideals is nilpotent?

  1. Feb 17, 2014 #1
    I can't wrap my head around this proof that the sum of two nilpotent ideals is nilpotent, I get stuck at one stage:

    http://imageshack.com/a/img706/5732/5wgq.png [Broken]


    I'm fine with every except showing by induction [itex] (I+J)^{N+k} = I^k \cap J + I \cap J^k [/itex]. Here's my attempt;

    Base case: k = 1,

    [itex] (I+J)^{N+1} = [I+J, (I+J)^N] \subseteq [I+J,I \cap J] = [I, I \cap J] + [J, I \cap J] \subseteq I \cap J + I \cap J [/itex]

    since [itex] [I, I \cap J], [J, I \cap J] \subseteq I \cap J [/itex] as [itex] I \cap J [/itex] is an ideal.

    Now inductive step;

    [itex] (I+J)^{N+k+1} = [I+J, (I+J)^{N+k}] = [I+J, I^k \cap J + I \cap J^k] = [I, I^k \cap J] + [J, I^k \cap J] + [I,I \cap J^k] + [J,I \cap J^k] [/itex]

    Now it's easy to see

    [itex] [I, I^k \cap J] \subseteq I^{k+1} \cap J [/itex]

    [itex] [J, I \cap J^k] \subseteq I \cap J^{k+1} [/itex]

    But I have no idea what I can do with the [itex] [J, I^k \cap J] + [I,I \cap J^k] [/itex] term so that it reduces to the form I want. Any help?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 17, 2014 #2

    jgens

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    Gold Member

    Are you stuck using this exact argument? Unless I have misunderstood something the proof should go through easily with a slight modification: Essentially after repeatedly multiplying I∩J by I+J we should end up with something like ∑Ij∩Jk-j and choosing k so large that either Ij or Jk-j vanishes for each choice of j should complete the argument. Maybe I have misunderstood something here though and this fails for some reason.
     
  4. Feb 17, 2014 #3
    You're quite right, and that's a different way of looking at it than I did. I think by this method it shows it is zero when [itex] k = 2N [/itex] while the one in my screenshot shows it for lower, at [itex] k = N [/itex]. It doesn't matter though since my goal is to show it's just nilpotent, thanks.
     
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