nyuuzyou/wikis huggingface dataset:
Ideal Quotient is Ideal

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathfrak a, \mathfrak b$ be ideals of $R$.
Then the ideal quotient $\ideal {\mathfrak a : \mathfrak b}$ is indeed an ideal.
== Proof ==
We shall check $(1)$-$(3)$ of Test for Ideal.
=== (1) ===
We have:
:$0 \mathfrak b = \set 0 \subseteq \mathfrak a$
That is:
:$0 \in \ideal {\mathfrak a : \mathfrak b}$
=== (2) ===
Let $x,y \in \ideal {\mathfrak a : \mathfrak b}$.
Then:
:$x \mathfrak b \subseteq a$
and:
:$\paren {-y} \mathfrak b \subseteq a$
Therefore:
:$\paren {x - y} \mathfrak b \subseteq x \mathfrak b + \paren {-y} \mathfrak b \subseteq a$
That is:
:$x - y \in \ideal {\mathfrak a : \mathfrak b}$
=== (3) ===
Let $x \in \ideal {\mathfrak a : \mathfrak b}$.
Let $ r \in R$.
Then:
:$\paren{r x} \mathfrak b \subseteq r \paren {x \mathfrak b} \subseteq r \mathfrak a \subseteq \mathfrak a$
That is:
:$r x \in \ideal {\mathfrak a : \mathfrak b}$
Category:Commutative Algebra
Category:Ideal Theory
Ideal induced by Congruence Relation defines that Congruence

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\EE$ be a congruence relation on $R$.
Let $J = \eqclass {0_R} \EE$ be the ideal induced by $\EE$.
Then the equivalence defined by the coset space $\struct {R, +} / \struct {J, +}$ is $\EE$ itself.
== Proof ==
Let $J = \eqclass {0_R} \EE$.
From Congruence Relation on Ring induces Ideal, we have that $J$ is an ideal of $R$.
From Ideal is Additive Normal Subgroup, we have that $\struct {J, +}$ is a normal subgroup of $\struct {R, +}$.
From Normal Subgroup induced by Congruence Relation defines that Congruence, the equivalence defined by the partition $\struct {R, +} / \struct {J, +}$ is $\EE$.
As $\EE$ was the congruence relation on $R$ that was originally posited, we already know that it is compatible with $\circ$.
Thus the equivalence defined by $J$ is the same congruence relation on $R$ that gave rise to $J$ to start with.
Hence the result.
== Also see ==
* Congruence Relation on Ring induces Ideal
* Quotient Ring is Ring
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorems $22.2$
Category:Congruence Relations
Category:Ideal Theory
Category:Quotient Rings
Ideal induces Congruence Relation on Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $J$ be an ideal of $R$
Then $J$ induces a congruence relation $\EE_J$ on $R$ such that $\struct {R / J, +, \circ}$ is a quotient ring.
== Proof ==
From Ideal is Additive Normal Subgroup, we have that $\struct {J, +}$ is a normal subgroup of $\struct {R, +}$.
Let $x \mathop {\EE_J} y$ denote that $x$ and $y$ are in the same coset, that is:
:$x \mathop {\EE_J} y \iff x + N = y + N$
From Congruence Modulo Normal Subgroup is Congruence Relation, $\EE_J$ is a congruence relation for $+$.
Now let $x \mathop {\EE_J} x', y \mathop {\EE_J} y'$.
By definition of congruence modulo $J$:
:$x + \paren {-x'} \in J$
:$y + \paren {-y'} \in J$
Then:
:$x \circ y + \paren {-x' \circ y'} = \paren {x + \paren {-x'} } \circ y + x' \circ \paren {y + \paren {-y'} } \in J$
demonstrating that $\EE_J$ is a congruence relation for $\circ$.
Hence the result by definition of quotient ring.
== Also see ==
* Congruence Relation on Ring induces Ideal
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorems $22.3$
Category:Congruence Relations
Category:Ideal Theory
Category:Quotient Rings
Ideal is Additive Normal Subgroup

== Theorem ==
Let $\left({R, +, \circ}\right)$ be a ring.
Let $J$ be an ideal of $R$.
Then $\left({J, +}\right)$ is a normal subgroup of $\left({R, +}\right)$.
== Proof ==
As $J$ is an ideal, $\left({J, +}\right)$ is a subgroup of $\left({R, +}\right)$.
By definition of a ring, $\left({R, +}\right)$ is abelian.
The result follows from Subgroup of Abelian Group is Normal.
Category:Ideal Theory
Ideal is Bimodule over Ring

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Let $J \subseteq R$ be an ideal of $R$.
Let $\circ_l : R \times J \to J$ be the restriction of $\times$ to $R \times J$.
Let $\circ_r : J \times R \to J$ be the restriction of $\times$ to $J \times R$.
Then $\struct {J, +, \circ_l, \circ_r}$ is a bimodule over $\struct {R, +, \times}$.
=== Corollary ===
== Proof ==
By definition of an ideal, $J$ is both a left ideal and a right ideal.
From Left Ideal is Left Module over Ring then $\struct {J, +, \circ_l}$ is a left module.
From Right Ideal is Right Module over Ring then $\struct {J, +, \circ_r}$ is a right module.
Then:
Hence $\struct {J, +, \circ_l, \circ_r}$ is a bimodule over $\struct {R, +, \times}$ by definition.
== Also see ==
* Ring is Bimodule over Ring
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Module Theory
Category:Ideal Theory
Ideal is Bimodule over Ring/Ring is Bimodule over Ring

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Then $\struct {R, +, \times, \times}$ is a bimodule over $\struct {R, +, \times}$.
== Proof ==
From Ring is Ideal of Itself and Ideal is Bimodule over Ring, $\struct {R, +, \times, \times}$ is a bimodule over $\struct {R, +, \times}$.
== Also see ==
* Ideal is Bimodule over Ring
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Module Theory
Category:Ring Theory
Ideal is Contained in Contraction of Extension

== Theorem ==
Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak a \subseteq A$ be an ideal.
Then $\mathfrak a$ is contained in the contraction of its extension by $f$:
:$\mathfrak a \subseteq \mathfrak a^{ec}$
== Proof ==
By definition of extension and generated ideal:
:$f \sqbrk {\mathfrak a} \subseteq \mathfrak a^e$
By Subset of Domain is Subset of Preimage of Image:
:$\mathfrak a \subseteq f^{-1} \sqbrk {f \sqbrk {\mathfrak a}}$
By Preimage of Subset is Subset of Preimage:
:$\mathfrak a \subseteq f^{-1} \sqbrk {f \sqbrk {\mathfrak a }} \subseteq f^{-1} \sqbrk {\mathfrak a^e}$
Category:Ideal Theory
Category:Commutative Algebra
Category:Ring Theory
Ideal is Dual of Filter (Order Theory)

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$.
The following are dual statements:
:$T$ is a filter of $\struct {S, \preceq}$
:$T$ is an ideal of $\struct {S, \preceq}$
== Proof ==
By definition of filter:
:$T$ is a filter of $\struct {S, \preceq}$
:
By the duality principle, the dual of this statement is:
By definition of ideal:
:$T$ is an ideal of $\struct {S, \preceq}$
Category:Dual Pairs (Order Theory)
Category:Order Theory
Ideal is Filter in Dual Ordered Set

== Theorem ==
Let $P = \struct {S, \preceq}$ be an ordered set.
Let $X$ be a subset of $S$.
Then
:$X$ is ideal in $P$
:$X$ is filter in $P^{-1}$
where $P^{-1} = \struct {S, \succeq}$ denotes the dual of $P$.
== Proof ==
=== Sufficient Condition ===
Let $X$ be ideal in $P$.
By definition of ideal in ordered set:
:$X$ is non-empty directed lower.
By definition of directed:
:$\forall x, y \in X: \exists z \in X: x \preceq z \land y \preceq z$
Then
:$\forall x, y \in X: \exists z \in X: z \succeq x \land z \succeq y$
By definition
:$X$ is filtered in $P^{-1}$
By definition of lower section:
:$\forall x \in X, y \in S: y \preceq x \implies y \in X$
Then
:$\forall x \in X, y \in S: x \succeq y \implies y \in X$
By definition
:$X$ is an upper section in $P^{-1}$.
Thus by definition of filter in ordered set
:$X$ is a filter in $P^{-1}$.
=== Necessary Condition ===
Let $X$ be a filter in $P^{-1}$.
By definition of filter in ordered set
:$X$ is non-empty, filtered in $P^{-1}$, and upper in $P^{-1}$.
By definition of filtered in $P^{-1}$: 
:$\forall x, y \in X: \exists z \in X: z \succeq x \land z \succeq y$
Then
:$\forall x, y \in X: \exists z \in X: x \preceq z \land y \preceq z$
By definition:
:$X$ is directed.
By definition of upper section in $P^{-1}$:
:$\forall x \in X, y \in S: x \succeq y \implies y \in X$
Then:
:$\forall x \in X, y \in S: y \preceq x \implies y \in X$
By definition:
:$X$ is a lower section.
Thus by definition of ideal in ordered set:
:$X$ is an ideal in $P$.
== Sources ==
* 
* 
Category:Order Theory
Category:Dual Orderings
Ideal is Subring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring, and let $J$ be an ideal of $R$.
Then $J$ is a subring of $R$.
== Proof ==
This follows directly from the definition of an ideal and Subring Test.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
Category:Ideal Theory
Ideal is Unit Ideal iff Includes Unity

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathfrak a$ be an ideal of $A$.
Then:
:$\mathfrak a = A \iff 1 \in \mathfrak a$
where $1$ denotes the unity of $A$.
== Proof ==
$\implies$ is trivial.
To see $\impliedby$, suppose $1 \in \mathfrak a$.
Let $a \in A$ be arbitrary.
Then:
Category:Commutative Algebra
Category:Ideal Theory
Ideal of Algebra over Field Embeds into Unitization as Ideal

== Theorem ==
Let $K$ be a field.
Let $A$ be an algebra over $K$ that is not unital. 
Let $I$ be an ideal of $A$.
Let $A_+$ be the unitization of $A$.
Let:
:$I_0 = \set {\tuple {x, 0_K} : x \in I}$
Then $I_0$ is an ideal of $A_+$. 
== Proof ==
From the definition of an ideal, $I$ is a vector subspace of $A$. 
Hence from Vector Subspace of Algebra over Field Embeds into Unitization as Vector Subspace, $I_0$ is a vector subspace of $A_+$. 
Let $\tuple {x, 0_K} \in I$ and $\tuple {y, \lambda} \in A_+$. 
Then we have:
:$\tuple {x, 0_K} \tuple {y, \lambda} = \tuple {x y + \lambda x, 0_K}$
Since $I$ is an ideal, we have $x y \in I$.
Since $I$ is a vector subspace of $A$, we have $x y + \lambda x \in I$.
Hence we have $\tuple {x, 0_K} \tuple {y, \lambda} \in I_0$.
Similarly we have:
:$\tuple {y, \lambda} \tuple {x, 0_K} = \tuple {y x + \lambda x, 0_K}$
Since $I$ is an ideal, we have $y x \in I$.
Since $I$ is a vector subspace of $A$, we have $y x + \lambda x \in I$.
Hence we have $\tuple {y, \lambda} \tuple {x, 0_K} \in I_0$.
Category:Unitizations of Algebras over Fields
Ideal of External Direct Sum of Rings

== Theorem ==
Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be rings.
Let 
:$\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$
be their direct product.
For each $k \in \closedint 1 n$, let:
:${R_k}' = \set {\tuple {x_1, \ldots, x_n} \in R: \forall j \ne k: x_j = 0}$
Then:
:$\forall k \in \closedint 1 n: {R_k}'$ is an ideal of $R$.
== Proof ==
Let $y = \tuple {y_1, y_2, \ldots, y_n} \in R$.
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in R_k$
By definition of direct product, we have:
:$x \circ y = \tuple {x_1 \circ y_1, x_2 \circ y_2, \ldots, x_n \circ y_n}$
:$y \circ x = \tuple {y_1 \circ x_1, y_2 \circ x_2, \ldots, y_n \circ x_n}$
But we have:
:$\forall j \ne k: x_j \circ y_j = 0 = y_j \circ x_j$
because $x_j = 0$.
So it follows that $x \circ y \in R_k$ and $y \circ x \in R_k$.
Hence ${R_k}'$ is an ideal of $R$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.8$
* : $\S 3.1$: Direct sums
Category:Ideal Theory
Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries

== Example of Ideal of Ring ==
Let $R$ be the set of all order $2$ square matrices of the form $\begin{pmatrix} x & y \\ 0 & z \end{pmatrix}$ with $x, y, z \in \R$.
Let $S$ be the set of all order $2$ square matrices of the form $\begin{pmatrix} x & y \\ 0 & 0 \end{pmatrix}$ with $x, y \in \R$.
Then $R$ is a ring and $S$ is an ideal of $R$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
Category:Examples of Ideals of Rings
Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries/Corollary

== Example of Ideal of Ring ==
Let $R$ be the set of all order $2$ square matrices of the form $\begin {pmatrix} x & y \\ 0 & z \end {pmatrix}$ with $x, y, z \in \R$.
Let $S$ be the set of all order $2$ square matrices of the form $\begin {pmatrix} x & y \\ 0 & 0 \end {pmatrix}$ with $x, y \in \R$.
Then:
:$R / S \cong \R$
where:
:$R / S$ is the quotient ring of $R$ by $S$
:$\cong$ denotes ring isomorphism.
== Proof ==
From Ideal of Ring: Order 2 Matrices with some Zero Entries:
:$S$ is an ideal of $R$
Having defined the ring homomorphism $\phi: R \to \R$:
:$\forall \mathbf A \in R: \map \phi {\begin {pmatrix} x & y \\ 0 & z \end {pmatrix} } = z$
from the First Ring Isomorphism Theorem:
:$\Img \phi \cong R / \map \ker \phi$
from which follows the result.
== Sources ==
* : Chapter $9$: Rings: Exercise $10$
Category:Examples of Ideals of Rings
Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries/Proof 1

== Example of Ideal of Ring ==
== Proof ==
Let $\begin{pmatrix} x_1 & y_1 \\ 0 & z_1 \end{pmatrix}, \begin{pmatrix} x_2 & y_2 \\ 0 & z_2 \end{pmatrix} \in R$.
Then:
Thus by the Subring Test $R$ is a subring of the ring of order $2$ matrices over $\R$.
We have that, for example, $\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in S$
Hence $S \ne \O$.
Let $\begin{pmatrix} x_1 & y_1 \\ 0 & z_1 \end{pmatrix}, \begin{pmatrix} x_2 & y_2 \\ 0 & z_2 \end{pmatrix} \in S$.
and so $S$ is closed under matrix subtraction.
Now let $\begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \in S$ for real $a, b$.
Let $\begin{pmatrix} x & y \\ 0 & z \end{pmatrix} \in R$.
Then we have:
and:
Thus, by the Test for Ideal, $S$ is an ideal of $R$.
== Sources ==
* : $\S 58.4$ Ideals: Worked example
Category:Examples of Ideals of Rings
Ideal of Ring/Examples/Order 2 Matrices with some Zero Entries/Proof 2

== Example of Ideal of Ring ==
== Proof ==
Consider the mapping $\phi: R \to \R$ defined as:
:$\forall \mathbf A \in R: \map \phi {\begin {pmatrix} x & y \\ 0 & z \end {pmatrix} } = z$
It is to be demonstrated that $\phi$ is a ring homomorphism whose kernel is $S$.
Thus:
and:
Thus by definition $\phi$ is a ring homomorphism.
By definition of $S$ itself, we have that:
:$S \subseteq \map \ker \phi$
Then we have that:
Hence:
:$\map \ker \phi = S$
From Kernel of Ring Homomorphism is Ideal:
:$S$ is an ideal of $R$.
== Sources ==
* : Chapter $9$: Rings: Exercise $10$
Category:Examples of Ideals of Rings
Ideal of Ring/Examples/Set of Even Integers

== Example of Ideal of Ring ==
The set $2 \Z$ of even integers forms an ideal of the ring of integers.
== Proof ==
Let $x \in 2 \Z$.
Then:
:$\forall y \in \Z: x y \in 2 \Z$
and:
:$\forall y \in \Z: y x \in 2 \Z$
Hence the result by definition of ideal.
== Sources ==
* : $\S 58$. Ideals
Category:Examples of Ideals of Rings
Category:Even Integers
Ideal of Ring is Contained in Radical

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathfrak a \subseteq A$ be an ideal.
Then $\mathfrak a$ is contained in its radical:
:$\mathfrak a \subseteq \operatorname{Rad} \left({\mathfrak a}\right)$
== Proof ==
Let $a \in \mathfrak a$.
By definition of power:
: $a^1 = a$
Thus:
: $a \in \operatorname{Rad} \left({\mathfrak a}\right)$
Category:Radical of Ideals
Ideal of Ring of Polynomials over Field has Unique Monic Polynomial forming Principal Ideal

== Theorem ==
Let $F$ be a field.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Let $J$ be a non-null ideal of $F \sqbrk X$.
Then there exists exactly one monic polynomial $f \in F \sqbrk X$ such that:
:$J = \ideal f$
where $\ideal f$ is the principal ideal generated by $f$ in $F \sqbrk X$.
== Proof ==
Let $f_1$ and $f_2$ be generators of $J$.
Then $f_1$ and $f_2$ are unit multiples of each other.
The units of $F \sqbrk X$ are the non-zero elements of $F$.
== Sources ==
* : Chapter $9$: Rings: Exercise $22$
Category:Principal Ideals of Rings
Category:Polynomial Theory
Ideal of Unit is Whole Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Let $J$ be an ideal of $R$.
If $J$ contains a unit of $R$, then $J = R$.
=== Corollary ===
== Proof ==
Let $u \in J$, where $u \in U_R$.
Also by definition, we have $u^{-1} \in U_R$.
Let $x \in R$.
Thus $R \subseteq J$.
As $J \subseteq R$ by definition, it follows that $J = R$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.2$
* : $\S 58.2$ Ideals
Category:Ideal Theory
Ideal of Unit is Whole Ring/Corollary

== Corollary to Ideal of Unit is Whole Ring ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Let $J$ be an ideal of $R$.
If $J$ contains the unity of $R$, then $J = R$.
== Proof ==
Follows directly from Ideal of Unit is Whole Ring and Unity is Unit.
== Sources ==
* : Chapter $5$: Rings: $\S 21$. Ideals: Theorem $35$
* : $\S 2$: Exercise $1$
* : $\S 58.2$ Ideals
Category:Ideal Theory
Ideal of Unital Algebra is Modular

== Theorem ==
Let $R$ be a ring. 
Let $A$ be a unital $R$-algebra.
Let $I$ be an ideal of $A$.
Then $I$ is a modular ideal of $A$. 
== Proof ==
Let ${\mathbf 1}_A$ be the identity element of $A$.
Let ${\mathbf 0}_A$ be the zero vector of $A$.
Then we have:
:$a {\mathbf 1}_A = {\mathbf 1}_A a$
for each $a \in A$. 
Hence we have:
:$a - {\mathbf 1}_A a = {\mathbf 0}_A$
and:
:$a - a {\mathbf 1}_A = {\mathbf 0}_A$
Because $I$ is a submodule of $A$:
:${\mathbf 0}_A \in A$
Hence:
:$a - u a \in I$
and so:
:$a - a u \in I$
for $u = {\mathbf 1}_A$.
Hence $I$ is a modular ideal of $A$ by definition.
== Also see ==
* Quotient Algebra is Unital iff Quotienting Ideal is Modular
== Sources ==
* : $1.1$: Banach Algebras
Category:Algebras
Category:Modular Ideals of Algebras
Category:Unital Algebras
Ideals Containing Ideal Form Lattice

== Theorem ==
Let $J$ be an ideal of a ring $R$.
Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$.
Then the ordered set $\struct {\mathbb L_J, \subseteq}$ is a lattice.
== Proof ==
Let $b_1, b_2 \in \mathbb L_J$.
Then from Set of Ideals forms Complete Lattice:
:$(1): \quad b_1 + b_2 \in \mathbb L_J$ and is the supremum of $\set {b_1, b_2}$
:$(2): \quad b_1 \cap b_2 \in \mathbb L_J$ and is the infimum of $\set {b_1, b_2}$
Thus $\struct {\mathbb L_J, \subseteq}$ is a lattice.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.7$
Category:Ideal Theory
Category:Lattice Theory
Ideals Containing Ideal Isomorphic to Quotient Ring

== Theorem ==
Let $J$ be an ideal of a ring $R$.
Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$.
Let the ordered set $\struct {\map {\mathbb L} {R / J}, \subseteq}$ be the set of all ideals of $R / J$.
Let the mapping $\Phi_J: \struct {\mathbb L_J, \subseteq} \to \struct {\map {\mathbb L} {R / J}, \subseteq}$ be defined as:
:$\forall a \in \mathbb L_J: \map {\Phi_J} a = \map {q_J} a$
where $q_J: a \to a / J$ is the quotient epimorphism from $a$ to $a / J$ from the definition of quotient ring.
Then $\Phi_J$ is an isomorphism.
== Proof ==
Let $b \in \mathbb L_J$.
From the way $\mathbb L_J$ is defined:
:$J \subseteq b$
Thus by Preimage of Image of Subring under Ring Homomorphism:
:$\map {q_J^{-1} } {\map {q_J} b} = b + J = b$
Let $c$ be an ideal of $R / J$.
Then, by Image of Preimage of Subring under Ring Epimorphism:
:$\map {q_J} {\map {q_J^{-1} } c} = c$
Thus by Bijection iff Left and Right Inverse, $\Phi_J$ is a bijection.
Hence:
:$\forall c \in \map {\mathbb L} {R / J}: \map {q_J^{-1} } {\Phi_J} c = \map {q_J^{-1} } c$
Now to show that $\Phi_J$ is an isomorphism.
Let $b_1, b_2 \in \mathbb L_J$.
Let $b_1 \subseteq b_2$.
Then from Subset Maps to Subset:
:$\map {q_J} {b_1} \subseteq \map {q_J} {b_2}$
Conversely, suppose $\map {q_J} {b_1} \subseteq \map {q_J} {b_2}$.
By what we have just proved:
:$b_1 = \map {q_J^{-1} } {\map {q_J} {b_1} } \subseteq \map {q_J^{-1} } {\map {q_J} {b_2} } = b_2$
Thus $\Phi_J$ is an isomorphism.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.7$
Category:Ideal Theory
Category:Quotient Rings
Ideals are Continuous Lattice Subframe of Power Set

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $I = \paren {\map {\operatorname{Ids} } L, \precsim}$ be an inclusion ordered set
where
:$\map {\operatorname{Ids} } L$ denotes the set of all ideals in $L$,
:$\mathord \precsim = \mathord \subseteq \cap \paren {\map {\operatorname{Ids} } L \times \map {\operatorname{Ids} } L}$
Let $P = \struct {\powerset S, \precsim'}$ be an inclusion ordered set
where
:$\powerset S$ denotes the power set of $S$,
:$\mathord \precsim' = \mathord \subseteq \cap \paren {\powerset S \times \powerset S}$
Then $I$ is continuous lattice subframe of $P$.
== Proof ==
By definition of subset:
:$\map {\operatorname{Ids} } L \subseteq \powerset S$
Then
:$\mathord \precsim = \mathord \precsim' \cap \paren {\map {\operatorname{Ids} } L \times \map {\operatorname{Ids} } L}$
Hence $I$ is ordered subset of $P$.
=== Infima Inheriting ===
Let $A$ be a subset of $\map {\operatorname{Ids} } L$ such that:
:$A$ admits an infimum in $P$.
By proof of Power Set is Complete Lattice:
:$\ds \inf_P A = \bigcap A$
By Intersection of Semilattice Ideals is Ideal/Set of Sets:
:$\ds \inf_P A \in \map {\operatorname{Ids} } L$
Thus by Infimum in Ordered Subset:
:$A$ admits an infimum in $I$ and $\ds \inf_I A = \inf_P A$
Hence $I$ inherits infima.
=== Directed Suprema Inheriting ===
Let $D$ be a directed subset of $\map {\operatorname{Ids} } L$ such that:
:$D$ admits a supremum in $P$.
By proof of Power Set is Complete Lattice:
:$\ds \sup_P D = \bigcup D$
We will prove that:
:$\ds \bigcup D$ is an ideal in $L$.
==== Directed ====
Let $\ds x, y \in \bigcup D$.
By definition of union:
:$\exists I_1 \in D: x \in I_1$
and
:$\exists I_2 \in D: y \in I_2$
By definition of directed:
:$\exists I \in D: I_1 \precsim I \land I_2 \precsim I$
By definition of $\precsim$:
:$I_1 \subseteq I$ and $I_2 \subseteq I$
By definition of subset:
:$x, y \in I$
By definition of directed:
:$\exists z \in I: x \preceq z \land y \preceq z$
Thus by definition of union:
:$\ds \exists z \in \bigcup D: x \preceq z \land y \preceq z$
==== Lower Section ====
Let $\ds x \in \bigcup D$, $y \in S$ such that:
:$y \preceq x$
By definition of union:
:$\exists I \in D: x \in I$
By definition of lower section:
:$y \in I$
Thus by definition of union:
:$\ds y \in \bigcup D$
==== Non-Empty Set ====
By definition of directed:
:$D$ is non-empty and $\forall I \in D: I$ is non-empty.
Thus by definitions of non-empty set and union:
:$\ds \bigcup D$ is non-empty.
By definition of $\operatorname{Ids}$:
:$\ds \bigcup D \in \map {\operatorname{Ids} } L$
Hence $I$ inherits directed suprema.
== Sources ==
* 
* 
Category:Join and Meet Semilattices
Ideals equal Filters in Dual Ordered Set

== Theorem ==
Let $L_1 = \struct {S, \preceq_1}$ be an ordered set.
Let $L_2 = \struct {S, \preceq_2}$ be a dual ordered set of $L_1$.
Then $\map {\mathit {Ids} } {L_1} = \map {\mathit {Filt} } {L_2}$
where
:$\map {\mathit {Filt} } {L_2}$ denotes the set of all filters of $L_2$
:$\map {\mathit {Ids} } {L_1}$ denotes the set of all ideals of $L_1$.
== Proof ==
Let $x$ be a set.
By definition of $\mathit {Filt}$:
:$x \in \map {\mathit {Filt} } {L_2} \iff x$ is a filter on $L_2$.
By Ideal is Filter in Dual Ordered Set:
:$x \in \map {\mathit {Filt} } {L_2} \iff x$ is an ideal in $L_1$.
By definition of $\mathit{Ids}$:
:$x \in \map {\mathit {Filt} } {L_2} \iff x \in \map {\mathit {Ids} } {L_1}$
Hence by definition of set equality:
:$\map {\mathit {Ids} } {L_1} = \map {\mathit {Filt} } {L_2}$
== Sources ==
* 
Category:Dual Orderings
Ideals form Algebraic Lattice

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $I = \struct {\map {\operatorname{Ids} } L, \precsim}$ be an inclusion ordered set
where
:$\map {\operatorname{Ids} } L$ denotes the set of all ideals in $L$
:$\mathord \precsim = \mathord \subseteq \cap \paren {\map {\operatorname{Ids} } L \times \map {\operatorname{Ids} } L}$
Then $I$ is an algebraic lattice.
== Proof ==
By definition of subset:
:$\map {\operatorname{Ids} } L \subseteq \powerset S$
where $\powerset S$ denotes the power set of $S$.
Define:
:$P = \struct {\powerset S, \precsim'}$
where:
:$\mathord \precsim' = \mathord\subseteq \cap \paren {\powerset S \times \powerset S}$
By Ideals are Continuous Lattice Subframe of Power Set:
:$I$ is an continuous lattice subframe of $P$.
By Lattice of Power Set is Algebraic:
:$P$ is an algebraic lattice.
Thus by Continuous Lattice Subframe of Algebraic Lattice is Algebraic Lattice:
:$I$ is an algebraic lattice.
== Sources ==
* 
* 
Category:Join and Meet Semilattices
Category:Continuous Lattices