Let \(R\) be a ring (with unit).
What is an ideal of \(R\)?
If \(I\) and \(J\) are ideals of \(R\), prove that there is a least (with respect to inclusion) ideal of \(R\) that contains both \(I\) and \(J\) as subsets. This new ideal is called the join of \(I\) with \(J\): \(I \vee J\).
If \(I\) and \(J\) are ideals of \(R\), let \[IJ = \left\{\sum_{i=1}^n x_i y_i : n > 0, x_1,\dots,x_n \in I, \text{ and } y_1,\dots,y_n \in J\right\}.\] Show that \(IJ\) is an ideal of \(R\) that is contained in both \(I\) and \(J\). If \(K\) is any ideal of \(R\) that is contained in both \(I\) and \(J\), must \(K\) be contained in \(IJ\)?
Solution
1. An ideal of \(R\) is a subset \(I \subseteq R\) such that \((I,+)\) is a subgroup of \((R,+)\), that is, \(0\in I\) and \(x-y\in I\) for all \(x,y\in I\), and such that \(rx\in I\) and \(xr\in I\) for all \(r \in R\) and \(x \in I\).This answers part (a).
2. The sum \(I + J = \{x + y : x \in I, y \in J\}\) is the least ideal containing both \(I\) and \(J\), so \(I\vee J=I+J\).
2.1. \(I + J\) is an ideal of \(R\).
We have \(0 = 0 + 0 \in I+J\). For \(x_1+y_1, x_2+y_2 \in I+J\), \((x_1+y_1) - (x_2+y_2) = (x_1-x_2) + (y_1-y_2) \in I+J\). For any \(r \in R\), \(r(x+y) = rx + ry \in I+J\) and \((x+y)r = xr + yr \in I+J\), because \(I\) and \(J\) are ideals.
2.2. \(I \subseteq I + J\) and \(J \subseteq I + J\).
For \(x \in I\), \(x = x + 0 \in I+J\) since \(0 \in J\); similarly, for \(y \in J\), \(y = 0 + y \in I+J\).
2.3. If \(M\) is an ideal of \(R\) containing both \(I\) and \(J\), then \(I + J \subseteq M\).
For \(x \in I \subseteq M\) and \(y \in J \subseteq M\), closure of \(M\) under addition gives \(x + y \in M\).
2.4. Q.E.D.
Steps 1, 2, and 3.
3. \(IJ\) is an ideal of \(R\) and \(IJ\subseteq I\cap J\).
3.1. \(IJ\) is an additive subgroup of \(R\).
Taking \(n=1\) and \(x_1=y_1=0\) gives \(0\in IJ\). The difference \(\sum_i x_i y_i - \sum_j x'_j y'_j=\sum_i x_iy_i+\sum_j(-x'_j)y'_j\) is again a finite sum of products with first factors in \(I\) and second factors in \(J\).
3.2. For \(r \in R\) and \(x = \sum_{i=1}^n x_i y_i \in IJ\), both \(rx\) and \(xr\) lie in \(IJ\).
We have \(rx = \sum_{i=1}^n (rx_i) y_i\) with \(rx_i \in I\), and \(xr = \sum_{i=1}^n x_i (y_i r)\) with \(y_i r \in J\), because \(I\) and \(J\) are two-sided ideals.
3.3. Every element of \(IJ\) lies in \(I \cap J\).
Each product \(x_iy_i\) lies in \(I\) because \(x_i\in I\) and \(I\) absorbs right multiplication, and lies in \(J\) because \(y_i\in J\) and \(J\) absorbs left multiplication. The ideal \(I\cap J\) is closed under addition, so it contains \(\sum_i x_iy_i\).
3.4. Q.E.D.
Steps 1 and 2 show that \(IJ\) is an ideal, and step 3 gives \(IJ\subseteq I\cap J\).
4. An ideal \(K\) contained in both \(I\) and \(J\) need not be contained in \(IJ\).
In \(R=\ZZ\), let \(I=J=K=2\ZZ\). Then \(IJ\) is the set of finite sums of products \((2a)(2b)=4ab\), which is \(4\ZZ\). The ideal \(K\) is contained in \(I\) and \(J\), but \(2\in K\) and \(2\notin4\ZZ=IJ\).
5. Q.E.D.
Step 1 answers part (a), step 2 proves part (b), and steps 3 and 4 answer part (c).