# Last step of the proof of the Chinese remainder theorem.

For the Chinese Remainder Theorem for rings we have:

$$A/(I\cap J) \cong A/I\times A/J$$

So far I have proven that there is a ring homomorphism from $\phi :A \rightarrow A/I\times A/J$ and the $ker(\phi)=I\cap J$. Additionally, since I and J are comaximal, $I + J = A$, implies $I\cap J=IJ$

My question is, how does this last fact prove isomorphism in the last step of the proof?

$I + J = 1\,\Rightarrow\, i + j = 1\,$ for some $\,i\in I,\, j\in J\,$ therefore $\,i\equiv 0\pmod{I},\ i\equiv 1\pmod{J}\,$ so $\,\phi(i) = (0,1).\,$ Similarly $\,\phi(j) = (1,0).\,$ Thus $\phi\,$ is onto: $\,{\rm Im}\,\phi = A/I\times A/J,\,$ hence applying the First Isomorphism Theorem yields the result.
Remark $\$ This generalizes to any number of pair-comaximal ideals $\, I + J_k = 1,\,$ since then there are elements $\ i_k+j_k = 1,\ \ i_k\in I,\, j_k\in J_k,\,$ therefore
$\qquad x := j_1\cdots j_n = (1-i_1)\cdot (1-i_n)\equiv 1^n\equiv 1\pmod{I},\,$ and $\ x\equiv 0\pmod {J_k}$
• But doesn't that just imply the $A/(I\cap J) \cong im(\phi) \leq A/I\times A/J$? – kslote1 Apr 5 '14 at 2:34