nyuuzyou/wikis huggingface dataset:
Ring of Continuous Real-Valued Functions is Ring

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $\struct {\map C {S, \R}, +, *}$ be the ring of continuous real-valued functions from $S$.
Then:
:$\struct {\map C {S, \R}, +, *}$ is a ring
== Proof ==
By definition of ring of continuous real-valued functions:
:$\struct {\map C {S, \R}, +, *}$ is the ring of continuous mappings from $S$ to $\R$.
From Ring of Continuous Mappings is Subring of All Mappings:
:$\struct {\map C {S, \R}, +, *}$ is a subring
By definition of subring:
:$\struct {\map C {S, \R}, +, *}$ is a ring
== Also see ==
* Zero of Ring of Continuous Real-Valued Functions
* Unity of Ring of Continuous Real-Valued Functions
* Ring of Continuous Real-Valued Functions is Commutative
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.3$
Category:Rings of Continuous Real-Valued Functions
Ring of Endomorphisms is Ring with Unity

== Theorem ==
Let $\struct {G, \oplus}$ be an abelian group.
Let $\mathbb G$ be the set of all group endomorphisms of $\struct {G, \oplus}$.
Let $\struct {\mathbb G, \oplus, *}$ denote the ring of endomorphisms on $\struct {G, \oplus}$.
Then $\struct {\mathbb G, \oplus, *}$ is a ring with unity.
== Proof ==
By Structure Induced by Group Operation is Group, $\struct {\mathbb G, \oplus}$ is an abelian group.
By Set of Homomorphisms to Abelian Group is Subgroup of All Mappings, it follows that $\struct {\mathbb G, \oplus}$ is a subgroup of $\struct {G^G, \oplus}$.
Next, we establish that $*$ is associative.
By definition, $\forall u, v \in \mathbb G: u * v = u \circ v$ where $u \circ v$ is defined as composition of mappings.
Associativity of $*$ follows directly from Composition of Mappings is Associative.
Next, we establish that $*$ is distributive over $\oplus$.
Let $u, v, w \in \mathbb G$.
Then:
:$\paren {u \oplus v} * w = \paren {u \oplus v} \circ w$
:$u * \paren {v \oplus w} = u \circ \paren {v \oplus w}$
So let $x \in G$.
Then:
So $\paren {u \oplus v} * w = \paren {u * w} \oplus \paren {v * w}$.
Similarly:
So:
:$u * \paren {v \oplus w} = \paren {u * v} \oplus \paren {u * w}$
Thus $*$ is distributive over $\oplus$.
The ring axioms are satisfied, and $\struct {\mathbb G, \oplus, *}$ is a ring.
The zero is easily checked to be the mapping which takes everything to the identity:
:$e: G \to \set {e_G}: \map e x = e_G$
The unity is easily checked to be the identity mapping, which is known to be an automorphism.
== Also see ==
* Ring of Endomorphisms is not necessarily Commutative Ring
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Example $21.2$
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $10$
Category:Rings of Endomorphisms
Ring of Endomorphisms is not necessarily Commutative Ring

== Theorem ==
Let $\struct {G, \oplus}$ be an abelian group.
Let $\mathbb G$ be the set of all group endomorphisms of $\struct {G, \oplus}$.
Let $\struct {\mathbb G, \oplus, *}$ denote the ring of endomorphisms on $\struct {G, \oplus}$.
Then $\struct {\mathbb G, \oplus, *}$ is not necessarily a commutative ring with unity.
== Proof ==
From Ring of Endomorphisms is Ring with Unity, we have that $\struct {\mathbb G, \oplus, *}$ is a ring with unity.
It remains to show that the operation $*$ is not necessarily commutative.
Category:Rings of Endomorphisms
Ring of Idempotents is Idempotent Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {A, \oplus, \circ}$ be its ring of idempotents.
Then $\struct {A, \oplus, \circ}$ is an idempotent ring.
== Proof ==
First, it is to be established that $\struct {A, \oplus, \circ}$ is a ring in the first place.
This we do by verifying the ring axioms.
=== ===
Let $x, y \in A$.
It is to be shown that $x \oplus y \in A$, i.e. that $x \oplus y$ is an idempotent element of $R$.
Compute as follows:
Hence $x \oplus y \in A$, as desired.
=== ===
Let $x, y, z \in A$.
It is to be shown that $\oplus$ is associative, that is:
:$\paren {x \oplus y} \oplus z = x \oplus \paren {y \oplus z}$
This is shown by the following computation:
=== ===
Let $x, y \in A$.
It is to be shown that $\oplus$ is commutative, that is:
:$x \oplus y = y \oplus x$
This is shown by the following computation:
=== ===
Let $x \in A$, and let $0_R$ be the zero of $R$.
By Ring Zero is Idempotent, $0_R \in A$.
It suffices to show that:
:$x \oplus 0_R = x$
since was already verified above.
Now:
as desired.
=== ===
Let $x \in A$.
It is to be shown that there exists $y \in A$ such that:
:$x \oplus y = y \oplus x = 0_R$
by $(A3)$ above.
In fact, one has:
so that each $x \in A$ is its own inverse for $\oplus$.
=== ===
Let $x, y \in A$.
It is to be shown that:
:$x \circ y \in A$
that is, that $x \circ y$ is idempotent.
We have the following computation:
=== ===
Immediate from Restriction of Associative Operation is Associative.
=== ===
By Restriction of Commutative Operation is Commutative, $\circ$ is commutative on $A$.
Thus to establish distributivity, it suffices to verify, for $x, y, z \in A$:
:$x \circ \paren {y \oplus z} = \paren {x \circ y} \oplus \paren {x \circ z}$
To this end, we compute as follows:
Therefore, having verified all ring axioms, we conclude $\struct {A, \oplus, \circ}$ is a ring.
By assumption all $x \in A$ are idempotent elements for $\circ$.
Thus $\circ$ is an idempotent operation on $A$.
Consequently, $\struct {A, \oplus, \circ}$ is an idempotent ring.
== Also see ==
* Ring of Idempotents of Commutative and Unitary Ring is Boolean Ring
== Sources ==
* : $\S 1$: Exercise $7$
Category:Ring Theory
Category:Idempotent Rings
Ring of Idempotents of Commutative and Unitary Ring is Boolean Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative and unitary ring.
Let $\struct {A, \oplus, \circ}$ be its ring of idempotents.
Then $\struct {A, \oplus, \circ}$ is a Boolean ring.
== Proof ==
From Ring of Idempotents is Idempotent Ring, $\struct {A, \oplus, \circ}$ is an idempotent ring.
By Unity is Unity in Ring of Idempotents, $\struct {A, \oplus, \circ}$ is also a unitary ring.
Hence, by definition, $\struct {A, \oplus, \circ}$ is a Boolean ring.
== Also see ==
* Ring of Idempotents is Idempotent Ring
== Sources ==
* : $\S 1$: Exercise $7$
Category:Ring Theory
Ring of Integers Modulo 2/Cayley Tables

== Cayley Tables for Ring of Integers Modulo $2$ ==
The Ring of Integers Modulo $2$:
:$\struct {\Z_2, +_2, \times_2}$
can be described completely by showing its Cayley tables:
:$\begin{array} {r|rr}
\struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\
\hline
\eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\
\eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\
\end{array}
\qquad
\begin{array}{r|rr}
\struct {\Z_2, \times_2} & \eqclass 0 2 & \eqclass 1 2 \\
\hline
\eqclass 0 2 & \eqclass 0 2 & \eqclass 0 2 \\
\eqclass 1 2 & \eqclass 0 2 & \eqclass 1 2 \\
\end{array}$
They can be presented more simply as:
:$\begin{array}{r|rr}
+ & 0 & 1 \\
\hline
0 & 0 & 1 \\
1 & 1 & 0 \\
\end{array}
\qquad
\begin{array}{r|rr}
\times & 0 & 1 \\
\hline
0 & 0 & 0 \\
1 & 0 & 1 \\
\end{array}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.1$
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Example $6.1$
Category:Examples of Cayley Tables
Ring of Integers Modulo Composite is not Integral Domain

== Theorem ==
Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$‎ be the ring of integers modulo $m$.
Let $m$ be a composite number.
Then $\struct {\Z_m, +, \times}$ is not an integral domain.
== Proof ==
Let $m \in \Z: m \ge 2$ be composite.
Then:
:$\exists k, l \in \N_{> 0}: 1 < k < m, 1 < l < m: m = k \times l$
Thus:
So $\struct {\Z_m, +, \times}$‎ is a ring with zero divisors.
So by definition $\struct {\Z_m, +, \times}$‎ is not an integral domain.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains
* : Chapter $5$: Rings: $\S 18$. Definition of a Ring: Example $29$
Category:Ring of Integers Modulo m
Category:Integral Domains
Ring of Integers Modulo Prime is Field

== Theorem ==
Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$‎ be the ring of integers modulo $m$.
Then:
:$m$ is prime
:
:$\struct {\Z_m, +, \times}$ is a field.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also see ==
* Definition:Field of Integers Modulo Prime
* Definition:Galois Field
== Sources ==
* : Introduction: Notation
* : $0$: Some Conventions and some Basic Facts
* : $\S 55$. Special types of ring and ring elements: $(4)$
Category:Ring of Integers Modulo m
Category:Galois Fields
Category:Ring of Integers Modulo Prime is Field
Ring of Integers Modulo Prime is Field/Proof 1

== Theorem ==
== Proof ==
=== Prime Modulus ===
$\struct {\Z_m, +, \times}$‎ is a commutative ring with unity by definition.
From Reduced Residue System under Multiplication forms Abelian Group, $\struct {\Z'_m, \times}$ is an abelian group.
$\Z'_m$ consists of all the elements of $\Z_m$ coprime to $m$.
Now when $m$ is prime, we have, from Reduced Residue System Modulo Prime:
:$\Z'_m = \set {\eqclass 1 m, \eqclass 2 m, \ldots, \eqclass {m - 1} m}$
That is:
:$\Z'_m = \Z_m \setminus \set {\eqclass 0 m}$
where $\setminus$ denotes set difference.
Hence the result.
=== Composite Modulus ===
Now suppose $m \in \Z: m \ge 2$ is composite.
From Ring of Integers Modulo Composite is not Integral Domain, $\struct {\Z_m, +, \times}$ is not an integral domain.
From Field is Integral Domain $\struct {\Z_m, +, \times}$ is not a field.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $5$
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 89$
* : $\S 19.2$: Properties of $\Z_m$ as an algebraic system
Category:Ring of Integers Modulo Prime is Field
Ring of Integers Modulo Prime is Field/Proof 2

== Theorem ==
== Proof ==
Let $p$ be prime.
From Irreducible Elements of Ring of Integers, we have that $p$ is irreducible in the ring of integers $\struct {\Z, +, \times}$.
From Ring of Integers is Principal Ideal Domain, $\struct {\Z, +, \times}$ is a principal ideal domain.
Let $\ideal p$ denote the principal ideal of $\struct {\Z, +, \times}$ generated by $p$.
By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\ideal p$ is a maximal ideal of $\struct {\Z, +, \times}$.
Let $\Z / \ideal p$ denote the quotient ring of $\Z$ by $\paren p$.
By Maximal Ideal iff Quotient Ring is Field, $\Z / \ideal p$, is a field.
But $\Z / \ideal p$ is exactly $\struct {\Z_p, +, \times}$.
Let $p$ be composite.
Then $p$ is not irreducible in $\struct {\Z, +, \times}$.
Thus by Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\ideal p$ is not a maximal ideal of $\struct {\Z, +, \times}$.
Hence by Maximal Ideal iff Quotient Ring is Field, $\Z / \ideal p$ is not a field.
== Sources ==
* : $\S 63$. Construction of fields as factor rings
Category:Ring of Integers Modulo Prime is Field
Ring of Integers Modulo Prime is Field/Proof 3

== Theorem ==
== Proof ==
Let $m$ be prime.
From Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain.
Let $\eqclass a m \ne \eqclass 0 m$ be a residue class modulo $m$.
We need to find a residue class modulo $m$ $\eqclass x m$ such that $\eqclass a m \eqclass x m = \eqclass 1 m$.
Because $m$ is prime and $m \nmid a$, we have:
:$\gcd \set {a, m} = 1$
Hence from Bézout's Identity:
:$1 = x a + y m$
for some $x, y \in \Z$.
Thus we have:
So every non-zero residue class modulo $m$ has an inverse.
So by definition $\struct {\Z_m, +, \times}$ is a field.
Now suppose $m \in \Z: m \ge 2$ is composite.
From Ring of Integers Modulo Composite is not Integral Domain, $\struct {\Z_m, +, \times}$ is not an integral domain.
From Field is Integral Domain $\struct {\Z_m, +, \times}$ is not a field.
== Sources ==
* : Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $20$
Category:Ring of Integers Modulo Prime is Field
Ring of Integers Modulo Prime is Field/Proof 4

== Theorem ==
== Proof ==
Let $m$ be prime.
From Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain.
From Finite Integral Domain is Galois Field, $\struct {\Z_m, +, \times}$ is a field.
Now suppose $m \in \Z: m \ge 2$ is composite.
From Ring of Integers Modulo Composite is not Integral Domain, $\struct {\Z_m, +, \times}$ is not an integral domain.
From Field is Integral Domain $\struct {\Z_m, +, \times}$ is not a field.
== Sources ==
* : Chapter $4$: Fields: $\S 15$. Examples of Fields
Category:Ring of Integers Modulo Prime is Field
Ring of Integers Modulo Prime is Integral Domain

== Corollary to Ring of Integers Modulo Prime is Field ==
Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$‎ be the ring of integers modulo $m$.
Then:
:$m$ is prime
:
:$\struct {\Z_m, +, \times}$ is an integral domain.
== Proof 1 ==
== Proof 2 ==
Category:Ring of Integers Modulo m
Category:Integral Domains
Category:Ring of Integers Modulo Prime is Field
Category:Ring of Integers Modulo Prime is Integral Domain
Ring of Integers Modulo Prime is Integral Domain/Proof 2

== Theorem ==
== Proof ==
From Ring of Integers Modulo m is Ring, $\struct {\Z_m, +, \times}$‎ is a ring.
It remains to be shown that $\struct {\Z_m, +, \times}$‎ has no proper zero divisors $m$ is prime.
=== $m$ Composite ===
Let $m$ be composite.
Then:
:$m = m_1 m_2$
where $0 < m_1 < m, 0 < m_2 < m$.
Then:
:$\eqclass {m_1} m \eqclass {m_2} m = \eqclass 0 m$
and so both $\eqclass {m_1} m$ and $\eqclass {m_2} m$ are proper zero divisors.
Hence if $m$ is not prime then $\struct {\Z_m, +, \times}$‎ is not an integral domain by definition.
=== $m$ Prime ===
Let $m$ be prime.
Let $\eqclass a m \eqclass b m = \eqclass 0 m$.
Then by Modulo Multiplication is Well-Defined:
:$\eqclass {a b} m = \eqclass 0 m$
Thus either
:$m \divides a$ or $m \divides b$
where $\divides$ denotes the divisibility relation.
If $m \divides a$ then $\eqclass a m = 0$.
If $m \divides b$ then $\eqclass b m = 0$.
Thus neither $a$ nor $b$ is a proper zero divisor.
Hence there are no proper zero divisors of $\struct {\Z_m, +, \times}$.
Hence, by definition, $\struct {\Z_m, +, \times}$‎ is an integral domain
== Sources ==
* : Chapter $1$: Integral Domains: $\S 6$. The Residue Classes: Theorem $6$
Category:Ring of Integers Modulo Prime is Integral Domain
Ring of Integers Modulo m cannot be Ordered Integral Domain

== Theorem ==
Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$‎ be the ring of integers modulo $m$.
Then $\struct {\Z_m, +, \times}$ cannot be an ordered integral domain.
== Proof ==
First note that from Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$‎ is an integral domain only when $m$ is prime.
So for $m$ composite the result holds.
If $m$ is prime, and $\struct {\Z_m, +, \times}$ is therefore an integral domain, its order is finite. 
The result follows from Finite Integral Domain cannot be Ordered.
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $11$
Category:Ring of Integers Modulo m
Category:Ordered Integral Domains
Ring of Integers Modulo m is Ring

== Theorem ==
For all $m \in \N: m \ge 2$, the ring of integers modulo $m$:
:$\struct {\Z_m, +_m, \times_m}$
is a commutative ring with unity $\eqclass 1 m$.
The zero of $\struct {\Z_m, +_m, \times_m}$ is $\eqclass 0 m$.
== Proof ==
First we check the ring axioms:
:
::The Integers Modulo $m$ under Addition form Abelian Group.
::From Modulo Addition has Identity, $\eqclass 0 m$ is the identity of the additive group $\struct {\Z_m, +_m}$.
From Integers Modulo m under Multiplication form Commutative Monoid:
::
::$\struct {\Z_m, \times_m}$ is closed.
::
::$\struct {\Z_m, \times_m}$ is associative.
::
::$\struct {\Z_m, \times_m}$ has an identity $\eqclass 1 m$.
::
::$\struct {\Z_m, \times_m}$ is commutative.
Then:
::
:: $\times_m$ distributes over $+_m$ in $\Z_m$.
== Also see ==
* Quotient Epimorphism from Integers by Principal Ideal
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $4$
* : Chapter $1$: Integral Domains: $\S 6$. The Residue Classes: Theorem $5$
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $2$
Category:Commutative Algebra
Category:Ring of Integers Modulo m
Ring of Integers has no Zero Divisors

== Theorem ==
The integers have no zero divisors:
:$\forall x, y, \in \Z: x \times y = 0 \implies x = 0 \lor y = 0$
This can equivalently be expressed:
:$\forall x, y, \in \Z: x \ne 0 \land y \ne 0 \implies x \times y \ne 0$
== Proof ==
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.
$\boxminus$ is the congruence relation defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested.
From the method of construction, $\eqclass {c, c} {}$, where $c$ is any element of the natural numbers $\N$, is the identity of $\struct {\Z, +}$.
To ease the algebra, we will take $\eqclass {0, 0} {}$ as a canonical instance of this equivalence class.
We need to show that:
:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {0, 0} {} \implies \eqclass {a, b} {} = \eqclass {0, 0} {} \lor \eqclass {c, d} {} = \eqclass {0, 0} {}$
From Natural Numbers form Commutative Semiring, we can take it for granted that:
:addition and multiplication are commutative and associative on the natural numbers $\N$
:natural number multiplication is distributive over natural number addition.
So:
We have to be careful here, and bear in mind that $a, b, c, d$ are natural numbers, and we have not defined (and, at this stage, will not define) subtraction on such entities.
, suppose that $\eqclass {c, d} {} \ne \eqclass {0, 0} {}$.
Then $c \ne d$.
, suppose also that $c > d$.
From Ordering in terms of Addition, $\exists p \in \N: d + p = c$ where $p > 0$.
Then:
Similarly for when $c < d$.
Thus:
:$\eqclass {c, d} {} \ne \eqclass {0, 0} {} \implies \eqclass {a, b} {} = \eqclass {0, 0} {}$
A similar argument shows that:
:$\eqclass {a, b} {} \ne \eqclass {0, 0} {} \implies \eqclass {c, d} {} = \eqclass {0, 0} {}$
The equivalence between the two forms of the statement of this theorem follows from De Morgan's Laws: Conjunction of Negations and the Rule of Transposition.
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf Z. \, 6$
Category:Integers
Ring of Integers is Principal Ideal Domain/Proof 1

== Theorem ==
== Proof ==
Let $J$ be an ideal of $\Z$.
Then $J$ is a subring of $\Z$, and so $\left({J, +}\right)$ is a subgroup of $\left({\Z, +}\right)$.
But by Integers under Addition form Infinite Cyclic Group, the group $\left({\Z, +}\right)$ is cyclic, generated by $1$.
Thus by Subgroup of Cyclic Group is Cyclic, $\left({J, +}\right)$ is cyclic, generated by some $m \in \Z$.
Therefore from the definition of principal ideal, $J = \left\{{k m: k \in \Z}\right\} = \left({m}\right)$, and is thus a principal ideal.
Category:Ring of Integers is Principal Ideal Domain
Ring of Integers is Principal Ideal Domain/Proof 3

== Theorem ==
== Proof ==
Let $U$ be an arbitrary ideal of $\Z$.
Let $c$ be a non-zero element of $U$.
Then both $c$ and $-c$ belong to $\ideal a$ and one of them is positive.
Thus $U$ contains strictly positive elements.
Let $b$ be the smallest strictly positive element of $U$.
By the Set of Integers Bounded Below by Integer has Smallest Element, $b$ is guaranteed to exist.
If $\ideal b$ denotes the ideal generated by $b$, then $\ideal b \subseteq U$ because $b\in U$ and $U$ is an ideal.
Let $a \in U$.
By the Division Theorem:
:$\exists q, r \in \Z, 0 \le r < b: a = b q + r$
As $a, b \in U$ it follows that so does $r = a - b q$.
By definition of $b$ it follows that $r = 0$.
Thus:
:$a = b q \in \ideal b$
and so:
:$U \subseteq \ideal b$
From the above:
:$U = \ideal b$
It follows by definition that $U$ is a principal ideal of $\Z$.
Recall that $U$ was an arbitrary ideal of $\Z$.
Hence by definition $\Z$ is a principal ideal domain.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.3$
* : Chapter $5$: Rings: $\S 21$. Ideals: Theorem $38$
Category:Ring of Integers is Principal Ideal Domain
Ring of Integers of Algebraic Number Field is UFD iff Class Number is 1

== Theorem ==
Let $K$ be a field of algebraic numbers.
Let $\OO_K$ be the ring of integers of $K$.
Then $\OO_K$ is a unique factorization domain (UFD) the class number of $K$ is $1$.
== Proof ==
== Sources ==
* 
* 
Category:Class Groups
Ring of Integers of Number Field is Free Z-Module

== Theorem ==
Let $K$ be an algebraic number field.
Let $\OO_K$ be its ring of integers.
Let $\sqbrk {K : \Q}$ denote the degree of field extension $K : \Q$.
Then $\OO_K$ is a free $\Z$-module of dimension $\sqbrk {K : \Q}$.
== Proof ==
== Sources ==
* : Chapter $2$ Ring of Integers: Rings of integers are finitely generated: Proposition $2.29$
Category:Rings of Integers of Number Fields
Category:Free Modules
Category:Algebraic Number Theory
Ring of Linear Operators is Ring

== Theorem ==
Let $R$ be a ring.
Let $\struct {G, +}$ be an abelian group..
Let $\struct {G, +, \circ}_R$ be an $R$-module.
Let $\struct {\map {\LL_R} G, +, \circ}$ be the ring of linear operators on $G$, where:
:$+$ denotes pointwise addition
:$\circ$ denotes composition of linear operators.
Then $\struct {\map {\LL_R} G, +, \circ}$ is a ring.
== Proof ==
Let $\phi$ and $\psi$ be elements of $\map {\LL_R} G$.
From Composite of R-Algebraic Structure Homomorphisms is Homomorphism, $\phi \circ \psi$ is also an element of $\map {\LL_R} G$.
That is, $\struct {\map {\LL_R} G, \circ}$ is closed.
From Set of Linear Transformations under Pointwise Addition forms Abelian Group, $\struct {\map {\LL_R} G, +}$ is an abelian group.
Hence by the One-Step Subgroup Test:
:$\forall \phi, \psi \in \map {\LL_R} G: \phi + \paren {-\psi} \in \map {\LL_R} G$
Then by the Subring Test:
:$\struct {\map {\LL_R} G, +, \circ}$ is a subring of the ring of endomorphisms of $\struct {G, +}$
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Ring of Linear Operators
Ring of Polynomial Forms is Commutative Ring with Unity

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $A = R \sqbrk {\set {X_j: j \in J} }$ be the set of all polynomial forms over $R$ in the indeterminates $\set {X_j: j \in J}$.
Then $\struct {A, +, \circ}$ is a commutative ring with unity.
== Proof ==
We must show that the commutative and unitary ring axioms are satisfied:
=== Proof of the additive axioms ===
A1:
This is shown by Polynomials Closed under Addition.
A2-A5:
According to the formal definition, a polynomial is a map from the free commutative monoid to $R$.
Now observe that addition of polynomial forms is induced by addition in $R$.
Therefore:
:A2 is shown by Structure Induced by Associative Operation is Associative
:A3 is shown by Induced Structure Identity
:A4 is shown by Pointwise Inverse in Induced Structure
:A5 is shown by Structure Induced by Commutative Operation is Commutative
=== Proof of the multiplicative axioms ===
M1:
This is shown by Polynomials Closed under Ring Product.
Multiplication of polynomial forms is not induced by multiplication in $R$, so we must show the multiplicative axioms by hand.
M2:
This is shown by Multiplication of Polynomials is Associative.
M3:
This is shown by Polynomials Contain Multiplicative Identity.
M4:
This is shown by Multiplication of Polynomials is Commutative.
D:
This is shown by Multiplication of Polynomials Distributes over Addition.
Therefore, all of the axioms of a commutative ring with unity are satisfied.
Category:Polynomial Rings