Proof

We will first show that the set $$I + J$$ is an ideal of the ring $$R$$.

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 $$R$$.

Now, we will show that $$I+J$$ is the greatest ideal dividing $$I$$ and $$J$$.

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 $$I+J=gcd(I,J)$$.

Next, we show that the set $$I \cap J$$ is an ideal of the ring $$R$$.

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 $$R$$.

Last, we show that the set $$I \cap J$$ is a least common multiple of the ideals $$I$$ and $$J$$.

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 $$I\cap J=lcm(I,J)$$.

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References

Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013