Unital Ring Homomorphism by Idempotent

Theorem

Let $A$ be a commutative ring with unity.

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 unital ring homomorphism from $\struct {A, +, \circ}$ to $\struct {\ideal e, +, \circ}$ with kernel the ideal $\ideal {1 - e}$ generated by $1 - e$.

Proof

By Ring Homomorphism by Idempotent $f$ is a surjective ring homomorphism with kernel $\ideal {1 - e}$.

By Ring by Idempotent $A$ is a commutative ring with unity whose unity is $e$.

Then

\(\ds \map f 1\)

\(=\)

\(\ds e \cdot 1\)

Definition of $f$

\(\ds \)

\(=\)

\(\ds e\)

$1$ is the unity of $A$.

Thus $f$ is a unital ring homomorphism.

$\blacksquare$