Let \(I\), \(J\) be two ideals in a commutative ring \(R\) (with unit).
Define \(K = \{x : x \in I \text{ and } x \in J\}\). Show that \(K\) is an ideal.
If \(R\) is a principal ideal domain, so \(I = (i)\), \(J = (j)\), give a formula for a generator \(k\) of \(K\).
Solution
1. The set \(K=I\cap J\) is an ideal of \(R\).
Because \(I\) and \(J\) are ideals, \(0\in I\) and \(0\in J\), hence \(0\in K\). If \(a,b\in K\), then \(a,b\in I\) and \(a,b\in J\). Therefore \[ a-b\in I\qquad\text{and}\qquad a-b\in J, \] so \(a-b\in K\). Thus \(K\) is an additive subgroup of \(R\).
For \(r\in R\) and \(a\in K\), one has \(a\in I\cap J\). Since both \(I\) and \(J\) are ideals, \[ ra\in I\qquad\text{and}\qquad ra\in J. \] Hence \(ra\in K\). Thus \(K\) is an ideal.
2. If \(R\) is a PID, then\[ I\cap J=(\operatorname{lcm}(i,j)), \]where the least common multiple is defined up to multiplication by a unit.
Write \(I=(i)\) and \(J=(j)\). For any \(x\in R\), \[ x\in I\cap J \iff i\mid x\text{ and }j\mid x. \] Thus the elements of \(I\cap J\) are exactly the common multiples of \(i\) and \(j\).
Since \(R\) is a PID, \(I\cap J\) is principal; write \(I\cap J=(k)\). Then \(k\) itself is a common multiple of \(i\) and \(j\). Conversely, every common multiple \(x\) lies in \((k)\), so \(k\mid x\). Hence \(k\) is a least common multiple of \(i\) and \(j\) in the divisibility sense. Therefore one may take \[ k=\operatorname{lcm}(i,j), \] up to a unit.
Equivalently, when \(i,j\ne0\), if \(d=\gcd(i,j)\) then one may choose \[ k=\frac{ij}{d} \] up to a unit. The ideal formula \(I\cap J=(\operatorname{lcm}(i,j))\) also covers \(I=(0)\) or \(J=(0)\), in which case the intersection is \((0)\).