most recent 30 from http://mathoverflow.net 2013-06-19T17:28:00Z http://mathoverflow.net/feeds/question/80871 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80871/how-many-idempotent-elements-are-in-z-m david 2011-11-14T06:39:23Z 2011-11-14T06:47:37Z

<p>How many idempotent elements are in Z_m or How many roots have this polynomial in Z_m</p> <p>f(x)=x^2 +x procedure of proof is important for me.</p>

http://mathoverflow.net/questions/80871/how-many-idempotent-elements-are-in-z-m/80872#80872 Greg Martin 2011-11-14T06:47:37Z 2011-11-14T06:47:37Z

<p>Idempotent elements are roots of $g(x)=x^2-x$; my answer will apply equally well to $f(x)=x^2+x$.</p> <p>The important step is the Chinese Remainder Theorem: one way of stating it is that if $m=p_1^{r_1}\times\cdots\times p_k^{r_k}$ is factored into powers of distinct primes, then the ring $Z_m$ is equal to the direct product of rings $Z_{p_1^{r_1}} \times \cdots \times Z_{p_k^{r_k}}$. So it suffices to count the number of roots in each ring $Z_{p_i^{r_i}}$ and then multiply those numbers together to obtain the number of roots in $Z_m$.</p> <p>I think it will be easy to convince yourself that the polynomial $g(x)=x^2-x$ has exactly two roots in any ring of the form $Z_{p^r}$.</p> <p>(By the way, the isomorphism between $Z_m$ and $Z_{p_1^{r_1}} \times \cdots \times Z_{p_k^{r_k}}$ is completely explicit, so this even gives a way to construct the idempotent elements, not just count them.)</p>