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I know I asked a similar question here. It helped me a lot and I understood it well. But I can't find my way through a "generalized" method to resolve it if I "split" my ring in, let's say, three other rings.

For example, I want to find the idempotent elements of $\Bbb Z_{540}$, so:

$$\Bbb Z_{540} = \Bbb Z_{27} \times \Bbb Z_4 \times\Bbb Z_5$$

I tried to resolve it as the one for $\Bbb Z_{36}$ (see the link), but it didn't work. For $(\widetilde0, \bar0, \mathring1)$ I had:

$$ \color{blue}1\cdot \color{red}{27\cdot a} + \color{blue}1\cdot\color{red}{4\cdot b} + \color{blue}0\cdot\color{red}{5\cdot c} = 1 $$

For $a = -1, b = 2$ and $c = 4$, the result is $521\equiv_{540}$ which is not good. How it should really be done? Thanks!

AcidMammoth
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2 Answers2

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Do it first for $\Bbb Z_{20} = \Bbb Z_4 \times\Bbb Z_5$ and then for $\Bbb Z_{540} = \Bbb Z_{27} \times \Bbb Z_{20}$.

lhf
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Hint: In the product ring ${\Bbb Z}_n =\prod_{i=1}^m {\Bbb Z}_{p_i^{n_i}}$ with $n=\prod_{i=1}^m p_i^{n_i}$, the idempotents $a=(a_1,\ldots,a_m)$ are given by idempotents $a_i\in{\Bbb Z}_{p_i^{n_i}}$, $1\leq i\leq m$, since the multiplication is component-wise. For instance, $0,1$ are always idempotent.

The local rings ${\Bbb Z}_{p_i^{n_i}}$ need to be checked individually. Units $a\ne 1$ are not idempotent, since $a^2=a$ implies $a=1$. The remaining nonzero elements of ${\Bbb Z}_{p_i^{n_i}}$ are all zero-divisors. My guess is that these elements are not idempotent either. I hand this question back to you.

Wuestenfux
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