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### Abstract Algebra: An Introduction

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

# If ${\mathbit{R}}$ is a ring with identity, and ${\mathbit{f}}{\mathbf{:}}{\mathbit{R}}{\mathbf{\to }}{\mathbit{S}}$ is a homomorphism from ${\mathbit{R}}$ to a ring ${\mathbit{S}}$ , prove that ${\mathbit{f}}\left({1}_{R}\right)$ is an idempotent in ${\mathbit{S}}$ . [Idempotent were defined in Exercise 3 of Section 3.2.]

Hence it is proved that $f\left({1}_{R}\right)$ is an idempotent in $S$ .

See the step by step solution

## Property of Rings

If any ring $R$ has elements such that, $a,b\in R$, then, addition and multiplication of the function of its elements is respectively given by:

$f\left(a+b\right)=f\left(a\right)+f\left(b\right)\phantom{\rule{0ex}{0ex}}f\left(ab\right)=f\left(a\right)f\left(b\right)$

## Homomorphism

We have a homomorphism of rings as$f:R\to S$:

In this case, we get:

$f\left({1}_{R}\right)=f\left({1}_{R}{1}_{R}\right)\phantom{\rule{0ex}{0ex}}=f\left({1}_{R}\right)f\left({1}_{R}\right)\phantom{\rule{0ex}{0ex}}=f{\left({1}_{R}\right)}^{2}$

Hence proved, $f\left({1}_{R}\right)$ is an idempotent in $S$ .