Show \(IJ\subseteq I\cap J\)Let \(x\in IJ\). Then \(x\) is a finite sum of elements of the form \(ab\), where \(a\in I\) and \(b\in J\). Since \(I\) is an ideal, \(ab\in I\). Since \(J\) is an ideal, \(ab\in J\). Thus, \(ab\in I\cap J\). Therefore, any sum of such elements is also in \(I\cap J\), so \(x\in I\cap J\). Show \(I\cap J\subseteq IJ\)Since \(I\) and \(J\) are coprime, \(I+J=R\). This means there exist \(i\in I\) and \(j\in J\) such that \(i+j=1\). Let \(x\in I\cap J\). Multiply \(x\) by \(1\): \(x=x\cdot 1\). Substitute \(1=i+j\): \(x=x(i+j)\). Distribute: \(x=xi+xj\). Since \(x\in J\) and \(i\in I\), \(xi\in JI\). Since \(x\in I\) and \(j\in J\), \(xj\in IJ\). As \(R\) is commutative, \(JI=IJ\). Therefore, \(x=xi+xj\in IJ+IJ=IJ\). | 
