Ring of Idempotents is Idempotent Ring

Theorem
Let $\struct {R, +, \circ}$ be a commutative ring.

Let $\struct {A, \oplus, \circ}$ be its ring of idempotents.

Then $\struct {A, \oplus, \circ}$ is an idempotent ring.

Proof
First, it is to be established that $\struct {A, \oplus, \circ}$ is a ring in the first place.

This we do by verifying the ring axioms.

Axiom $(\text A 0)$: Closure for $\oplus$
Let $x, y \in A$.

It is to be shown that $x \oplus y \in A$, i.e. that $x \oplus y$ is an idempotent element of $R$.

Compute as follows:

Hence $x \oplus y \in A$, as desired.

Axiom $(\text A 1)$: Associativity of $\oplus$
Let $x, y, z \in A$.

It is to be shown that $\oplus$ is associative, that is:


 * $\paren {x \oplus y} \oplus z = x \oplus \paren {y \oplus z}$

This is shown by the following computation:

Axiom $(\text A 2)$: Commutativity of $\oplus$
Let $x, y \in A$.

It is to be shown that $\oplus$ is commutative, that is:


 * $x \oplus y = y \oplus x$

This is shown by the following computation:

Axiom $(\text A 3)$: Identity for $\oplus$
Let $x \in A$, and let $0_R$ be the zero of $R$.

By Ring Zero is Idempotent, $0_R \in A$.

It suffices to show that:


 * $x \oplus 0_R = x$

since was already verified above.

Now:

as desired.

Axiom $(\text A 4)$: Inverses for $\oplus$
Let $x \in A$.

It is to be shown that there exists $y \in A$ such that:


 * $x \oplus y = y \oplus x = 0_R$

by $(A3)$ above.

In fact, one has:

so that each $x \in A$ is its own inverse for $\oplus$.

Axiom $(\text M 0)$: Closure for $\circ$
Let $x, y \in A$.

It is to be shown that:


 * $x \circ y \in A$

that is, that $x \circ y$ is idempotent.

We have the following computation:

Axiom $(\text M 1)$: Associativity of $\circ$
Immediate from Restriction of Associative Operation is Associative.

Axiom $(\text D)$: Distributivity
By Restriction of Commutative Operation is Commutative, $\circ$ is commutative on $A$.

Thus to establish distributivity, it suffices to verify, for $x, y, z \in A$:


 * $x \circ \paren {y \oplus z} = \paren {x \circ y} \oplus \paren {x \circ z}$

To this end, we compute as follows:

Therefore, having verified all ring axioms, we conclude $\struct {A, \oplus, \circ}$ is a ring.

By assumption all $x \in A$ are idempotent elements for $\circ$.

Thus $\circ$ is an idempotent operation on $A$.

Consequently, $\struct {A, \oplus, \circ}$ is an idempotent ring.

Also see

 * Ring of Idempotents of Commutative and Unitary Ring is Boolean Ring