Let \(R\) be a principal ideal domain and let \(I,J\) be nonzero ideals. Prove that \[ IJ=I\cap J \] if and only if \[ I+J=R. \]
Solution
1. For any two ideals \(I,J\) in a commutative ring,\[ IJ\subseteq I\cap J. \]
Every generator \(xy\) of the product ideal \(IJ\), with \(x\in I\) and \(y\in J\), lies in both \(I\) and \(J\). Hence every finite sum of such products lies in \(I\cap J\).
2. If\[ I+J=R, \]then\[ IJ=I\cap J. \]
Choose \(u\in I\) and \(v\in J\) with \[ u+v=1. \] If \(x\in I\cap J\), then \[ x = xu+xv. \] Since \(x\in J\) and \(u\in I\), one has \(xu\in IJ\); since \(x\in I\) and \(v\in J\), one has \(xv\in IJ\). Thus \(x\in IJ\), proving \(I\cap J\subseteq IJ\). Step 1 gives the reverse inclusion.
3. Suppose\[ IJ=I\cap J. \]Write\[ I=(a), \qquad J=(b), \qquad I+J=(d), \]with \(a,b,d\neq0\). Then \(d\) is a unit.
Since \((a)+(b)=(d)\), the element \(d\) divides both \(a\) and \(b\). Write \[ a=da', \qquad b=db'. \] The element \[ c\coloneqq da'b' \] satisfies \[ c=ab' \in (a)=I \] and \[ c=ba' \in (b)=J. \] Hence \[ c\in I\cap J=IJ=(ab). \] Thus for some \(r\in R\), \[ da'b' = r ab = r d^2a'b'. \] Because \(R\) is a domain and \(da'b'\neq0\), cancellation gives \[ 1=rd. \] Therefore \(d\) is a unit.
4. If\[ IJ=I\cap J, \]then\[ I+J=R. \]
By step 3, \[ I+J=(d) \] with \(d\) a unit. Hence \((d)=R\).
5. Consequently,\[ \boxed{ IJ=I\cap J \quad\Longleftrightarrow\quad I+J=R }. \]
The forward implication is step 4, and the reverse implication is step 2.
6. Q.E.D.
Step 5 is the asserted equivalence.