Question

Prove Theorem 6.3

Short answer

It can be proved$I=\left\{r_1{c}_1+r_2{c}_2+......r_n{c}_n\overline{)r_1,r_2.....,r_n\in R}\right\}$ is an ideal in$R$

Step-by-step solution

Statement

Let $R$be a commutative ring with identity and$c_1,c_2,.....c_n\in R$ Then the set$I=\left\{r_1{c}_1+r_2{c}_2+....r_n{c}_n\overline{)r_1,r_2,....r_n\in R}\right\}$ is an ideal in $R$

Results used

We shall use Theorem to prove the above result.

Theorem of a non-empty subset$I$ of a ring is an ideal if and only if it has the following 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$ar\in I$ .

Supposition

According to Theorem , suppose that $a,b\in I$and $r\in R${where $I=\left\{r_1{c}_1+r_2{c}_2+....+r_n{c}_n\overline{)r_1,r_2,...,r_n\in R}\right\}$

To prove: $a-b\in I$and $ra\in I$

{Note that $ar\in I$automatically by commutativity}

 Using hypothesis

$a,b\in I$

Therefore,

$$\begin{gathered} a=s_1{c}_1+s_2{c}_2+....s_n{c}_n \\ b=t_1{c}_1+t_2{c}_2+.....t_n{c}_n \\ {s}_1,s_2,...s_n\in Rand{t}_1,t_2,.....t_n\in R \end{gathered}$$

Proving the first condition 

Now,

$$\begin{gathered} a-b=s_1{c}_1+s_2{c}_2+.....s_n{c}_n-\left(t_1{c}_1+t_2{c}_2+....t_n{c}_n\right) \\ a-b=s_1{c}_1-t_1{c}_1+s_2{c}_2-t_2{c}_2+......+s_n{c}_n-t_n{c}_n \\ a-b=\left(s_1-t_1\right)c_1+\left(s_2-t_2\right)c_2+.....+\left(s_n-t_n\right)c_n \end{gathered}$$

Hence,

$$\begin{gathered} s_1,s_2,......s_n\in R \\ {t}_1,t_2,........t_n\in R \end{gathered}$$

So,$s_i-t_i\in R$ .

Hence, $a-b\in I$.

Proving the second condition

$$\begin{gathered} ra=\left(r{s}_1\right)c_1+\left(r{s}_2\right)c_2+......+\left(r{s}_n\right)c_n \\ ra\in I \end{gathered}$$

Conclusion  

Then the set $I=\left\{r_1{c}_1+r_2{c}_2+.....+r_n{c}_n\overline{)r_1,r_2,....r_n\in R}\right\}$is an ideal in $R$