Product of rings: If $K$ is an ideal of $R\times S$, then there exists $I$ ideal of $R$, $J$ ideal of $S$ such that $K=I\times J$. Let $R$ and $S$ be two rings. We consider the product $R\times S$.
It is a ring with operations of sum and product defined coordinate by coordinate, i.e.
$$(r_1, s_1) + (r_2, s_2) = (r_1 + r_2, s_1 + s_2) \text{  and  } (r_1, s_1) · (r_2, s_2) = (r_1 · r_2, s_1 · s_2)$$
The element $1$ of the ring $R \times S$ is $(1, 1)$ and the element $0$ is $(0, 0)$.
$(a)$ Let $I$ be a two-sided ideal of $R$ and let $J$ be a two-sided ideal of $J$. Show that $I \times J$ is a two-sided ideal of $R \times S$.
$(b)$ Show that the converse holds. If $K$ is an ideal of $R\times S$, then there exists $I$ ideal of $R$, $J$ ideal of $S$ such that $K=I\times J$.
I have been able to show part $(a)$ by

*

*$I$ is nonempty, $J$ nonempty so $I\times J$ is nonempty

*If $(i,j)$, $(i',j') \in I\times J$, then $(i,j) + (i',j') \in I\times J$

*If $(i,j) \in  I\times J$, $(a,b)\in R\times S$ then $(i,j)\cdot (a,b) \in I\times J$
However I am not sure how to prove $(b)$. Any help would be appreciated.

For (a) you should also note that $(a,b)(i,j) \in I \times J$ (you want $I \times J$ to be two-sided).
For (b): Let $I := \{i \in R \mid \exists s \in S : (i,s) \in K\}$ and $J := \{j \in S \mid \exists r \in R: (r,j) \in K\}$. Then $I$ is an ideal: $I$ is non-empty, if $i,i' \in I$, say $(i,s), (i',s') \in K$, then $(i+i', s+s') \in K$, hence $i+i' \in I$. For $i \in I$, $r \in R$, say $(i, s) \in K$, we have $(ir,s), (ri,s) \in K$, hence $ir, ri \in I$, so $I$ in a two-sided ideal, same for $J$.
It remains to show that $I \times J = K$. On one hand, if $(i,j) \in K$, then $i \in I$, $j \in J$ by definition of $I$ and $J$, that is $K \subseteq I \times J$. Now suppose $i \in I$, $j \in J$. Then for some $r \in R$, $s \in S$, we have $(i,s), (r,j) \in K$. That is
$$ (i,j) = (i,0) + (0,j) = (i,s) \cdot (1,0) + (r,j) \cdot (0,1) \in K $$
Hence $I \times J \subseteq K$.
