State the Chinese remainder theorem in any form you like. Prove it.
Solution
Let \(R\) be a commutative ring and let \(I_1,\dots,I_n\) be pairwise comaximal ideals. The Chinese remainder theorem states that the natural map \[ \phi:R\to\prod_{i=1}^n R/I_i, \qquad r\mapsto(r+I_1,\dots,r+I_n) \] is surjective and has kernel \[ \ker\phi=\bigcap_{i=1}^n I_i=I_1\cdots I_n. \] Hence \[ R/(I_1\cdots I_n)\cong\prod_{i=1}^nR/I_i. \]
1. The kernel is \(\bigcap_i I_i\).
By definition, \(\phi(r)=0\) exactly when \(r\in I_i\) for every \(i\).
2. For pairwise comaximal ideals,\[ \bigcap_i I_i=I_1\cdots I_n. \]
It suffices to prove the two-ideal case and iterate. If \(I+J=R\), choose \(a\in I\) and \(b\in J\) with \(a+b=1\). For \(x\in I\cap J\), \[ x=x(a+b)=xa+xb\in IJ. \] The reverse inclusion \(IJ\subseteq I\cap J\) is automatic.
3. The map \(\phi\) is surjective.
For each \(i\), pairwise comaximality gives \[ I_i+\prod_{j\ne i}I_j=R. \] Choose \(e_i\in\prod_{j\ne i}I_j\) with \(e_i\equiv1\pmod{I_i}\). Then \(e_i\equiv0\pmod{I_j}\) for \(j\ne i\). Given residues \(r_i+I_i\), the element \[ r=\sum_i r_i e_i \] has \(r\equiv r_i\pmod{I_i}\) for every \(i\).
The first isomorphism theorem now gives the claimed product decomposition.