## How many idempotent elements are in Z_m [closed]

How many idempotent elements are in Z_m or How many roots have this polynomial in Z_m

f(x)=x^2 +x procedure of proof is important for me.

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 That looks too much like homework: -1. – Julien Puydt Nov 14 2011 at 6:43 maybe you meant x^2-x? Also, by Z_m do you mean Z/mZ or p-adics?(or something else entirely? – B. Bischof Nov 14 2011 at 6:43

## closed as too localized by Gjergji Zaimi, Andres Caicedo, Ryan Budney, Torsten Ekedahl, Bruce WestburyNov 14 2011 at 8:40

Idempotent elements are roots of $g(x)=x^2-x$; my answer will apply equally well to $f(x)=x^2+x$.
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 $Z_m$.
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 $Z_{p^r}$.
(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.)