Show that intersections, products, and sums of ideals are ideals.
Solution
1. The intersection of any nonempty family of ideals of \(R\) is an ideal.
Let \(\{I_\alpha\}_{\alpha\in A}\) be ideals and set \(I=\bigcap_{\alpha\in A}I_\alpha\). Since \(0\in I_\alpha\) for every \(\alpha\), \(0\in I\). If \(x,y\in I\), then \(x-y\in I_\alpha\) for every \(\alpha\), hence \(x-y\in I\). If \(r\in R\) and \(x\in I\), then \(rx,xr\in I_\alpha\) for every \(\alpha\), so \(rx,xr\in I\). Thus \(I\) is a two-sided ideal.
2. If \(I,J\trianglelefteq R\), then\[ I+J=\{a+b:a\in I,\ b\in J\} \]is an ideal.
We have \(0=0+0\in I+J\). If \(a+b,c+d\in I+J\), with \(a,c\in I\) and \(b,d\in J\), then \[ (a+b)-(c+d)=(a-c)+(b-d)\in I+J. \] For \(r\in R\), \[ r(a+b)=ra+rb\in I+J, \qquad (a+b)r=ar+br\in I+J, \] because \(I\) and \(J\) are ideals. Hence \(I+J\trianglelefteq R\).
3. If \(I,J\trianglelefteq R\), their product\[ IJ=\left\{\sum_{k=1}^n a_kb_k:n\ge0,\ a_k\in I,\ b_k\in J\right\} \]is an ideal.
The empty sum is \(0\), so \(0\in IJ\). Finite sums of products are closed under addition and additive inverses. If \[ x=\sum_{k=1}^n a_kb_k\in IJ \] and \(r\in R\), then \[ rx=\sum_{k=1}^n (ra_k)b_k\in IJ \] because \(ra_k\in I\), while \[ xr=\sum_{k=1}^n a_k(b_kr)\in IJ \] because \(b_kr\in J\). Thus \(IJ\trianglelefteq R\).
4. Therefore intersections, sums, and products of ideals are ideals.
This is exactly <1>1–<1>3. Finite sums of more than two ideals follow by induction from 2, and finite products from 3.