How to find the nilpotent elements of $\mathbb{Z}/(\prod p_i^{n_i})$? I've been following MIT's old opencourseware class on commutative algebra. For one problem, I want to find the nilpotent and idempotent elements of $\mathbb{Z}/(n)$, where $n=\prod p_i^{n_i}$.
I know $\mathbb{Z}/(n)\cong\prod\mathbb{Z}/(p_i^{n_i})$. Also, $\text{nilrad}(\mathbb{Z}/(p_i^{n_i}))=\bigcap P$ where the intersection ranges over $\text{Spec}(\mathbb{Z}/(p_i^{n_i}))$. But the prime ideals of $\mathbb{Z}/(p_i^{n_i})$ are in bijection with the primes of $\mathbb{Z}$ containing $(p_i^{n_i})$, which is just $(p_i)$. So the nilpotent elements is the image $(p_i)/(p_i^{n_i})$ under the canonical projection. With that, I could theoretically find the nilpotent elements in each factor, and then find the nilpotent elements of $\mathbb{Z}/(n)$ by the Chinese remainder theorem. Is there a more concrete way to express the set of nilpotenet elements, or is this the best we can do?
 A: This answer is about the idempotents (I gather from your comment above, @hmIII, that you want to know also about idempotents). As you write, they are got from the CRT, i.e. the isomorphism
$$\mathbb{Z}/ n \cong \prod \mathbb{Z}/ p_i^{n_i}.$$ Let $e_i$ be the element of $\mathbb{Z}/ n$ that is $1$ module $p_i^{n_i}$ and $0$ modulo every other $p_j^{n_j}$. These give the primitive idempotents; it is a general principle that every idempotent $e$ is a sum $$e=\sum_{i \in S} e_i$$ (the set $S$ might be empty) of these primitive idempotents. 
To write down particular representatives for the $e_i$ one might proceed as follows: let $m_i=\prod_{j \neq i } p_i^{n_i}$ and compute (using the Euclidean algorithm) a representative $l_i \in \mathbb{Z}$ for the inverse $m_i^{-1}$ of $m_i$ modulo $p_i^{n_i}$. Then $e_i=l_i m_i \ \mathrm{mod} \ n$ is the $i$th primitive idempotent. This amounts to a constructive proof of the CRT; you write "is that all you can say", and the answer is that constructing these idempotents is essentially equivalent to proving the CRT, for the isomorphism can be specified explicitly as $(a_i)_{i \in I} \mapsto \sum a_i e_i \in \mathbb{Z} / n$, where $I$ is an index set for the prime factors of $n$ and $(a_i) \in \prod_{i \in I} \mathbb{Z} / p_i^{n_i}$.
Worked examples: (1) Let $n=80=2^4 5$. We have $13=5^{-1}$ mod $16$ and $1=16^{-1}$ mod $5$ so the primitive idempotents are $e_1=13*5=65$ and $e_2=1*16=16$. Together with $0$ and $1$ these are the only idempotents. 
(2) Let $n=60=2^2*3*5$. We have $(3*5)^{-1}=3$ mod $4$, $(4*5)^{-1}=2$ mod $3$, and $(4*3)^{-1}=3$ mod $5$ so the primitive idempotents are $e_1=3*15=45$, $e_2=2*20=40$, and $e_3=3*12=36$. There are $8$ idempotents total, obtained by summing over all possible subsets of $\{e_1,e_2,e_3\}$.
A: Perhaps this is not the way you want to go but the nilpotent elements mod $n$ correspond to the numbers divisible by all the prime factors of $n$. In other words, the multiples of the radical of $n$.
