I've been working with the Chinese Remainder Theorem for rings:
Let $R$ be a ring and let ${A}_{1}, {A}_{2}, ..., {A}_{n}$ be ideals in $R$ such that for each $ i ,j \ \epsilon \ {1,2,...,n}, {A}_{i}$ is comaximal to ${A}_{j} $ for each $i \neq j$. Then $R/({A}_{1}{A}_{2}...{A}_{n}) \simeq R/{A}_{1} \times R/{A}_{2} \times ... \times R/{A}_{n}$.
I have no problem with the proof but, the book that I'm using says that this Theorem gives the following isomorphisms:
$(Z/nmZ)^{ \times }\simeq (Z/nZ)^{ \times }\times(Z/mZ)^{ \times }$
If $ n=({p}_{1})^{{\alpha}_{1}} ({p}_{2})^{{\alpha}_{2}}...({p}_{k})^{{\alpha}_{k}}\ $ then $Z/nZ\quad \simeq \quad Z/{ { p }_{ 1 } }^{ { \alpha }_{ 1 } }Z\quad \times \quad Z/{ { p }_{ 2 } }^{ { \alpha }_{ 2 } }Z\quad \times \quad ...\quad \times \quad Z/{ { p }_{ k } }^{ { \alpha }_{ k } }Z $
I have no idea how to prove this with the Chinese Remainder Theorem. Isn't it necessary that $ Z/{ p }_{ i } ^{ { \alpha }_{ i }} Z$ is ideal, and comaximal to the others $ Z/{ p }_{ j} ^{ { \alpha }_{ j }} Z$? How can i prove this?