Ring of Idempotents is Idempotent Ring

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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 $(A0)$: 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:

\(\displaystyle \paren {x \oplus y} \circ \paren {x \oplus y}\) \(=\) \(\displaystyle \paren {x + y - 2 x \circ y} \circ \paren {x + y - 2 x \circ y}\) $\quad$ Definition of $\oplus$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {x + y - 2 x \circ y} \circ x + \paren {x + y - 2 x \circ y} \circ y - \paren {x + y - 2 x \circ y} \circ \paren {2 x \circ y}\) $\quad$ $\circ$ distributes over $+$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ x + y \circ x - \paren {2 x \circ y} \circ x + x \circ y + y \circ y - \paren {2 x \circ y} \circ y\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle x \circ \paren {2 x \circ y} - y \circ \paren {2 x \circ y} + \paren {2 x \circ y} \circ \paren {2 x \circ y}\) $\quad$ $\circ$ distributes over $+$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ x + y \circ x - 2 x \circ y \circ x + x \circ y + y \circ y - 2 x \circ y \circ y\) $\quad$ $\quad$
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle 2 x \circ x \circ y - 2 y \circ x \circ y + 4 x \circ y \circ x \circ y\) $\quad$ Product of Integral Multiples $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x + x \circ y - 2 x \circ y + x \circ y + y - 2 x \circ y - 2 x \circ y - 2 x \circ y + 4 x \circ y\) $\quad$ $\circ$ is commutative, $x, y$ are idempotent $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x + y - 2 x \circ y\) $\quad$ Integral Multiple Distributes over Ring Addition $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \oplus y\) $\quad$ Definition of $\oplus$ $\quad$

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

$\Box$


Axiom $(A1)$: Associativity of $\oplus$

Let $x, y, z \in A$.

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

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

This is shown by the following computation:

\(\displaystyle \paren {x \oplus y} \oplus z\) \(=\) \(\displaystyle \paren {x \oplus y} + z - 2 \paren {x \oplus y} \circ z\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x + y - 2 x \circ y + z - 2 \paren {x + y - 2 x \circ y} \circ z\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x + y + z - 2 x \circ y - 2 x \circ z - 2 y \circ z + 4 x \circ y \circ z\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x + \paren {y + z - 2 y \circ z} - 2 x \circ \paren {y + z - 2 y \circ z}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x + \paren {y \oplus z} - 2 x \circ \paren {y \oplus z}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \oplus \paren {y \oplus z}\) $\quad$ $\quad$


$\Box$


Axiom $(A2)$: Commutativity of $\oplus$

Let $x, y \in A$.

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

$x \oplus y = y \oplus x$

This is shown by the following computation:

\(\displaystyle x \oplus y\) \(=\) \(\displaystyle x + y - 2 x \circ y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle y + x - 2 y \circ x\) $\quad$ $+, \circ$ are commutative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle y \oplus x\) $\quad$ $\quad$

$\Box$


Axiom $(A3)$: 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 $(A2)$ was already verified above.


Now:

\(\displaystyle x \oplus 0_R\) \(=\) \(\displaystyle x + 0_R - 2 x \circ 0_R\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x\) $\quad$ Ring Product with Zero $\quad$

as desired.

$\Box$


Axiom $(A4)$: 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:

\(\displaystyle x \oplus x\) \(=\) \(\displaystyle x + x - 2 x \circ x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2 x - 2 x\) $\quad$ $x$ is an idempotent element $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0_R\) $\quad$ $\quad$

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

$\Box$


Axiom $(M0)$: Closure for $\circ$

Let $x, y \in A$.

It is to be shown that:

$x \circ y \in A$

i.e. that $x \circ y$ is idempotent.

We have the following computation:

\(\displaystyle \paren {x \circ y} \circ \paren {x \circ y}\) \(=\) \(\displaystyle x \circ x \circ y \circ y\) $\quad$ $\circ$ is associative and commutative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ y\) $\quad$ $x, y$ are idempotent elements $\quad$

$\Box$


Axiom $(M1)$: Associativity of $\circ$

Immediate from Restriction of Associative Operation is Associative.

$\Box$


Axiom $(D)$: 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:

\(\displaystyle x \circ \paren {y \oplus z}\) \(=\) \(\displaystyle x \circ \paren {y + z - 2 y \circ z}\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ y + x \circ z - 2 x \circ y \circ z\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ y + x \circ z - 2 x \circ x \circ y \circ z\) $\quad$ $x$ is an idempotent element $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ y + x \circ z - 2 x \circ y \circ x \circ z\) $\quad$ $\circ$ is commutative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {x \circ y} \oplus \paren {x \circ z}\) $\quad$ $\quad$

$\Box$


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.

$\blacksquare$


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