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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?

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

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  • $\begingroup$ In $\Bbb Z_{51}$ $51=2p$ so how did you find that? Using CRT? $\endgroup$ – Baby Elephant Jul 14 '19 at 1:25
  • $\begingroup$ 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. $\endgroup$ – TMO Jul 14 '19 at 8:18

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