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

So $f$ is surjective.

Let $x, y \in A$.

Then

and:

So $f$ is a ring homomorphism.

We have

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

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

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

Thus:


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

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

Also see

 * Unital Ring Homomorphism by Idempotent