nyuuzyou/wikis huggingface dataset:
Right Product with Degenerate Linear Transformation is Degenerate

== Theorem ==
Let $U, V, W$ be vector spaces over a field $K$.
Let $G: U \to V$ be a degenerate linear transformation.
Let $M: V \to W$ be a linear transformation.
Then $M \circ G$ is degenerate.
== Proof ==
Let $\set {s_1, \ldots, s_n}$ be a generator of $\Img G$.
Then $\set {\map M {s_1}, \ldots, \map M {s_n} }$ is a generator of $\Img {M \circ G}$.
By Cardinality of Generator of Vector Space is not Less than Dimension:
:$\map \dim {\Img {M \circ G}} \le n$
== Sources ==
* : $2.2$: Index of a Linear Map
Category:Linear Algebra
Right Quasigroup if (1-3) Parastrophe of Magma is Magma

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let the $(1-3)$ parastrophe of $\struct {S, \circ}$ be a magma.
Then $\struct {S, \circ}$ is a right quasigroup.
== Proof ==
By the definition of a right quasigroup it must be shown that:
:$\forall a, b \in S: \exists ! x \in S: x \circ a = b$
 there exists $a, b \in S$ such that $x \circ a = b$ does not have a unique solution for $x$.
Then in the $(1-3)$ parastrophe of $\struct {S, \circ}$ we see that $\circ$ as a mapping either fails to be left-total or many-to-one for $b \circ a$.
So $\struct {S, \circ}$ is not a magma which contradicts our assumption.
Category:Parastrophes
Category:Quasigroups
Category:Magmas
Right Regular Representation by Inverse is Group Action

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*: G \times G \to G$ be the operation:
:$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$
where $\rho_g$ is the right regular representation of $G$ with respect to $g$.
Then $*$ is a group action.
== Proof ==
The group action axioms are investigated in turn.
Let $g, h, a \in G$.
Thus:
demonstrating that holds.
Then:
demonstrating that holds.
== Also defined as ==
Some sources denote a mapping by placing the symbol defining that mapping on the left of its operand.
Thus under such a convention:
:$\map {\rho_g} a$ is written $a \rho_g$
and:
:$g * a$ is written $a * g$ (or even $a g$ in sources which do not place a high regard on clarity).
Thus the right regular representation itself can directly be defined as being a group action without the need to take the inverse:
Such a treatment can be found in .
As it runs contrary to the conventions used on , beyond its mention here it will not be used.
However, this example indicates how the arbitrary nature of notational conventions can cause the details of results to be equally arbitrarily dependent upon the convention used.
== Also see ==
* Left Regular Representation is Group Action
* Group Acts on Itself
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(iv)}$
Category:Group Actions
Right Regular Representation by Inverse is Transitive Group Action

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*: G \times G \to G$ be the group action:
:$\forall g, h \in G: g * h = \map {\rho_{g^{-1} } } h$
where $\rho_g$ is the right regular representation of $G$ with respect to $g$.
Then $*$ is a transitive group action.
== Proof ==
Let $g, h \in G$.
Then:
$h$ is arbitrary, therefore the above holds for all $h \in G$.
By definition of orbit:
:$\Orb h = \set {t \in G: \exists g \in G: g * h = t}$
That is:
:$\Orb h = G$
Hence the result by definition of transitive group action.
== Also see ==
* Left Regular Representation is Transitive Group Action
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(iv)}$
Category:Regular Representations
Category:Transitive Group Actions
Right Regular Representation of 0 is Bijection in B-Algebra

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-algebra.
Then the right regular representation of $\struct {X, \circ}$ $0$ is a bijection.
== Proof ==
 states:
:$\forall x \in X: x \circ 0 = x$
and so, for all $x \in X$:
:$\map {\rho_0} x = x$
That is:
:$\rho_0 = I_X$
which is the identity mapping on $X$.
The result follows from Identity Mapping is Bijection.
Category:B-Algebras
Category:Regular Representations
Right Regular Representation of Subset Product

== Theorem ==
Let $\struct {S, \circ}$ be a magma.
Let $T \subseteq S$ be a subset of $S$.
Let $\rho_a: S \to S$ be the right regular representation of $S$ with respect to $a$.
Then:
:$\rho_a \sqbrk T = T \circ \set a = T \circ a$
where $T \circ a$ denotes subset product with a singleton.
== Proof ==
== Also see ==
* Left Regular Representation of Subset Product
== Sources ==
* : $\S 41$. Multiplication of subsets of a group
Category:Regular Representations
Right Regular Representation wrt Right Cancellable Element on Finite Semigroup is Bijection

== Theorem ==
Let $\struct {S, \circ}$ be a finite semigroup.
Let $a \in S$ be right cancellable.
Then the right regular representation $\rho_a$ of $\struct {S, \circ}$ with respect to $a$ is a bijection.
== Proof ==
By Right Cancellable iff Right Regular Representation Injective, $\rho_a$ is an injection.
By hypothesis, $S$ is finite.
From Injection from Finite Set to Itself is Surjection, $\rho_a$ is a surjection.
Thus $\rho_a$ is injective and surjective, and therefore a bijection.
== Also see ==
* Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection
Category:Semigroups
Category:Regular Representations
Category:Cancellability
Right Self-Distributive Operation with Left Identity is Idempotent

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be right self-distributive.
Let $\struct {S, \circ}$ have a left identity.
Then $\circ$ is an idempotent operation.
== Proof ==
Let the left identity of $\struct {S, \circ}$ be $e_L$.
We have:
The result follows by definition of idempotent operation.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.23 \ \text{(c)}$
Category:Idempotence
Category:Self-Distributive Operations
Category:Identity Elements
Right Shift Operator is Linear Mapping

== Theorem ==
Let $X = Y = \ell^2$ be 2-sequence spaces over real numbers.
Let $R : X \to Y$ be the right shift operator.
Then $R$ is a linear mapping.
== Proof ==
Let $x = \tuple {x_1, x_2,x_3, \ldots}, y = \tuple {y_1, y_2, y_3, \ldots} \in \ell^2$
Let $\alpha \in \R$.
=== Distributivity ===
=== Positive homogenity ===
== Sources ==
* : Chapter $\S 2.1$: Continuous and linear maps. Linear transformations
Category:Linear Transformations
Right Shift Operator on 2-Sequence Space is Continuous

== Theorem ==
Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the $2$-sequence space with $2$-norm.
Let $R : \ell^2 \to \ell^2$ be the right shift operator.
Then $R$ is continuous on $\struct {\ell^2, \norm {\, \cdot \,}_2}$.
== Proof ==
Let $\sequence {a_n}_{n \mathop \in \N} = \tuple {a_1, a_2, a_3, \ldots}$ be a $2$-sequence.
By Continuity of Linear Transformation between Normed Vector Spaces, $R$ is continuous in $\struct {\ell^2, \norm {\, \cdot \,}_2}$.
== Sources ==
* : Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$
Category:Operator Theory
Category:Continuous Mappings
Right Zero Divisor of Commutative Ring is Left Zero Divisor

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $x \in R$ be a right zero divisor of $R$.
Then $x$ is also a left zero divisor of $R$.
== Proof ==
Let $x \in R$ be a right zero divisor of $R$.
Let $0_R$ denote the zero of $R$.
Then:
Hence $x$ is a left zero divisor of $R$ by definition.
== Sources ==
* 
* 
Category:Zero Divisors
Ring Direct Product of Modulo Integers is Isomorphic to Ring Modulo Product iff Coprime

== Theorem ==
Let $m, n \in \Z_{>1}$.
Let $\struct {\Z_m, +_m, \times_m}$ and $\struct {\Z_n, +_n, \times_n}$ be the rings of integers modulo $m$ and $n$ respectively.
Let $\struct {\Z_m \times \Z_n}$ be the direct product of $\Z_m$ and $\Z_n$.
Let $\struct {\Z_{m n}, +_{m n}, \times_{m n} }$ be the ring of integers modulo $mn$.
Then $\struct {\Z_m \times \Z_n}$ is isomorphic to $\struct {\Z_{m n}, +_{m n}, \times_{m n} }$ $m$ and $n$ are coprime.
== Proof ==
== Sources ==
* : $\S 2.3$: Congruences
Category:Commutative Algebra
Category:Modulo Arithmetic
Ring Element is Unit iff Unit in Integral Extension

== Theorem ==
let $A$ be a commutative ring with unity.
Let $a \in A$.
Let $B$ be an integral ring extension of $A$.
:$(1): \quad a$ is a unit of $A$
:$(2): \quad a$ is a unit of $B$
== Proof ==
=== 1 implies 2 ===
Follows from Ring Homomorphism Preserves Units.
=== 2 implies 1 ===
Let $a$ be a unit of $B$.
Let $P \in A \sqbrk x$ be a monic polynomial with $\map P {1 / a} = 0$.
Let $n$ be its degree and $\map P x = x^n + \map Q x$.
Then $1 + a^n \map Q {1 / a} = 0$.
Note that $a^{n - 1} \map Q {1 / a} \in A$.
Thus $a$ is a unit of $A$, with inverse $-a^{n - 1} \map Q {1 / a}$.
Category:Integral Ring Extensions
Ring Element is Zero Divisor iff not Cancellable

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring which is not null.
Let $z \in R^*$.
Then $z$ is a zero divisor $z$ is not cancellable for $\circ$.
== Proof ==
=== Sufficient Condition ===
Let $z$ be a zero divisor.
Then either $z \circ x = 0_R$ or $x \circ z = 0_R$ for some $x \in R^*$.
Then:
:$z \circ 0_R = 0_R = 0_R \circ z$
and so $z$ is not cancellable.
=== Necessary Condition ===
Let $z$ not be cancellable in $R$.
Then there exists $x, y \in R$ such that $x \ne y$ and:
:$z \circ x = z \circ y$
Then:
But $x \ne y$, so $x + \paren {-y} \ne 0$.
Thus $z$ is a zero divisor.
Similarly if $x \circ z = y \circ z$ where $x \ne y$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $21$. Rings and Integral Domains: Theorem $21.2$
* : Chapter $1$: Rings - Definitions and Examples: $3$: Some special classes of rings: Lemma $1.4$
Category:Zero Divisors
Ring Epimorphism Preserves Ideals

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $J$ be an ideal of $R_1$.
Then $\phi \sqbrk J$ is an ideal of $R_2$.
== Proof ==
$J$ is an ideal of $R_1$, so it is also a subring of $R_1$.
From Ring Homomorphism Preserves Subrings, it follows that $\phi \sqbrk J$ is a subring of $R_2$.
Now suppose $u \in \phi \sqbrk J$.
Let $v \in R_2$.
Then $\exists x \in J, y \in R_1$ such that $\map \phi x = u, \map \phi y = v$.
Thus, by the morphism property:
:$u \circ_2 v = \map \phi x \circ_2 \map \phi y = \map \phi {x \circ_1 y}$
So $u \circ_2 v \in \phi \sqbrk J$ because $x \circ_1 y \in J$.
Similarly $v \circ_2 u \in \phi \sqbrk J$ also.
The result follows.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 2^\circ$
Category:Ideal Theory
Category:Ring Epimorphisms
Ring Epimorphism Preserves Subrings

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $S$ be a subring of $R_1$.
Then $\phi \sqbrk S$ is a subring of $R_2$.
== Proof ==
A direct application of Ring Homomorphism Preserves Subrings.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 2^\circ$
Category:Ring Epimorphisms
Category:Subrings
Ring Epimorphism Preserves Unity

== Theorem ==
Let $A$ be a ring with unity $1$.
Let $B$ be a ring.
Let $f: A \to B$ be a ring epimorphism.
Then $\map f 1$ is a unity of $B$.
== Proof ==
By definition, $f$ is a semigroup homomorphism between multiplicative semigroups.
A unity of a ring is by definition an identity element of its multiplicative semigroup.
Thus the result follows from Epimorphism Preserves Identity.
Category:Ring Epimorphisms
Ring Epimorphism from Integers to Integers Modulo m

== Theorem ==
Let $\struct {\Z, +, \times}$ be the ring of integers.
Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.
Let $\phi: \struct {\Z, +, \times} \to \struct {\Z_m, +_m, \times_m}$ be the mapping defined as:
:$\forall x \in \Z: \map \phi x = \eqclass x m$
where $\eqclass x m$ is the residue class modulo $m$.
Then $\phi$ is a ring epimorphism, but specifically not a ring monomorphism.
The image of $\phi$ is $\struct {\Z_m, +_m, \times_m}$.
The kernel of $\phi$ is $m \Z$, the set of integer multiples of $m$.
== Proof ==
Let $a, b \in \Z$.
Then:
Hence $\phi$ is a ring homomorphism.
Now let $\eqclass a m \in \Z_m$.
By definition of residue class modulo $m$:
:$\eqclass a m = \set {x \in \Z: \exists k \in \Z: z = a + k m}$
Setting $k = 0$:
:$\map \phi a = \eqclass a m$
and so:
:$\map {\phi^{-1} } {\eqclass a m} \ne \O$
Thus $\phi$ is a surjection.
Now setting $k = 1$, for example, we have that:
:$\map \phi {a + m} = \eqclass a m$
and so:
:$\map \phi a = \map \phi {a + m}$.
So $\phi$ is specifically not an injection.
It follows by definition that $\phi$ is a ring epimorphism, but specifically not a ring monomorphism.
Next we note that:
:$\forall x \in \Z: \map \phi x \in \Z_m$
and so:
:$\Img \phi = \Z_m$
Finally, we have that the kernel of $\phi$ is:
:$\map \ker \phi = \set {x \in \Z: \map \phi x = \eqclass 0 m}$
Let $\map \phi x = \eqclass 0 m$
Then $x = 0 + k m$ for some $k \in \Z$.
That is, $x \in m \Z$ and so:
:$\map \ker \phi \subseteq m \Z$
Now let $x \in m \Z$.
Then:
:$\exists k \in \Z: x = 0 + k m$
and so by definition:
:$\map \phi x = \eqclass 0 m$
So:
:$m \Z \subseteq \map \ker \phi$
Hence:
:$\map \ker \phi = m \Z$
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Example $3$
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Example $45$
Category:Ring Epimorphisms
Category:Modulo Arithmetic
Category:Integers
Ring Epimorphism with Trivial Kernel is Isomorphism

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $K = \map \ker \phi$ be the kernel of $\phi$.
$\phi$ is an isomorphism $K = \set {0_{R_1} }$.
== Proof ==
From Kernel is Trivial iff Monomorphism, $\phi$ is a ring monomorphism $K = \set {0_{R_1} }$.
As $\phi$ is also an epimorphism, the result follows.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 1^\circ$
Category:Ring Epimorphisms
Category:Ring Isomorphisms
Ring Homomorphism Preserves Negatives

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then:
:$\forall x \in R_1: \map \phi {-x} = -\paren {\map \phi x}$
== Proof ==
We have that Ring Homomorphism of Addition is Group Homomorphism.
The result follows from Group Homomorphism Preserves Inverses.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Theorem $44 \ \text{(ii)}$
* : $\S 2.2$: Homomorphisms: Definition $2.4$
* : $\S 57.1$ Ring homomorphisms: $\text{(ii)}$
Category:Ring Homomorphisms
Ring Homomorphism Preserves Subrings

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let $S$ be a subring of $R_1$.
Then $\phi \sqbrk S$ is a subring of $R_2$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Ring Homomorphism Preserves Subrings
Category:Ring Homomorphisms
Category:Subrings
Ring Homomorphism Preserves Subrings/Corollary

== Corollary to Ring Homomorphism Preserves Subrings ==
Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be rings.
The image of a ring homomorphism $\phi: R_1 \to R_2$ is a subring of $R_2$.
== Proof ==
From Ring is Subring of Itself, $R_1$ is a subring of itself.
The result then follows from Ring Homomorphism Preserves Subrings.
== Sources ==
* : $\S 2.2$: Homomorphisms: Lemma $2.6 \ \text{(ii)}$
* : $\S 57.3$ Ring homomorphisms: $\text{(i)}$
Category:Ring Homomorphism Preserves Subrings
Category:Ring Homomorphisms
Category:Subrings
Ring Homomorphism Preserves Subrings/Proof 1

== Theorem ==
== Proof ==
Since $S \ne \O$, $\phi \sqbrk S \ne \O$.
From Group Homomorphism Preserves Subgroups, $\struct {\phi \sqbrk S, +_2}$ is a subgroup of $\struct {R_2, +_2}$.
From Homomorphism Preserves Subsemigroups, $\struct {\phi \sqbrk S, \circ_2}$ is a subsemigroup of $\struct {R_2, \circ_2}$.
Thus, as $\struct {R_2, +_2}$ is a group and $\struct {R_2, \circ_2}$ is a semigroup, the result follows.
Category:Ring Homomorphism Preserves Subrings
Ring Homomorphism Preserves Subrings/Proof 2

== Theorem ==
== Proof ==
From Morphism Property Preserves Closure, $\phi \sqbrk {R_1}$ is a closed algebraic structure.
From Epimorphism Preserves Rings, $\phi \sqbrk S$ is a ring.
Hence the result, from the definition of subring.
Category:Ring Homomorphism Preserves Subrings
Ring Homomorphism Preserves Subrings/Proof 3

== Theorem ==
== Proof ==
Let $S$ be a subring of $R_1$.
Since $S \ne \O$ it follows that $\phi \sqbrk S \ne \O$.
Let $x, y \in \phi \sqbrk S$.
Then:
:$\exists s, t \in S: x = \map \phi s, y = \map \phi t$
So:
As $S$ is a subring of $R_1$, it is closed under $+_1$ and the taking of negatives.
Thus $s +_1 \paren {-t} \in S$ and so $x +_2 \paren {-y} \in \phi \sqbrk S$.
Similarly:
Because $S$ is a subring of $R_1$, it is closed under $\circ_1$.
Thus $s \circ_1 t \in S$ and so $x \circ_2 y \in \phi \sqbrk S$.
The result follows from Subring Test.
== Sources ==
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Theorem $46 \ \text{(i)}$
* : $\S 57.3$ Ring homomorphisms: $\text{(i)}$
Category:Ring Homomorphism Preserves Subrings
Ring Homomorphism Preserves Zero

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Let:
:$0_{R_1}$ be the zero of $R_1$
:$0_{R_2}$ be the zero of $R_2$.
Then:
:$\map \phi {0_{R_1} } = 0_{R_2}$
== Proof ==
By definition, if $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ are rings then $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ are groups.
Again by definition:
:the zero of $\struct {R_1, +_1, \circ_1}$ is the identity of $\struct {R_1, +_1}$
:the zero of $\struct {R_2, +_2, \circ_2}$ is the identity of $\struct {R_2, +_2}$.
The result follows from Group Homomorphism Preserves Identity.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms
* : Chapter $5$: Rings: $\S 24$. Homomorphisms: Theorem $44 \ \text{(i)}$
* : $\S 2.2$: Homomorphisms: Definition $2.4$
* : $\S 57.1$ Ring homomorphisms: $\text{(i)}$
Category:Ring Homomorphisms
Ring Homomorphism by Idempotent

== Theorem ==
Let $A$ be a commutative ring.
Let $e \in A$ be an idempotent element.
Let $\ideal e$ be the ideal of $A$ generated by $e$.
Then the mapping:
:$f: A \to \ideal e: a \mapsto e a$ 
is a surjective ring homomorphism with kernel the ideal $\ideal {1 - e}$ generated by $1 - e$.
== Proof ==
Let $e a \in \ideal e$ for some arbitrary $a \in A$.
Then
So $f$ is surjective.
Let $x, y \in A$.
Then 
and:
So $f$ is a ring homomorphism.
We have
Thus $1 - e \in \map \ker f$ and $\ideal {1 - e} \subseteq \map \ker f$.
Suppose $a \in \map \ker f$.
Then $e a = \map f a = 0$.
Thus:
:$\map \ker f \subseteq \ideal {1 - e}$
so:
:$\map \ker f = \ideal {1 - e}$
== Also see ==
* Unital Ring Homomorphism by Idempotent
Category:Ring Theory
Category:Commutative Rings
Category:Ring Homomorphisms
Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism

== Theorem ==
Let $\struct {R, +_R, \circ}$ and $\struct {S, +_S, *}$ be rings whose zeros are $0_R$ and $0_S$ respectively.
Let $\phi: R \to S$ be a ring homomorphism.
If $R$ is a division ring, then either:
:$(1): \quad \phi$ is a monomorphism (that is, $\phi$ is injective)
:$(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in R: \map \phi a = 0_S$).
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Ring Homomorphism from Field is Monomorphism or Zero Homomorphism
Category:Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism
Category:Division Rings
Category:Monomorphisms (Abstract Algebra)
Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism/Proof 1

== Theorem ==
== Proof ==
We have that:
:The kernel of a homomorphism is an ideal of $R$
:the only ideals of a division ring are trivial.
So $\map \ker \phi = \set {0_R}$ or $R$. 
If $\map \ker \phi = \set {0_R}$, then $\phi$ is injective by Kernel is Trivial iff Monomorphism.
If $\map \ker \phi = R$, $\phi$ is the zero homomorphism by definition.
Category:Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism
Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism/Proof 2

== Theorem ==
== Proof ==
From Surjection by Restriction of Codomain, we can restrict the codomain of $\phi$ and consider the mapping $\phi': R \to \Img R$
As $\phi'$ is now a surjective homomorphism, it is by definition an epimorphism.
Then an Epimorphism from Division Ring to Ring is either null or an isomorphism.
As an isomorphism is by definition injective, the result follows.
Category:Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism
Ring Homomorphism from Field is Monomorphism or Zero Homomorphism

== Theorem ==
Let $\struct {F, +_F, \circ}$ be a field whose zero is $0_F$.
Let $\struct {S, +_S, *}$ be a ring whose zero is $0_S$.
Let $\phi: F \to S$ be a ring homomorphism.
Then either:
:$(1): \quad \phi$ is a monomorphism (that is, $\phi$ is injective)
or
:$(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in F: \map \phi a = 0_S$).
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism
== Sources ==
* : Chapter $9$: Rings: Exercise $8$
Category:Ring Homomorphism from Field is Monomorphism or Zero Homomorphism
Category:Field Theory
Category:Monomorphisms (Abstract Algebra)
Ring Homomorphism from Field is Monomorphism or Zero Homomorphism/Proof 2

== Theorem ==
== Proof ==
Let $\phi: F \to S$ be a ring homomorphism.
Suppose $\phi$ is not a monomorphism.
By definition, $\phi$ is not an injection.
So there must exist $a, b \in F: \map \phi a = \map \phi b$.
Let $k = a +_F \paren {-b}$.
Then:
As $a \ne b$ then $k \ne 0$ and so has a product inverse $\exists k^{-1} \in F$.
So for any $x \in F$ we can write $x = k \circ \paren {k^{-1} \circ x}$ and so:
So if $\phi$ is not a monomorphism, it is the zero homomorphism.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Theorem $3.1$
Category:Ring Homomorphism from Field is Monomorphism or Zero Homomorphism
Ring Homomorphism from Ring with Unity to Integral Domain Preserves Unity

== Theorem ==
Let $\struct {R, +_R, \circ_R}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {D, +_D, \circ_D}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\phi: R \to D$ be a ring homomorphism such that:
:$\map \ker \phi \ne R$
where $\map \ker \phi$ denotes the kernel of $\phi$.
Then $\map \phi {1_R} = 1_D$.
== Proof ==
 $\map \phi {1_R} = 0_D$.
Let $x \in R$ be arbitrary.
Then:
But this contradicts the assertion that $\map \ker \phi \ne R$.
It follows that $\map \phi {1_R} \ne 0_D$.
Then we have:
Hence the result.
== Sources ==
* : Chapter $9$: Rings: Exercise $9$
Category:Ring Homomorphisms
Category:Rings with Unity
Category:Integral Domains
Ring Homomorphism of Addition is Group Homomorphism

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then $\phi: \struct {R_1, +_1} \to \struct {R_2, +_2}$ is a group homomorphism.
== Proof ==
From the definition of a ring, both $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ are abelian groups.
The result follows from the definitions of ring homomorphism and group homomorphism.
Category:Ring Homomorphisms
Category:Group Homomorphisms