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 $\ x\equiv 0\pmod {J_k}$


Just apply the first isomorphism theorem to complete the proof.

  • 1
    $\begingroup$ But doesn't that just imply the $A/(I\cap J) \cong im(\phi) \leq A/I\times A/J$? $\endgroup$ – kslote1 Apr 5 '14 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.