Ring Homomorphism by Idempotent
Jump to navigation
Jump to search
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:
- $\map \ker f = \ideal {1 - e}$
$\blacksquare$