# Show that subrings of a nilpotent ring are nilpotent.

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 $a - b$ $\epsilon$ $S$?

• The thing you're trying to show is not the thing you're being asked to show. – Qiaochu Yuan May 27 '14 at 4:22
• 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 $(a−b)^n=0$? – user153580 May 27 '14 at 5:34
• 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. – Qiaochu Yuan May 27 '14 at 8:05
• What is your definition of a nilpotent ring? – Luiz Cordeiro May 27 '14 at 19:52
• Just that every element in R is nilpotent, so for every $a \epsilon R$ there is a positive integer $n$ such that $a^n = 0$ – user153580 May 28 '14 at 6:26