3
$\begingroup$

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.

$\endgroup$

2 Answers 2

6
$\begingroup$

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 $I \times J \subseteq K$.

$\endgroup$
2
  • $\begingroup$ How to prove $I \times J \subseteq K$ in a ring without multiplicative identity? $\endgroup$ Sep 28, 2022 at 19:23
  • $\begingroup$ @Paulo Sigiani The statement need not be true if there is no multiplicative identity. $\endgroup$
    – Allen Bell
    Nov 30, 2023 at 6:10
0
$\begingroup$

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 $K = I \times J$. $\blacksquare$

$\endgroup$
1
  • $\begingroup$ 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 $(r,0)=(r,s)(1,0)$. $\endgroup$
    – Allen Bell
    Nov 30, 2023 at 16:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .