$M=I\times J $ for some $I,J$

Question

Let $R,S$ be two rings with identity. Prove that every ideal of $R\times S$ is of the form $I \times J$ where $I$ is an ideal of $R$ and $J$ is an ideal of $S$ .

Obviously $I \times J$ is an ideal of $R\times S$. Conversely let us assume that $M$ is an ideal of $R\times S$.

To show $M=I\times J $. How to proceed?

Answer 1

Let $I=\{x\in R:(x,0)\in M\}$ and $J=\{y\in S:(0,y)\in M\}$.

If $(x,y)\in M$, then $(x,0)=(x,y)(1,0)\in M$ and $(0,y)=(x,y)(0,1)\in M$. This can be used to show one of the inclusions. The other one is easy.