Let $p_1,\cdots,p_k$ be distinct primes, and let $n=p_1\cdots p_k$. If $R$ is the ring of integers modulo $n$, show that there are exactly $2^k$ elements $a\in R$ such that $a^2=a$.
Observation: this exercise appears on Herstein's book.
I guess this will work with induction on $k$. For $k=1$, if $a^2=a$, then $a(a-1)=0$ and since $R$ is domain, it follows that $a=0$ or $a=1$. OK
Now how can I work with induction hyphotesis? Thanks in advance.
I don't think induction simplifies matters here.
By Chinese remainder theorem, $\mathbb{Z}/n\mathbb{Z} \cong \prod\limits_{i=1}^{k} \mathbb{Z}/p_i\mathbb{Z}$
Notice that an element is idempotent in the right-hand-side product iff all of its components are idempotent. In our product, each component is a field, which possesses precisely the idempotents $0$ and $1$.