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Please excuse my sketchy presentation/ formatting, I am completely new to the site and LaTeX.

  1. Show that subrings of a nilpotent ring are nilpotent

So I know that we need to use the subring criterion, i.e:

Let R be a ring and S $ \subseteq $ R such that S $ \not= \emptyset $. Then S $ \le $ R iff:

  • $ \forall a, b $ $ \epsilon $ $ S $, $ a - b $ $ \epsilon $ $S$ holds
  • $ \forall a, b $ $ \epsilon $ $ S $, $ ab $ $ \epsilon $ $S$ holds

The second condition is easy to show: $ a^n = 0, b^n = 0 $ therefore $ (ab)^n=a^nb^n =0 $.

But the first condition confuses me; I know that we need to show that $ (a+b)^n = \sum\limits_{i=0}^n \binom {n} {i} a^ib^{n-i} = 0 $ and I can do that, but I don't understand why we're not trying to show that $ (a-b)^n = 0 $ to satisfy $ a - b $ $\epsilon $ $ S $?

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  • $\begingroup$ The thing you're trying to show is not the thing you're being asked to show. $\endgroup$ – Qiaochu Yuan May 27 '14 at 4:22
  • $\begingroup$ Do you mean I shouldn't be trying to show that $ (a+b)^n = \sum\limits_{i=0}^n \binom {n} {i} a^ib^{n-i} = 0 $ or that $ (a−b)^n=0 $? $\endgroup$ – user153580 May 27 '14 at 5:34
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    $\begingroup$ No, I mean you shouldn't be talking about $(a + b)^n$ at all. The problem is not to prove that anything is a subring, but to assume that something already is a subring and prove something else about it. $\endgroup$ – Qiaochu Yuan May 27 '14 at 8:05
  • $\begingroup$ What is your definition of a nilpotent ring? $\endgroup$ – Luiz Cordeiro May 27 '14 at 19:52
  • $\begingroup$ Just that every element in R is nilpotent, so for every $ a \epsilon R $ there is a positive integer $ n $ such that $ a^n = 0 $ $\endgroup$ – user153580 May 28 '14 at 6:26

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