I am studying for my algebra qualifying exam in January, and this was one of the questions on a recent exam.
Let $R$ be a commutative ring with identity, and let $I$ and $J$ be ideals of $R$. If $I+J=R$, show that $IJ=I \cap J$.
Here is my approach, and I just want to know if it is correct or not.
Suppose $I$ and $J$ are ideals of a commutative ring $R$ such that $I+J=R$. We wants to show that $IJ=I \cap J$. So, we have the following:
$IJ \subset I \cap J$:
Let $x \in IJ$. Then $x=ab$ for some $a \in I$ and $b \in J$. Moreover, $a+b \in I+J$ and so $a+b \in R$. Since $IJ$ can be shown to be an ideal of $R$, then $x(a+b) \in IJ \implies x \in I$, and, since $R$ is commutative, $(a+b)x \in IJ \implies x \in J$. Hence, $x \in I \cap J$, and so $IJ \subset I \cap J$.
$I \cap J \subset IJ$:
Let $x \in I \cap J$. Then, $x \in I$ and $x \in J$. Hence, $xx \in IJ$, and so $x \in IJ$ since $IJ$ is an ideal. Thus, $I \cap J \subset IJ$.
Hence, we have $IJ=I \cap J$.