# 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 $I \times J \subseteq K$.