 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! 14

Expert-verified Found in: Page 149 ### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624 # Prove Theorem 6.3

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 $R$

See the 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,

$a={s}_{1}{c}_{1}+{s}_{2}{c}_{2}+....{s}_{n}{c}_{n}\phantom{\rule{0ex}{0ex}}b={t}_{1}{c}_{1}+{t}_{2}{c}_{2}+.....{t}_{n}{c}_{n}\phantom{\rule{0ex}{0ex}}{s}_{1},{s}_{2},...{s}_{n}\in Rand{t}_{1},{t}_{2},.....{t}_{n}\in R$

## Proving the first condition

Now,

$a-b={s}_{1}{c}_{1}+{s}_{2}{c}_{2}+.....{s}_{n}{c}_{n}-\left({t}_{1}{c}_{1}+{t}_{2}{c}_{2}+....{t}_{n}{c}_{n}\right)\phantom{\rule{0ex}{0ex}}a-b={s}_{1}{c}_{1}-{t}_{1}{c}_{1}+{s}_{2}{c}_{2}-{t}_{2}{c}_{2}+......+{s}_{n}{c}_{n}-{t}_{n}{c}_{n}\phantom{\rule{0ex}{0ex}}a-b=\left({s}_{1}-{t}_{1}\right){c}_{1}+\left({s}_{2}-{t}_{2}\right){c}_{2}+.....+\left({s}_{n}-{t}_{n}\right){c}_{n}$

Hence,

${s}_{1},{s}_{2},......{s}_{n}\in R\phantom{\rule{0ex}{0ex}}{t}_{1},{t}_{2},........{t}_{n}\in R\phantom{\rule{0ex}{0ex}}$

So, ${s}_{i}-{t}_{i}\in R$ .

Hence, $a-b\in I$ .

## Proving the second condition

$ra=\left(r{s}_{1}\right){c}_{1}+\left(r{s}_{2}\right){c}_{2}+......+\left(r{s}_{n}\right){c}_{n}\phantom{\rule{0ex}{0ex}}ra\in I$

## 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 $R$ ### Want to see more solutions like these? 