Ring Homomorphism by Idempotent

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:

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

$\blacksquare$