Lie theory - sum of nilpotent ideals is nilpotent?

1. Feb 17, 2014

Silversonic

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 $(I+J)^{N+k} = I^k \cap J + I \cap J^k$. 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?

Last edited by a moderator: May 6, 2017
2. Feb 17, 2014

jgens

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.

3. Feb 17, 2014

Silversonic

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.