# Ring Homomorphism by Idempotent

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## Theorem

Let $A$ be a commutative ring.

Let $e \in A$ be an idempotent element.

Let $\ideal e$ be the ideal of $A$ generated by $e$.

Then the mapping:

- $f: A \to \ideal e: a \mapsto e a$

is a surjective ring homomorphism with kernel the ideal $\ideal {1 - e}$ generated by $1 - e$.

## Proof

Let $e a \in \ideal e$ for some arbitrary $a \in A$.

This article, or a section of it, needs explaining.In particular: Technically, one needs to prove that a general element of $\ideal e$ has this form. --Wandynsky (talk) 15:38, 30 July 2021 (UTC)
Go to it.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

Then

\(\ds \map f {e a}\) | \(=\) | \(\ds e e a\) | Definition of $f$ | |||||||||||

\(\ds \) | \(=\) | \(\ds ea\) | $e$ is idempotent |

So $f$ is surjective.

Let $x, y \in A$.

Then

\(\ds \map f {x + y}\) | \(=\) | \(\ds e \paren {x + y}\) | Definition of $f$ | |||||||||||

\(\ds \) | \(=\) | \(\ds e x + e y\) | Ring Axiom $\text D$: Distributivity of Product over Addition | |||||||||||

\(\ds \) | \(=\) | \(\ds \map f x + \map f y\) | Definition of $f$ |

and:

\(\ds \map f {x y}\) | \(=\) | \(\ds e x y\) | Definition of $f$ | |||||||||||

\(\ds \) | \(=\) | \(\ds e^2 x y\) | $e$ is idempotent | |||||||||||

\(\ds \) | \(=\) | \(\ds e x e y\) | $A$ is commutative | |||||||||||

\(\ds \) | \(=\) | \(\ds \map f x \map f y\) | Definition of $f$ |

So $f$ is a ring homomorphism.

We have

\(\ds \map f {1 - e}\) | \(=\) | \(\ds e \paren {1 - e}\) | Definition of $f$ | |||||||||||

\(\ds \) | \(=\) | \(\ds e^2 - e\) | Ring Axiom $\text D$: Distributivity of Product over Addition | |||||||||||

\(\ds \) | \(=\) | \(\ds 0\) | $e$ is idempotent |

Thus $1 - e \in \map \ker f$ and $\ideal {1 - e} \subseteq \map \ker f$.

This article, or a section of it, needs explaining.In particular: Technically, one needs a result of the form: 'set contained in ideal' implies 'ideal generated by set contained in ideal' --Wandynsky (talk) 15:38, 30 July 2021 (UTC)
Go to it.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

Suppose $a \in \map \ker f$.

Then $e a = \map f a = 0$.

\(\ds a\) | \(=\) | \(\ds e a + \paren {1 - e} a\) | Ring Axiom $\text D$: Distributivity of Product over Addition | |||||||||||

\(\ds \) | \(=\) | \(\ds \paren {1 - e} a\) | $e a = 0$ |

Thus:

- $\map \ker f \subseteq \ideal {1 - e}$

so:

- $\map \ker f = \ideal {1 - e}$

$\blacksquare$