Let \(R\) be a commutative ring and let \(I, J\subseteq R\) be ideals.
\(I + J = \theset{i+j \suchthat i\in I, j\in J}\) is the smallest ideal containing \(I\) and \(J\).
\(IJ \coloneqq \theset{\sum_{k=1}^N x_k y_k \suchthat N\geq 0,\ x_k\in I,\ y_k \in J}\) is an ideal, the ideal generated by the products \(xy\) with \(x\in I\) and \(y\in J\), and \(IJ\subseteq I\cap J\).
\(I \cap J\) is an ideal.
If \(I+J = R\), then \(I \cap J = IJ\).
(a)–(c) Each set contains \(0\) and is closed under subtraction and under multiplication by elements of \(R\); any ideal containing \(I\) and \(J\) contains all sums \(i+j\), and any ideal containing all products \(xy\) contains their finite sums. Each product \(xy\) lies in \(I\) and in \(J\), so \(IJ\subseteq I\cap J\).
- Write \(1=i+j\) with \(i\in I\) and \(j\in J\). For \(x\in I\cap J\), \(x=xi+xj\), where \(xi\in JI=IJ\) and \(xj\in IJ\), so \(x\in IJ\).
A union of ideals need not be an ideal: in \(\ZZ\), \(2\in 2\ZZ\cup3\ZZ\) and \(3\in2\ZZ\cup3\ZZ\), but \(2+3=5\notin2\ZZ\cup3\ZZ\).