Ring of Idempotents is Idempotent Ring

From ProofWiki
Jump to navigation Jump to search

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.


Ring Axiom $\text A0$: Closure under Addition

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:

\(\ds \paren {x \oplus y} \circ \paren {x \oplus y}\) \(=\) \(\ds \paren {x + y - 2 x \circ y} \circ \paren {x + y - 2 x \circ y}\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds \paren {x + y - 2 x \circ y} \circ x + \paren {x + y - 2 x \circ y} \circ y - \paren {x + y - 2 x \circ y} \circ \paren {2 x \circ y}\) Ring Axiom $\text D$: Distributivity of Product over Addition
\(\ds \) \(=\) \(\ds x \circ x + y \circ x - \paren {2 x \circ y} \circ x + x \circ y + y \circ y - \paren {2 x \circ y} \circ y\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds x \circ \paren {2 x \circ y} - y \circ \paren {2 x \circ y} + \paren {2 x \circ y} \circ \paren {2 x \circ y}\) Ring Axiom $\text D$: Distributivity of Product over Addition
\(\ds \) \(=\) \(\ds x \circ x + y \circ x - 2 x \circ y \circ x + x \circ y + y \circ y - 2 x \circ y \circ y\)
\(\ds \) \(\) \(\, \ds - \, \) \(\ds 2 x \circ x \circ y - 2 y \circ x \circ y + 4 x \circ y \circ x \circ y\) Product of Integral Multiples
\(\ds \) \(=\) \(\ds x + x \circ y - 2 x \circ y + x \circ y + y - 2 x \circ y - 2 x \circ y - 2 x \circ y + 4 x \circ y\) $\circ$ is commutative, $x, y$ are idempotent
\(\ds \) \(=\) \(\ds x + y - 2 x \circ y\) Integral Multiple Distributes over Ring Addition
\(\ds \) \(=\) \(\ds x \oplus y\) Definition of $\oplus$

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

$\Box$


Ring Axiom $\text A1$: Associativity of Addition

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:

\(\ds \paren {x \oplus y} \oplus z\) \(=\) \(\ds \paren {x \oplus y} + z - 2 \paren {x \oplus y} \circ z\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds x + y - 2 x \circ y + z - 2 \paren {x + y - 2 x \circ y} \circ z\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds x + y - 2 x \circ y + z - 2 x \circ z - 2 y \circ z + 4 x \circ y \circ z\) Ring Axiom $\text D$: Distributivity of Product over Addition
\(\ds \) \(=\) \(\ds x + y + z - 2 y \circ z - 2 x \circ y - 2 x \circ z + 4 x \circ y \circ z\) Ring Axiom $\text A2$: Commutativity of Addition
\(\ds \) \(=\) \(\ds x + \paren { y + z - 2 y \circ z} - 2 x \circ y - 2 x \circ z + 4 x \circ y \circ z\) Ring Axiom $\text A1$: Associativity of Addition
\(\ds \) \(=\) \(\ds x + \paren {y + z - 2 y \circ z} - 2 x \circ \paren {y + z - 2 y \circ z}\) Ring Axiom $\text D$: Distributivity of Product over Addition
\(\ds \) \(=\) \(\ds x + \paren {y \oplus z} - 2 x \circ \paren {y \oplus z}\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds x \oplus \paren {y \oplus z}\) Definition of $\oplus$

$\Box$


Ring Axiom $\text A2$: Commutativity of Addition

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:

\(\ds x \oplus y\) \(=\) \(\ds x + y - 2 x \circ y\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds y + x - 2 y \circ x\) Ring Axiom $\text A2$: Commutativity of Addition and \circ is commutative
\(\ds \) \(=\) \(\ds y \oplus x\) Definition of $\oplus$

$\Box$


Ring Axiom $\text A3$: Identity for Addition

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 Ring Axiom $\text A2$: Commutativity of Addition was already verified above.


Now:

\(\ds x \oplus 0_R\) \(=\) \(\ds x + 0_R - 2 x \circ 0_R\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds x\) Ring Product with Zero

as desired.

$\Box$


Ring Axiom $\text A4$: Inverses for Addition

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:

\(\ds x \oplus x\) \(=\) \(\ds x + x - 2 x \circ x\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds 2 x - 2 x\) $x$ is an idempotent element
\(\ds \) \(=\) \(\ds 0_R\) Ring Subtraction equals Zero iff Elements are Equal

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

$\Box$


Ring Axiom $\text M0$: Closure under Product

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:

\(\ds \paren {x \circ y} \circ \paren {x \circ y}\) \(=\) \(\ds x \circ \paren {y \circ x} \circ y\) Ring Axiom $\text M1$: Associativity of Product
\(\ds \) \(=\) \(\ds x \circ \paren {x \circ y} \circ y\) $\circ$ is commutative
\(\ds \) \(=\) \(\ds \paren {x \circ x} \circ \paren {y \circ y}\) Ring Axiom $\text M1$: Associativity of Product
\(\ds \) \(=\) \(\ds x \circ y\) $x, y$ are idempotent elements

$\Box$


Ring Axiom $\text M1$: Associativity of Product

Immediate from Restriction of Associative Operation is Associative.

$\Box$


Ring Axiom $\text D$: Distributivity of Product over Addition

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:

\(\ds x \circ \paren {y \oplus z}\) \(=\) \(\ds x \circ \paren {y + z - 2 y \circ z}\) Definition of $\oplus$
\(\ds \) \(=\) \(\ds x \circ y + x \circ z - 2 x \circ y \circ z\) Ring Axiom $\text D$: Distributivity of Product over Addition
\(\ds \) \(=\) \(\ds x \circ y + x \circ z - 2 x \circ x \circ y \circ z\) $x$ is an idempotent element
\(\ds \) \(=\) \(\ds x \circ y + x \circ z - 2 x \circ y \circ x \circ z\) $\circ$ is commutative
\(\ds \) \(=\) \(\ds \paren {x \circ y} \oplus \paren {x \circ z}\) Definition of $\oplus$

$\Box$


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.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Ring_of_Idempotents_is_Idempotent_Ring&oldid=628037"