Question

If $I$is an ideal in $R$and $J$is an ideal in the ring $S$, prove that $I\times J$is an ideal inthe ring$R\times S$ .

Short answer

It is proved$I\times J$ is an ideal in the ring$R\times S$ .

Step-by-step solution

Determine theorem 6.1 (i) condition 

Consider that $I$and $J$are non-empty and $I\times J$is empty using Theorem 6.1 and applying conditions.

Now, taking $\left(r_1,s_1\right),\left(r_2,s_2\right)\in I\times J$. It has $r_1,r_2\in I$and $s_1,s_2\in J$.

Solving furthermore,

$\left(r_1,s_1\right)-\left(r_2,s_2\right)=\left(r_1-r_2,s_1-s_2\right)\in I\times J$

Here, $r_1-r_2\in I$and$s_1-s_2\in J$

Here,$I$ and $J$are ideals. Therefore, condition (i) of Theorem 6.1 is satisfied.

Determine I×J is an ideal in the ring R×S 

Now consider $\left(r,s\right)\in R\times S$, where $r\in R$and $s\in S$.

$\left(r,s\right)\left(r_1,s_1\right)=\left(r{r}_1,s{s}_1\right)\in I\times J$

$r{r}_1\in I$and $s{s}_1\in J$, where, $I$and $J$are ideals. Similarly,

$\left(r_1,s_1\right)\left(r,s\right)=\left(r_{1}r,s_{1}s\right)\in I\times J$

$r_{1}r\in I$and $s_{1}s\in J$.

Here, $I$and $J$are ideals. Therefore, condition (ii) of Theorem 6.1 is satisfied.

Hence,$I\times J$is an ideal in the ring$R\times S$ .