# The only idempotent element in $\Bbb Z_{51}$ are $0$ and $1$. [duplicate]

True or False: The only idempotent elements in $$\Bbb Z_{51}$$ are $$0$$ and $$1$$.

Here $$\Bbb Z_{51}$$ is not an integral domain, but I am guessing that it is tricky so how to approach this problem?

• Well, why not try a smaller number to warm up. What are the idempotents in $\mathbb Z_6$, say? – lulu Jul 13 '19 at 22:00

We have $$18^2 = 18 \text{ mod } 51$$ and $$34^2 = 34 \text{ mod } 51$$. These are the only non-trivial idempotents.
Consider for example the ring $$\mathbb{Z}/n\mathbb{Z}$$, where $$n = 2 \cdot p$$ for some odd prime $$p$$. Then we have $$2p \mid p(p-1) = p^2 - p$$ as $$p-1$$ is even. In other words $$p^2 = p \text{ mod } 2p$$.
If $$e \in R$$ is idempotent, then $$1- e \in R$$ is also idempotent since $$(1-e)^2 = 1 - 2e + e^2 = 1 - 2e + e = 1 - e.$$
This means that in the above example we get that $$1-p = 1 - p +2p = p + 1 \in \mathbb{Z}/n\mathbb{Z}$$ is also idempotent. One can show that these are the only non-trivial ones as I will do now.
The more general approach would be to use the chinese remainder theorem and the fact that the only idempotents in $$\mathbb{Z}/p^l\mathbb{Z}$$ are the trivial ones. By that you can count the idempotents in $$\mathbb{Z}/n\mathbb{Z}$$, as the idempotents will exactly be the tuples consisting of $$0$$ and $$1$$ entries. What I just explained is that if $$n = p_1^{\nu_1} \dots p_r^{\nu_r}$$, then $$\mathbb{Z}/n\mathbb{Z}$$ has $$2^r$$ idempotents.
• In $\Bbb Z_{51}$ $51=2p$ so how did you find that? Using CRT? – Baby Elephant Jul 14 '19 at 1:25
• Yes, but you do not calculate the order modulo something. We have $51 = 3 \cdot p$ for $p = 17$. Yes, exactly. I found these by my "more general approach" via the chinese remainder theorem. – TMO Jul 14 '19 at 8:18