So I have to prove that prove that $[0]$ and $[1]$ are the only two idempotent elements of $\mathbb{Z}_p$.

Here is how my proof goes:

Suppose there is another idempotent element of $\mathbb{Z}_p$ which we call $[a]$, where $[a]\neq [0] $ and $[1]$. Further $1<[a]<[p]$. But since $p\geq 2$ it follows that $2\leq[a]<[p]$, a contradiction when $p=2$.

Not sure if this is correct or not. I don't think its correct if not a small hint would help. Thanks

Edit: New Proof:

Suppose that there exists another idempotent element call it $[a]$ such that ${[a]}^2=[a]$. Also $a \neq 0$ and $a \neq 1$. Let $x\in[a]$ then it follows $x \equiv a \pmod{p} \iff p|(x-a) \iff ps=x-a$ for some $s\in\mathbb{Z}$ and $x \equiv a^2 \pmod{p} \iff p|(x-a^2) \iff pt=x-a^2$ for some $t\in\mathbb{Z}$. It follows that $ps-pt= a^2-a \implies p|(a^2-a) \iff a^2-a \equiv 0 \pmod {p}$. It follows that since $p|(a^2-a)$ then by Euclid's lemma we get $p|a$ or $p|(a-1)$. In other words $a \equiv 0 \pmod{p}$ or $a \equiv 1 \pmod{p}$. A contradiction since $a<p$ ( since $a$ make up the congruence classes of $\mathbb{Z}_p$ and since $a\neq 1$ and $a \neq 0$ .Thus $a=0$ or $a=1$ are the only two idempotent elements in $\mathbb{Z}_p$

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    $\begingroup$ what exactly is $\mathbb{Z_P}$? $\endgroup$ – Robert Lewis Aug 11 '13 at 5:07
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    $\begingroup$ < is something that makes sense for integers. It is not something that makes sense for $\mathbb{Z}_p$. Note, for example, $[p] = [0]$ in $\mathbb{Z}_p$. $\endgroup$ – user14972 Aug 11 '13 at 5:08
  • $\begingroup$ It means integers mod p $\endgroup$ – user60887 Aug 11 '13 at 5:09
  • $\begingroup$ Ask yourself: Why was it necessary for $p$ to be a prime? If it is not used in an essential way, then the argument simply has to be wrong. Notice that in $\mathbb{Z}_6$ also $[3]$ and $[4]$ are idempotents, so the fact has to depend on $p$ being prime in an essential way. See for example this answer for more about the non-prime case. $\endgroup$ – Jyrki Lahtonen Aug 11 '13 at 16:33
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    $\begingroup$ Your new proof works, except you don't need to do all that work just to get $p|a^2-a$. This is, essentially, what was given. $\endgroup$ – Jonathan Y. Aug 11 '13 at 23:17

If $F$ is any field, and $e \in F$ is idempotent, then by definition $e^2 = e$. Now if $e \ne 0$, multiplying through be $e^{-1}$ yields $e = 1$. QED.

It's actually considerably more general:

If $F$ is an integral domain, write $e^2 - e = e(e - 1) = 0$; now if $e \ne 0$, again we have $e = 1$.

If $F$ is a division ring, the same arguments apply.

Presumably the list of kinds of rings for which $0, 1$ are the only idempotents goes on . . .

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    $\begingroup$ Noncommutative domains capture all those cases listed, and then a new case would be (noncommutative) local rings. Uniform rings and hollow rings also don't permit any idempotents... $\endgroup$ – rschwieb Aug 11 '13 at 23:34
  • $\begingroup$ @rschwieb: Wow! Sounds like your list of "kinds of rings" is a lot longer than mine! OK, I'd best hit wikipedia. Thanks for the lesson! Cheers, BobLewis $\endgroup$ – Robert Lewis Aug 11 '13 at 23:38
  • $\begingroup$ Yeah, the notes I produced to organize what rings I know has about 100 types. It's was a fun exercise! $\endgroup$ – rschwieb Aug 11 '13 at 23:43
  • $\begingroup$ @rscwieb: sounds cool, love to see it! $\endgroup$ – Robert Lewis Aug 12 '13 at 0:15

Hint: Show that the only solutions of the congruence $x(x-1)\equiv 0\pmod{p}$ are given by $x\equiv 0\pmod{p}$ and $x-1\equiv 0\pmod{p}$. If a prime divides a product $\dots$.

  • $\begingroup$ Would this work too if I say $a\neq 0$ and $a \neq 1$. Further $1<a<p$. But since $p \geq 2$ it follows that $2 \leq a<p$. A contradiction when p=2. $\endgroup$ – user60887 Aug 11 '13 at 5:22
  • $\begingroup$ It would show that $[0]$ and $[1]$ are the only idempotents when the prime is $2$. But you want to show that they are the only idempotents whatever the prime is. $\endgroup$ – André Nicolas Aug 11 '13 at 5:26
  • $\begingroup$ oh ok I get it. $\endgroup$ – user60887 Aug 11 '13 at 5:27

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