Let $R$ and $S$ be two rings, then $K$ is an ideal in $R\times S$ if and only if there are $K_1$ and $K_2$ such that $K_1$ is an ideal of $R$ and $K_2$ is an ideal of $S$ such that $K = K_1 \times K_2$.
How do you show that $K_1 = \{a \in R \mid (a,0) \in K\}$ is an ideal of $R$ and $K_2 = \{b \in S \mid (0,b) \in K\}$ is an ideal of S, and the converse too.
Thank you very much.
For the inclusion $K_1 \times K_2 \subset K$, let $(a,b) \in K_1 \times K_2$, then $(a,0) \in K$ and $(0,b) \in K$.
For the reverse inclusion, let $(a,b) \in K$, then $(1,0) \cdot (a,b)=(a,0) \in K$ and $(0,1) \cdot (a,b)=(0,b) \in K$. Note that I assume $1 \in R,S$.