# 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 $I \times J \subseteq K$.

• How to prove $I \times J \subseteq K$ in a ring without multiplicative identity? Sep 28, 2022 at 19:23
• @Paulo Sigiani The statement need not be true if there is no multiplicative identity. Nov 30, 2023 at 6:10

I am posting my solution here in case a wandering math student is stuck on this problem like I was. Credit to @martini for helping me to understand the solution.

Let $$K$$ be an ideal in $$R \times S$$. Define \begin{align*} I &= \{r \in R : (r,s) \in K \text{ for some } s \in S\}, \\ J &= \{s \in S : (r,s) \in K \text{ for some } r \in R\}. \end{align*} We show that $$I$$ is an ideal in $$R$$. Firstly, $$I$$ is not empty, $$0 \in I$$ since $$(0,0) \in K$$. Let $$i_1,i_2 \in I$$. Then, there are some $$s_1,s_2 \in S$$ such that $$(i_1,s_1)$$ and $$(i_2,s_2)$$ are in $$K$$. Since $$K$$ is an ideal, $$(i_1,s_1) - (i_2,s_2) = (i_1 - i_2,s_1 - s_2) \in K.$$ Clearly, $$s_1 - s_2 \in S$$ and so $$i_1 - i_2 \in I$$. I.e., $$I$$ is closed under subtraction. Similarly, $$(i_1,s_1) \cdot (i_2,s_2) = (i_1 i_2,s_1 s_2) \in K.$$ Again, $$s_1 s_2 \in S$$ and so $$i_1 i_2 \in I$$. Thus, $$I$$ is closed under multiplication. Hence $$I$$ is a subrng of $$R$$. Now, let $$i \in I$$ and $$r \in R$$. Then, there exists some $$s \in S$$ such that $$(i,s) \in K$$. Notice that $$(r,1) \in R \times S$$. Hence, \begin{align*} (i,s) \cdot (r,1) &= (ir,s) \in K, \\ (r,1) \cdot (i,s) &= (ri,s) \in K \end{align*} Thus, $$ir, ri \in I$$. Therefore, $$I$$ is an ideal in $$R$$. A symmetric argument guarantees that $$J$$ is an ideal in $$S$$.

Next, we show that $$K = I \times J$$. Let $$(k_1,k_2) \in K$$. By definition, $$k_1 \in I$$ and $$k_2 \in J$$. Therefore, $$K \subseteq I \times J$$. Let $$(i,j) \in I \times J$$. By definition, there exist $$r \in R$$ and $$s \in S$$ such that $$(i,s),(r,j) \in K$$. Observe that $$(i,j) = (i,0) + (0,j) = (i,s) \cdot (1,0) + (r,j) \cdot (0,1) \in K.$$ Therefore, $$I \times J \subseteq K$$ and so $$K = I \times J$$. $$\blacksquare$$

• In both your solution and martini's, you can define $I=\{\,r\in R\mid (r,0)\in K\,\}$ and likewise for $J$. This defines the same $I$, since $(r,0)=(r,s)(1,0)$. Nov 30, 2023 at 16:30