-1
$\begingroup$

Can we find the idempotents in $(\mathbb{Z}_n,+,\cdot)$ for any $n$?

Is there a general rule?

Note: Trying to consider the prime factors!

| cite | improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ Hint: If $(m.n)=1$, $\mathbb Z_{mn}\cong \mathbb Z_{m}\times \mathbb Z_n$. FInd the idemopotents for each of those rings and use Chinese Remainder Theorem to find them all. $\endgroup$ – Thomas Andrews Aug 11 '13 at 3:36
  • $\begingroup$ Essentially, it reduces to finding the idempotents of $\mathbb Z_{p^k}$ for $p$ prime. $\endgroup$ – Thomas Andrews Aug 11 '13 at 3:36
  • $\begingroup$ I intend to find the elements not their number! Is there a general rule? Note: Number theory can be used! $\endgroup$ – UNM Aug 11 '13 at 3:37
  • $\begingroup$ This question is NOT a duplicate! the OP is asking about the idempotent elements in $\mathbb Z_n$, not just their number. $\endgroup$ – Matemáticos Chibchas Aug 11 '13 at 9:10
2
$\begingroup$
  1. Understand the Chinese Remainder Theorem.

  2. Use Thomas' hints to understand why your problem becomes that of finding the idempotents of $\mathbb{Z}/p^m\mathbb{Z}.$

  3. See why any zero divisors of $\mathbb{Z}/p^m\mathbb{Z}$ must be nilpotent and find the nilpotent elements.

  4. Idempotents satisfy $x(x-1)=0,$ so using 3 see why the only idempotents are the trivial ones.

  5. Use the isomorphism in the Chinese Remainder Theorem to tell you about idempotents in $\mathbb{Z}/n\mathbb{Z}.$

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Unfortunately, Chinese remainder theorem states that the natural mapping $\mathbb Z_{mn}\to\mathbb Z_m\times\mathbb Z_n$ (with GCD$(m,n)=1$) is a ring monomorphism, so in particular (by cardinality considerations) it is an isomorphism, but it says nothing about the nature of the inverse map, which is precisely what is needed in order to solve the OP question. $\endgroup$ – Matemáticos Chibchas Aug 11 '13 at 9:13

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