nyuuzyou/wikis huggingface dataset:
Ideals form Arithmetic Lattice

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below lattice.
Let $I = \struct {\map {\mathit {Ids} } L, \precsim}$ be an inclusion ordered set, where:
:$\map {\mathit {Ids} } L$ denotes the set of all ideals in $L$
:$\mathord \precsim = \mathord \subseteq \cap \paren {\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$
Then $I$ is an arithmetic lattice.
== Proof ==
We will prove that:
:$\forall x, y \in \map K I: \exists z \in \map K I: z \precsim x \land z \precsim y \land \forall v \in \map K I: v \precsim x \land v \precsim y \implies v \precsim z$
Let $x, y \in \map K I$.
By definition of compact subset:
:$x$ and $y$ are compact elements in $I$.
By Compact Element iff Principal Ideal and definition of principal ideal:
:$\exists a \in S: x = a^\preceq$
and
:$\exists b \in S: y = b^\preceq$
By Intersection of Semilattice Ideals is Ideal:
:$x \cap y \in \map {\mathit {Ids} } L$
Define $z = x \cap y$.
By Meet in Set of Ideals:
:$z = x \wedge_I y$
We will prove that
:$\paren {a \wedge b}^\preceq \subseteq z$
Let $c \in \paren {a \wedge b}^\preceq$
By definition of lower closure of element:
:$c \preceq a \wedge b$
By Meet Precedes Operands:
:$a \wedge b \preceq a$ and $a \wedge b \preceq b$
By definition of transitivity:
:$c \preceq a$ and $c \preceq b$
By definition of lower closure of element:
:$c \in x$ and $c \in y$
Thus by definition of intersection:
:$c \in z$
We will prove that:
:$z \subseteq \paren {a \wedge b}^\preceq$
Let $c \in z$.
By definition of intersection:
:$c \in x$ and $c \in y$
By definition of lower closure of element:
:$c \preceq a$ and $c \preceq b$
By definitions of infimum and lower bound:
:$c \preceq a \wedge b$
Thus by definition of lower closure of element:
:$c \in \paren {a \wedge b}^\preceq$
By definition of set equality:
:$z = \paren {a \wedge b}^\preceq$
By definition:
:$z$ is a principal ideal.
By Compact Element iff Principal Ideal:
:$z$ is a compact element in $I$.
Thus by definition of compact subset:
:$z \in \map K I$
Thus by Meet Precedes Operands:
:$z \precsim x$ and $z \precsim y$
Let $v \in \map K I$ such that
:$v \precsim x$ and $v \precsim y$
By definition of $\precsim$:
:$v \subseteq x$ and $v \subseteq y$
By Intersection is Largest Subset:
:$v \subseteq z$
Thus by definition of $\precsim$:
:$v \precsim z$
By definition:
:$\map K I$ forms a meet semilattice.
By Compact Subset is Join Subsemilattice:
:$\map K I$ forms a join semilattice.
By definition:
:$\map K I$ forms a lattice.
By Ideals form Algebraic Lattice:
:$I$ is an algebraic lattice.
Thus by Arithmetic iff Compact Subset form Lattice in Algebraic Lattice:
:$I$ is an arithmetic lattice.
== Sources ==
* 
* 
Category:Lattice Theory
Category:Continuous Lattices
Ideals form Complete Lattice

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\II = \struct {\map {\operatorname{Ids} } L, \subseteq}$ be an inclusion ordered set,
where $\map {\operatorname{Ids} } L$ denotes the set of all ideals in $L$.
Then $\II$ is complete lattice.
== Proof ==
Let $X \subseteq \map {\operatorname{Ids} } L$.
By Intersection of Semilattice Ideals is Ideal/Set of Sets:
:$\ds \bigcap X$ is an ideal.
By Intersection is Largest Subset/General Result:
:$\ds \forall A \in \map {\operatorname{Ids} } L: \paren {\forall I \in X: A \subseteq I} \iff A \subseteq \bigcap X$
Thus by definition:
:$X$ admits an infimum.
Thus by dual to Lattice is Complete iff it Admits All Suprema:
:$\II$ is complete lattice.
== Sources ==
* 
Category:Complete Lattices
Ideals of Division Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a division ring whose zero is $0_R$.
The only ideals of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself.
That is, $\struct {R, +, \circ}$ has no non-null proper ideals.
== Proof ==
From Null Ring is Ideal, $\set {0_R}$ is an ideal of $\struct {R, +, \circ}$, as $\struct {R, +, \circ}$, being a division ring, is also a ring.
By definition, every non-zero element of a division ring is a unit.
So $S \ne \set {0_R} \implies \exists x \in S: x \ne 0_R$ such that $r$ is a unit of $R$.
The result follows from Ideal of Unit is Whole Ring.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.2$: Corollary
Category:Ideal Theory
Ideals of Field

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Then $\struct {R, +, \circ}$ is a field the only ideals of $\struct {R, +, \circ}$ are $\struct {R, +, \circ}$ and $\set {0_R}$.
== Proof ==
=== Necessary Condition ===
Let $\struct {R, +, \circ}$ be a field.
The result follows from Field has 2 Ideals.
=== Sufficient Condition ===
Suppose that the only ideals of $\struct {R, +, \circ}$ are $\struct {R, +, \circ}$ and $\set {0_R}$.
The result follows from Commutative and Unitary Ring with 2 Ideals is Field
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.5$
* : $\S 2$: Exercise $12$
Category:Field Theory
Category:Ideal Theory
Category:Commutative Algebra
Ideals of P-adic Integers

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Then the ideals of $\Z_p$ are the principal ideals:
:$\text a) \quad \set 0$
:$\text b) \quad \forall k \in \N: p^k \Z_p$
=== Corollary ===
== Proof ==
Let $\nu_p: \Q_p \to \Z \cup \set {+\infty}$ denote the $p$-adic valuation on the $p$-adic numbers.
Let $\Z_p^\times$ denote the $p$-adic units.
Let $I \ne \set 0$ be a non-null ideal of $\Z_p$.
Hence:
:$\exists j \in I : \map {\nu_p} j < \infty$
Let:
:$k = \inf \set {\map {\nu_p} i : i \in I}$
Hence:
:$k \le j < \infty$
Let:
:$a \in I : a \ne 0 \land \map {\nu_p} a = k$
From P-adic Number is Power of p Times P-adic Unit:
:$\exists u \in \Z_p^\times : a = p^k u$
We have:
Let $b \in I$.
=== Case 1 : $b \ne 0$ ===
Let:
:$w = \map {\nu_p} b$
Then:
:$k \le w < \infty$
From P-adic Number is Power of p Times P-adic Unit:
:$\exists u' \in \Z_p^\times : b = p^w u'$
We have:
=== Case 2 : $b = 0$ ===
We have:
In either case:
$b \in p^k \Z_p$
It follows that:
:$I \subseteq p^k \Z_p$ 
By the definition of set equality:
:$I = p^k \Z_p$
Category:P-adic Integers
Category:Ideals of P-adic Integers
Ideals of P-adic Integers/Corollary

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p$.
Then:
:$\Z_p$ is a principal ideal domain
== Proof ==
From Ideals of P-adic Integers, all ideals of $\Z_p$ are the principal ideals:
:* $\quad\set 0$
:* $\quad\forall n \in \N : p^n \Z_p$
Hence $\Z_p$ is a principal ideal domain by definition.
== Sources ==
* : $\S 1.8$ Algebraic properties of $p$-adic integers: Proposition $1.46$
Category:Ideals of P-adic Integers
Ideals of Ring of Integers Modulo m

== Theorem ==
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {\Z_m, +, \times}$ denote the ring of integers modulo $m$.
The ideals of $\struct {\Z_m, +, \times}$ are of the form:
:$d \Z / m \Z$
where $d$ is a divisor of $m$.
== Proof ==
Let $J$ be an ideal of $\struct {\Z_m, +, \times}$.
Recall from Ring of Integers Modulo m is Ring that $\struct {\Z_m, +}$ is the additive group of integers modulo $m$.
From Integers Modulo $m$ under Addition form Cyclic Group, $\struct {\Z_m, +}$ is a cyclic group.
By definition of ideal, $\struct {J, +}$ is a subgroup of $\struct {\Z_m, +}$.
Let $\struct {G, +}$ be a subgroup of $\struct {\Z_m, +}$.
From Subgroup of Cyclic Group is Cyclic $\struct {G, +}$ is a cyclic group.
Hence by definition of cyclic group, $\struct {G, +}$ can be generated by $d$ and denoted $\gen d$, where $d \divides m$.
By Lagrange's Theorem, for a finite cyclic group of order $k$, the order of every subgroup is a divisor of $k$.
By Bijection from Divisors to Subgroups of Cyclic Group there is exactly one subgroup for each divisor.
It follows that all ideals of $\struct {\Z_m, +, \times}$ are of form $\gen d$, where $d$ is a positive divisor of $m$.
== Examples ==
Category:Ring of Integers Modulo m
Category:Ideal Theory
Category:Ideals of Ring of Integers Modulo m
Ideals of Ring of Integers Modulo m/Examples/Order 12

== Examples of Ideals of Ring of Integers Modulo m ==
Let $\struct {\Z_{12}, +, \times}$ denote the ring of integers modulo $12$.
The underlying sets of the ideals of $\struct {\Z_{12}, +, \times}$ are:
:$\set {0, 2, 4, 6, 8, 10}$
:$\set {0, 3, 6, 9}$
:$\set {0, 4, 8}$
:$\set {0, 6}$
== Sources==
* : Chapter $5$: Rings: $\S 21$. Ideals: Example $36$
Category:Ideals of Ring of Integers Modulo m
Ideals with Coprime Radicals are Coprime

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathfrak a, \mathfrak b \subseteq A$ be ideals.
Let their radicals be coprime:
:$\map \Rad {\mathfrak a} + \map \Rad {\mathfrak b} = \ideal 1$
Then $\mathfrak a$ and $\mathfrak b$ are coprime:
:$\mathfrak a + \mathfrak b = \ideal 1$
== Proof ==
We have:
By Unit Ideal iff Radical is Unit Ideal:
:$\mathfrak a + \mathfrak b = \ideal 1$
That is, $\mathfrak a$ and $\mathfrak b$ are coprime.
Category:Radical of Ideals
Idempotent Element/Examples/One is Idempotent for Multiplication

== Example of Idempotent Element ==
$1$ is idempotent under the operation of multiplication in the set of real numbers $\R$.
== Proof ==
From Non-Zero Real Numbers under Multiplication form Abelian Group, $\struct {\R_{\ne 0}, \times}$ forms an abelian group whose identity element is $1$.
The result follows from Identity Element is Idempotent.
== Sources ==
* 
* 
Category:Examples of Idempotence
Idempotent Element/Examples/Unit Matrix

== Example of Idempotent Element ==
Let $\mathbf I_n$ be the unit matrix of order $n$.
Then $\mathbf I_n$ is idempotent under the operation of (conventional) matrix multiplication.
== Proof ==
From Unit Matrix is Identity for Matrix Multiplication, $\mathbf I_n$ forms an identity element for matrix multiplication.
The result follows from Identity Element is Idempotent.
== Sources ==
* 
* 
Category:Examples of Idempotence
Idempotent Element/Examples/Zero is Idempotent for Addition

== Example of Idempotent Element ==
$0$ is idempotent under the operation of addition in the set of integers $\Z$, but no other element of $\Z$ is so.
== Proof ==
From Integers under Addition form Abelian Group, $\struct {\Z, +}$ forms an abelian group whose identity element is $0$.
The result follows from Identity is only Idempotent Element in Group.
Category:Examples of Idempotence
Idempotent Elements for Integer Multiplication

== Theorem ==
There are exactly two integers which are idempotent with respect to multiplication:
:$0 \times 0 = 0$
:$1 \times 1 = 1$
== Proof ==
The integers $\struct {\Z, +, \times}$ form an integral domain.
By definition of integral domain, therefore, the integers form a ring with no (proper) zero divisors.
The result follows from Idempotent Elements of Ring with No Proper Zero Divisors.
== Sources ==
* : $\S 1.5$
Category:Integer Multiplication
Idempotent Elements form Submonoid of Commutative Monoid

== Theorem ==
Let $\struct {S, \circ}$ be a commutative monoid.
Let $e \in S$ be the identity element of $\struct {S, \circ}$.
Let $I$ be the set of all elements of $S$ that are idempotent under $\circ$.
That is:
:$I = \set {x \in S: x \circ x = x}$
Then $\struct {I, \circ}$ is a submonoid of $\struct {S, \circ}$ with identity $e$.
== Proof ==
By Idempotent Elements form Subsemigroup of Commutative Semigroup, $\struct {I, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
By Identity Element is Idempotent:
:$e \in I$
By Identity of Algebraic Structure is Preserved in Substructure, $e$ is an identity of $\struct {I, \circ}$.
Since $\struct {T, \circ}$ is a semigroup and has an identity, $\struct {T, \circ}$ is a monoid.
Since $T \subseteq S$, $\struct {T, \circ}$ is a submonoid of $\struct {S, \circ}$ with identity $e$.
Category:Commutative Monoids
Idempotent Elements form Subsemigroup of Commutative Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup such that $\circ$ is commutative.
Let $I$ be the set of all elements of $S$ that are idempotent under $\circ$.
That is:
:$I = \set {x \in S: x \circ x = x}$
Then $\struct {I, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
== Proof 1 ==
== Proof 2 ==
Category:Subsemigroups
Category:Idempotence
Category:Idempotent Elements form Subsemigroup of Commutative Semigroup
Idempotent Elements form Subsemigroup of Commutative Semigroup/Proof 1

== Theorem ==
== Proof ==
By Subsemigroup Closure Test we need only show that:
:For all $x, y \in I$: $x \circ y \in I$.
That is:
:$\paren {x \circ y} \circ \paren {x \circ y} = x \circ y$
We reason as follows:
Hence the result.
Category:Idempotent Elements form Subsemigroup of Commutative Semigroup
Idempotent Elements form Subsemigroup of Commutative Semigroup/Proof 2

== Theorem ==
== Proof ==
By Subsemigroup Closure Test we need only show that:
: For all $x, y \in I$: $x \circ y \in I$.
As $x, y \in I$, they are idempotent.
We have that $\circ$ is commutative.
Thus, by definition, $x$ and $y$ commute.
From Product of Commuting Idempotent Elements is Idempotent, $\left({x \circ y}\right)$ is idempotent.
That is:
:$x \circ y \in I$
Category:Idempotent Elements form Subsemigroup of Commutative Semigroup
Idempotent Elements of Ring with No Proper Zero Divisors

== Theorem ==
Let $\struct {R, +, \circ}$ be a non-null ring with no (proper) zero divisors.
Let $x \in R$.
Then:
:$x \circ x = x \iff x \in \set {0_R, 1_R}$
That is, the only elements of $\struct {R, \circ}$ that are idempotent are zero and unity.
== Proof ==
We have $0_R \circ 0_R = 0_R$, so that sorts out one element.
Let $R^*$ be the ring $R$ without the zero: $R^* = R \setminus \set {0_R}$.
By Ring Element is Zero Divisor iff not Cancellable, all elements of $R^*$ that are not zero divisors are cancellable.
Therefore all elements of $R^*$ are cancellable.
Suppose $x \circ x = x$ where $x \ne 0_R$.
Then $x \circ x = x = x \circ 1_R$.
As $x$ is cancellable, the result follows.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.3$
Category:Ring Theory
Idempotent Magma Element forms Singleton Submagma

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $x \in S$ be an idempotent element of $\struct {S, \circ}$.
Then $\struct {\set x, \circ}$ is a submagma of $\struct {S, \circ}$.
== Proof ==
By Singleton of Element is Subset:
:$x \in S \iff \set x \subseteq S$
By the definition of idempotence:
:$x \circ x = x \in \set x$
Thus $\set x$ is a subset of $S$ which is closed under $\circ$.
By the definition of submagma, the result follows.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.3$
Category:Magmas
Idempotent Non-Trivial Quasigroup is Not a Loop

== Theorem ==
Let $\left({S, \circ}\right)$ be an idempotent quasigroup whose underlying set $S$ comprises more than one element.
Then $\left({S, \circ}\right)$ is not an algebra loop, that is, it has no identity element.
== Proof ==
 $\left({S, \circ}\right)$ has an identity element $e$.
Then by Identity Element is Idempotent:
: $e \circ e = e$
Consider $e' \in S$ where $e' \ne e$.
Since $e$ is an identity element:
:$e' \circ e = e'$
Also, by assumption, $\circ$ is idempotent, so:
:$e' \circ e' = e'$
Then by the definition of a quasigroup, the left regular representation:
:$\lambda_{e'}: S \to S, x \mapsto e' \circ x$
is a permutation, and in particular, an injection.
Hence $e' = e$, which contradicts our assumption that $e' \ne e$.
So $\left({S, \circ}\right)$ has no identity element and is not an algebra loop.
Category:Quasigroups
Category:Algebra Loops
Category:Idempotence
Idempotent Ring has Characteristic Two

== Theorem ==
Let $\struct {R, +, \circ}$ be an idempotent non-null ring.
Denote with $0_R$ the zero of $R$.
Then $\struct {R, +, \circ}$ has characteristic $2$.
=== Corollary ===
== Proof ==
Let $x \in R$.
Then:
Hence the result.
== Sources ==
* : $\S 1$
Category:Idempotent Rings
Idempotent Ring has Characteristic Two/Corollary

== Theorem ==
Let $\struct {R, +, \circ}$ be an idempotent non-null ring.
Then:
:$\forall x \in R: -x = x$
== Proof ==
Let $0_R$ denote the zero of $R$.
Let $x \in R$.
Then:
Hence the result.
== Sources ==
* : Chapter $9$: Rings: Exercise $1$
Category:Idempotent Rings
Idempotent Ring is Commutative

== Theorem ==
Let $\struct {R, +, \circ}$ be an idempotent ring.
Denote with $0_R$ the zero of $R$.
Then $\struct {R, +, \circ}$ is a commutative ring.
== Proof ==
Let $x, y \in R$.
Then:
Subtracting $x + y$ from both sides yields:
Finally, subtracting $y \circ x$ from both sides, we obtain:
:$x \circ y = y \circ x$
and conclude $R$ is a commutative ring.
== Sources ==
* : $\S 1$
* : Chapter $9$: Rings: Exercise $1$
Category:Idempotent Rings
Idempotent Semigroup/Examples/Relation induced by Inverse Element

== Example of Idempotent Semigroup ==
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $\RR$ be the relation on $S$ defined as:
:$\forall a, b \in S: a \mathrel \RR b \iff \paren {a \circ b \circ a = a \land b \circ a \circ b = b}$
That is, such that $a$ is the inverse of $b$ and $b$ is the inverse of $a$.
=== Properties ===
The following properties can be deduced:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19$
Category:Idempotent Semigroups
Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties

== Example of Idempotent Semigroup ==
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $\RR$ be the relation on $S$ defined as:
:$\forall a, b \in S: a \mathrel \RR b \iff \paren {a \circ b \circ a = b \land b \circ a \circ b = a}$
That is, such that $a$ is the inverse of $b$ and $b$ is the inverse of $a$.
Then the following properties can be deduced:
=== Property $1$ ===
=== Property $2$ ===
=== Property $3$ ===
=== Property $4$ ===
=== Property $5$ ===
=== Property $6$ ===
Category:Idempotent Semigroups
Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties/1

== Example of Idempotent Semigroup ==
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $\RR$ be the relation on $S$ defined as:
:$\forall a, b \in S: a \mathrel \RR b \iff \paren {a \circ b \circ a = a \land b \circ a \circ b = b}$
That is, such that $a$ is the inverse of $b$ and $b$ is the inverse of $a$.
Let $x \circ y = y$ and $y \circ x = x$.
Then for all $z \in S$:
:$\paren {z \circ x} \mathrel \RR \paren {z \circ y}$
and:
:$\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
== Proof ==
From we take it for granted that $\struct {S, \circ}$ is closed under $\circ$.
From we take it for granted that $\circ$ is associative.
Hence parentheses will be used whenever it makes groupings of operations more clear.
We have:
Similarly:
Hence by definition of $\RR$:
:$\forall z \in S: \paren {z \circ x} \mathrel \RR \paren {z \circ y}$
Then:
and:
Hence by definition of $\RR$:
:$\forall z \in S: \paren {x \circ z} \mathrel \RR \paren {y \circ z}$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19 \ \text {(b)}$
Category:Idempotent Semigroups
Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties/2

== Example of Idempotent Semigroup ==
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $\RR$ be the relation on $S$ defined as:
:$\forall a, b \in S: a \mathrel \RR b \iff \paren {a \circ b \circ a = a \land b \circ a \circ b = b}$
That is, such that $a$ is the inverse of $b$ and $b$ is the inverse of $a$.
Let $x \circ y = x$ and $y \circ x = y$.
Then for all $z \in S$:
:$\paren {x \circ z} \mathrel \RR \paren {y \circ z}$
and:
:$\paren {z \circ x} \mathrel \RR \paren {z \circ y}$
== Proof ==
From we take it for granted that $\circ$ is associative.
Hence parentheses will be employed whenever it makes groupings of operations more clear.
We have:
Similarly:
Hence by definition of $\RR$:
:$\forall z \in S: \paren {x \circ z} \mathrel \RR \paren {y \circ z}$
Then:
and:
Hence by definition of $\RR$:
:$\forall z \in S: \paren {z \circ x} \mathrel \RR \paren {z \circ y}$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19 \ \text {(d)}$
Category:Idempotent Semigroups
Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties/3

== Example of Idempotent Semigroup ==
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $\RR$ be the relation on $S$ defined as:
:$\forall a, b \in S: a \mathrel \RR b \iff \paren {a \circ b \circ a = a \land b \circ a \circ b = b}$
That is, such that $a$ is the inverse of $b$ and $b$ is the inverse of $a$.
Let:
Then:
:$x \mathrel \RR z$
== Proof ==
From we take it for granted that $\circ$ is associative.
Hence parentheses will be employed whenever it makes groupings of operations more clear.
Recall:
From $x \circ y = y$ and $y \circ x = x$
:$(1): \quad \forall z \in S: z \circ x \circ z \circ y = z \circ y$
using Properties of Idempotent Semigroup: $1$
From $y \circ z = y$ and $z \circ y = z$ 
:$(2): \quad \forall x \in S: y \circ x \circ z \circ x = y \circ x$
using Properties of Idempotent Semigroup: $2$, exchanging $z$ and $x$
We have:
and:
Hence the result, by definition of $\RR$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19 \ \text {(e)}$
Category:Idempotent Semigroups
Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties/4

== Example of Idempotent Semigroup ==
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $\RR$ be the relation on $S$ defined as:
:$\forall a, b \in S: a \mathrel \RR b \iff \paren {a \circ b \circ a = a \land b \circ a \circ b = b}$
That is, such that $a$ is the inverse of $b$ and $b$ is the inverse of $a$.
$\RR$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Thus $\RR$ is seen to be reflexive.
=== Symmetry ===
Thus $\RR$ is seen to be symmetric.
=== Transitivity ===
Let $a \mathrel \RR b$ and $b \mathrel \RR c$.
Thus we have:
and:
Let:
We have:
and:
Hence using Properties of Idempotent Semigroup: $1$:
Then we have:
and:
Hence using Properties of Idempotent Semigroup: $2$:
Thus we have:
Hence by Idempotent Semigroup: Relation induced by Inverse Element: $3$:
:$x \mathrel \RR z$
That is:
Thus $\RR$ is seen to be transitive.
$\RR$ has been shown to be reflexive, symmetric and transitive.
Hence by definition $\RR$ is an equivalence relation.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19 \ \text {(f)}$
Category:Idempotent Semigroups
Idempotent Semigroup/Examples/Relation induced by Inverse Element/Properties/5

== Example of Idempotent Semigroup ==
Let $\struct {S, \circ}$ be an idempotent semigroup.
Let $\RR$ be the relation on $S$ defined as:
:$\forall a, b \in S: a \mathrel \RR b \iff \paren {a \circ b \circ a = a \land b \circ a \circ b = b}$
That is, such that $a$ is the inverse of $b$ and $b$ is the inverse of $a$.
$\RR$ is a congruence relation for $\circ$.
== Proof ==
From Idempotent Semigroup: Relation induced by Inverse Element: $4$:
:$\RR$ is an equivalence relation.
Let $a, b \in S$ be arbitrary such that $a \mathrel \RR b$.
Then:
Then we have:
Then:
As $a$ and $b$ are arbitrary, the result follows by Equivalence Relation is Congruence iff Compatible with Operation.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.19 \ \text {(g)}$
Category:Idempotent Semigroups