avewright/proofwiki-math huggingface dataset:
Theorem: Element of Principal Ideal Domain is Finite Product of Irreducible Elements

Let $R$ be a principal ideal domain.
Let $p \in R$ such that $p \ne 0$ and $p$ is not a unit.
Then there exist irreducible elements $p_1, \ldots, p_n$ such that $p = p_1 \cdots p_n$.

Proof:
Let $S = \set {\ideal p: p \in R \setminus \set 0, p \text{ is not a unit, and } p \text{ is not a product of finitely many irreducible elements} }$.
Define a relation $\RR$ on $S$ by $\ideal a \mathrel \RR \ideal b$ {{iff}} $\ideal a \subsetneq \ideal b$.
For any $\ideal p \in S$, since $p$ is not irreducible, we can write $p = r_1 s_1$ where neither $r_1$ nor $s_1$ is zero or a unit.
At least one of $r_1$ and $s_1$ is not a finite product of irreducible elements, or else $p$ would be, contradicting $\ideal p \in S$.
{{WLOG}}, if $r_1$ is such an element, then $\ideal p \subsetneq \ideal {r_1}$ is a proper inclusion of ideals, as $r_1$ is not a unit.
Hence, $\ideal {r_1} \in S$, and $\ideal p \mathrel \RR \ideal {r_1}$, which means that $\RR$ is a left-total relation on $S$.
By the Axiom of Dependent Choice, such a relation establishes an infinite strictly ascending chain of ideals:
:$\ideal p \subsetneq \ideal {r_1} \subsetneq \ideal {r_2} \subsetneq \ideal {r_3} \subsetneq \cdots$
By Principal Ideal Domain fulfills Ascending Chain Condition, $S = \O$, or else such an ascending chain would exist, leading to a contradiction.
{{qed}}
Theorem: Principal Ideal Domain fulfills Ascending Chain Condition

Let $R$ be a principal ideal domain.
Then $R$ fulfils the ascending chain condition.

Proof:
Let $I_1 \subseteq I_2 \subseteq I_3 \subseteq \dotsb$ be an ascending chain of ideals.
Build $\ds I = \bigcup_{i \mathop = 1}^\infty I_i$. 
$I$ is an ideal.
Since $R$ is a principal ideal domain, $I = \ideal a$ for some $a \in R$.
Now, since $a \in I$, there is some $n$ such that $a \in I_n$.
Thus:
:$\ideal a \subseteq I_n$
By definition $I_n \subset I = \ideal a$, and so $I_n = I$.
Thus:
:$\forall m \ge n: I_m = I$
{{MissingLinks|Still some links to be made, I'll get to it later.}}
{{qed}}
{{Proofread}}
Category:Ideal Theory
Category:Principal Ideal Domains
9ii5qo3qx0b0y56doy4xwvz4euapzbe
Theorem: Subspace of Product Space is Homeomorphic to Factor Space

Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Suppose that $X$ is non-empty.
Then for each $i \in I$ there is a subspace $Y_i \subseteq X$ which is homeomorphic to $\struct {X_i, \tau_i}$.
Specifically, for any $z \in X$, let:
:$Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$
and let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
Then $\struct {Y_i, \upsilon_i}$ is homeomorphic to $\struct {X_i, \tau_i}$, where the homeomorphism is the restriction of the projection $\pr_i$ to $Y_i$.

Proof:
From Finite Cartesian Product of Non-Empty Sets is Non-Empty both $T_1 \times T_2$ and $T_2 \times T_1$ are both non-empty.
The conclusions follow immediately from Subspace of Product Space is Homeomorphic to Factor Space.
{{qed}}
Theorem: Subspace of Product Space is Homeomorphic to Factor Space

Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Suppose that $X$ is non-empty.
Then for each $i \in I$ there is a subspace $Y_i \subseteq X$ which is homeomorphic to $\struct {X_i, \tau_i}$.
Specifically, for any $z \in X$, let:
:$Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$
and let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
Then $\struct {Y_i, \upsilon_i}$ is homeomorphic to $\struct {X_i, \tau_i}$, where the homeomorphism is the restriction of the projection $\pr_i$ to $Y_i$.

Proof:
The conclusions are symmetrical.
{{WLOG}}, therefore, it will be shown that $T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$.
Let $f: T_1 \to T_1 \times \set b$ be defined as:
:$\map f x = \tuple {x, b}$
=== Lemma ===
{{:Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Lemma}}{{qed|lemma}}
$f^{-1}$ is a restriction to the subspace $T_1 \times \set b$ of the projection $\pr_1$ of $T_1 \times T_2$ onto $T_1$.
From Projection from Product Topology is Continuous , $\pr_1$ is a continuous.
From Restriction of Continuous Mapping is Continuous, $f^{-1}$ is a continuous.
It is to be shown that $f$ is continuous.
Let $x \in T_1$.
Let $U$ be an open set in $T_1 \times \set b$ such that:
:$\map f x \in U$
By definition of the subspace topology then for some $U'$ open in $T_1 \times T_2$:
:$U' \cap \paren {T_1 \times \set b} = U$
By the definition of the Natural Basis of Product Topology, there exist open sets $V_1$ and $V_2$ in $T_1$ and $T_2$, respectively, such that:
:$\map f x = \tuple {x, b} \in V_1 \times V_2 \subseteq U'$
Then for any $y \in V_1$:
:$\map f y = \tuple {y, b} \in V_1 \times V_2 \subseteq U'$
But:
:$\tuple {y, b} \in T_1 \times \set b$
so:
:$\tuple {y, b} \in U$
Thus if $y \in V_1$, it follows that $\map f y \in U$.
Then $f$ is continuous by definition.
{{qed}}
Theorem: Subspace of Product Space is Homeomorphic to Factor Space

Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Suppose that $X$ is non-empty.
Then for each $i \in I$ there is a subspace $Y_i \subseteq X$ which is homeomorphic to $\struct {X_i, \tau_i}$.
Specifically, for any $z \in X$, let:
:$Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$
and let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
Then $\struct {Y_i, \upsilon_i}$ is homeomorphic to $\struct {X_i, \tau_i}$, where the homeomorphism is the restriction of the projection $\pr_i$ to $Y_i$.

Proof:
Let $z \in X$.
Let $i \in i$.
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$. 
For all $j \in I$ let:
::<nowiki>$Z_j = \begin{cases} X_i & i = j \\
\set{z_j} & j \ne i \end{cases}$</nowiki>
=== Lemma 1 ===
{{:Subspace of Product Space is Homeomorphic to Factor Space/Proof 1/Lemma 1}}{{qed|lemma}}
From Product Space of Subspaces is Subspace of Product Space , $\struct {Y_i, \upsilon_i}$ is a product space.
Consider the projection:
:$p_i: \struct {Y_i, \upsilon_i} \to \struct {X_i, \tau_i}$
From Projection from Product of Family is Injection iff Other Factors are Singletons, $p_i$ is injective.
From Projection from Product of Family is Surjective, $p_i$ is surjective.
From  Projection from Product Topology is Continuous:General Result, $p_i$ is continuous.
From  Projection from Product Topology is Open:General Result, $p_i$ is open.
Thus, by definition, we have that $p_i$ is a homeomorphism.
Consider the projection:
:$\pr_i: \struct {X, \tau} \to \struct {X_i, \tau_i}$
and the restriction:
:$\pr_i {\restriction_{Y_i} }: \struct {Y_i, \upsilon_i} \to \struct {X_i, \tau_i}$
=== Lemma 2 ===
{{:Subspace of Product Space is Homeomorphic to Factor Space/Proof 1/Lemma 2}}{{qed|lemma}}
Thus $\pr_i {\restriction_{Y_i} }: Y_i \to X_i$ is a homeomorphism.
{{qed}}
Theorem: Subspace of Product Space is Homeomorphic to Factor Space

Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.
Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.
Suppose that $X$ is non-empty.
Then for each $i \in I$ there is a subspace $Y_i \subseteq X$ which is homeomorphic to $\struct {X_i, \tau_i}$.
Specifically, for any $z \in X$, let:
:$Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$
and let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.
Then $\struct {Y_i, \upsilon_i}$ is homeomorphic to $\struct {X_i, \tau_i}$, where the homeomorphism is the restriction of the projection $\pr_i$ to $Y_i$.

Proof:
Let $z \in X$.
Let $i \in I$. 
Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.
Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$. 
Let $p_i = \pr_i {\restriction_{Y_i}}$.
Note that by definitions of a restriction and a projection then:
:$\forall y \in Y_i: \map {p_i} y = y_i$ 
=== $p_i$ is an injection ===
{{:Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Injection}}{{qed|lemma}}
=== $p_i$ is a surjection ===
{{:Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Surjection}}{{qed|lemma}}
=== $p_i$ is a continuous mapping ===
{{:Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Continuous Mapping}}{{qed|lemma}}
=== $p_i$ is an open mapping ===
{{:Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Open Mapping}}{{qed}}
Theorem: Closed Set of Countable Fort Space is G-Delta

Let $T = \struct {S, \tau_p}$ be a Fort space on a countably infinite set $S$.
Let $H \subseteq S$ be closed in $T$.
Then $H$ is a $G_\delta$ set.

Proof:
By definition of Fort space, $H$ is finite or contains $p$.
Consider the set of sets defined by:
:$\DD = \set {S \setminus \set z: z \notin H}$
Because $H$ is finite, $S \setminus H$ is countably infinite.
From its method of construction, $\DD$ has the same cardinality as $S \setminus H$ and so is countably infinite.
We have that the elements of $\DD$ are all open sets.
Thus, by definition, any set of the form $\ds H \subseteq \bigcap_{V \mathop \in \DD} V$ is a $G_\delta$ set.
Let $r \in H$.
Then:
:$\forall z \in S: z \notin H: r \in S \setminus \set z$
Thus by definition of $\DD$:
:$\ds H \subseteq \bigcap_{V \mathop \in \DD} V$
Let $r \notin H$.
Then:
:$\exists z \in S: r \notin S \setminus \set z \in \DD$
So:
:$\ds \relcomp S H \subseteq \relcomp S {\bigcap_{V \mathop \in \DD} V}$
and so:
:$\ds H = \bigcap_{V \mathop \in \DD} V$
Hence the result.
{{qed}}
Category:Countable Fort Spaces
Category:Examples of G-Delta Sets
664o6pc2cu1l3j908q1374651cy3j55
Theorem: Ring of Algebraic Integers

Let $K / \Q$ be a number field.
Let $\Z \sqbrk x$ denote the polynomial ring in one variable over $\Z$.
Let $\OO_K$ denote the set of all elements of $K / \Q$ which are a root of some monic polynomial $P \in \Z \sqbrk x$.
That is, let $\OO_K$ denote the algebraic integers over $K$.
Then $\OO_K$ is a ring, called the '''Ring of Algebraic Integers'''.

Proof:
This is a special case of Integral Closure is Subring.
We have an extension of commutative rings with unity, $\Z \subseteq K$, and $\OO_K$ is the integral closure of $\Z$ in $K$.
The theorem says that $\OO_K$ is a subring of $K$.
{{qed}}
Category:Algebraic Number Theory
oo93c5tvo6i3bunkms20vt61z8taawl
Theorem: Subtraction of Divisors obeys Distributive Law

{{:Euclid:Proposition/VII/7}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a - c = \dfrac 1 n \paren {b - d}$

Proof:
Let $AB$ be that aliquot part of the (natural) number $CD$ which $AE$ subtracted is of $CF$ subtracted.
We need to show that the remainder $EB$ is also the same part of the number $CD$ which $AE$ subtracted is of $CF$ subtracted.
:500px
Whatever part $AE$ is of $CF$, let the same part $EB$ be of $CG$.
Then from {{EuclidPropLink|book = VII|prop = 5|title = Divisors obey Distributive Law}}, whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $AB$ of $GF$.
But whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $AB$ of $CD$, by hypothesis.
Therefore, whatever aliquot part $AB$ is of $GF$, the same aliquot part is it of $CD$ also.
Therefore $GF = CD$.
Let $CF$ be subtracted from each.
Therefore $GC = FD$.
We have that whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $EB$ of $CG$
Therefore whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $EB$ of $FD$.
But whatever aliquot part $AE$ is of $CF$, the same aliquot part also is $AB$ of $CD$.
Therefore the remainder $EB$ is the same aliquot part of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
{{qed}}
Theorem: Subtraction of Divisors obeys Distributive Law

{{:Euclid:Proposition/VII/7}}
In modern algebraic language:
:$a = \dfrac 1 n b, c = \dfrac 1 n d \implies a - c = \dfrac 1 n \paren {b - d}$

Proof:
A direct application of the Distributive Property:
{{begin-eqn}}
{{eqn | l = \frac 1 n b - \frac 1 n d
      | r = \frac 1 n b + \frac 1 n \paren {-d}
      | c = 
}}
{{eqn | r = \frac 1 n \paren {b + \paren {-d} }
      | c = 
}}
{{eqn | r = \frac 1 n \paren {b - d}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Minimal Polynomial is Unique

Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then the minimal polynomial of $\alpha$ over $K$ is unique.

Proof:
Let $f$ be a minimal polynomial of $\alpha$ over $K$ according to definition 1.
By Minimal Polynomial is Irreducible, we have that $f$ is irreducible over $K$. 
Let $g$ be another polynomial in $K \sqbrk x$ such that $\map g \alpha = 0$.
By the definition of minimal polynomial, $\map \deg f \le \map \deg g$, where $\deg$ denotes degree.
By Division Theorem for Polynomial Forms over Field, there exist polynomials $q, r \in K \sqbrk x$ such that:
:$g = q f + r$ and $\map \deg r < \map \deg f$.
Suppose $\map \deg r > 0$.
Then evaluating both sides of the equation above at $\alpha$, we obtain $\map r \alpha = 0$.
This contradicts the minimality of the degree of $f$.
Thus, $r$ is constant and equal to $0$.
We have now shown that $f$ divides all polynomials in $K \sqbrk x$ which vanish at $\alpha$.
By the monic restriction, it then follows that $f$ is unique.
{{qed}}
Theorem: Minimal Polynomial is Unique

Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then the minimal polynomial of $\alpha$ over $K$ is unique.

Proof:
Let $f, g \in K \sqbrk x$ be minimal polynomials for $\alpha$ according to definition 2.
That is, let $f$ and $g$ be irreducible monic polynomials in $K \sqbrk x$ with $\map f \alpha = \map g \alpha = 0$.
Suppose $f$ and $g$ are distinct.
Then $f$ and $g$ are coprime.
Thus there exist polynomials $a, b \in K \sqbrk x$ with $a f + b g = 1$.
Taking the evaluation homomorphism in $\alpha$, we obtain the contradiction that $0=1$.
{{qed}}
Theorem: Minimal Polynomial is Irreducible

Let $L / K$ be a field extension.
Let $\alpha \in L$ be algebraic over $K$.
Then the minimal polynomial in $\alpha$ over $K$ is unique and irreducible.

Proof:
Let $\map f x$ be a minimal polynomial of $\alpha$ over $K$ of degree $n$.
From Minimal Polynomial is Unique we have that $\map f x$ is unique.
Since $K$ is a field, we may assume that the coefficient of $x^n$ is $1$.
{{AimForCont}} that $f$ is not irreducible.
Then there exist non-constant polynomials $g, h \in K \sqbrk x$ such that $f = g h$.
By definition of $f$ as the minimal polynomial in $\alpha$:
:$0 = \map f \alpha = \map g \alpha \, \map h \alpha$
From Field is Integral Domain, $L$ is an integral domain.
Therefore, as $\map g \alpha \, \map h \alpha \in L$, either $\map g \alpha = 0$ or $\map h \alpha = 0$.
This contradicts the minimality of the degree of $f$.
Hence the result, by Proof by Contradiction.
{{qed}}
Theorem: Equivalence of Definitions of Normal Extension

Let $L / K$ be an algebraic field extension.
{{TFAE|def = Normal Extension}}

Proof:
=== Definition $1$ implies Definition $2$ ===
Let $\alpha \in L$ be an arbitrary element.
Let $\sigma: L \mapsto \overline K$ be an arbitrary embedding of $L$ fixing $K$.
We wish to show that $\map \sigma \alpha \in L$.
Let $m_\alpha$ be the minimal polynomial of $\alpha$ over $K$, which exists because $L / K$ is algebraic.
Since $\sigma$ fixes $K$, $\map \sigma \alpha$ must also be a root of $m_\alpha$.
By our assumption, $\alpha \in L$ implies that all roots of $m_\alpha$ are in $L$ and consequently $\map \sigma \alpha \in L$.
{{qed|lemma}}
Theorem: Abstract Model of Algebraic Extensions

Let $K$ be a field and $\alpha \in \overline{K}$ be an element of the algebraic closure of $K$ which is algebraic over $K$.
Let $m_\alpha$ be the minimal polynomial of $\alpha$ over $K$.
Then:
:$K[\alpha] \cong K[x]/ \langle m_\alpha\rangle$

Proof:
Let $\phi: K[x] \to K [\alpha]$ be the homomorphism which fixes $K$ and maps $x$ to $\alpha$.
The kernel of $\phi$ consists of all polynomials which vanish at $\alpha$, which is precisely the ideal $\langle m_\alpha\rangle$.
The result follows by the First Fundamental Theorem on Ring Homomorphisms.
{{qed}}
Category:Field Extensions
av97dw9oef35ouryrxbxz5e0nxfkjas
Theorem: Union of Local Bases is Basis

Let $T = \struct {X, \tau}$ be a topological space.
For each $x \in X$, let $\BB_x$ be a local basis at $x$ which consists entirely of open sets.
Then $\ds \BB = \bigcup_{x \mathop \in X} \BB_x$ is a basis for the topology $\tau$.

Proof:
Let $U \in \tau$ be any open set of $T$.
Consider any $x \in U$.
Then, by definition of local basis, there exists a $B_x \in \BB_x$ such that $B_x \subseteq U$.
Also by definition of local basis:
:$x \in B_x$
So by Set Union Preserves Subsets:
:$U = \ds \bigcup_{x \mathop \in U} \set x \subseteq \bigcup_{x \mathop \in U} B_x$
Also, since $B_x \subseteq U$ for all $x$, by Union is Smallest Superset: General Result:
:$\ds \bigcup_{x \mathop \in U} B_x \subseteq U$
Thus by definition of set equality:
:$\ds U = \bigcup_{x \mathop \in U} B_x$
So, by definition, $\ds \BB = \bigcup_{x \mathop \in X} \BB_x$ is a basis for the topology $\tau$.
{{qed}}
Theorem: Subtraction of Multiples of Divisors obeys Distributive Law

{{:Euclid:Proposition/VII/8}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a - c = \dfrac m n \paren {b - d}$

Proof:
Let the (natural) number $AB$ be the same aliquant part of the (natural) number $CD$ that $AE$ subtracted is of $CF$ subtracted.
We need to show that $EB$ is also the same aliquant part of the remainder $FD$ that the whole $AB$ is of the whole $CD$.
:400px
Let $GH = AB$.
Then whatever aliquant part $GH$ is of $CD$, the same aliquant part also is $AE$ of $CF$.
Let $GH$ be divided into the aliquant parts of $CD$, namely $GK + KH$, and $AE$ into the aliquant parts of $CF$, namely $AL + LE$.
Thus the multitude of $GK, KH$ will be equal to the multitude of $AL, LE$.
We have that whatever aliquot part $GK$ is of $CD$, the same aliquot part also is $AL$ of $CF$.
We also have that $CD > CF$.
Therefore $GK > AL$.
Now let $GM = AL$.
Then whatever aliquot part $GK$ is of $CD$, the same aliquot part also is $GM$ of $CF$.
Therefore from Subtraction of Divisors obeys Distributive Law the remainder $MK$ is of the same aliquot part of the remainder $FD$ that the whole $GK$ is of the whole $CD$.
Again, we have that whatever aliquot part $KH$ is of $CD$, the same aliquot part also is $EL$ of $CF$.
We also have that $CD > CF$.
Therefore $HK > EL$.
Let $KN = EL$.
Then whatever aliquot part $KH$ is of $CD$, the same aliquot part also is $KN$ of $CF$.
Therefore from Subtraction of Divisors obeys Distributive Law the remainder $NH$ is of the same aliquot part of the remainder $FD$ that the whole $KH$ is of the whole $CD$.
But $MK$ was proved to be the same aliquot part of $FD$ that $GK$ is of $CD$.
Therefore $MK + NH$ is the same aliquant part of $DF$ that $HG$ is of $CD$.
But $MK + NH = EB$ and $HG = BA$.
Therefore $EB$ is the same aliquant part of $FD$ that $AB$ is of $CD$.
{{Qed}}
Theorem: Subtraction of Multiples of Divisors obeys Distributive Law

{{:Euclid:Proposition/VII/8}}
In modern algebraic language:
:$a = \dfrac m n b, c = \dfrac m n d \implies a - c = \dfrac m n \paren {b - d}$

Proof:
A direct application of the Distributive Property:
{{begin-eqn}}
{{eqn | l = \frac m n b - \frac m n d
      | r = \frac m n b + \frac m n \paren {-d}
      | c = 
}}
{{eqn | r = \frac m n \paren {b + \paren {-d} }
      | c = 
}}
{{eqn | r = \frac m n \paren {b - d}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Proportional Numbers have Proportional Differences

{{:Euclid:Proposition/VII/11}}
That is:
:$a : b = c : d \implies \left({a - c}\right) : \left({b - d}\right) = a : b$
where $a : b$ denotes the ratio of $a$ to $b$.

Proof:
As the whole $AB$ is to the whole $CD$, so let the $AE$ subtracted be to $CF$ subtracted.
We need to show that $EB : FD = AB : CD$.
:200px
We have that :$AB : CD = AE : CF$.
So from {{EuclidDefLink|VII|20|Proportional}} we have that whatever aliquot part or aliquant part $AB$ is of $CD$, the same aliquot part or aliquant part is $AE$ of $CF$.
So from:
: {{EuclidPropLink|book = VII|prop = 7|title = Subtraction of Divisors obeys Distributive Law}}
and:
: {{EuclidPropLink|book = VII|prop = 8|title = Subtraction of Multiples of Divisors obeys Distributive Law}}
$EB$ is the same aliquot part or aliquant part of $FD$ that $AB$ is of $CD$.
So by {{EuclidDefLink|VII|20|Proportional}} $EB : FD = AB : CD$.
{{qed}}
{{Euclid Note|11|VII}}
Theorem: Galois Group is Group

Let $L / K$ be a normal extension.
Let $\Gal {L / K}$ be the Galois group of $L / K$.
Then $\Gal {L / K}$ forms a group under the operation of composition of mappings.

Proof:
The Galois group $\Gal {L / K}$ of $L / K$ is defined as:
:$\Gal {L / K} = \leftset{\sigma: L \to L: \sigma}$ is an automorphism of $L$ such that $\sigma$ fixes $K$ point-wise$\rightset{}$
We have that $\Gal {L / K}$ is a subset of the automorphism group on $L / K$.
We initially note that the Identity Mapping is Automorphism which trivially fixes $K$.
Thus $\Gal {L / K}$ is not the empty set.
First, to show that $\Gal {L / K}$ is closed under composition of mappings.
Let $\sigma, \tau \in \Gal {L / K}$.
Let $\rho = \sigma \circ \tau$.
If $k \in K$, then:
:$\map \rho k = \map \sigma {\map \tau k} = \map \sigma k = k$
and so $\rho$ fixes $K$.
By Composite of Isomorphisms in Algebraic Structure is Isomorphism, $\rho$ is an automorphism.
Thus, $\Gal {L / K}$ is closed under composition.
Next, by Inverse of Algebraic Structure Isomorphism is Isomorphism, $\sigma^{-1}$ is an automorphism, and:
:$\forall k \in K: \map \sigma k = k \implies \map {\sigma^{-1} } k = k$
Thus, $\sigma^{-1} \in \Gal {L / K}$.
The result follows from the Two-Step Subgroup Test.
That is, $\Gal {L / K}$ is a group.
{{qed}}
Category:Galois Theory
f71qu2ra82nvdtzi8zei20o2ue5rpi3
Theorem: Relation of Ratios to Products

{{:Euclid:Proposition/VII/19}}
That is:
:$a : b = c : d \iff ad = bc$

Proof:
Let $A, B, C, D$ be four (natural) numbers in proportion, so that $A : B = C : D$.
Let $A \times D = E$ and $B \times C = F$.
We need to show that $E = F$.
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Let $A \times C = G$.
Then $A \times C = G$ and $A \times D = E$.
So from {{EuclidPropLink||book=VII|prop=17|title=Multiples of Ratios of Numbers}}:
: $C : D = G : E$
But we have:
: $C : D = A : B$
: $A : B = G : E$
: $A \times C = G$
: $B \times C = F$
Then from {{EuclidPropLink|book=VII|prop=18|title=Ratios of Multiples of Numbers}}:
: $A : B = G : F$
Further, we have that:
: $A : B = G : E$
Thus:
: $G : F = G : E$
So from {{EuclidPropLink|book=V|prop=9|title=Magnitudes with Same Ratios are Equal}}:
: $E = F$
{{qed|lemma}}
Now suppose that $E = F$.
We need to show that:
: $A : B = C : D$
Using the same construction, from {{EuclidPropLink|book=V|prop=7|title=Ratios of Equal Magnitudes}}:
: $G : E = G : F$
But from {{EuclidPropLink|book=VII|prop=17|title=Multiples of Ratios of Numbers}}:
: $G : E = C : D$
Then from {{EuclidPropLink|book=VII|prop=18|title=Ratios of Multiples of Numbers}}:
: $G : F = A : B$
So:
: $A : B = C : D$
{{qed}}
{{Euclid Note|19|VII}}