(related to Lemma: Greatest Common Divisor and Least Common Multiple of Ideals)
According to the definition of ideals, if \(I\lhd R\) and \(J\lhd R\) are ideals by hypothesis, then \((I, + )\) and \((J, + )\) are subgroups of the additive group \((R,+ )\). According to the definition of the set \(I+J\) and for an \(i_1,i_2\in I\) and \(j_1,j_2\in J\) we therefore have for each element \((i_1+j_1)+(i_2+j_2)\) contained in \(I+J\) that also the element \((i_1+j_1) - (i_2+j_2)\) is contained in \(I+J\). Thus, it is verified by the first subgroup criterion, that the set \((I + J, + )\) is a subgroup of the additive group \((R,+)\). Moreover, if \(r\in R\) and \(i+j\in I+J\), then we have \[r(i+j)=ri+rj\in I+J.\] Therefore, \(I + J\) is an ideal of the ring \(R\).
First of all, we observe that \(I+J\) divides both, \(I\) and \(J\), since \(I+J\supseteq I\) and \(I+J\supseteq J\). Now, for any other ideal \(K\lhd R\) dividing both, \(I\) and \(J\) we have either \(I+J\supseteq K\) or \(K\supseteq I+ J\). In the case \(I+J=K\) there is nothing to show. Suppose \(I+J\supset K\). Then, there would exist an \(x\in I+J\) with \(x\not\in K\). But this contradicts \(K\supseteq I\owns x\), (analogously \(K\supseteq J\owns x\)), since \(K\) divides both ideals by assumption. Thus, we must have \(K\supseteq I + J\). This demonstrates that for any other ideal \(K\) dividing both, \(I\) and \(J\), it also divides \(I+J\). This means that \(I + J\) is the greatest common divisor, or \(I+J=gcd(I,J)\).
Since \((I, + )\) and \((J, + )\) are subgroups of the additive group \((R,+ )\), also \((I\cap J, + ) \) is a subgroup by second subgroup criterion. Moreover, if \(r\in R\) and \(x\in I\cap J\), then \(x\in I\) and \(x\in J\), so \(rx\in I\) and \(rx\in J\) since \(I\) and \(J\) are ideals. But this means that \(rx\in I\cap J\). Therefore, \(I \cap J\) is an ideal of the ring \(R\).
First of all, we observe that \(I\supseteq I\cap J\) and \(J\supseteq I\cap J\), thus \(I\cap J\) is a multiple of both ideals (in other words both ideals divide \(I\cap J\) simultaneously). For any other ideal \(K\lhd R\) being the multiple of both ideals (\(I\supseteq K\) and \(J\supseteq K\)) we have either \(I\cap J\supseteq K\) or \(K\supseteq I\cap J\). In the case \(I\cap J=K\) there is nothing to show. Suppose \(K\supset I\cap J\). Then, there would exist an \(x\in K\) with \(x\not\in I\cap J\). This contradicts \(x\in I\cap J\), since \(I\subseteq K\owns x\) and \(J\subseteq K\owns x\) simultaneously. Thus, we must have \(I\cap J\supseteq K\). This demonstrates that for any other ideal \(K\) being a multiple of both, \(I\) and \(J\), it is also a multiple of \(I\cap J\). This means that \(I\cap J\) is the least common multiple, or \(I\cap J=lcm(I,J)\).
