Question

Let $\mathit{I}$and $\mathit{J}$be ideals in $\mathit{R}$. Let $\mathit{I}\mathit{J}$denote the set of all possible finite sums of elements of the form $\mathit{a}\mathit{b}$(with $\mathit{a}\mathbf{\in }\mathit{I}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{J}$ ), that is,

$\mathit{I}\mathit{J}\mathbf{=}\left\{a_1{b}_1+a_2{b}_2+.....+a_n{b}_n\overline{)n}\ge 1,a_k\in I,b_k\in J\right\}$

Prove that $\mathit{I}\mathit{J}$is an ideal, $\mathit{I}\mathit{J}$is called the product of$\mathit{I}$ and $\mathit{J}$.

Short answer

It is proved that $IJ$is an ideal.

Step-by-step solution

Write the conditions for theorem 6.1

A non-empty subset$I$ of a ring$R$ is an ideal if and only if it follows the listed properties:

1. If $a,b\in I$then $a-b\in I$
2. If $r\in R$and $a\in I$, then $ra\in I$and$ar\in I$

Show that either I=3 or I=c

Now, verify the above conditions as follows:

1. If$\sum _{i=1}^m{a}_i{b}_i,\sum _{j=1}^n{c}_j{d}_j\in IJ$ then we have,

$\left(\sum _{i=1}^m{a}_i{b}_i\right)‐\left(\sum _{j=1}^n{a}_j{b}_j\right)=\sum _{k=1}^{m+n}\left\{\begin{array}{ll}{a}_k{b}_k& k\le m\\ -c_{k-m}{d}_{k-m}& k>m\end{array}\right.\in IJ$

1. If$\sum _{i=1}^m{a}_i{b}_i\in IJ$ and$r\in R$ then we have,

role="math" localid="1653727939872" $$\begin{gathered} r\left(\sum _{i=1}^m{a}_i{b}_i\right)=\sum _{i=1}^m\left(r{a}_i\right)b_i\in IJ \\ \left(\sum _{i=1}^m{a}_i{b}_i\right)r=\sum _{i=1}^m{a}_i\left(b_{i}r\right)\in IJ \end{gathered}$$

Thus,

Hence, the statement is proved.