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I want to count the number of idempotent and nilpotent elements in a ring. Is there a method to do this? I'm particularly interested in $\mathbb{Z}_n$.

HSN
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kadir
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    Idempotents: http://math.stackexchange.com/questions/45747/how-many-idempotent-elements-does-the-ring-mathbb-z-n-contain – Exodd May 12 '16 at 13:00
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    Nilpotents: http://math.stackexchange.com/questions/1179602/to-find-the-nilpotent-elements-of-bbb-z-n-and-also-the-number-of-nilpotent-el?rq=1 – Exodd May 12 '16 at 13:02

1 Answers1

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Idempotents

Because $x$ and $x-1$ are coprime, the idempotents mod $p^e$ are only $0$ and $1$.

If $n=p_1^{e_1} \cdots p_k^{e_k}$, then every idempotent mod $n$ is an idempotents mod $p_i^{e_i}$.

Conversely, by the Chinese remainder theorem, idempotents mod $p_i^{e_i}$ can be combined uniquely into an idempotent mod $n$. More precisely, for each $k$-tuple $(\varepsilon_1, \ldots, \varepsilon_k)$ with $\varepsilon_i = 0, 1$, there is exactly one idempotent mod $n$.

Therefore, there are $2^k$ idempotents mod $n$.

Nilpotents

$a$ is nilpotent mod $n$ iff $a$ contains all the prime factors of $n$.

In other words, the nilpotents are the multiples of $rad(n) = p_1 \cdots p_k$.

Therefore, there are $\frac{n}{rad(n)}=p_1^{e_1-1} \cdots p_k^{e_k-1}$ nilpotents mod $n$.

lhf
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