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Found in: Page 149

Abstract Algebra: An Introduction

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

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If $I$ is an ideal in a field $F$ , prove that $I=\left({0}_{F}\right)$ or $I=F$ .

It can be proved $I=\left({0}_{F}\right)$ or $I=F$ .

See the step by step solution

Previous results used

We shall use the following result to prove this.

Given $R$ is a ring with identity and $I$ is an ideal in $R$.

If ${1}_{R}\in I$ , then $I=R$ .

Proving the first condition

If $I=\left({0}_{F}\right)$, there is nothing to prove.

Proving the second condition

If $I\ne \left({0}_{F}\right)$, there exists $a\in I$ such that $a\ne {0}_{F}$ .

Now, we know that $F$ is a field.

Therefore, $a$ must be a unit.

Hence, from the above result, we will conclude that $I=F$ .

Conclusion

Hence,$I=\left({0}_{F}\right)$ or $I=F$ .

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