Book edition 3rd

Author(s) Thomas W Hungerford, David Leep

Pages 608 pages

ISBN 9781111569624

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Short Answer

Prove Theorem 6.3

It can be proved $I = \{r_{1} c_{1} + r_{2} c_{2} + . . . . . . r_{n} c_{n} r_{1}, r_{2} . . . . ., r_{n} \in R\}$ is an ideal in $R$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Statement

Let $R$ be a commutative ring with identity and $c_{1}, c_{2}, . . . . . c_{n} \in R$ Then the set $I = \{r_{1} c_{1} + r_{2} c_{2} + . . . . r_{n} c_{n} r_{1}, r_{2}, . . . . r_{n} \in R\}$ 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:

If $a, b \in I$ , then $a - b \in I$ .
If $r \in R$ and $a \in I$ , then $r a \in I$ and $a r \in I$ .

Supposition

According to Theorem , suppose that $a, b \in I$ and $r \in R$ {where $I = \{r_{1} c_{1} + r_{2} c_{2} + . . . . + r_{n} c_{n} r_{1}, r_{2}, . . ., r_{n} \in R\}$

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

{Note that $a r \in I$ automatically by commutativity}

Using hypothesis

$a, b \in I$

Therefore,

$a = s_{1} c_{1} + s_{2} c_{2} + . . . . s_{n} c_{n} b = t_{1} c_{1} + t_{2} c_{2} + . . . . . t_{n} c_{n} s_{1}, s_{2}, . . . s_{n} \in R a n d t_{1}, t_{2}, . . . . . t_{n} \in R$

Proving the first condition

Now,

$a - b = s_{1} c_{1} + s_{2} c_{2} + . . . . . s_{n} c_{n} - (t_{1} c_{1} + t_{2} c_{2} + . . . . t_{n} c_{n}) a - b = s_{1} c_{1} - t_{1} c_{1} + s_{2} c_{2} - t_{2} c_{2} + . . . . . . + s_{n} c_{n} - t_{n} c_{n} a - b = (s_{1} - t_{1}) c_{1} + (s_{2} - t_{2}) c_{2} + . . . . . + (s_{n} - t_{n}) c_{n}$

Hence,

$s_{1}, s_{2}, . . . . . . s_{n} \in R t_{1}, t_{2}, . . . . . . . . t_{n} \in R$

So, $s_{i} - t_{i} \in R$ .

Hence, $a - b \in I$ .

Proving the second condition

$r a = (r s_{1}) c_{1} + (r s_{2}) c_{2} + . . . . . . + (r s_{n}) c_{n} r a \in I$

Conclusion

Then the set $I = \{r_{1} c_{1} + r_{2} c_{2} + . . . . . + r_{n} c_{n} r_{1}, r_{2}, . . . . r_{n} \in R\}$ is an ideal in $R$

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Abstract Algebra: An Introduction

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Short Answer

Prove Theorem 6.3

Step by Step Solution

Statement

Results used

Supposition

Using hypothesis

Proving the first condition

Proving the second condition

Conclusion

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Abstract Algebra: An Introduction

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Prove Theorem 6.3

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