Idempotent elements in a modulo n ring

Question

I'm trying to find the idempotent elements of the ring ($\Bbb Z_{36} $, +, $ \cdot $) so I "split" it into $ \operatorname{Idemp}(\Bbb Z_4 \times \Bbb Z_9) $, meaning $\operatorname{Idemp}(\Bbb Z_4) \times \operatorname{Idemp}(\Bbb Z_9) $.

I know the idempotent elements of $\Bbb Z_4$ and $\Bbb Z_9$, but I don't understand how to "transform" them back to $\Bbb Z_{36}$. For example, I know that pair $(\widetilde{0}, \bar{1})$ in $\Bbb Z_4 \times \Bbb Z_9$ corresponds to $\widehat{28}$ in $\Bbb Z_{36}$, and that $(\widetilde{1}, \bar{0})$ corresponds to $\widehat{9}$, but why?

If there are other ways to find the idempotents for a large $n$, it's OK, but I wish to understand this particular case I just asked. Thanks!

Answer 1

It's easier to transform $28$ from $\Bbb Z_{36}$ to $\Bbb Z_4\times \Bbb Z_9$: Then we just reduce the normal way for modular arithmetic:

• In $\Bbb Z_4$, $28$ reduces to $0$
• In $\Bbb Z_9$, $28$ reduces to $1$

Translating the other way (from $(0,1)$ to $28$) is a bit more tricky, and for small examples, I usually brute force it. In this case:

I know that whatever the answer is, it reduces to $1$ in $\Bbb Z_9$, so my options are $1, 10, 19$ and $28$. Of these, only $28$ reduces to $0$ in $\Bbb Z_4$, so that's the answer.

This is the second approach suggested by the Wikipedia article on the Chinese remainder theorem. The third approach uses the actual proof of the theorem to find the correct remainder, and it's more constructive, albeit somewhat longer for this small example:

First, find $a, b$ such that $4a + 9b = 1$. This may be done by the extended Euclidean algorithm, or in a simple case like this, just immediately spotting that $9-8 = 1$: We get $a = -2, b = 1$. We now get a solution as $$ \color{blue}0\cdot \color{red}{1\cdot 9} + \color{blue}1\cdot\color{red}{(-2)\cdot 4} = -8\equiv_{36}28 $$ where the blue numbers are the components of our element in $\Bbb Z_4\times\Bbb Z_9$, and the red parts are the terms from $-2\cdot 4 + 1\cdot 9 = 1$. Note that the component from $\Bbb Z_4$ is multiplied by the $9$ term from $-2\cdot 4 + 1\cdot 9$, and vice versa.

The idea of this third approach is that from $4\cdot (-2) + 9\cdot 1 = 1$, we get that $4\cdot (-2)\in \Bbb Z_{36}$ corresponds to $(0,1)\in \Bbb Z_4\times\Bbb Z_9$ and $9\cdot 1\in \Bbb Z_{36}$ corresponds to $(1,0)$. So we just add however many we need of each, and we get our answer.

Answer 2

HINT: For integers $m,n\geq 2$ with $\gcd(m,n)=1$, the mapping $${\Bbb Z}_{mn}\rightarrow{\Bbb Z}_m\times {\Bbb Z}_n:x\mapsto (x\mod m,x\mod n)$$ is a ring isomorphism.

The surjectivity is usually proved by the Chinese remainder theorem, i.e., the simultaneaous congruences $$x\equiv a\mod m\quad\mbox{and}\quad x\equiv b\mod n$$ have a unique solution $x$ (an integer $x$ with $0\leq x <mn$).

SOLUTION: The solution is $x=b-yn(b-a)$ where $1=yn+zm$.