How do I show that for the qoutient ring of a ring $R$ and an ideal $N=\{a\in R| a^{n}=0\}$, the only nilpotent elements in the qoutient is $0$?
But a logic question, isn't it not true, because N is in the qoutient ring and N is all the nilpotent elements?
Otherwise, let's take an item from the quotient and look when he is niplotent: $$\left(a+N\right)^n=N$$ $\left(a+N\right)^n=a^n+N=N$ that is true if and only if $a\in N$ or $a=0$, and that's why I asked the qeuestion, because $a$ in R (and in particular we can choose him from N and get the neutral cosset, N itself, so it seems to be that all the nilpotent items are in the quotient, isn't it?).
Take $x$ in $R/N$ such that $x^n=0$. Now $x=a+N$ for some $a\in R$, and so we know that $a^n\in N$, that is for some $m$ $a^{nm}=0$. In other words $a\in N$, so that $x=a+N=0$ in $R/N$.