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 $a \in A$.



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



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:

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

$\blacksquare$


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