nyuuzyou/wikis huggingface dataset:
Ring Homomorphism whose Kernel contains Ideal

== Theorem ==
Let $R$ be a ring.
Let $J$ be an ideal of $R$.
Let $\nu: R \to R / J$ be the quotient epimorphism.
Let $\phi: R \to S$ be a ring homomorphism such that:
:$J \subseteq \map \ker \phi$
where $\map \ker \phi$ is the kernel of $\phi$.
Then there exists a unique ring homomorphism $\psi: R / J \to S$ such that:
:$\phi = \psi \circ \nu$
where $\circ$ denotes composition of mappings.
:$\begin {xy} \xymatrix@L + 2mu@ + 1em {
 R \ar[rr]^*{\nu}
 \ar[rdrd]_*{\phi}
& & R / J \ar[dd]_*{\psi} \\
\\
& & S 
} \end {xy}$
Also:
:$\map \ker \psi = \map \ker \phi / J$
== Proof ==
Suppose $\phi = \psi \circ \nu$.
Let $J + x$ be an arbitrary element of $R / J$.
Then:
:$(1) \quad \map \psi {J + x} = \psi \circ \map \nu x = \map \phi x$
So there is only one possible way to define $\psi$.
Now suppose $J + x = J + x'$.
Then $x + \paren {-x'} \in J$.
So $x + \paren {-x'} \in \map \ker \phi$ as $J \subseteq \map \ker \phi$.
That is, $\map \phi {x + \paren {-x'} } = 0_S$.
So $\map \phi x = \map \phi {x'}$ and so $\psi$ as defined in $(1)$ is well-defined.
Now suppose $J + x, J + y \in R / J$.
We have:
So $\psi$ preserves ring addition.
Then:
So $\psi$ preserves ring product, and so $\psi$ is a ring homomorphism.
Finally:
So:
:$\map \ker \psi = \map \ker \phi / J$
== Sources ==
* : $\S 2.2$: Homomorphisms: Theorem $2.8$
Category:Ideal Theory
Ring Isomorphic to Polynomial Ring is Polynomial Ring/One Variable

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $R \sqbrk X$ be a polynomial ring in one variable $X$ over $R$.
Let $\iota : R \to R \sqbrk X$ denote the canonical embedding.
Let $S$ be a commutative ring with unity and $f: R \sqbrk X \to S$ be a ring isomorphism.
Then $\struct {S, f \circ \iota, \map f X}$ is a polynomial ring in one variable $\map f X$ over $R$.
== Proof ==
Category:Polynomial Rings
Ring Less Zero is Semigroup for Product iff No Proper Zero Divisors

== Theorem ==
Let $\struct {R, +, \circ}$ be a non-null ring.
Then $R$ has no zero divisors $\struct {R^*, \circ}$ is a semigroup.
== Proof ==
=== Necessary Condition ===
Let $\struct {R, +, \circ}$ be a non-null ring with no zero divisors.
The set $R^* = R \setminus \set {0_R} \ne \O$ as $\struct {R, +, \circ}$ is non-null.
All elements of $R$ are not zero divisors, and therefore are cancellable.
$\struct {R, +, \circ}$ is closed under $\circ$, from the fact that there are no zero divisors, and also that $\struct {R, \circ}$ is also closed.
From Restriction of Associative Operation is Associative, ring product is associative on $\struct {R^*, +, \circ}$, as it is associative on $\struct {R, +, \circ}$.
Thus $\struct {R^*, \circ}$ is a semigroup.
=== Sufficient Condition ===
Let $\struct {R^*, \circ}$ be a semigroup.
Then:
:$\neg \exists x, y \in R^*: x \circ y \notin R^*$
Thus:
:$\neg \exists x, y \in R^*: x \circ y = 0_R$
and the result follows.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.2$: Corollary
* : $\S 55.2$ Special types of ring and ring elements
Category:Ring Theory
Category:Semigroups
Category:Zero Divisors
Ring Monomorphism from Integers to Rationals

== Theorem ==
Let $\phi: \Z \to \Q$ be the mapping from the integers $\Z$ to the rational numbers $\Q$ defined as:
:$\forall x \in \Z: \map \phi x = \dfrac x 1$
Then $\phi$ is a (ring) monomorphism, but specifically not an epimorphism.
== Proof ==
First note that:
:$\forall a, b \in \Z: a \ne b \implies \dfrac a 1 = \map \phi a \ne \map \phi b = \dfrac b 1$
and so clearly $\phi$ is an injection.
However, take for example $\dfrac 1 2$:
:$\not \exists a \in \Z: \map \phi a = \dfrac 1 2$
as $\dfrac 1 2 \notin \Img \phi$.
So $\phi$ is not a surjection.
Next let $a, b \in \Z$.
So $\phi$ can be seen to be an injective, but not surjective, ring homomorphism.
Hence the result by definition of ring monomorphism and ring epimorphism.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Example $4$
Category:Ring Monomorphisms
Category:Integers
Category:Rational Numbers
Ring Negative is Unique

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $a \in R$.
Then the ring negative $-a$ of $a$ is unique.
== Proof ==
The ring negative is, by definition of a ring, the inverse element of $a$ in the additive group $\struct {R, +}$.
The result then follows from Inverse in Group is Unique.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Ring Theory
Ring Operations on Coset Space of Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\powerset R$ be the power set of $R$.
Let $J$ be an ideal of $R$.
Let $X$ and $Y$ be cosets of $J$.
Let $X +_\PP Y$ be the sum of $X$ and $Y$, where $+_\PP$ is the operation induced on $\powerset R$ by $+$.
Similarly, let $X \circ_\PP Y$ be the product of $X$ and $Y$, where $\circ_\PP$ is the operation induced on $\powerset R$ by $\circ$.
=== Sum of Cosets of Ideals is Sum in Quotient Ring ===
=== Product of Cosets of Ideals is Subset of Product in Quotient Ring ===
== Examples ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
Category:Ideal Theory
Category:Ring Operations on Coset Space of Ideal
Ring Operations on Coset Space of Ideal/Examples/Integer Multiples of 5

== Example of Use of Ring Operations on Coset Space of Ideal ==
Let $\ideal 5$ denote the set of all integer multiples of $5$.
Then their product $\ideal 5 \circ_\PP \ideal 5$ in $\powerset \Z$ is a proper subset of their product in $\Z / \ideal 5$.
== Proof ==
$\ideal 5$ is a principal ideal of the ring $\Z$.
In the ring $\Z / \ideal 5$ we have:
:$\ideal 5 \circ \ideal 5 = \ideal 5$
However, in $\powerset \Z$, we have $\ideal 5 \circ_\PP \ideal 5 = \ideal {25}$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
Category:Ring Operations on Coset Space of Ideal
Ring Product preserves Inequalities on Positive Elements

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be an ordered ring.
Let $x, y, z, w \in R$.
Let $0 < x < y$ and $0 < z < w$.
Then:
:$0 < z \circ x < w \circ y$
== Proof ==
By Properties of Ordered Ring $(6)$:
:$z \circ x < z \circ y$
:$z \circ y < w \circ y$
Then by transitivity of $\circ$:
:$z \circ x < w \circ y$
Also by Properties of Ordered Ring $(6)$:
:$z \circ 0 < z \circ x$
Hence by Ring Product with Zero:
:$0 < z \circ x$
Category:Ordered Rings
Category:Inequalities
Ring Product with Zero

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Then:
:$\forall x \in R: 0_R \circ x = 0_R = x \circ 0_R$
That is, the zero is a zero element for the ring product, thereby justifying its name.
== Proof ==
Because $\struct {R, +, \circ}$ is a ring, $\struct {R, +}$ is a group.
Since $0_R$ is the identity in $\struct {R, +}$, we have $0_R + 0_R = 0_R$.
From the Cancellation Laws, all group elements are cancellable, so every element of $\struct {R, +}$ is cancellable for $+$.
Thus:
Next:
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.9$
* : Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $2 \ \text{(iii)}$
* : Chapter $1$: Rings - Definitions and Examples: $3$: Some special classes of rings: Lemma $1.2 \ \text{(i)}$
* : $\S 54.1$ The definition of a ring and its elementary consequences
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
* 
* : $\S 1$: Exercise $6$
Category:Ring Theory
Ring Subtraction equals Zero iff Elements are Equal

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$
Then:
:$\forall a, b \in R: a - b = 0_R \iff a = b$
where $a - b$ denotes ring subtraction.
== Proof ==
== Sources ==
* : Chapter $1$: Integral Domains: $\S 4$. Elementary Properties: Theorem $2 \ \text{(ii)}$
Category:Ring Theory
Ring Without Unity may have Quotient Ring with Unity

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $I$ be an ideal of $R$.
Let $\struct {R / I, +, \circ}$ be the associated quotient ring.
Then $\struct {R / I, +, \circ}$ may have a unity even if $\struct {R, +, \circ}$ has not.
== Proof ==
Consider the external direct product of rings $\Z \oplus 2 \Z$.
From Integer Multiples form Commutative Ring, $2 \Z$ does not admit a unity.
By Unity of External Direct Sum of Rings, neither does $\Z \oplus 2 \Z$.
Now consider the ideal $\set 0 \times 2 \Z$ of $\Z \oplus 2 \Z$.
We have for all $a \in \Z$ and $b, c \in 2 \Z$ that:
:$\tuple {0, b} - \tuple {0, c} = \tuple {0, b - c}$
:$\tuple {a, b} \cdot \tuple {0, c} = \tuple {0, b \cdot c}$
so by the Test for Ideal, indeed $\set 0 \times 2 \Z$ is an ideal in $\Z \oplus 2 \Z$.
By Quotient Ring of External Direct Sum of Rings, we have:
:$\paren {\Z \oplus 2 \Z} / \paren {\set 0 \times 2 \Z} \cong \paren {Z / \set 0} \oplus \paren {2 \Z / 2 \Z}$
By Quotient Ring by Null Ideal and Quotient Ring Defined by Ring Itself is Null Ring, this last ring is isomorphic to $\Z \times \set 0 \cong \Z$.
Since $\Z$ has a unity, this construction provides an example of the required kind.
== Also see ==
* Quotient Ring of Ring with Unity is Ring with Unity - this result demonstrates that its converse does not hold.
Category:Rings with Unity
Category:Quotient Rings
Ring Zero is Idempotent

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose ring zero is $0_R$.
Then $0_R$ is an idempotent element of $R$ under the ring product $\circ$:
:$0_R \circ 0_R = 0_R$
== Proof ==
By Ring Product with Zero (applied to $0_R$):
:$0_R \circ 0_R = 0_R$
which was to be proven. 
Category:Ring Theory
Ring Zero is Unique/Proof 3

== Theorem ==
== Proof ==
Suppose $0$ and $0'$ are both ring zeroes of $\struct {R, +, \circ}$.
Then by Ring Product with Zero:
:$0' \circ 0 = 0$ by dint of $0$ being a ring zero
:$0' \circ 0 = 0'$ by dint of $0'$ being a ring zero.
So $0 = 0' \circ 0 = 0'$.
So $0 = 0'$ and there is only one ring zero of $\struct {R, +, \circ}$ after all.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Ring Zero is Unique
Ring Zero is not Cancellable

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring which is not null.
Let $0$ be the ring zero of $R$.
Then $0$ is not a cancellable element for the ring product $\circ$.
== Proof ==
 $0$ is cancellable.
Let $a, b \in R$ such that $a \ne b$.
By definition of ring zero:
:$0 \circ a = 0 = 0 \circ b$
By our supposition that $0$ is cancellable:
:$a = b$
The result follows by Proof by Contradiction.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains
Category:Ring Theory
Ring by Idempotent

== Theorem ==
Let $\struct {A, +, \circ}$ be a commutative ring.
Let $e$ be an idempotent element of $A$.
Then the ideal $I := \ideal e$ generated by $e$ is a commutative ring with unity $\struct {I, +, \circ}$ with unity $e$.
== Proof ==
Because $\struct {I, +}$ is an ideal of $\struct {A, +, \circ}$, it follows that $\struct {I, +, \circ}$ is a ring (not necessarily unital).
By definition of generated ideal, $I$ is the intersection of all ideals of $A$ containing $e$.
Thus $e \mathop \in I$.
Let $e a \in I$ for some arbitrary $a \in A$.
Then
It follows that $e$ is a unity of $\struct {I, +, \circ}$.
Hence $\struct {I, +, \circ}$ is a commutative ring with unity with unity $e$.
Category:Ring Theory
Ring is Commutative iff Opposite Ring is Itself

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Let $\struct {R, +, *}$ be the opposite ring of $\struct {R, +, \times}$.
Then $\struct {R, +, \times}$ is a commutative ring :
:$\struct {R, +, \times} = \struct {R, +, *}$
== Proof ==
By definition of the opposite ring:
:$\forall x, y \in R: x * y = y \times x$
=== Necessary Condition ===
Let $\struct {R, +, \times}$ be a commutative ring. 
Then:
=== Sufficient Condition ===
Let $\struct {R, +, \times} = \struct {R, +, *}$.
Then:
Thus by definition $\struct {R, +, \times}$ is a commutative ring.
== Sources ==
* : Chapter $4$: Rings and Modules: $\S 4.1$: The Definitions Recalled
Category:Examples of Rings
Ring is Ideal of Itself

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Then $R$ is an ideal of $R$.
== Proof ==
From Ring is Subring of Itself, $\struct {R, +, \circ}$ is a subring of $\struct {R, +, \circ}$.
Also:
:$\forall x, y \in \struct {R, +, \circ}: x \circ y \in R$
thus fulfilling the condition for $R$ to be an ideal of $R$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
* : Chapter $5$: Rings: $\S 22$. Quotient Rings: Example $40$
* : $\S 2.2$: Homomorphisms: Definition $2.5$
* : $\S 58.1$ Ideals
Category:Ideal Theory
Ring is Module over Itself/Proof 1

== Theorem ==
== Proof ==
Note that:
$\struct {R, +, \circ}$ is a ring by assumption.
$\struct {R, +}$ is an abelian group by the definition of a ring.
Let us verify the module axioms:
Axiom $(1)$ and $(2)$ follow from distributivity of $\circ$.
Axiom $(3)$ follows from associativity of $\circ$. 
Category:Ring is Module over Itself
Ring is Subring of Itself

== Theorem ==
Let $R$ be a ring.
Then $R$ is a subring of itself.
== Proof ==
$R$ is a ring and $R \subseteq R$.
It follows by definition that $R$ is a subring of $R$.
== Sources ==
* : $\S 56.1$ Subrings and subfields
Category:Subrings
Ring is not Empty

== Theorem ==
A ring cannot be empty.
== Proof ==
In a ring $\struct {R, +, \circ}$, $\struct {R, +}$ forms a group.
From Group is not Empty, the group $\struct {R, +}$ contains at least the identity, so cannot be empty.
So every ring $\struct {R, +, \circ}$ contains at least the identity for ring addition.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $1$: The definition of a ring: Definitions $1.1 \ \text{(c)}$
Category:Ring Theory
Ring of Algebraic Integers

== Theorem ==
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 $K$.
Category:Algebraic Number Theory
Ring of Arithmetic Functions is Ring with Unity

== Theorem ==
Let $\AA$ be the set of all arithmetic functions.
Let $*$ denote Dirichlet convolution, and $+$ the pointwise sum of functions.
The ring of arithmetic functions $\struct {\AA, +, *}$ is a commutative ring with unity.
== Proof ==
By Structure Induced by Abelian Group Operation is Abelian Group, $\struct {\AA, +}$ is an abelian group.
By Properties of Dirichlet Convolution, $*$ is commutative, associative and has a unity.
Therefore $\struct {\AA, +, *}$ is a commutative ring with unity.
Category:Arithmetic Functions
Category:Examples of Rings
Ring of Bounded Continuous Functions is Ring of Continuous Functions for Compact Space

== Theorem ==
Let $\struct {K, \tau}$ be a compact space.
Let $\R$ denote the real number line.
Let $\struct {\map C {S, \R}, +, *}$ be the ring of continuous real-valued functions from $S$.
Let $\struct {\map {C^*} {S, \R}, +, *}$ be the ring of bounded continuous real-valued functions from $S$.
Then:
:$\struct {\map {C^*} {S, \R}, +, *} = \struct {\map C {S, \R}, +, *}$
== Proof ==
Follows immediately from:
* Compact Space is Pseudocompact
* Ring of Bounded Continuous Functions is Ring of Continuous Functions for Pseudocompact Space
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.4$
Category:Rings of Continuous Real-Valued Functions
Category:Rings of Bounded Continuous Real-Valued Functions
Category:Compact Spaces
Ring of Bounded Continuous Functions is Ring of Continuous Functions for Pseudocompact Space

== Theorem ==
Let $\struct {K, \tau}$ be a pseudocompact space.
Let $\R$ denote the real number line.
Let $\struct {\map C {S, \R}, +, *}$ be the ring of continuous real-valued functions from $S$.
Let $\struct {\map {C^*} {S, \R}, +, *}$ be the ring of bounded continuous real-valued functions from $S$.
Then:
:$\struct {\map {C^*} {S, \R}, +, *} = \struct {\map C {S, \R}, +, *}$
== Proof ==
Follows immediately from the definitions of:
* Definition:Pseudocompact Space
* Definition:Ring of Continuous Real-Valued Functions
* Definition:Ring of Bounded Continuous Real-Valued Functions
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.4$
Category:Rings of Continuous Real-Valued Functions
Category:Rings of Bounded Continuous Real-Valued Functions
Category:Pseudocompact Spaces
Ring of Bounded Continuous Functions is Subring of Continuous Real-Valued Functions

== 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 $S$.
Let $\struct {\map {C^*} {S, \R}, +, *}$ be the ring of bounded continuous real-valued functions from $S$.
Then:
:$\struct {\map {C^*} {S, \R}, +, *}$ is a subring of $\struct {\map C {S, \R}, +, *}$
== Proof ==
From Ring of Continuous Real-Valued Functions is Ring:
:$\struct {\map C {S, \R}, +, *}$ is a ring.
From Additive Inverse in Ring of Continuous Real-Valued Functions:
:$\forall f \in R^S :$ the additive inverse of $f$ is the pointwise negation $-f$, defined by:
::$\forall s \in S: \map {\paren {-f} } s := - \map f s$
From the Subring Test:
:$\struct {\map {C^*} {S, \R}, +, *}$ is a subring of $\struct {\map C {S, \R}, +, *}$
:$(1) \quad \map {C^*} {S, R} \ne \O$
:$(2) \quad \forall f, g \in \map {C^*} {S, R} : f + \paren{-g} \in \map {C^*} {S, R}$
:$(3) \quad \forall f, g \in \map {C^*} {S, R} : f * g \in \map {C^*} {S, R}$
=== $(1) \quad \map {C^*} {S, \R} \ne \O$ ===
Let $0_{\R^S}: S \to \R$ denote the constant mapping defined by:
:$\forall s \in S : \map {0_{\R^S}} s = 0$
From Constant Mapping is Continuous and Constant Real-Valued Function is Bounded:
:$0_{\R^S} \in \map {C^*} {S, \R}$
It follows that:
:$\map {C^*} {S, \R} \ne \O$
=== $(2) \quad \forall f, g \in \map {C^*} {S, \R} : f + \paren{-g} \in \map {C^*} {S, \R}$ ===
Let $f, g \in \map {C^*} {S, \R}$.
From Negation Rule for Bounded Continuous Real-Valued Function:
:$-g \in \map {C^*} {S, \R}$
From Sum Rule for Bounded Continuous Real-Valued Functions:
:$f + \paren{-g} \in \map {C^*} {S, \R}$
It follows that:
:$\forall f, g \in \map {C^*} {S, \R} : f + \paren{-g} \in \map C {S, R}$
=== $(3) \quad \forall f, g \in \map {C^*} {S, \R} : f * g \in \map {C^*} {S, \R}$ ===
Let $f, g \in \map {C^*} {S, \R}$.
From Product Rule for Bounded Continuous Real-Valued Functions:
:$f * g \in \map {C^*} {S, \R}$
It follows that:
:$\forall f, g \in \map {C^*} {S, \R} : f * g \in \map {C^*} {S, \R}$
From Subring Test:
:$\struct{\map {C^*} {S, \R}, +, *}$ is a subring of $\struct {\map C {S, \R}, +, *}$.
== Also see ==
* Zero of Ring of Bounded Continuous Real-Valued Functions
* Additive Inverse in Ring of Bounded Continuous Real-Valued Functions
* Unity of Ring of Bounded Continuous Real-Valued Functions
* Ring of Bounded Continuous Real-Valued Functions is Commutative
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.4$
Category:Rings of Bounded Continuous Real-Valued Functions
Ring of Bounded Continuous Real-Valued Functions is Commutative

== 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 bounded continuous real-valued functions from $S$.
Then:
:$\struct{\map {C^*} {S, \R}, +, *}$ is a commutative ring
== Proof ==
Let $\struct{\map C {S, \R}, +, *}$ be the ring of continuous real-valued functions from $S$.
From Ring of Bounded Continuous Functions is Subring of Continuous Real-Valued Functions:
:$\struct{\map {C^*} {S, \R}, +, *}$ is a subring of $\struct{\map C {S, \R}, +, *}$
From Ring of Continuous Real-Valued Functions is Commutative:
:$\struct{\map C {S, \R}, +, *}$ is a commutative ring
From Subring of Commutative Ring is Commutative:
:$\struct{\map {C^*} {S, \R}, +, *}$ is a commutative ring
== Also see ==
* Ring of Bounded Continuous Functions is Subring of Continuous Real-Valued Functions
* Zero of Ring of Bounded Continuous Real-Valued Functions
* Additive Inverse in Ring of Bounded Continuous Real-Valued Functions
* Unity of Ring of Bounded Continuous Real-Valued Functions
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.4$
Category:Rings of Bounded Continuous Real-Valued Functions
Ring of Continuous Mappings is Subring of All Mappings

== Theorem ==
Let $\struct {S, \tau_{_S} }$ be a topological space.
Let $\struct {R, +, *, \tau_{_R} }$ be a topological ring.
Let $\struct {R^S, +, *}$ be the ring of mappings from $S$ to $R$.
Let $\struct{\map C {S, R}, +, *}$ be the ring of continuous mappings from $S$ to $R$.
Then:
:$\struct{\map C {S, R}, +, *}$ is a subring of $\struct {R^S, +, *}$
== Proof ==
From Structure Induced by Ring Operations is Ring:
:$\struct {R^S, +, *}$ is a ring.
From Structure Induced by Ring Operations is Ring:
:$\forall f \in R^S :$ the additive inverse of $f$ is the pointwise negation $-f$, defined by:
::$\forall s \in S: \map {\paren {-f} } s := - \map f s$
From the Subring Test:
:$\struct{\map C {S, R}, +, *}$ is a subring of $\struct {R^S, +, *}$
:$(1) \quad \map C {S, R} \ne \O$
:$(2) \quad \forall f, g \in \map C {S, R} : f + \paren{-g} \in \map C {S, R}$
:$(3) \quad \forall f, g \in \map C {S, R} : f * g \in \map C {S, R}$
=== $(1) \quad \map C {S, R} \ne \O$ ===
Let $0_R$ denote the zero of $\struct {R, +, *, \tau_{_R} }$.
Let $0_{R^S}: S \to R$ denote the constant mapping defined by:
:$\forall s \in S : \map {0_{R^S}} s = 0_R$
From Constant Mapping is Continuous:
:$0_{R^S} \in \map C {S, R}$
It follows that:
:$\map C {S, R} \ne \O$
=== $(2) \quad \forall f, g \in \map C {S, R} : f + \paren{-g} \in \map C {S, R}$ ===
Let $f, g \in \map C {S, R}$.
From Negation Rule for Continuous Mappings to Topological Ring:
:$-g \in \map C {S, R}$
From Sum Rule for Continuous Mappings into Topological Ring:
:$f + \paren{-g} \in \map C {S, R}$
It follows that:
:$\forall f, g \in \map C {S, R} : f + \paren{-g} \in \map C {S, R}$
=== $(3) \quad \forall f, g \in \map C {S, R} : f * g \in \map C {S, R}$ ===
Let $f, g \in \map C {S, R}$.
From Product Rule for Continuous Mappings to Topological Ring:
:$f * g \in \map C {S, R}$
It follows that:
:$\forall f, g \in \map C {S, R} : f * g \in \map C {S, R}$
From Subring Test:
:$\struct{\map C {S, R}, +, *}$ is a subring of $\struct {R^S, +, *}$.
== Also see ==
* Zero of Ring of Continuous Mappings
* Additive Inverse in Ring of Continuous Mappings
* Unity of Ring of Continuous Mappings
* Commutativity of Ring of Continuous Mappings
Category:Rings of Continuous Mappings
Ring of Continuous Real-Valued Functions is Commutative

== 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 commutative 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 Commutativity of Ring of Continuous Mappings:
:$\struct{\map C {S, \R}, +, *}$ is a commutative ring
== Also see ==
* Ring of Continuous Real-Valued Functions is Ring
* Zero of Ring of Continuous Real-Valued Functions
* Unity of Ring of Continuous Real-Valued Functions
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.3$
Category:Rings of Continuous Real-Valued Functions