Suppose that \(R\) is a commutative ring with identity and \(I\) and \(J\) are ideals of \(R\). Suppose that \(I + J = R\).
Prove that \(IJ = I \cap J\).
Prove that \(R/(IJ)\) is isomorphic to \(R/I \oplus R/J\). (This is the Chinese Remainder Theorem).
Solution
Choose \(u\in I\) and \(v\in J\) such that \(u+v=1\). The finite direct sum in the statement carries componentwise addition and multiplication, so it is the product ring \(R/I\times R/J\).
1. The ideals \(IJ\) and \(I\cap J\) are equal.
Every product \(ij\), with \(i\in I\) and \(j\in J\), belongs to both ideals. Their intersection is closed under finite sums, so \(IJ\subseteq I\cap J\).
Conversely, for \(z\in I\cap J\) one has \(z=zu+zv\). Since \(u\in I\), \(z\in J\), and \(R\) is commutative, \(zu\in IJ\). Since \(z\in I\) and \(v\in J\), also \(zv\in IJ\). Thus \(z\in IJ\), proving the reverse inclusion.
2. The map\[ \phi:R\longrightarrow R/I\times R/J,\qquad r\longmapsto(r+I,r+J) \]is a surjective ring homomorphism with kernel \(IJ\).
Both components are quotient homomorphisms, so \(\phi\) preserves addition, multiplication, and the identity. Its kernel is \(I\cap J\), which equals \(IJ\) by step 1.
For arbitrary representatives \(a,b\in R\), put \(r=av+bu\). Then \(r-a=u(b-a)\in I\) and \(r-b=v(a-b)\in J\). Consequently \(\phi(r)=(a+I,b+J)\), proving surjectivity.
3. The required isomorphism and its inverse are\[ \begin{aligned} \overline\phi:R/(IJ)&\longrightarrow R/I\times R/J, &r+IJ&\longmapsto(r+I,r+J),\\ \psi:R/I\times R/J&\longrightarrow R/(IJ), &(a+I,b+J)&\longmapsto av+bu+IJ. \end{aligned} \]
The map \(\overline\phi\) is well defined because \(IJ\subseteq I\cap J\). It is a ring homomorphism. Changing \(a\) by an element \(i\in I\) changes \(av+bu\) by \(iv\in IJ\); changing \(b\) by \(j\in J\) changes it by \(ju\in IJ\). Thus \(\psi\) is also well defined.
The residue computation in step 2 gives \(\overline\phi\psi(a+I,b+J)=(a+I,b+J)\). In the other direction, \(\psi\overline\phi(r+IJ)=r(v+u)+IJ=r+IJ\). Therefore \(\overline\phi\) is a bijective ring homomorphism and hence a ring isomorphism.