4
$\begingroup$

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 $\mathbb{Z}_p$

$\endgroup$
  • 1
    $\begingroup$ what exactly is $\mathbb{Z_P}$? $\endgroup$ – Robert Lewis Aug 11 '13 at 5:07
  • 3
    $\begingroup$ < 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
  • 1
    $\begingroup$ 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
7
$\begingroup$

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 . . .

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $\dots$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.