Lie theory - sum of nilpotent ideals is nilpotent?

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Silversonic
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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


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

Base case: k = 1,

(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

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

Now inductive step;

(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]

Now it's easy to see

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

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

But I have no idea what I can do with the [J, I^k \cap J] + [I,I \cap J^k] term so that it reduces to the form I want. Any help?
 
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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.
 
jgens said:
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.

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 k = 2N while the one in my screenshot shows it for lower, at k = N. It doesn't matter though since my goal is to show it's just nilpotent, thanks.
 
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