$\mathbb{Z}/n\mathbb{Z}$ is a semisimple ring provided n is square-free

Question

Show that $\mathbb{Z}/n\mathbb{Z}$ is a semisimple ring if n is square-free..

Some articles suggests that we can use Chinese remainder theorem. But I can't exactly figure out how to do that.

Thanks in advance!!!

Answer 1

Take the prime factorization $n=p_1^{e_1}\cdots p_n^{e_n}$. Then $\phi:{\Bbb Z}/n{\Bbb Z}\rightarrow \prod_i {\Bbb Z}/p_i^{e_i}$ defined by $m\mapsto (m\;{\rm mod}\;p_i^{e_i})$ is a ring isomorphism. If $n$ is square-free, then $e_1=\ldots=e_n=1$ and ${\Bbb Z}/p_i{\Bbb Z}$ is a field and so simple. Thus your ring is a product of simple rings and so semisimple. The proof of the ring isomorphy (surjectivity) requires the Chinese remainder theorem.