nilpotent ideals 
Possible Duplicate:
The set of all nilpotent element is an ideal of R 

An element $a$ of a ring $R$ is nilpotent if $a^n = 0$ for some positive integer $n$. Let $R$ be a
commutative ring, and let $N$ be the set of all nilpotent elements of $R$.
(a) I'm trying to show that $N$ is an ideal, and that the only nilpotent element of $R/N$ is the zero element.
(c) What are the nilpotent elements of $R = \mathbb{Z}_{24}$? And what is the quotient ring $R/N$ in that
case? (what known ring is it isomorphic to?)

The first question on why the set of all nilpotent elements in a commutative ring $R$ is an ideal (also called the nilradical of $R$, denoted by $\mathfrak{R}$) has already been answered numerous times on this site. I will tell you why $R/\mathfrak{R}$ has no nilpotent elements. 
Suppose in the quotient we have an element $\overline{a}$ such that $\overline{a}^n = 0$. But then multiplication in the quotient is well defined so that $\overline{a}^n = \overline{a^n}$. This means that $a^n$ must be in $\mathfrak{R}$, so that there exists a $k$ such that $(a^n)^k =0$. But then this means that $a^{nk} = 0$ so that $a \in \mathfrak{R}$. In otherwords, in the quotient $\overline{a} = 0$ proving that the quotient ring $R/ \mathfrak{R}$ has no nilpotent elements.
For question (c) I think you can do it by bashing out the algebra, but let me tell you that $\Bbb{Z}_{24}$ is guaranteed to have nilpotent elements because $24$ is not square free (it's prime factorisation is $24 = 2^{3} \cdot 3$).
