avewright/proofwiki-math huggingface dataset:
Theorem: Ordinal is Proper Subset of Successor

Let $\alpha$ be an ordinal.
Then:
:$\alpha \subsetneqq \alpha^+$
where $\alpha^+$ denotes the successor set of $\alpha$.
That is:
:$\alpha$ is a proper subset of $\alpha^+$.

Proof:
{{AimForCont}} $\alpha = \alpha^+$.
By definition:
:$\alpha^+ = \alpha \cup \set \alpha$
and so:
:$\alpha \subseteq \alpha^+$
and:
:$\alpha \in \alpha^+$
which leads to:
:$\alpha \in \alpha$
But from Ordinal is not Element of Itself:
:$\alpha \notin \alpha$
Hence by Proof by Contradiction:
:$\alpha \ne \alpha^+$
But we have:
:$\alpha \subseteq \alpha^+$
Hence:
:$\alpha \subsetneqq \alpha^+$
{{qed}}
Theorem: Successor Mapping on Ordinals is Strictly Progressing

Let $\On$ denote the class of all ordinals.
Let $s$ denote the successor mapping on $\On$:
:$\forall \alpha \in \On: \map s \alpha := \alpha \cup \set \alpha$
where $\alpha$ is an ordinal.
Then $s$ is a strictly progressing mapping.

Proof:
From Ordinal is Proper Subset of Successor:
:$\alpha \subsetneqq \alpha^+$
where $\alpha^+$ is identified as $\map s \alpha$.
The result follows by definition of strictly progressing mapping.
{{qed}}
Theorem: Class of All Ordinals is Proper Class

Let $\On$ denote the class of all ordinals.
Then $\On$ is a proper class.
That is, $\On$ is not a set.

Proof:
We have that Successor Mapping on Ordinals is Strictly Progressing.
The result follows from Superinductive Class under Strictly Progressing Mapping is Proper Class.
{{qed}}
Theorem: Nilpotent Element is Contained in Prime Ideals

Let $R$ be a commutative ring with unity.
Let $\mathfrak p \subset R$ be a prime ideal of $R$.
Let $x \in R$ be a nilpotent element of $R$.
Then $x \in \mathfrak p$.

Proof:
Let $x$ be nilpotent in $R$ as asserted.
Then by definition:
:$\exists n \in \Z_{>0}: x^n = 0$
But $0 \in \mathfrak p$ so:
:$x^n \in \mathfrak p$
{{begin-eqn}}
{{eqn | l = x
      | r = \map \Rad {\mathfrak p}
      | o = \in
      | c = {{Defof|Radical of Ideal of Ring|index=1}} }}
{{eqn | r = \mathfrak p
      | c = Radical of Prime Ideal}}
{{end-eqn}}
{{qed}}
Category:Prime Ideals of Rings
sae3q6gwfhmfjox0lz80luxzxdhlj3p
Theorem: Well-Ordering of Class of All Ordinals under Subset Relation

Let $\On$ denote the class of all ordinals.
$\On$ is well-ordered by the subset relation such that the following $3$ conditions hold:
{{begin-axiom}}
{{axiom | n = 1
        | lc= the smallest ordinal is $0$
}}
{{axiom | n = 2
        | lc= for $\alpha \in \On$, the immediate successor of $\alpha$ is its successor set $\alpha^+$
}}
{{axiom | n = 3
        | lc= every limit ordinal is the union of the set of smaller ordinals.
}}
{{end-axiom}}

Proof:
We have that Class of All Ordinals is $g$-Tower.
By Zero is Smallest Ordinal, $0$ is the smallest element of $\On$.
We identify the natural number $0$ via the von Neumann construction of the natural numbers as:
:$0 := \O$
The result then follows directly from $g$-Tower is Well-Ordered under Subset Relation.
{{qed}}
Theorem: Strict Ordering of Ordinals is Equivalent to Membership Relation

Let $\On$ denote the class of all ordinals.
Let $<$ denote the (strict) usual ordering of $\On$.
Then:
:$\forall \alpha, \beta \in \On: \alpha < \beta \iff \alpha \in \beta$

Proof:
=== Necessary Condition ===
Let $\alpha \in \beta$.
Then from Ordinal is Transitive:
:$\alpha \subseteq \beta$
But if $\alpha = \beta$ we would have $\alpha \in \alpha$.
This is contrary to Ordinal is not Element of Itself.
Hence we have:
:$\alpha \subseteq \beta$
and:
:$\alpha \ne \beta$
That is:
:$\alpha \subsetneqq \beta$
Hence by definition of the (strict) usual ordering of $\On$:
:$\alpha < \beta$
{{qed|lemma}}
Theorem: Ordinal Membership is Transitive

Let $\On$ denote the class of all ordinals.
Then:
:$\forall \alpha, \beta, \gamma \in \On: \paren {\alpha \in \beta} \land \paren {\beta \in \gamma} \implies \alpha \in \gamma$

Proof:
By Strict Ordering of Ordinals is Equivalent to Membership Relation the statement to be proved is equivalent to:
:$\forall \alpha, \beta, \gamma \in \On: \paren {\alpha \subsetneqq \beta} \land \paren {\beta \subsetneqq \gamma} \implies \alpha \subsetneqq \gamma$
which follows (indirectly) from Subset Relation is Transitive.
{{qed}}
Theorem: Equality of Successors implies Equality of Ordinals

Let $\On$ denote the class of all ordinals.
Then:
:$\forall \alpha, \beta \in \On: \alpha^+ = \beta^+ \implies \alpha = \beta$

Proof:
From Class of All Ordinals is Well-Ordered by Subset Relation:
:$\alpha^+$ is the immediate successor of $\alpha$
:$\beta^+$ is the immediate successor of $\beta$
and no two distinct elements of $\On$ can have the same immediate successor.
The result follows.
{{qed}}
Theorem: Exists Ordinal Greater than Set of Ordinals

Let $S$ be a set of ordinals.
Then there exists an ordinal greater than every element of $S$:
:If $S$ contains a greatest ordinal $\alpha$, then $\alpha^+$ is greater than every element of $S$
:If $S$ does not contain a greatest ordinal, then $\bigcup S$ is greater than every element of $S$.

Proof:
Recall that Class of All Ordinals is Well-Ordered by Subset Relation.
Suppose $S$ contains a greatest ordinal $\alpha$.
Because $\alpha^+$ is greater than $\alpha$ by definition, it follows {{apriori}} that $\alpha^+$ is greater than every element of $S$.
{{qed|lemma}}
Suppose $S$ does not contain a greatest ordinal.
Consider the union $\bigcup S$ of $S$.
If $\bigcup S \in S$ it would be the greatest element of $S$.
Hence $\bigcup S \notin S$.
Let $\alpha \in S$.
Then:
:$\alpha \ne \bigcup S$
But:
:$\alpha \subseteq \bigcup S$
and so:
:$\alpha \subsetneqq S$
From Union of Set of Ordinals is Ordinal, $\bigcup S$ is an ordinal.
Hence by definition of the usual ordering on ordinals:
:$a < \bigcup S$
Thus $\bigcup S$ is greater than every element of $S$.
Hence the result.
{{qed}}
Theorem: Successor of Ordinal Smaller than Limit Ordinal is also Smaller

Let $\On$ denote the class of all ordinals.
Let $\lambda \in \On$ be a limit ordinal.
Then:
:$\forall \alpha \in \On: \alpha < \lambda \implies \alpha^+ < \lambda$

Proof:
Let $\lambda$ be a limit ordinal such that $\alpha < \lambda$.
From Successor of Element of Ordinal is Subset
Then as $\alpha^+$ is the successor set of $\alpha$ it follows that:
:$\alpha^+ \le \lambda$
{{explain|Find the result that says $\alpha < \beta \implies \alpha^+ \le \beta$}}
But as $\lambda$ is not a successor ordinal:
:$\alpha^+ \ne \lambda$
Hence:
:$\alpha^+ < \lambda$
{{qed}}
Theorem: Unit Ideal is Principal Ideal Generated by Unity

Let $A$ be a commutative ring with unity.
Then:
:$A = \ideal 1$
where:
:$A$ is called the unit ideal of $A$
:$\ideal 1$ denotes the principal ideal generated by the unity of $A$

Proof:
$\ideal 1 \subseteq A$ is clear by definition of principal ideal.
To see $A \subseteq \ideal 1$, let $a \in A$ be an arbitrary element.
Then:
{{begin-eqn}}
{{eqn | l = a
      | r = a 1
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \ideal 1
      | o = \in
      | c = {{Defof|Principal Ideal of Ring}}
}}
{{end-eqn}}
{{qed}}
Category:Commutative Algebra
Category:Ideal Theory
54lr7uq7u7nzjixg8hve0xosi7twhfk
Theorem: Union of Ordinal is Subset of Itself

Let $\alpha$ be an ordinal.
Then:
:$\bigcup \alpha \subseteq \alpha$
where $\bigcup \alpha$ denotes the union of $\alpha$.

Proof:
Let $x \in \bigcup \alpha$.
Then:
:$\exists \beta \in \alpha: x \in \beta$
By Element of Ordinal is Ordinal, $\beta$ is an ordinal.
Thus:
:$x \in \beta$ and $\beta \in \alpha$
Hence by Ordinal Membership is Transitive:
:$x \in \alpha$
{{qed}}
Theorem: Ideal is Unit Ideal iff Includes Unity

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:
{{begin-eqn}}
{{eqn | l = a
      | r = a 1
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \mathfrak a
      | o = \in
      | c = since $1 \in \mathfrak a$, by {{Defof|Ideal of Ring}}
}}
{{end-eqn}}
{{qed}}
Category:Commutative Algebra
Category:Ideal Theory
1td4musqjlq40tibu5yfksdvrjmdqmr
Theorem: Unit Ideal iff Radical is Unit Ideal

Let $A$ be a commutative ring with unity.
Let $\mathfrak a$ be an ideal of $A$.
Then:
:$ \mathfrak a = A \iff \map \Rad {\mathfrak a} = A$
where:
:$A$ is called the unit ideal of $A$
:$\map \Rad {\mathfrak a}$ denotes the radical of $\mathfrak a$

Proof:
{{begin-eqn}}
{{eqn | l = \mathfrak a 
      | r = A
}}
{{eqn | ll= \leadstoandfrom
      | l = 1
      | r = A
      | o = \in
      | c = Ideal is Unit Ideal iff Includes Unity
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists n \in \Z_{>0}
      | l = 1^n
      | r = A
      | o = \in
      | c = since $ 1^n = 1$ by {{Defof|Unity of Ring}}
}}
{{eqn | ll= \leadstoandfrom
      | l = 1
      | r = \map \Rad {\mathfrak a}
      | o = \in
      | c = {{Defof|Radical of Ideal of Ring|index = 1}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \Rad {\mathfrak a}
      | r = A
      | c = Ideal is Unit Ideal iff Includes Unity
}}
{{end-eqn}}
{{qed}}
Category:Radical of Ideals
mrabwu8dunaa27yj94jnovwe2orjq82
Theorem: Ordinal equals Successor of its Union

Let $\alpha$ be an ordinal.
Then:
:$\bigcup \alpha^+ = \alpha$
where:
:$\alpha^+$ denotes the successor set of $\alpha$
:$\bigcup \alpha^+$ denotes the union of $\alpha^+$.

Proof:
Let $x \in \bigcup \alpha^+$.
Then there exists an ordinal $\beta$ such that:
:$x \in \beta$
and:
:$\beta \in \alpha^+$
By definition of the usual ordering of ordinals:
:$\beta < \alpha^+$
Thus:
:$\beta \le \alpha$
Hence because $x < \beta$:
:$x < \alpha$
and thus:
:$x \in \alpha$
That is:
:$\bigcup \alpha^+ \subseteq \alpha$
Now suppose $x \in \alpha$.
Since we have:
:$\alpha \in \alpha^+$
we have:
:$x \in \bigcup \alpha^+$
Thus:
:$a \subseteq \bigcup \alpha^+$
Hence by set equality:
:$a = \bigcup \alpha^+$
{{qed}}
Theorem: Transitive Class of Ordinals is Subset of Ordinal not in it

Let $A$ be a transitive class of ordinals.
Let $\alpha$ be an ordinal which is not an element of $A$.
Then:
:$A \subseteq \alpha$

Proof:
Let $A$ and $\alpha$ be {{hypothesis}}.
Let $\beta \in A$ be arbitrary.
Because $\beta \in A$ and $\alpha \notin A$ we have that:
:$\beta \ne \alpha$
Because $\beta \in A$ and $A$ is transitive:
:all elements of $\beta$ are in $A$
But because $\alpha \notin A$:
:$\alpha \notin \beta$
Thus we have:
:$\alpha \notin \beta$
and:
:$\alpha \ne \beta$
Hence from Ordinal Membership is Trichotomy:
:$\beta \in \alpha$
As $\beta$ is arbitrary, it follows that:
:$A \subseteq \alpha$
{{qed}}
Theorem: Transitive Set of Ordinals is Ordinal

Let $x$ be a transitive set of ordinals.
Then $x$ is itself an ordinal.

Proof:
Let $x$ be a transitive set of ordinals according to the statement of the theorem.
We have from Class of All Ordinals is Well-Ordered by Subset Relation that $\On$ is well-ordered by $\subseteq$.
By Exists Ordinal Greater than Set of Ordinals there exists $\alpha$ such that $\alpha \notin x$.
Hence let $\alpha$ be the smallest ordinal not in $x$.
So all ordinals smaller than $\alpha$ are in $x$.
Hence from Ordinal equals its Initial Segment:
:$\alpha \subseteq x$
From Transitive Class of Ordinals is Subset of Ordinal not in it:
:$x \subseteq \alpha$
Hence by set equality:
:$x = \alpha$
Thus $x$ is an ordinal.
{{qed}}
Theorem: Class of All Ordinals is Only Proper Class of Ordinals

Let $A$ be a transitive proper class of ordinals.
Then $A$ is the class of all ordinals $\On$.

Proof:
Let $A$ be a transitive class of ordinals.
Let there exist $\alpha \in \On$ such that $\alpha \notin A$.
Then by Transitive Class of Ordinals is Subset of Ordinal not in it:
:$A \subseteq \alpha$
But that makes $A$ a set.
So if $A$ is a proper class, it must contain all ordinals.
That is:
:$A = \On$
{{qed}}
Theorem: Set is Ordinal iff Every Transitive Proper Subset is Element of it

Let $x$ be a set.
Then:
:$x$ is an ordinal
{{iff}}
:every transitive proper subset of $x$ is an element of $x$.

Proof:
=== Necessary Condition ===
Let $\alpha$ be an arbitrary ordinal.
Let $y$ be a proper subset of $\alpha$ such that $y$ is transitive.
From Transitive Set of Ordinals is Ordinal it follows that $y$ is an ordinal.
By definition of usual ordering of ordinals:
:$y < \alpha$
Hence from Strict Ordering of Ordinals is Equivalent to Membership Relation:
:$y \in \alpha$
As $\alpha$ is arbitrary, it follows that every transitive proper subset of $\alpha$ is an element of $\alpha$.
{{qed|lemma}}
Theorem: Class such that Every Transitive Subset is Element of it Contains All Ordinals

Let $K$ be a class.
Let $K$ be such that every transitive subset of $K$ is an element of $K$.
Then every ordinal is an element of $K$.

Proof:
Let us assume the hypothesis.
Let $\On$ denote the class of all ordinals.
From Well-Ordering of Class of All Ordinals under Subset Relation, $\On$ is well-ordered under $\subseteq$.
Hence we can use the First Principle of Transfinite Induction.
Let $\alpha$ be an ordinal such that every ordinal less than $\alpha$ is an element of $K$.
Thus $\alpha \subseteq K$.
As Ordinal is Transitive, it follows {{hypothesis}} that:
:$\alpha \in K$
So if all ordinals less than $\alpha$ are elements of $K$, then so is $\alpha$.
Hence by the First Principle of Transfinite Induction, all ordinals are in $K$.
{{qed}}
Theorem: Element of Every Transitive-Closed Class is Ordinal

Let $x$ be a set such that $x$ is an element of every transitive-closed class.
Then $x$ is an ordinal.

Proof:
{{ProofWanted|Not even sure if it actually is true yet. The exercise as set in S&F asks the question whether it is or not. If it turns out not true, we rename this page.}}
Theorem: Ideal Quotient is Ideal

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.
Theorem: Ordinal Addition/Examples/Ordinal Addition by Two

Let $2$ denote the successor of the ordinal $1$.
Then:
:$x + 2 = x^{++}$

Proof:
{{begin-eqn}}
{{eqn | l = x + 2
      | r = x + 1^+
      | c = 
}}
{{eqn | r = \paren {x + 1}^+
      | c = {{Defof|Ordinal Addition}}
}}
{{eqn | r = x^{++}
      | c = Ordinal Addition by One
}}
{{end-eqn}}
{{qed}}
Theorem: Ordinal Addition/Examples/Ordinal Addition by Natural Number

Let $n$ be a natural number.
Then:
:$x + \paren {n + 1} = \paren {x + n}^+$

Proof:
{{begin-eqn}}
{{eqn | l = x + \paren {n + 1}
      | r = x + n^+
      | c = Ordinal Addition by One
}}
{{eqn | r = \paren {x + n}^+
      | c = {{Defof|Ordinal Addition}}
}}
{{end-eqn}}
{{qed}}
Theorem: Set of Natural Numbers is Ordinal

The set of natural numbers $\N$ is an ordinal.

Proof:
From Natural Number is Ordinal, every element of $\N$ is an ordinal.
From Union of Set of Ordinals is Ordinal, $\bigcup \N$ is therefore itself an ordinal.
From Set of Natural Numbers Equals its Union:
:$\bigcup \N = \N$
Hence the result.
{{qed}}
Theorem: Set of Natural Numbers is Smallest Ordinal Greater than All Natural Numbers

The set of natural numbers $\N$ is the smallest ordinal which is greater than all natural numbers.

Proof:
From Set of Natural Numbers is Ordinal, $\N$ is an ordinal.
{{finish}}
Theorem: Set of Natural Numbers is Limit Ordinal

The set of natural numbers $\N$ is a limit ordinal.

Proof:
From Set of Natural Numbers is Ordinal, $\N$ is an ordinal.
{{finish}}
Theorem: Radical of Primary Ideal is Smallest Larger Prime Ideal

Let $R$ be a commutative ring with unity.
Let $\mathfrak q$ be a primary ideal of $R$.
Let $\map \Rad {\mathfrak q}$ be the radical of $\mathfrak q$.
Then $\map \Rad {\mathfrak q}$ is the smallest prime ideal containing $\mathfrak q$.

Proof:
First, we show that $\map \Rad {\mathfrak q}$ is a prime ideal.
Let $x y \in \map \Rad {\mathfrak q}$.
Then by definition of radical of ideal:
:$\exists n \in \N_{>0} : \paren {xy}^n \in \mathfrak q$
By Commutativity $(\text M 2)$:
:$x^n y^n \in \mathfrak q$
By definition of primary ideal:
:$x^n \in \mathfrak q \lor \exists m \in \N_{>0} : \paren {y^n}^m = y^{n m} \in \mathfrak q$
Thus, by definition of radical of ideal:
:$x \in \map \Rad {\mathfrak q} \lor y \in \map \Rad {\mathfrak q}$
So, $\map \Rad {\mathfrak q}$ is a prime ideal.
Finally, by Prime Ideal Including Ideal Includes Radical, every prime ideal $\mathfrak p$ satisfies:
:$\mathfrak q \subseteq \mathfrak p \implies \map \Rad {\mathfrak q} \subseteq \mathfrak p$
That is, $\map \Rad {\mathfrak q}$ is the smallest prime ideal including $\mathfrak q$.
{{qed}}
Category:Radical of Ideals
Category:Primary Ideals
Category:Prime Ideals of Rings
mr24lipzg8sa9uzd5sk5li4137tex4m
Theorem: Image of Class under Mapping is Image of Restriction of Mapping to Class

Let $V$ be a basic universe
Let $f: V \to V$ be a mapping.
Let $A$ be a class.
Let $f \sqbrk A$ denote the image of $A$ under $f$.
Then $f \sqbrk A$ is the image of the restriction of $f$ to $A$:
:$f \sqbrk A = \Img {f {\restriction} A}$

Proof:
By definition, $f {\restriction} A$ is class of all ordered pairs $\tuple {a, \map f a}$, where $a \in A$.
The result follows from Restriction of Mapping is Subset of Cartesian Product.
{{qed}}
Theorem: Image of Set under Mapping is Set iff Restriction is Set

Let $V$ be a basic universe
Let $f: V \to V$ be a mapping.
Let $x$ be a set.
Let $f \sqbrk x$ denote the image of $x$ under $f$.
Let $f {\restriction} x$ denote the restriction of $f$ to $x$.
Then $f \sqbrk x$ is a set {{iff}} $f {\restriction} x$ is a set.

Proof:
Let $f {\restriction} x$ be a set.
Then its image is also a set.
But then:
:$\Img {f {\restriction} x} = f \sqbrk x$
Conversely let $f \sqbrk x$ be a set.
From Cartesian Product of Sets is Set:
:$x \times f \sqbrk x$ is a set.
From Restriction of Mapping is Subclass of Cartesian Product:
:$f {\restriction} A \subseteq A \times f \sqbrk A$
The result follows from Subclass of Set is Set.
{{qed}}
Theorem: Restriction of Mapping is Subclass of Cartesian Product

Let $V$ be a basic universe
Let $f: V \to V$ be a mapping.
Let $A$ be a class.
Let $f \sqbrk A$ denote the image of $A$ under $f$.
Let $f {\restriction} A$ denote the restriction of $f$ to $A$.
Then $f {\restriction} A$ is a subclass of the cartesian product of $A$ with its image:
:$f {\restriction} A \subseteq A \times f \sqbrk A$

Proof:
{{Proofread|Missing explicit mention of Cartesian Product in the proof}}
Follows directly from:
:the definition of restriction
:the definition of mapping.
{{qed}}
Theorem: Radical of Prime Ideal

Let $R$ be a commutative ring with unity.
Let $\mathfrak p$ be a prime ideal of $R$.
Let $\map \Rad {\mathfrak p}$ be the radical of $\mathfrak p$.
Then:
:$\map \Rad {\mathfrak p} = \mathfrak p$

Proof:
=== $\supseteq$ ===
Let $x \in \mathfrak p$.
Since $x = x^1$, by {{Defof|Radical of Ideal of Ring|index=1}}:
:$x \in \map \Rad {\mathfrak p}$
{{qed|lemma}}
Theorem: Prime Ideal Including Ideal Includes Radical

Let $R$ be a commutative ring with unity.
Let $\mathfrak p$ be a prime ideal.
Let $\mathfrak a$ be an ideal of $R$ such that:
:$\mathfrak a \subseteq \mathfrak p$
Let $\map \Rad {\mathfrak a}$ be the radical of $\mathfrak a$.
Then:
:$\map \Rad {\mathfrak a} \subseteq \mathfrak p$

Proof:
Let $x \in \relcomp R {\mathfrak p}$.
By {{Defof|Prime Ideal of Ring/Commutative and Unitary Ring|Prime Ideal|index=3}}:
{{begin-eqn}}
{{eqn | ll = \forall n \in \N_{>0} :
      | l = x^n
      | r = \mathfrak p
      | o = \not\in
}}
{{eqn | ll = \leadsto
      | l = x^n
      | r = \mathfrak a
      | o = \not\in
      | c = since $\mathfrak a \subseteq \mathfrak p$
}}
{{end-eqn}}
Therefore, by {{Defof|Radical of Ideal of Ring|Radical|index=2}}:
:$x \not\in \map \Rad {\mathfrak a}$
Thus:
:$\relcomp R {\mathfrak p} \subseteq \relcomp R {\map \Rad {\mathfrak a} }$
Therefore, by Relative Complement inverts Subsets:
:$\map \Rad {\mathfrak a} \subseteq \mathfrak p$
{{qed}}
Category:Radical of Ideals
Category:Ideal Theory
Category:Commutative Algebra
Category:Ring Theory
g7ijpddc5aue2wng6f949eaxus9n1pz
Theorem: Composite of Mapping with Inverse of Another is Identity implies Mappings are Equal

Let $A$ and $B$ be classes.
Let $f$ and $g$ be mappings on $A \times B$.
Let $f$ and $g$ be such that:
:$f \circ g^{-1} = I_B$
where:
:$g^{-1}$ denotes the inverse of $g$
:$I_B$ denotes the identity mapping on $B$
:$\circ$ denotes composition of mappings.
Then:
:$f = g$
{{questionable|Each of the definitions inverse, identity mapping and composition of mappings is defined on sets only. They need to be expanded to classes.}}

Proof:
Let $\tuple {a, b} \in f$.
Then because $\tuple {b, b} \in I_B$ we must have:
:$\tuple {b, a} \in g^{-1}$
and so by definition of inverse of mapping:
:$\tuple {a, b} \in g$
Hence:
:$f \subseteq g$
Let $\tuple {a, b} \in g$.
Then by definition of inverse of mapping:
:$\tuple {b, a} \in g^{-1}$
Then because $\tuple {b, b} \in I_B$ we must have:
:$\tuple {a, b} \in f$
Hence:
:$g \subseteq f$
Hence by definition of set equality:
:$f = g$
{{qed}}
Theorem: Ideal Contains Extension of Contraction

Let $A$ and $B$ be commutative rings with unity.
Let $f : A \to B$ be a ring homomorphism.
Let $\mathfrak b \subseteq B$ be an ideal.
Then $\mathfrak b$ contains the extension of its contraction by $f$:
:$\mathfrak b^{ce} \subseteq \mathfrak b$

Proof:
{{begin-eqn}}
{{eqn | l = f \sqbrk {\mathfrak b ^c}
      | r = f \sqbrk {f^{-1} \sqbrk {\mathfrak b} }
      | c = {{Defof|Contraction of Ideal}}
}}
{{eqn | r = \mathfrak b
      | o = \subseteq
      | c = Image of Preimage under Mapping
}}
{{end-eqn}}
Since $\mathfrak b^{ce}$ is generated by $f \sqbrk {\mathfrak b ^c}$ according to {{Defof|Extension of Ideal}}, it is contained in $\mathfrak b$
{{qed}}
Category:Ideal Theory
Category:Commutative Algebra
Category:Ring Theory
hmdhgsrmjm1cobaxfj1pmthjdzm321f
Theorem: Strict Lower Closure of Element is Proper Lower Section

Let $A$ be a class under an ordering $\preccurlyeq$.
Let $a \in A$.
Let $a^\prec$ denote the strict lower closure of $a$ in $A$.
Then $a^\prec$ is a proper lower section of $A$.

Proof:
By definition of strict lower closure:
:$a^\prec := \set {b \in S: b \prec a}$
By definition of $\prec$:
:$a \not \prec a$
and so:
:$a \notin a^\prec$
and so:
:$a \in A \setminus a^\prec$
Let $x \in a^\prec$.
Then by definition of $a^\prec$:
:$x \prec a$
and so:
:$\exists a \in A \setminus a^\prec: x \prec a$
and so $x$ is in a proper lower section of $A$.
As $x$ is arbitrary, it follows that every element of $a^\prec$ is in a proper lower section of $A$.
Hence $a^\prec$ ''is'' a proper lower section of $A$.
{{qed}}
Theorem: Proper Lower Section under Well-Ordering is Initial Segment

Let $A$ be a class under a well-ordering $\preccurlyeq$.
Let $B$ be a proper lower section of $A$.
Let $b$ be the smallest element of $A \setminus B$.
Then $B$ is the initial segment of $b$ in $A$.

Proof:
By definition of proper lower section of $A$:
:$\forall x \in B: \forall a \in A \setminus B: x \preccurlyeq a$
while:
:$B \ne \O$
:$B \ne A$
Because $B \ne A$ we have that $A \setminus B \ne \O$.
Because $\preccurlyeq$ is a well-ordering, $A \setminus B$ does indeed have a smallest element, which we can indeed call $b$.
We have {{apriori}} that:
:$\forall x \in B: \forall a \in A \setminus B: x \preccurlyeq a$
This applies in particular to $b \in A \setminus B$:
:$\forall x \in B: \exists b \in A \setminus B: x \preccurlyeq b$
That is, $B$ is the initial segment of $b$ in $A$ by definition.
{{qed}}
Theorem: Well-Ordering on Set is Proper Well-Ordering

Let $\struct {S, \preccurlyeq}$ be a well-ordered set.
Then $\preccurlyeq$ is a proper well-ordering.

Proof:
By definition, a proper well-ordering is a well-ordering on a class such that:
:every proper lower section of $S$ is a set.
We have {{afortiori}} that a proper lower section of $S$ is a subclass of $S$.
But here we have that $S$ is a set.
The result follows from Subclass of Set is Set.
{{qed}}