id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-600 | Identity is only Idempotent Element in Group | Every group has exactly one idempotent element: the identity. | The Identity Element is Idempotent.
From the Cancellation Laws, all group elements are cancellable.
The result follows from Identity is only Idempotent Cancellable Element.
{{qed}} | Every [[Definition:Group|group]] has [[Definition:Exactly One|exactly one]] [[Definition:Idempotent Element|idempotent element]]: the [[Definition:Identity Element|identity]]. | The [[Identity Element is Idempotent]].
From the [[Cancellation Laws]], all [[Definition:Group|group]] [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]].
The result follows from [[Identity is only Idempotent Cancellable Element]].
{{qed}} | Identity is only Idempotent Element in Group/Proof 1 | https://proofwiki.org/wiki/Identity_is_only_Idempotent_Element_in_Group | https://proofwiki.org/wiki/Identity_is_only_Idempotent_Element_in_Group/Proof_1 | [
"Group Theory",
"Identities are Idempotent",
"Identity is only Idempotent Element in Group"
] | [
"Definition:Group",
"Definition:Unique",
"Definition:Idempotence/Element",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Identity Element is Idempotent",
"Cancellation Laws",
"Definition:Group",
"Definition:Element",
"Definition:Cancellable Element",
"Identity is only Idempotent Cancellable Element"
] |
proofwiki-601 | Identity is only Idempotent Element in Group | Every group has exactly one idempotent element: the identity. | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $x \in G$ such that $x \circ x = x$.
{{begin-eqn}}
{{eqn | l = e
| r = x \circ x^{-1}
| c = {{Group-axiom|3}}
}}
{{eqn | r = \paren {x \circ x} \circ x^{-1}
| c = {{hypothesis}}: $x \circ x = x$
}}
{{eqn | r = x \circ \paren {x \circ x^{-1... | Every [[Definition:Group|group]] has [[Definition:Exactly One|exactly one]] [[Definition:Idempotent Element|idempotent element]]: the [[Definition:Identity Element|identity]]. | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $x \in G$ such that $x \circ x = x$.
{{begin-eqn}}
{{eqn | l = e
| r = x \circ x^{-1}
| c = {{Group-axiom|3}}
}}
{{eqn | r = \paren {x \circ x} \circ x^{-1}
| c = {{hypothesis}}: $x \c... | Identity is only Idempotent Element in Group/Proof 2 | https://proofwiki.org/wiki/Identity_is_only_Idempotent_Element_in_Group | https://proofwiki.org/wiki/Identity_is_only_Idempotent_Element_in_Group/Proof_2 | [
"Group Theory",
"Identities are Idempotent",
"Identity is only Idempotent Element in Group"
] | [
"Definition:Group",
"Definition:Unique",
"Definition:Idempotence/Element",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-602 | Group Product Identity therefore Inverses | Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$.
Then if either:
:$g h = e$
or:
:$h g = e$
it follows that:
:$g = h^{-1}$
and:
:$h = g^{-1}$ | From the Division Laws for Groups:
:$g h = e \implies g = e h^{-1} = h^{-1}$
Also by the Division Laws for Groups:
:$g h = e \implies h = g^{-1} e = g^{-1}$
{{qed}} | Let $g$ and $h$ be [[Definition:Element|elements]] of a [[Definition:Group|group]] $G$ whose [[Definition:Identity Element|identity element]] is $e$.
Then if either:
:$g h = e$
or:
:$h g = e$
it follows that:
:$g = h^{-1}$
and:
:$h = g^{-1}$ | From the [[Division Laws for Groups]]:
:$g h = e \implies g = e h^{-1} = h^{-1}$
Also by the [[Division Laws for Groups]]:
:$g h = e \implies h = g^{-1} e = g^{-1}$
{{qed}} | Group Product Identity therefore Inverses/Part 1/Proof 1 | https://proofwiki.org/wiki/Group_Product_Identity_therefore_Inverses | https://proofwiki.org/wiki/Group_Product_Identity_therefore_Inverses/Part_1/Proof_1 | [
"Group Theory",
"Group Product Identity therefore Inverses"
] | [
"Definition:Element",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Division Laws for Groups",
"Division Laws for Groups"
] |
proofwiki-603 | Group Product Identity therefore Inverses | Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$.
Then if either:
:$g h = e$
or:
:$h g = e$
it follows that:
:$g = h^{-1}$
and:
:$h = g^{-1}$ | Let $g h = e$.
Then:
{{begin-eqn}}
{{eqn | l = h
| r = e h
| c = {{Group-axiom|2}}
}}
{{eqn | r = \paren {g^{-1} g} h
| c = {{Group-axiom|3}}
}}
{{eqn | r = g^{-1} \paren {g h}
| c = {{Group-axiom|1}}
}}
{{eqn | r = g^{-1} e
| c = {{hypothesis}}
}}
{{eqn | r = g^{-1}
| c = {{Group-ax... | Let $g$ and $h$ be [[Definition:Element|elements]] of a [[Definition:Group|group]] $G$ whose [[Definition:Identity Element|identity element]] is $e$.
Then if either:
:$g h = e$
or:
:$h g = e$
it follows that:
:$g = h^{-1}$
and:
:$h = g^{-1}$ | Let $g h = e$.
Then:
{{begin-eqn}}
{{eqn | l = h
| r = e h
| c = {{Group-axiom|2}}
}}
{{eqn | r = \paren {g^{-1} g} h
| c = {{Group-axiom|3}}
}}
{{eqn | r = g^{-1} \paren {g h}
| c = {{Group-axiom|1}}
}}
{{eqn | r = g^{-1} e
| c = {{hypothesis}}
}}
{{eqn | r = g^{-1}
| c = {{Group-a... | Group Product Identity therefore Inverses/Part 1/Proof 2 | https://proofwiki.org/wiki/Group_Product_Identity_therefore_Inverses | https://proofwiki.org/wiki/Group_Product_Identity_therefore_Inverses/Part_1/Proof_2 | [
"Group Theory",
"Group Product Identity therefore Inverses"
] | [
"Definition:Element",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [] |
proofwiki-604 | Group Product Identity therefore Inverses | Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$.
Then if either:
:$g h = e$
or:
:$h g = e$
it follows that:
:$g = h^{-1}$
and:
:$h = g^{-1}$ | From the Division Laws for Groups:
:$h g = e \implies g = e h^{-1} = h^{-1}$
Also by the Division Laws for Groups:
:$h g = e \implies h = g^{-1} e = g^{-1}$
{{qed}} | Let $g$ and $h$ be [[Definition:Element|elements]] of a [[Definition:Group|group]] $G$ whose [[Definition:Identity Element|identity element]] is $e$.
Then if either:
:$g h = e$
or:
:$h g = e$
it follows that:
:$g = h^{-1}$
and:
:$h = g^{-1}$ | From the [[Division Laws for Groups]]:
:$h g = e \implies g = e h^{-1} = h^{-1}$
Also by the [[Division Laws for Groups]]:
:$h g = e \implies h = g^{-1} e = g^{-1}$
{{qed}} | Group Product Identity therefore Inverses/Part 2/Proof 1 | https://proofwiki.org/wiki/Group_Product_Identity_therefore_Inverses | https://proofwiki.org/wiki/Group_Product_Identity_therefore_Inverses/Part_2/Proof_1 | [
"Group Theory",
"Group Product Identity therefore Inverses"
] | [
"Definition:Element",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Division Laws for Groups",
"Division Laws for Groups"
] |
proofwiki-605 | Group Product Identity therefore Inverses | Let $g$ and $h$ be elements of a group $G$ whose identity element is $e$.
Then if either:
:$g h = e$
or:
:$h g = e$
it follows that:
:$g = h^{-1}$
and:
:$h = g^{-1}$ | Let $h g = e$.
Then:
{{begin-eqn}}
{{eqn | l = g
| r = e g
| c = {{Group-axiom|2}}
}}
{{eqn | r = \paren {h^{-1} h} g
| c = {{Group-axiom|3}}
}}
{{eqn | r = h^{-1} \paren {h g}
| c = {{Group-axiom|1}}
}}
{{eqn | r = h^{-1} e
| c = {{hypothesis}}
}}
{{eqn | r = h^{-1}
| c = {{Group-ax... | Let $g$ and $h$ be [[Definition:Element|elements]] of a [[Definition:Group|group]] $G$ whose [[Definition:Identity Element|identity element]] is $e$.
Then if either:
:$g h = e$
or:
:$h g = e$
it follows that:
:$g = h^{-1}$
and:
:$h = g^{-1}$ | Let $h g = e$.
Then:
{{begin-eqn}}
{{eqn | l = g
| r = e g
| c = {{Group-axiom|2}}
}}
{{eqn | r = \paren {h^{-1} h} g
| c = {{Group-axiom|3}}
}}
{{eqn | r = h^{-1} \paren {h g}
| c = {{Group-axiom|1}}
}}
{{eqn | r = h^{-1} e
| c = {{hypothesis}}
}}
{{eqn | r = h^{-1}
| c = {{Group-a... | Group Product Identity therefore Inverses/Part 2/Proof 2 | https://proofwiki.org/wiki/Group_Product_Identity_therefore_Inverses | https://proofwiki.org/wiki/Group_Product_Identity_therefore_Inverses/Part_2/Proof_2 | [
"Group Theory",
"Group Product Identity therefore Inverses"
] | [
"Definition:Element",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [] |
proofwiki-606 | Self-Inverse Elements Commute iff Product is Self-Inverse | Let $\struct {G, \circ}$ be a group.
Let $x, y \in \struct {G, \circ}$, such that $x$ and $y$ are self-inverse.
Then $x$ and $y$ commute {{iff}} $x \circ y$ is also self-inverse. | Let the identity element of $\struct {G, \circ}$ be $e_G$. | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $x, y \in \struct {G, \circ}$, such that $x$ and $y$ are [[Definition:Self-Inverse Element|self-inverse]].
Then $x$ and $y$ [[Definition:Commute|commute]] {{iff}} $x \circ y$ is also [[Definition:Self-Inverse Element|self-inverse]]. | Let the [[Definition:Identity Element|identity element]] of $\struct {G, \circ}$ be $e_G$. | Self-Inverse Elements Commute iff Product is Self-Inverse | https://proofwiki.org/wiki/Self-Inverse_Elements_Commute_iff_Product_is_Self-Inverse | https://proofwiki.org/wiki/Self-Inverse_Elements_Commute_iff_Product_is_Self-Inverse | [
"Group Theory",
"Commutativity"
] | [
"Definition:Group",
"Definition:Self-Inverse Element",
"Definition:Commutative/Elements",
"Definition:Self-Inverse Element"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-607 | Power Structure of Group is Semigroup | Let $\struct {G, \circ}$ be a group.
Let $\struct {\powerset G, \circ_\PP}$ be the power structure of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a semigroup. | We need to prove closure and associativity.
=== Closure ===
Let $\struct {G, \circ}$ be a group, and let $A, B \subseteq G$.
{{begin-eqn}}
{{eqn | o =
| r = \forall a \in A, b \in B: a \circ b \in G
| c =
}}
{{eqn | o = \leadsto
| r = A \circ B \subseteq G
| c = {{Defof|Subset Product}}
}}
{{e... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\struct {\powerset G, \circ_\PP}$ be the [[Definition:Power Structure|power structure]] of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a [[Definition:Semigroup|semigroup]]. | We need to prove [[Definition:Closed Algebraic Structure|closure]] and [[Definition:Associative Operation|associativity]].
=== Closure ===
Let $\struct {G, \circ}$ be a [[Definition:Group|group]], and let $A, B \subseteq G$.
{{begin-eqn}}
{{eqn | o =
| r = \forall a \in A, b \in B: a \circ b \in G
| c ... | Power Structure of Group is Semigroup/Proof 1 | https://proofwiki.org/wiki/Power_Structure_of_Group_is_Semigroup | https://proofwiki.org/wiki/Power_Structure_of_Group_is_Semigroup/Proof_1 | [
"Group Theory",
"Semigroups",
"Power Structures",
"Power Structure of Group is Semigroup"
] | [
"Definition:Group",
"Definition:Power Structure",
"Definition:Semigroup"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Associative Operation",
"Definition:Group",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Subset Product within Semigroup is Associative",
"Definition:Associative Operation",
"Definition:Semigroup"
] |
proofwiki-608 | Power Structure of Group is Semigroup | Let $\struct {G, \circ}$ be a group.
Let $\struct {\powerset G, \circ_\PP}$ be the power structure of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a semigroup. | By definition a group is also a semigroup.
The result then follows from Power Structure of Semigroup is Semigroup.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\struct {\powerset G, \circ_\PP}$ be the [[Definition:Power Structure|power structure]] of $\struct {G, \circ}$.
Then $\struct {\powerset G, \circ_\PP}$ is a [[Definition:Semigroup|semigroup]]. | By definition a [[Definition:Group|group]] is also a [[Definition:Semigroup|semigroup]].
The result then follows from [[Power Structure of Semigroup is Semigroup]].
{{qed}} | Power Structure of Group is Semigroup/Proof 2 | https://proofwiki.org/wiki/Power_Structure_of_Group_is_Semigroup | https://proofwiki.org/wiki/Power_Structure_of_Group_is_Semigroup/Proof_2 | [
"Group Theory",
"Semigroups",
"Power Structures",
"Power Structure of Group is Semigroup"
] | [
"Definition:Group",
"Definition:Power Structure",
"Definition:Semigroup"
] | [
"Definition:Group",
"Definition:Semigroup",
"Power Structure of Semigroup is Semigroup"
] |
proofwiki-609 | Inverse of Product of Subsets of Group | Let $\struct {G, \circ}$ be a group.
Let $X, Y \subseteq G$.
Then:
:$\paren {X \circ Y}^{-1} = Y^{-1} \circ X^{-1}$
where $X^{-1}$ is the inverse of $X$. | First, note that:
{{begin-eqn}}
{{eqn | o =
| r = x \in X, y \in Y
| c =
}}
{{eqn | o = \leadsto
| r = x^{-1} \in X^{-1}, y^{-1} \in Y^{-1}
| c = {{Defof|Inverse of Subset of Group}}
}}
{{eqn | o = \leadsto
| r = y^{-1} \circ x^{-1} \in Y^{-1} \circ X^{-1}
| c = {{Defof|Subset Prod... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $X, Y \subseteq G$.
Then:
:$\paren {X \circ Y}^{-1} = Y^{-1} \circ X^{-1}$
where $X^{-1}$ is the [[Definition:Inverse of Subset of Group|inverse]] of $X$. | First, note that:
{{begin-eqn}}
{{eqn | o =
| r = x \in X, y \in Y
| c =
}}
{{eqn | o = \leadsto
| r = x^{-1} \in X^{-1}, y^{-1} \in Y^{-1}
| c = {{Defof|Inverse of Subset of Group}}
}}
{{eqn | o = \leadsto
| r = y^{-1} \circ x^{-1} \in Y^{-1} \circ X^{-1}
| c = {{Defof|Subset Pro... | Inverse of Product of Subsets of Group/Proof 1 | https://proofwiki.org/wiki/Inverse_of_Product_of_Subsets_of_Group | https://proofwiki.org/wiki/Inverse_of_Product_of_Subsets_of_Group/Proof_1 | [
"Group Theory",
"Subset Products",
"Inverse of Product"
] | [
"Definition:Group",
"Definition:Inverse of Subset/Group"
] | [
"Inverse of Group Product"
] |
proofwiki-610 | Inverse of Product of Subsets of Group | Let $\struct {G, \circ}$ be a group.
Let $X, Y \subseteq G$.
Then:
:$\paren {X \circ Y}^{-1} = Y^{-1} \circ X^{-1}$
where $X^{-1}$ is the inverse of $X$. | === Superset ===
We will show that $\forall z \in Y^{-1} \circ X^{-1}: z \in \paren {X \circ Y}^{-1}$, from which:
:$Y^{-1} \circ X^{-1} \subseteq \paren {X \circ Y}^{-1}$
Let $z \in Y^{-1} \circ X^{-1}$.
By the definition of subset product:
:$\exists x' \in X^{-1}, y' \in Y^{-1}: z = y' \circ x'$
Then by Inverse of Gr... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $X, Y \subseteq G$.
Then:
:$\paren {X \circ Y}^{-1} = Y^{-1} \circ X^{-1}$
where $X^{-1}$ is the [[Definition:Inverse of Subset of Group|inverse]] of $X$. | === Superset ===
We will show that $\forall z \in Y^{-1} \circ X^{-1}: z \in \paren {X \circ Y}^{-1}$, from which:
:$Y^{-1} \circ X^{-1} \subseteq \paren {X \circ Y}^{-1}$
Let $z \in Y^{-1} \circ X^{-1}$.
By the definition of [[Definition:Subset Product|subset product]]:
:$\exists x' \in X^{-1}, y' \in Y^{-1}: z = ... | Inverse of Product of Subsets of Group/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Product_of_Subsets_of_Group | https://proofwiki.org/wiki/Inverse_of_Product_of_Subsets_of_Group/Proof_2 | [
"Group Theory",
"Subset Products",
"Inverse of Product"
] | [
"Definition:Group",
"Definition:Inverse of Subset/Group"
] | [
"Definition:Subset Product",
"Inverse of Group Product",
"Definition:Inverse of Subset/Group",
"Definition:Subset Product",
"Definition:Inverse of Subset/Group",
"Definition:Inverse of Subset/Group",
"Inverse of Inverse of Subset of Group",
"Definition:Inverse of Subset/Group"
] |
proofwiki-611 | Regular Representations in Group are Permutations | Let $\struct {G, \circ}$ be a group.
Let $a \in G$ be any element of $G$.
Then the left regular representation $\lambda_a$ and the right regular representation $\rho_a$ are permutations of $G$. | This follows directly from the fact that all elements of a group are by definition invertible.
Therefore the result Regular Representation of Invertible Element is Permutation applies.
{{Qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $a \in G$ be any [[Definition:Element|element]] of $G$.
Then the [[Definition:Left Regular Representation|left regular representation]] $\lambda_a$ and the [[Definition:Right Regular Representation|right regular representation]] $\rho_a$ are [[Definition:... | This follows directly from the fact that all [[Definition:Element|elements]] of a [[Definition:Group|group]] are by definition [[Definition:Invertible Element|invertible]].
Therefore the result [[Regular Representation of Invertible Element is Permutation]] applies.
{{Qed}} | Regular Representations in Group are Permutations | https://proofwiki.org/wiki/Regular_Representations_in_Group_are_Permutations | https://proofwiki.org/wiki/Regular_Representations_in_Group_are_Permutations | [
"Group Theory",
"Regular Representations"
] | [
"Definition:Group",
"Definition:Element",
"Definition:Regular Representations/Left Regular Representation",
"Definition:Regular Representations/Right Regular Representation",
"Definition:Permutation"
] | [
"Definition:Element",
"Definition:Group",
"Definition:Invertible Element",
"Regular Representation of Invertible Element is Permutation"
] |
proofwiki-612 | Set Equivalence of Regular Representations | If $S$ is a finite subset of a group $G$, then:
:$\card {a \circ S} = \card S = \left|{S \circ a}\right|$
That is, $a \circ S$, $S$ and $S \circ a$ are equivalent: $a \circ S \sim S \sim S \circ a$. | Follows immediately from the fact that both the left and right regular representation are permutations, and therefore bijections.
{{qed}} | If $S$ is a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of a [[Definition:Group|group]] $G$, then:
:$\card {a \circ S} = \card S = \left|{S \circ a}\right|$
That is, $a \circ S$, $S$ and $S \circ a$ are [[Definition:Set Equivalence|equivalent]]: $a \circ S \sim S \sim S \circ a$. | Follows immediately from the fact that both the [[Definition:Left Regular Representation|left]] and [[Definition:Right Regular Representation|right regular representation]] are [[Regular Representations in Group are Permutations|permutations]], and therefore [[Definition:Bijection|bijections]].
{{qed}} | Set Equivalence of Regular Representations | https://proofwiki.org/wiki/Set_Equivalence_of_Regular_Representations | https://proofwiki.org/wiki/Set_Equivalence_of_Regular_Representations | [
"Group Theory",
"Regular Representations"
] | [
"Definition:Finite Set",
"Definition:Subset",
"Definition:Group",
"Definition:Set Equivalence"
] | [
"Definition:Regular Representations/Left Regular Representation",
"Definition:Regular Representations/Right Regular Representation",
"Regular Representations in Group are Permutations",
"Definition:Bijection"
] |
proofwiki-613 | All Elements Self-Inverse then Abelian | Let $\struct {G, \circ}$ be a group.
Suppose that every element of $G$ is self-inverse.
Then $G$ is abelian. | Every element of $G$ is self-inverse, that is:
:$\forall x \in G: x \circ x = e$
In particular, for all $x, y \in G$:
:$\paren {x \circ y} \circ \paren {x \circ y} = e$
that is, $x \circ y$ is also self-inverse.
From Self-Inverse Elements Commute iff Product is Self-Inverse, it follows that:
:$\forall x, y \in G: y \ci... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Suppose that every [[Definition:Element|element]] of $G$ is [[Definition:Self-Inverse Element|self-inverse]].
Then $G$ is [[Definition:Abelian Group|abelian]]. | Every [[Definition:Element|element]] of $G$ is [[Definition:Self-Inverse Element|self-inverse]], that is:
:$\forall x \in G: x \circ x = e$
In particular, for all $x, y \in G$:
:$\paren {x \circ y} \circ \paren {x \circ y} = e$
that is, $x \circ y$ is also [[Definition:Self-Inverse Element|self-inverse]].
From [[... | All Elements Self-Inverse then Abelian | https://proofwiki.org/wiki/All_Elements_Self-Inverse_then_Abelian | https://proofwiki.org/wiki/All_Elements_Self-Inverse_then_Abelian | [
"Abelian Groups"
] | [
"Definition:Group",
"Definition:Element",
"Definition:Self-Inverse Element",
"Definition:Abelian Group"
] | [
"Definition:Element",
"Definition:Self-Inverse Element",
"Definition:Self-Inverse Element",
"Self-Inverse Elements Commute iff Product is Self-Inverse",
"Definition:Commutative/Elements",
"Definition:Abelian Group"
] |
proofwiki-614 | Commutation Property in Group | Let $\struct {G, \circ}$ be a group.
Then $x$ and $y$ commute {{iff}} $x \circ y \circ x^{-1} = y$. | {{begin-eqn}}
{{eqn | l = x \circ y
| r = y \circ x
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {x \circ y} \circ x^{-1}
| r = \paren {y \circ x} \circ x^{-1}
| c = Cancellation Laws
}}
{{eqn | ll= \leadstoandfrom
| l = x \circ y \circ x^{-1}
| r = y \circ \paren {x \circ x^{-1} }
... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then $x$ and $y$ [[Definition:Commute|commute]] {{iff}} $x \circ y \circ x^{-1} = y$. | {{begin-eqn}}
{{eqn | l = x \circ y
| r = y \circ x
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {x \circ y} \circ x^{-1}
| r = \paren {y \circ x} \circ x^{-1}
| c = [[Cancellation Laws]]
}}
{{eqn | ll= \leadstoandfrom
| l = x \circ y \circ x^{-1}
| r = y \circ \paren {x \circ x^{-1} ... | Commutation Property in Group | https://proofwiki.org/wiki/Commutation_Property_in_Group | https://proofwiki.org/wiki/Commutation_Property_in_Group | [
"Group Theory",
"Commutativity"
] | [
"Definition:Group",
"Definition:Commutative/Elements"
] | [
"Cancellation Laws"
] |
proofwiki-615 | Identity Mapping is Automorphism | The identity mapping $I_S: \struct {S, \circ} \to \struct {S, \circ}$ on the algebraic structure $\struct {S, \circ}$ is an automorphism.
Its image is $S$. | By definition, an automorphism is an isomorphism from an algebraic structure onto itself.
An isomorphism, in turn, is a bijective homomorphism.
From Identity Mapping is Bijection, the identity mapping $I_S: S \to S$ on the set $S$ is a bijection from $S$ onto itself.
Now we need to show it is a homomorphism.
Let $x, y ... | The [[Definition:Identity Mapping|identity mapping]] $I_S: \struct {S, \circ} \to \struct {S, \circ}$ on the [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {S, \circ}$ is an [[Definition:Automorphism (Abstract Algebra)|automorphism]].
Its [[Definition:Image of Mapping|image]] is $S$... | By definition, an [[Definition:Automorphism (Abstract Algebra)|automorphism]] is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from an [[Definition:Algebraic Structure|algebraic structure]] onto itself.
An [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], in turn, is a [[Definition:Bijection|bi... | Identity Mapping is Automorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Automorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Automorphism | [
"Automorphisms (Abstract Algebra)",
"Identity Mappings"
] | [
"Definition:Identity Mapping",
"Definition:Algebraic Structure/One Operation",
"Definition:Automorphism (Abstract Algebra)",
"Definition:Image (Set Theory)/Mapping/Mapping"
] | [
"Definition:Automorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Algebraic Structure",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Bijection",
"Definition:Homomorphism (Abstract Algebra)",
"Identity Mapping is Bijection",
"Definition:Identity Mapping"... |
proofwiki-616 | Group Homomorphism of Product with Inverse | Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.
Then:
{{begin-eqn}}
{{eqn | n = 1
| q = \forall x, y \in G
| l = \map \phi {x \circ y^{-1} }
| r = \map \phi x * \paren {\map \phi y}^{-1}
}}
{{eqn | n = 2
| q = \f... | Let $e_G$ and $e_H$ be the identities of $\struct {G, \circ}$ and $\struct {H, *}$ respectively.
By {{Group-axiom|0}}:
:$\forall x, y \in G: x \circ y^{-1} \in G, y^{-1} \circ x \in G$
;Result $(1)$:
{{begin-eqn}}
{{eqn | l = \map \phi {x \circ y^{-1} } * \map \phi y
| r = \map \phi { \paren {x \circ y^{-1} } \ci... | Let $\struct {G, \circ}$ and $\struct {H, *}$ be [[Definition:Group|groups]].
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Homomorphism|group homomorphism]].
Then:
{{begin-eqn}}
{{eqn | n = 1
| q = \forall x, y \in G
| l = \map \phi {x \circ y^{-1} }
| r = \map \phi x *... | Let $e_G$ and $e_H$ be the [[Definition:Identity Element|identities]] of $\struct {G, \circ}$ and $\struct {H, *}$ respectively.
By {{Group-axiom|0}}:
:$\forall x, y \in G: x \circ y^{-1} \in G, y^{-1} \circ x \in G$
;Result $(1)$:
{{begin-eqn}}
{{eqn | l = \map \phi {x \circ y^{-1} } * \map \phi y
| r = \map... | Group Homomorphism of Product with Inverse | https://proofwiki.org/wiki/Group_Homomorphism_of_Product_with_Inverse | https://proofwiki.org/wiki/Group_Homomorphism_of_Product_with_Inverse | [
"Group Homomorphisms"
] | [
"Definition:Group",
"Definition:Group Homomorphism",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-617 | Mapping to Square is Endomorphism iff Abelian | Let $\struct {G, \circ}$ be a group.
Let $\phi: G \to G$ be defined as:
:$\forall g \in G: \map \phi g = g \circ g$
Then $\struct {G, \circ}$ is abelian {{iff}} $\phi$ is a (group) endomorphism. | === Necessary Condition ===
Let $\struct {G, \circ}$ be an abelian group.
Let $a, b \in G$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {a \circ b}
| r = \paren {a \circ b} \circ \paren {a \circ b}
| c = Definition of $\phi$
}}
{{eqn | r = a \circ \paren {b \circ a} \circ b
| c = {{Group-ax... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\phi: G \to G$ be defined as:
:$\forall g \in G: \map \phi g = g \circ g$
Then $\struct {G, \circ}$ is [[Definition:Abelian Group|abelian]] {{iff}} $\phi$ is a [[Definition:Group Endomorphism|(group) endomorphism]]. | === Necessary Condition ===
Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]].
Let $a, b \in G$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi {a \circ b}
| r = \paren {a \circ b} \circ \paren {a \circ b}
| c = Definition of $\phi$
}}
{{eqn | r = a \circ \paren {b \circ a... | Mapping to Square is Endomorphism iff Abelian | https://proofwiki.org/wiki/Mapping_to_Square_is_Endomorphism_iff_Abelian | https://proofwiki.org/wiki/Mapping_to_Square_is_Endomorphism_iff_Abelian | [
"Abelian Groups",
"Group Endomorphisms"
] | [
"Definition:Group",
"Definition:Abelian Group",
"Definition:Group Endomorphism"
] | [
"Definition:Abelian Group",
"Definition:Commutative/Operation",
"Definition:Group Homomorphism",
"Definition:Group Endomorphism",
"Definition:Group Endomorphism",
"Definition:Abelian Group"
] |
proofwiki-618 | Induced Group Product is Homomorphism iff Commutative | Let $\struct {G, \circ}$ be a group.
Let $H_1, H_2$ be subgroups of $G$.
Let $\phi: H_1 \times H_2 \to G$ be defined such that:
:$\forall \tuple {h_1, h_2} \in H_1 \times H_2: \map \phi {h_1, h_2} = h_1 \circ h_2$
Then $\phi$ is a homomorphism {{iff}} every element of $H_1$ commutes with every element of $H_2$. | We have $\tuple {h_1, h_2} \circ \tuple {k_1, k_2} = \tuple {h_1 \circ k_1, h_2 \circ k_2}$ by definition of group direct product. | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H_1, H_2$ be [[Definition:Subgroup|subgroups]] of $G$.
Let $\phi: H_1 \times H_2 \to G$ be defined such that:
:$\forall \tuple {h_1, h_2} \in H_1 \times H_2: \map \phi {h_1, h_2} = h_1 \circ h_2$
Then $\phi$ is a [[Definition:Group Homomorphism|homomorp... | We have $\tuple {h_1, h_2} \circ \tuple {k_1, k_2} = \tuple {h_1 \circ k_1, h_2 \circ k_2}$ by definition of [[Definition:Group Direct Product|group direct product]]. | Induced Group Product is Homomorphism iff Commutative | https://proofwiki.org/wiki/Induced_Group_Product_is_Homomorphism_iff_Commutative | https://proofwiki.org/wiki/Induced_Group_Product_is_Homomorphism_iff_Commutative | [
"Group Homomorphisms"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Group Homomorphism",
"Definition:Commutative/Elements"
] | [
"Definition:Group Direct Product"
] |
proofwiki-619 | Isomorphism of Abelian Groups | Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group isomorphism.
Then $\struct {G, \circ}$ is abelian {{iff}} $\struct {H, *}$ is abelian. | We have that Isomorphism Preserves Commutativity.
Thus:
:$\forall x, y \in G: x \circ y = y \circ x \implies \map \phi x * \map \phi y = \map \phi y * \map \phi x$
Thus if $G$ is abelian, so is $H$.
As $\phi^{-1}: H \to G$ is also an isomorphism, it is clear that if $H$ is abelian, then so is $G$.
{{qed}} | Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Isomorphism|group isomorphism]].
Then $\struct {G, \circ}$ is [[Definition:Abelian Group|abelian]] {{iff}} $\struct {H, *}$ is [[Definition:Abelian Group|abelian]]. | We have that [[Isomorphism Preserves Commutativity]].
Thus:
:$\forall x, y \in G: x \circ y = y \circ x \implies \map \phi x * \map \phi y = \map \phi y * \map \phi x$
Thus if $G$ is [[Definition:Abelian Group|abelian]], so is $H$.
As $\phi^{-1}: H \to G$ is [[Inverse of Algebraic Structure Isomorphism is Isomorph... | Isomorphism of Abelian Groups | https://proofwiki.org/wiki/Isomorphism_of_Abelian_Groups | https://proofwiki.org/wiki/Isomorphism_of_Abelian_Groups | [
"Abelian Groups",
"Group Isomorphisms"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Abelian Group",
"Definition:Abelian Group"
] | [
"Isomorphism Preserves Commutativity",
"Definition:Abelian Group",
"Inverse of Algebraic Structure Isomorphism is Isomorphism",
"Definition:Abelian Group"
] |
proofwiki-620 | Opposite Group is Group | Let $\struct {G, \circ}$ be a group.
Let $\struct {G, *}$ be the opposite group to $G$.
Then $\struct {G, *}$ is a group. | === {{Group-axiom|0|nolink}} ===
$\struct {G, *}$ is closed:
:$b \circ a \in G \implies a * b \in G$
{{qed|lemma}}
=== {{Group-axiom|1|nolink}} ===
$*$ is associative on $G$:
{{begin-eqn}}
{{eqn | l = a * \paren {b * c}
| r = \paren {c \circ b} \circ a
| c = Definition of $*$
}}
{{eqn | r = c \circ \paren {... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\struct {G, *}$ be the [[Definition:Opposite Group|opposite group]] to $G$.
Then $\struct {G, *}$ is a [[Definition:Group|group]]. | === {{Group-axiom|0|nolink}} ===
$\struct {G, *}$ is [[Definition:Closed Algebraic Structure|closed]]:
:$b \circ a \in G \implies a * b \in G$
{{qed|lemma}}
=== {{Group-axiom|1|nolink}} ===
$*$ is [[Definition:Associative Operation|associative]] on $G$:
{{begin-eqn}}
{{eqn | l = a * \paren {b * c}
| r = \par... | Opposite Group is Group/Proof 1 | https://proofwiki.org/wiki/Opposite_Group_is_Group | https://proofwiki.org/wiki/Opposite_Group_is_Group/Proof_1 | [
"Opposite Group is Group",
"Opposite Groups"
] | [
"Definition:Group",
"Definition:Opposite Group",
"Definition:Group"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
... |
proofwiki-621 | Group Example: x inv c y | Let $\struct {G, \circ}$ be a group.
Let $c \in G$.
We define a new operation $*$ on $G$ as:
:$\forall x, y \in G: x * y = x \circ c^{-1} \circ y$
Then $\struct {G, *}$ is a group. | === {{Group-axiom|0|nolink}} ===
Let $x, y \in G$.
Then:
:$\forall x * y = x \circ c^{-1} \circ y \in G$ as $c^{-1} \in G$
thus demonstrating that $\struct {G, *}$ is closed.
{{qed|lemma}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $c \in G$.
We define a new [[Definition:Group Operation|operation]] $*$ on $G$ as:
:$\forall x, y \in G: x * y = x \circ c^{-1} \circ y$
Then $\struct {G, *}$ is a [[Definition:Group|group]]. | === {{Group-axiom|0|nolink}} ===
Let $x, y \in G$.
Then:
:$\forall x * y = x \circ c^{-1} \circ y \in G$ as $c^{-1} \in G$
thus demonstrating that $\struct {G, *}$ is [[Definition:Closed Algebraic Structure|closed]].
{{qed|lemma}} | Group Example: x inv c y | https://proofwiki.org/wiki/Group_Example:_x_inv_c_y | https://proofwiki.org/wiki/Group_Example:_x_inv_c_y | [
"Examples of Groups"
] | [
"Definition:Group",
"Definition:Group Product/Group Law",
"Definition:Group"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-622 | Symmetric Difference on Power Set forms Abelian Group | Let $S$ be a set such that $\O \subset S$ (that is, $S$ is non-empty).
Let $A \symdif B$ be defined as the symmetric difference between $A$ and $B$.
Let $\powerset S$ denote the power set of $S$.
Then the algebraic structure $\struct {\powerset S, \symdif}$ is an abelian group. | From Power Set is Closed under Symmetric Difference, we have that $\struct {\powerset S, \symdif}$ is closed.
The result follows directly from Set System Closed under Symmetric Difference is Abelian Group.
{{Qed}} | Let $S$ be a [[Definition:Set|set]] such that $\O \subset S$ (that is, $S$ is [[Definition:Non-Empty Set|non-empty]]).
Let $A \symdif B$ be defined as the [[Definition:Symmetric Difference|symmetric difference]] between $A$ and $B$.
Let $\powerset S$ denote the [[Definition:Power Set|power set]] of $S$.
Then the [[... | From [[Power Set is Closed under Symmetric Difference]], we have that $\struct {\powerset S, \symdif}$ is [[Definition:Closed Algebraic Structure|closed]].
The result follows directly from [[Set System Closed under Symmetric Difference is Abelian Group]].
{{Qed}} | Symmetric Difference on Power Set forms Abelian Group | https://proofwiki.org/wiki/Symmetric_Difference_on_Power_Set_forms_Abelian_Group | https://proofwiki.org/wiki/Symmetric_Difference_on_Power_Set_forms_Abelian_Group | [
"Abelian Groups",
"Symmetric Difference",
"Power Set"
] | [
"Definition:Set",
"Definition:Non-Empty Set",
"Definition:Symmetric Difference",
"Definition:Power Set",
"Definition:Algebraic Structure",
"Definition:Abelian Group"
] | [
"Power Set is Closed under Symmetric Difference",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Set System Closed under Symmetric Difference is Abelian Group"
] |
proofwiki-623 | Group is Subgroup of Itself | Let $\struct {G, \circ}$ be a group.
Then:
:$\struct {G, \circ} \le \struct {G, \circ}$
That is, a group is always a subgroup of itself. | By Set is Subset of Itself, we have that:
:$G \subseteq G$
Thus $\struct {G, \circ}$ is a group which is a subset of $\struct {G, \circ}$, and therefore a subgroup of $\struct {G, \circ}$.
{{Qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then:
:$\struct {G, \circ} \le \struct {G, \circ}$
That is, a [[Definition:Group|group]] is always a [[Definition:Subgroup|subgroup]] of itself. | By [[Set is Subset of Itself]], we have that:
:$G \subseteq G$
Thus $\struct {G, \circ}$ is a [[Definition:Group|group]] which is a [[Definition:Subset|subset]] of $\struct {G, \circ}$, and therefore a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$.
{{Qed}} | Group is Subgroup of Itself | https://proofwiki.org/wiki/Group_is_Subgroup_of_Itself | https://proofwiki.org/wiki/Group_is_Subgroup_of_Itself | [
"Subgroups"
] | [
"Definition:Group",
"Definition:Group",
"Definition:Subgroup"
] | [
"Set is Subset of Itself",
"Definition:Group",
"Definition:Subset",
"Definition:Subgroup"
] |
proofwiki-624 | Identity of Subgroup | Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of group $G$.
Then the identity of $H$ is also $e$. | A group is {{afortiori}} a monoid.
From the Cancellation Laws, all of its elements are cancellable.
The result then follows from Identity of Cancellable Monoid is Identity of Submonoid.
{{qed}} | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H$ be a [[Definition:Subgroup|subgroup]] of [[Definition:Group|group]] $G$.
Then the [[Definition:Identity Element|identity]] of $H$ is also $e$. | A [[Definition:Group|group]] is {{afortiori}} a [[Definition:Monoid|monoid]].
From the [[Cancellation Laws]], all of its [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]].
The result then follows from [[Identity of Cancellable Monoid is Identity of Submonoid]].
{{qed}} | Identity of Subgroup | https://proofwiki.org/wiki/Identity_of_Subgroup | https://proofwiki.org/wiki/Identity_of_Subgroup | [
"Subgroups",
"Identity Elements"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Group",
"Definition:Monoid",
"Cancellation Laws",
"Definition:Element",
"Definition:Cancellable Element",
"Identity of Cancellable Monoid is Identity of Submonoid"
] |
proofwiki-625 | Inverses in Subgroup | Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Then for each $h \in H$, the inverse of $h$ in $H$ is the same as the inverse of $h$ in $G$. | Let $h \in H$.
Let:
:$h'$ be the inverse of $h$ in $H$
:$h^{-1}$ be the inverse of $h$ in $G$.
From Identity of Subgroup:
:$h \circ h' = e$
From Inverse in Group is Unique, it follows that $h' = h^{-1}$.
{{qed}} | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then for each $h \in H$, the [[Definition:Inverse Element|inverse]] of $h$ in $H$ is the same as the [[Definition:Inverse Element|inverse]] of $h$ in $G$. | Let $h \in H$.
Let:
:$h'$ be the [[Definition:Inverse Element|inverse]] of $h$ in $H$
:$h^{-1}$ be the [[Definition:Inverse Element|inverse]] of $h$ in $G$.
From [[Identity of Subgroup]]:
:$h \circ h' = e$
From [[Inverse in Group is Unique]], it follows that $h' = h^{-1}$.
{{qed}} | Inverses in Subgroup | https://proofwiki.org/wiki/Inverses_in_Subgroup | https://proofwiki.org/wiki/Inverses_in_Subgroup | [
"Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Identity of Subgroup",
"Inverse in Group is Unique"
] |
proofwiki-626 | Subgroup of Abelian Group is Abelian | A subgroup of an abelian group is itself abelian. | Follows directly from Restriction of Commutative Operation is Commutative.
{{qed}} | A [[Definition:Subgroup|subgroup]] of an [[Definition:Abelian Group|abelian group]] is itself [[Definition:Abelian Group|abelian]]. | Follows directly from [[Restriction of Commutative Operation is Commutative]].
{{qed}} | Subgroup of Abelian Group is Abelian | https://proofwiki.org/wiki/Subgroup_of_Abelian_Group_is_Abelian | https://proofwiki.org/wiki/Subgroup_of_Abelian_Group_is_Abelian | [
"Abelian Groups",
"Subgroups"
] | [
"Definition:Subgroup",
"Definition:Abelian Group",
"Definition:Abelian Group"
] | [
"Restriction of Commutative Operation is Commutative"
] |
proofwiki-627 | Intersection of Subgroups is Subgroup | The intersection of two subgroups of a group is itself a subgroup of that group:
:$\forall H_1, H_2 \le \struct {G, \circ}: H_1 \cap H_2 \le G$
It also follows that $H_1 \cap H_2 \le H_1$ and $H_1 \cap H_2 \le H_2$. | Let $H = H_1 \cap H_2$ where $H_1, H_2 \le \struct {G, \circ}$.
Then:
{{begin-eqn}}
{{eqn | o =
| r = a, b \in H
| c =
}}
{{eqn | o = \leadsto
| r = a, b \in H_1 \land a, b \in H_2
| c = {{Defof|Set Intersection}}
}}
{{eqn | o = \leadsto
| r = a \circ b^{-1} \in H_1 \land a \circ b^{-1} ... | The [[Definition:Set Intersection|intersection]] of two [[Definition:Subgroup|subgroups]] of a [[Definition:Group|group]] is itself a [[Definition:Subgroup|subgroup]] of that [[Definition:Group|group]]:
:$\forall H_1, H_2 \le \struct {G, \circ}: H_1 \cap H_2 \le G$
It also follows that $H_1 \cap H_2 \le H_1$ and $H_... | Let $H = H_1 \cap H_2$ where $H_1, H_2 \le \struct {G, \circ}$.
Then:
{{begin-eqn}}
{{eqn | o =
| r = a, b \in H
| c =
}}
{{eqn | o = \leadsto
| r = a, b \in H_1 \land a, b \in H_2
| c = {{Defof|Set Intersection}}
}}
{{eqn | o = \leadsto
| r = a \circ b^{-1} \in H_1 \land a \circ b^{-1... | Intersection of Subgroups is Subgroup | https://proofwiki.org/wiki/Intersection_of_Subgroups_is_Subgroup | https://proofwiki.org/wiki/Intersection_of_Subgroups_is_Subgroup | [
"Subgroups",
"Set Intersection"
] | [
"Definition:Set Intersection",
"Definition:Subgroup",
"Definition:Group",
"Definition:Subgroup",
"Definition:Group"
] | [
"Definition:Group",
"One-Step Subgroup Test"
] |
proofwiki-628 | Union of Subgroups | Let $\struct {G, \circ}$ be a group.
Let $H, K \le G$ be subgroups of $G$.
Let neither $H \subseteq K$ nor $K \subseteq H$.
Then $H \cup K$ is ''not'' a subgroup of $G$. | As neither $H \subseteq K$ nor $K \subseteq H$, it follows from Set Difference with Superset is Empty Set that neither $H \setminus K = \O$ nor $K \setminus H = \O$.
So, let $h \in H \setminus K, k \in K \setminus H$.
Thus, $h \notin K, k \notin H$.
If $\struct {H \cup K, \circ}$ is a group, then it must be closed.
If ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H, K \le G$ be [[Definition:Subgroup|subgroups]] of $G$.
Let neither $H \subseteq K$ nor $K \subseteq H$.
Then $H \cup K$ is ''not'' a [[Definition:Subgroup|subgroup]] of $G$. | As neither $H \subseteq K$ nor $K \subseteq H$, it follows from [[Set Difference with Superset is Empty Set]] that neither $H \setminus K = \O$ nor $K \setminus H = \O$.
So, let $h \in H \setminus K, k \in K \setminus H$.
Thus, $h \notin K, k \notin H$.
If $\struct {H \cup K, \circ}$ is a [[Definition:Group|group]]... | Union of Subgroups | https://proofwiki.org/wiki/Union_of_Subgroups | https://proofwiki.org/wiki/Union_of_Subgroups | [
"Subgroups",
"Set Union",
"Union of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Subgroup"
] | [
"Set Difference with Superset is Empty Set",
"Definition:Group",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Subgroup"
] |
proofwiki-629 | Elements of Group with Equal Images under Homomorphisms form Subgroup | Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $f: G \to H$ and $g: G \to H$ be group homomorphisms.
Then the set:
:$S = \set {x \in G: \map f x = \map g x}$
is a subgroup of $G$. | Let the identities of $\struct {G, \circ}$ and $\struct {H, *}$ be $e_G$ and $e_H$ respectively.
By Homomorphism to Group Preserves Identity:
:$\map f {e_G} = \map g {e_G} = e_H$
Thus $e_G \in S$, and so $S \ne \O$.
Similarly, from Homomorphism to Group Preserves Inverses, $x \in S \implies x^{-1} \in S$.
Let $x, y \in... | Let $\struct {G, \circ}$ and $\struct {H, *}$ be [[Definition:Group|groups]].
Let $f: G \to H$ and $g: G \to H$ be [[Definition:Group Homomorphism|group homomorphisms]].
Then the [[Definition:Set|set]]:
:$S = \set {x \in G: \map f x = \map g x}$
is a [[Definition:Subgroup|subgroup]] of $G$. | Let the [[Definition:Identity Element|identities]] of $\struct {G, \circ}$ and $\struct {H, *}$ be $e_G$ and $e_H$ respectively.
By [[Homomorphism to Group Preserves Identity]]:
:$\map f {e_G} = \map g {e_G} = e_H$
Thus $e_G \in S$, and so $S \ne \O$.
Similarly, from [[Homomorphism to Group Preserves Inverses]], $x... | Elements of Group with Equal Images under Homomorphisms form Subgroup | https://proofwiki.org/wiki/Elements_of_Group_with_Equal_Images_under_Homomorphisms_form_Subgroup | https://proofwiki.org/wiki/Elements_of_Group_with_Equal_Images_under_Homomorphisms_form_Subgroup | [
"Group Homomorphisms"
] | [
"Definition:Group",
"Definition:Group Homomorphism",
"Definition:Set",
"Definition:Subgroup"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Homomorphism to Group Preserves Identity",
"Homomorphism to Group Preserves Inverses",
"Definition:Morphism Property",
"Definition:Morphism Property",
"Two-Step Subgroup Test"
] |
proofwiki-630 | Product of Subgroup with Itself | Let $\struct {G, \circ}$ be a group.
Then:
:$\forall H \le G: H \circ H = H$
where:
:$H \circ H$ denotes the subset product of $H$ with $H$
:$\le$ denotes the subgroup relation. | From Magma Subset Product with Self, we have:
:$H \circ H \subseteq H$
Let $e$ be the identity of $G$.
By Identity of Subgroup, it is also the identity of $H$.
So:
{{begin-eqn}}
{{eqn | l = h
| o = \in
| r = H
| c =
}}
{{eqn | ll= \leadsto
| l = e \circ h
| o = \in
| r = H \circ H
... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then:
:$\forall H \le G: H \circ H = H$
where:
:$H \circ H$ denotes the [[Definition:Subset Product|subset product]] of $H$ with $H$
:$\le$ denotes the [[Definition:Subgroup|subgroup relation]]. | From [[Magma Subset Product with Self]], we have:
:$H \circ H \subseteq H$
Let $e$ be the [[Definition:Identity Element|identity]] of $G$.
By [[Identity of Subgroup]], it is also the [[Definition:Identity Element|identity]] of $H$.
So:
{{begin-eqn}}
{{eqn | l = h
| o = \in
| r = H
| c =
}}
{{eq... | Product of Subgroup with Itself | https://proofwiki.org/wiki/Product_of_Subgroup_with_Itself | https://proofwiki.org/wiki/Product_of_Subgroup_with_Itself | [
"Subgroups",
"Subset Products"
] | [
"Definition:Group",
"Definition:Subset Product",
"Definition:Subgroup"
] | [
"Magma Subset Product with Self",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Identity of Subgroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Set Equality"
] |
proofwiki-631 | Inverse of Subgroup | Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Then:
:$H^{-1} = H$
where $H^{-1}$ is the inverse of $H$. | As $H$ is a subgroup of $G$:
:$\forall h \in H: h^{-1} \in H$
The result follows.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then:
:$H^{-1} = H$
where $H^{-1}$ is the [[Definition:Inverse of Subset of Group|inverse]] of $H$. | As $H$ is a [[Definition:Subgroup|subgroup]] of $G$:
:$\forall h \in H: h^{-1} \in H$
The result follows.
{{qed}} | Inverse of Subgroup | https://proofwiki.org/wiki/Inverse_of_Subgroup | https://proofwiki.org/wiki/Inverse_of_Subgroup | [
"Subgroups",
"Subset Products"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Inverse of Subset/Group"
] | [
"Definition:Subgroup"
] |
proofwiki-632 | Subset Product of Subgroups | Let $\struct {G, \circ}$ be a group.
Let $H, K$ be subgroups of $G$.
Then $H \circ K$ is a subgroup of $G$ {{iff}} $H$ and $K$ are permutable.
That is:
$H \circ K$ is a subgroup of $G$ {{iff}}:
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product. | Suppose $H \circ K$ is a subgroup of $G$.
Let $h \circ k \in H \circ K$.
Then $h \circ k$ is the inverse of some element $g$ of $H \circ K$.
Thus we can write $g = h' \circ k'$ for some $h' \in H$ and $k' \in K$.
So:
{{begin-eqn}}
{{eqn | l = h \circ k
| r = g^{-1}
| c = Inverse of Group Inverse: $g$ is the... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H, K$ be [[Definition:Subgroup|subgroups]] of $G$.
Then $H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$ {{iff}} $H$ and $K$ are [[Definition:Permutable Subgroups|permutable]].
That is:
$H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$ ... | Suppose $H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$.
Let $h \circ k \in H \circ K$.
Then $h \circ k$ is the [[Definition:Inverse Element|inverse]] of some [[Definition:Element|element]] $g$ of $H \circ K$.
Thus we can write $g = h' \circ k'$ for some $h' \in H$ and $k' \in K$.
So:
{{begin-eqn}}
{{eqn |... | Subset Product of Subgroups/Necessary Condition/Proof 1 | https://proofwiki.org/wiki/Subset_Product_of_Subgroups | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Necessary_Condition/Proof_1 | [
"Subgroups",
"Subset Products",
"Subset Product of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Permutable Subgroups",
"Definition:Subgroup",
"Definition:Subset Product"
] | [
"Definition:Subgroup",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Element",
"Inverse of Group Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Inverse of Group Product",
"Definition:Group",
"Definition:Subset",
"Inverse of Group Inverse",
"Inverse of Group Product",
... |
proofwiki-633 | Subset Product of Subgroups | Let $\struct {G, \circ}$ be a group.
Let $H, K$ be subgroups of $G$.
Then $H \circ K$ is a subgroup of $G$ {{iff}} $H$ and $K$ are permutable.
That is:
$H \circ K$ is a subgroup of $G$ {{iff}}:
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product. | Suppose $H \circ K$ is a subgroup of $G$.
Then:
{{begin-eqn}}
{{eqn | l = H \circ K
| r = \paren {H \circ K}^{-1}
| c = Inverse of Subgroup
}}
{{eqn | r = K^{-1} \circ H^{-1}
| c = Inverse of Product of Subsets of Group
}}
{{eqn | r = K \circ H
| c = Inverse of Subgroup
}}
{{end-eqn}}
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H, K$ be [[Definition:Subgroup|subgroups]] of $G$.
Then $H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$ {{iff}} $H$ and $K$ are [[Definition:Permutable Subgroups|permutable]].
That is:
$H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$ ... | Suppose $H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$.
Then:
{{begin-eqn}}
{{eqn | l = H \circ K
| r = \paren {H \circ K}^{-1}
| c = [[Inverse of Subgroup]]
}}
{{eqn | r = K^{-1} \circ H^{-1}
| c = [[Inverse of Product of Subsets of Group]]
}}
{{eqn | r = K \circ H
| c = [[Inverse of... | Subset Product of Subgroups/Necessary Condition/Proof 2 | https://proofwiki.org/wiki/Subset_Product_of_Subgroups | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Necessary_Condition/Proof_2 | [
"Subgroups",
"Subset Products",
"Subset Product of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Permutable Subgroups",
"Definition:Subgroup",
"Definition:Subset Product"
] | [
"Definition:Subgroup",
"Inverse of Subgroup",
"Inverse of Product of Subsets of Group",
"Inverse of Subgroup"
] |
proofwiki-634 | Subset Product of Subgroups | Let $\struct {G, \circ}$ be a group.
Let $H, K$ be subgroups of $G$.
Then $H \circ K$ is a subgroup of $G$ {{iff}} $H$ and $K$ are permutable.
That is:
$H \circ K$ is a subgroup of $G$ {{iff}}:
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product. | Suppose $H \circ K = K \circ H$.
First note that $H \circ K \ne \O$, as $e_G = e_G \circ e_G \in H \circ K$, from Identity of Subgroup.
Suppose $a_1, a_2 \in H, b_1, b_2 \in K$.
Then:
:$\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = a_1 \circ \paren {b_1 \circ a_2} \circ b_2$.
Since $H \circ K = K \circ H$, we s... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H, K$ be [[Definition:Subgroup|subgroups]] of $G$.
Then $H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$ {{iff}} $H$ and $K$ are [[Definition:Permutable Subgroups|permutable]].
That is:
$H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$ ... | Suppose $H \circ K = K \circ H$.
First note that $H \circ K \ne \O$, as $e_G = e_G \circ e_G \in H \circ K$, from [[Identity of Subgroup]].
Suppose $a_1, a_2 \in H, b_1, b_2 \in K$.
Then:
:$\paren {a_1 \circ b_1} \circ \paren {a_2 \circ b_2} = a_1 \circ \paren {b_1 \circ a_2} \circ b_2$.
Since $H \circ K = K \cir... | Subset Product of Subgroups/Sufficient Condition/Proof 1 | https://proofwiki.org/wiki/Subset_Product_of_Subgroups | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Sufficient_Condition/Proof_1 | [
"Subgroups",
"Subset Products",
"Subset Product of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Permutable Subgroups",
"Definition:Subgroup",
"Definition:Subset Product"
] | [
"Identity of Subgroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Inverse of Group Product",
"Two-Step Subgroup Test",
"Definition:Subgroup"
] |
proofwiki-635 | Subset Product of Subgroups | Let $\struct {G, \circ}$ be a group.
Let $H, K$ be subgroups of $G$.
Then $H \circ K$ is a subgroup of $G$ {{iff}} $H$ and $K$ are permutable.
That is:
$H \circ K$ is a subgroup of $G$ {{iff}}:
:$H \circ K = K \circ H$
where $H \circ K$ denotes subset product. | Suppose $H \circ K = K \circ H$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {H \circ K} \circ \paren {H \circ K}^{-1}
| r = H \circ K \circ K^{-1} \circ H^{-1}
| c = Inverse of Product of Subsets of Group
}}
{{eqn | r = H \circ K \circ K \circ H
| c = Inverse of Subgroup
}}
{{eqn | r = H \circ K \circ H
... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H, K$ be [[Definition:Subgroup|subgroups]] of $G$.
Then $H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$ {{iff}} $H$ and $K$ are [[Definition:Permutable Subgroups|permutable]].
That is:
$H \circ K$ is a [[Definition:Subgroup|subgroup]] of $G$ ... | Suppose $H \circ K = K \circ H$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {H \circ K} \circ \paren {H \circ K}^{-1}
| r = H \circ K \circ K^{-1} \circ H^{-1}
| c = [[Inverse of Product of Subsets of Group]]
}}
{{eqn | r = H \circ K \circ K \circ H
| c = [[Inverse of Subgroup]]
}}
{{eqn | r = H \circ K... | Subset Product of Subgroups/Sufficient Condition/Proof 2 | https://proofwiki.org/wiki/Subset_Product_of_Subgroups | https://proofwiki.org/wiki/Subset_Product_of_Subgroups/Sufficient_Condition/Proof_2 | [
"Subgroups",
"Subset Products",
"Subset Product of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Permutable Subgroups",
"Definition:Subgroup",
"Definition:Subset Product"
] | [
"Inverse of Product of Subsets of Group",
"Inverse of Subgroup",
"Product of Subgroup with Itself",
"Definition:By Hypothesis",
"Product of Subgroup with Itself",
"Definition:Set Equality/Definition 2",
"One-Step Subgroup Test using Subset Product"
] |
proofwiki-636 | Group Homomorphism Preserves Subgroups | Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups.
Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a group homomorphism.
Then:
:$H \le G_1 \implies \phi \sqbrk H \le G_2$
where:
:$\phi \sqbrk H$ denotes the image of $H$ under $\phi$
:$\le$ denotes subgroup.
That is, group homomorphism preserves subg... | Let $H \le G_1$.
First note that from Null Relation is Mapping iff Domain is Empty Set:
:$H \ne \O \implies \phi \sqbrk H \ne \O$
and so $\phi \sqbrk H$ is not empty.
Next, let $x, y \in \phi \sqbrk H$.
Then:
:$\exists h_1, h_2 \in H: x = \map \phi {h_1}, y = \map \phi {h_2}$
From the definition of Group Homomorphism, ... | Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be [[Definition:Group|groups]].
Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a [[Definition:Group Homomorphism|group homomorphism]].
Then:
:$H \le G_1 \implies \phi \sqbrk H \le G_2$
where:
:$\phi \sqbrk H$ denotes the [[Definition:Image of Subset under ... | Let $H \le G_1$.
First note that from [[Null Relation is Mapping iff Domain is Empty Set]]:
:$H \ne \O \implies \phi \sqbrk H \ne \O$
and so $\phi \sqbrk H$ is not [[Definition:Empty Set|empty]].
Next, let $x, y \in \phi \sqbrk H$.
Then:
:$\exists h_1, h_2 \in H: x = \map \phi {h_1}, y = \map \phi {h_2}$
From the... | Group Homomorphism Preserves Subgroups | https://proofwiki.org/wiki/Group_Homomorphism_Preserves_Subgroups | https://proofwiki.org/wiki/Group_Homomorphism_Preserves_Subgroups | [
"Subgroups",
"Group Homomorphisms"
] | [
"Definition:Group",
"Definition:Group Homomorphism",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Subgroup",
"Definition:Group Homomorphism",
"Definition:Subgroup"
] | [
"Null Relation is Mapping iff Domain is Empty Set",
"Definition:Empty Set",
"Definition:Group Homomorphism",
"Definition:Subgroup",
"One-Step Subgroup Test"
] |
proofwiki-637 | Image of Group Homomorphism is Subgroup | Let $\phi: G_1 \to G_2$ be a group homomorphism.
Then:
:$\Img \phi \le G_2$
where $\le$ denotes the relation of being a subgroup. | This is a special case of Group Homomorphism Preserves Subgroups, where we set $H = G_1$.
{{Qed}} | Let $\phi: G_1 \to G_2$ be a [[Definition:Group Homomorphism|group homomorphism]].
Then:
:$\Img \phi \le G_2$
where $\le$ denotes the relation of being a [[Definition:Subgroup|subgroup]]. | This is a special case of [[Group Homomorphism Preserves Subgroups]], where we set $H = G_1$.
{{Qed}} | Image of Group Homomorphism is Subgroup | https://proofwiki.org/wiki/Image_of_Group_Homomorphism_is_Subgroup | https://proofwiki.org/wiki/Image_of_Group_Homomorphism_is_Subgroup | [
"Group Homomorphisms",
"Subgroups"
] | [
"Definition:Group Homomorphism",
"Definition:Subgroup"
] | [
"Group Homomorphism Preserves Subgroups"
] |
proofwiki-638 | Conjugacy is Equivalence Relation | Conjugacy of group elements is an equivalence relation. | Checking each of the criteria for an equivalence relation in turn: | [[Definition:Conjugate of Group Element|Conjugacy of group elements]] is an [[Definition:Equivalence Relation|equivalence relation]]. | Checking each of the criteria for an [[Definition:Equivalence Relation|equivalence relation]] in turn: | Conjugacy is Equivalence Relation | https://proofwiki.org/wiki/Conjugacy_is_Equivalence_Relation | https://proofwiki.org/wiki/Conjugacy_is_Equivalence_Relation | [
"Conjugacy",
"Examples of Equivalence Relations"
] | [
"Definition:Conjugate (Group Theory)/Element",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-639 | Kernel of Group Homomorphism is Subgroup | The kernel of a group homomorphism is a subgroup of its domain:
:$\map \ker \phi \le \Dom \phi$ | Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.
From Homomorphism to Group Preserves Identity, $\map \phi {e_G} = e_H$, so $e_G \in \map \ker \phi$.
Therefore $\map \ker \phi \ne \O$.
Let $x, y \in \map \ker \phi$, so that $\map \phi x = \map \phi y = e_H$.
Then:
{{begin-eqn}}
{{eqn | l = \ma... | The [[Definition:Kernel of Group Homomorphism|kernel of a group homomorphism]] is a [[Definition:Subgroup|subgroup]] of its [[Definition:Domain of Mapping|domain]]:
:$\map \ker \phi \le \Dom \phi$ | Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Homomorphism|group homomorphism]].
From [[Homomorphism to Group Preserves Identity]], $\map \phi {e_G} = e_H$, so $e_G \in \map \ker \phi$.
Therefore $\map \ker \phi \ne \O$.
Let $x, y \in \map \ker \phi$, so that $\map \phi x = \map \phi y ... | Kernel of Group Homomorphism is Subgroup | https://proofwiki.org/wiki/Kernel_of_Group_Homomorphism_is_Subgroup | https://proofwiki.org/wiki/Kernel_of_Group_Homomorphism_is_Subgroup | [
"Subgroups",
"Kernels of Group Homomorphisms"
] | [
"Definition:Kernel of Group Homomorphism",
"Definition:Subgroup",
"Definition:Domain (Set Theory)/Mapping"
] | [
"Definition:Group Homomorphism",
"Homomorphism to Group Preserves Identity",
"Homomorphism with Identity Preserves Inverses",
"One-Step Subgroup Test"
] |
proofwiki-640 | Commutative Semigroup is Entropic Structure | A commutative semigroup is an entropic structure. | Let $\struct {S, \circ}$ be a commutative semigroup.
Let $a, b, c, d \in S$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {c \circ d}
| r = a \circ \paren {b \circ \paren {c \circ d} }
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = a \circ \paren {\paren {b \circ c} \circ d}
| c = {{Sem... | A [[Definition:Commutative Semigroup|commutative semigroup]] is an [[Definition:Entropic Structure|entropic structure]]. | Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
Let $a, b, c, d \in S$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {c \circ d}
| r = a \circ \paren {b \circ \paren {c \circ d} }
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = a \circ \paren {\pare... | Commutative Semigroup is Entropic Structure | https://proofwiki.org/wiki/Commutative_Semigroup_is_Entropic_Structure | https://proofwiki.org/wiki/Commutative_Semigroup_is_Entropic_Structure | [
"Semigroups",
"Entropic Structures",
"Commutativity"
] | [
"Definition:Commutative Semigroup",
"Definition:Entropic Structure"
] | [
"Definition:Commutative Semigroup",
"Definition:Commutative/Operation"
] |
proofwiki-641 | Abelian Group Induces Entropic Structure | Let $\struct {G, \circ}$ be an abelian group.
Let the operation $*$ be defined on $G$ such that:
:$\forall x, y \in G: x * y = x \circ y^{-1}$
Then $\struct {G, *}$ is an entropic structure. | We need to prove that $\forall a, b, c, d \in G: \paren {a * b} * \paren {c * d} = \paren {a * c} * \paren {b * d}$.
So:
{{begin-eqn}}
{{eqn | l = \paren {a * b} * \paren {c * d}
| r = \paren {a \circ b^{-1} } \circ \paren {c \circ d^{-1} }^{-1}
| c = Definition of $*$
}}
{{eqn | r = \paren {a \circ b^{-1} ... | Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]].
Let the operation $*$ be defined on $G$ such that:
:$\forall x, y \in G: x * y = x \circ y^{-1}$
Then $\struct {G, *}$ is an [[Definition:Entropic Structure|entropic structure]]. | We need to prove that $\forall a, b, c, d \in G: \paren {a * b} * \paren {c * d} = \paren {a * c} * \paren {b * d}$.
So:
{{begin-eqn}}
{{eqn | l = \paren {a * b} * \paren {c * d}
| r = \paren {a \circ b^{-1} } \circ \paren {c \circ d^{-1} }^{-1}
| c = Definition of $*$
}}
{{eqn | r = \paren {a \circ b^{-1... | Abelian Group Induces Entropic Structure | https://proofwiki.org/wiki/Abelian_Group_Induces_Entropic_Structure | https://proofwiki.org/wiki/Abelian_Group_Induces_Entropic_Structure | [
"Abelian Groups",
"Entropic Structures"
] | [
"Definition:Abelian Group",
"Definition:Entropic Structure"
] | [
"Inverse of Group Product",
"Inverse of Group Inverse",
"Inverse of Group Inverse",
"Inverse of Group Product"
] |
proofwiki-642 | Cancellable Semiring with Unity is Additive Semiring | Let $\struct {S, *, \circ}$ be a cancellable semiring with unity $1_S$.
Then the distributand $*$ is commutative.
That is to say, $\struct {S, *, \circ}$ is also an additive semiring. | Let $\struct {S, *, \circ}$ be a semiring, all of whose elements of $S$ are cancellable for $*$.
We expand the expression $\paren {a * b} \circ \paren {c * d}$ using the distributive law in two ways:
{{begin-eqn}}
{{eqn | l = \paren {a * b} \circ \paren {c * d}
| r = \paren {\paren {a * b} \circ c} * \paren {\par... | Let $\struct {S, *, \circ}$ be a [[Definition:Cancellable Semiring|cancellable semiring]] with [[Definition:Unity of Semiring|unity]] $1_S$.
Then the [[Definition:Distributand|distributand]] $*$ is [[Definition:Commutative Operation|commutative]].
That is to say, $\struct {S, *, \circ}$ is also an [[Definition:Addit... | Let $\struct {S, *, \circ}$ be a [[Definition:Semiring (Abstract Algebra)|semiring]], all of whose [[Definition:Element|elements]] of $S$ are [[Definition:Cancellable Element|cancellable]] for $*$.
We expand the expression $\paren {a * b} \circ \paren {c * d}$ using the [[Definition:Distributive Operation|distributive... | Cancellable Semiring with Unity is Additive Semiring | https://proofwiki.org/wiki/Cancellable_Semiring_with_Unity_is_Additive_Semiring | https://proofwiki.org/wiki/Cancellable_Semiring_with_Unity_is_Additive_Semiring | [
"Cancellable Semirings",
"Additive Semirings"
] | [
"Definition:Cancellable Semiring",
"Definition:Unity of Semiring",
"Definition:Distributive Operation/Distributand",
"Definition:Commutative/Operation",
"Definition:Additive Semiring"
] | [
"Definition:Semiring (Abstract Algebra)",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Distributive Operation",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:A Fortiori",
"Definition:Commutative/Operation",
"Definition:Additive Semiring",
"Category:Canc... |
proofwiki-643 | Ring is not Empty | A ring cannot be empty. | 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.
{{qed}} | A [[Definition:Ring (Abstract Algebra)|ring]] cannot be [[Definition:Empty Set|empty]]. | In a ring $\struct {R, +, \circ}$, $\struct {R, +}$ forms a [[Definition:Group|group]].
From [[Group is not Empty]], the [[Definition:Group|group]] $\struct {R, +}$ contains at least the [[Definition:Identity Element|identity]], so cannot be [[Definition:Empty Set|empty]].
So every ring $\struct {R, +, \circ}$ contai... | Ring is not Empty | https://proofwiki.org/wiki/Ring_is_not_Empty | https://proofwiki.org/wiki/Ring_is_not_Empty | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Empty Set"
] | [
"Definition:Group",
"Group is not Empty",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Empty Set",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Ring (Abstract Algebra)/Addition"
] |
proofwiki-644 | Ring Product with Zero | 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. | 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:
{{begin-eqn}}
{{eqn | l = x \circ \paren ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Then:
:$\forall x \in R: 0_R \circ x = 0_R = x \circ 0_R$
That is, the [[Definition:Ring Zero|zero]] is a [[Definition:Zero Element|zero element]] for the [[Definition:Ring Product|ring produc... | Because $\struct {R, +, \circ}$ is a [[Definition:Ring (Abstract Algebra)|ring]], $\struct {R, +}$ is a [[Definition:Group|group]].
Since $0_R$ is the [[Definition:Identity Element|identity]] in $\struct {R, +}$, we have $0_R + 0_R = 0_R$.
From the [[Cancellation Laws]], all [[Definition:Group|group]] [[Definition:El... | Ring Product with Zero | https://proofwiki.org/wiki/Ring_Product_with_Zero | https://proofwiki.org/wiki/Ring_Product_with_Zero | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Ring Zero",
"Definition:Zero Element",
"Definition:Ring (Abstract Algebra)/Product"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Cancellation Laws",
"Definition:Group",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Element",
"Definition:Cancellable Element",
"Cancellation Laws",
"Ca... |
proofwiki-645 | Product with Ring Negative | Let $\struct {R, +, \circ}$ be a ring.
Then:
:$\forall x, y \in \struct {R, +, \circ}: \paren {-x} \circ y = -\paren {x \circ y} = x \circ \paren {-y}$
where $\paren {-x}$ denotes the negative of $x$. | We have:
{{begin-eqn}}
{{eqn | l = \paren {x + \paren {-x} } \circ y
| r = 0_R \circ y
| c = {{Defof|Ring Zero}}
}}
{{eqn | r = 0_R
| c = Ring Product with Zero
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {x \circ y} + \paren {\paren {-x} \circ y}
| r = 0_R
| c = {{Ring-axiom|D}}
}}
... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Then:
:$\forall x, y \in \struct {R, +, \circ}: \paren {-x} \circ y = -\paren {x \circ y} = x \circ \paren {-y}$
where $\paren {-x}$ denotes the [[Definition:Ring Negative|negative]] of $x$. | We have:
{{begin-eqn}}
{{eqn | l = \paren {x + \paren {-x} } \circ y
| r = 0_R \circ y
| c = {{Defof|Ring Zero}}
}}
{{eqn | r = 0_R
| c = [[Ring Product with Zero]]
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {x \circ y} + \paren {\paren {-x} \circ y}
| r = 0_R
| c = {{Ring-axiom|D}... | Product with Ring Negative | https://proofwiki.org/wiki/Product_with_Ring_Negative | https://proofwiki.org/wiki/Product_with_Ring_Negative | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Negative"
] | [
"Ring Product with Zero"
] |
proofwiki-646 | Unity of Ring is Unique | A ring can have no more than one unity. | Let $\struct {R, +, \circ}$ be a ring.
If $\struct {R, \circ}$ has an identity, then it is a monoid.
From Identity of Monoid is Unique, it follows that such an identity is unique.
{{qed}} | A [[Definition:Ring (Abstract Algebra)|ring]] can have no more than one [[Definition:Unity of Ring|unity]]. | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
If $\struct {R, \circ}$ has an [[Definition:Identity (Abstract Algebra)|identity]], then it is a [[Definition:Monoid|monoid]].
From [[Identity of Monoid is Unique]], it follows that such an identity is unique.
{{qed}} | Unity of Ring is Unique | https://proofwiki.org/wiki/Unity_of_Ring_is_Unique | https://proofwiki.org/wiki/Unity_of_Ring_is_Unique | [
"Rings with Unity"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Unity (Abstract Algebra)/Ring"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)",
"Definition:Monoid",
"Identity of Monoid is Unique"
] |
proofwiki-647 | Trivial Ring from Abelian Group | An abelian group $\struct {G, +}$ may be turned into a trivial ring by defining the ring product to be:
:$\forall x, y \in G: x \circ y = e_G$ | Follows directly from the definition of a trivial ring.
{{qed}} | An [[Definition:Abelian Group|abelian group]] $\struct {G, +}$ may be turned into a [[Definition:Trivial Ring|trivial ring]] by defining the [[Definition:Ring Product|ring product]] to be:
:$\forall x, y \in G: x \circ y = e_G$ | Follows directly from the definition of a [[Definition:Trivial Ring|trivial ring]].
{{qed}} | Trivial Ring from Abelian Group | https://proofwiki.org/wiki/Trivial_Ring_from_Abelian_Group | https://proofwiki.org/wiki/Trivial_Ring_from_Abelian_Group | [
"Trivial Rings",
"Abelian Groups"
] | [
"Definition:Abelian Group",
"Definition:Trivial Ring",
"Definition:Ring (Abstract Algebra)/Product"
] | [
"Definition:Trivial Ring"
] |
proofwiki-648 | Null Ring iff Zero and Unity Coincide | The null ring is the only ring in which the unity and zero coincide. | The single element of the null ring serves as an identity for both of the operations.
So, in this particular ring, the unity and the zero are the same element.
A non-null ring contains some non-zero element $a$.
Since $1_R \circ a = a \ne 0_R = 0_R \circ a$, then $1_R \ne 0_R$.
So if a ring is non-null, its unity canno... | The [[Definition:Null Ring|null ring]] is the only [[Definition:Ring (Abstract Algebra)|ring]] in which the [[Definition:Unity of Ring|unity]] and [[Definition:Ring Zero|zero]] coincide. | The single [[Definition:Element|element]] of the [[Definition:Null Ring|null ring]] serves as an [[Definition:Identity Element|identity]] for both of the operations.
So, in this particular [[Definition:Ring (Abstract Algebra)|ring]], the [[Definition:Unity of Ring|unity]] and the [[Definition:Ring Zero|zero]] are the ... | Null Ring iff Zero and Unity Coincide | https://proofwiki.org/wiki/Null_Ring_iff_Zero_and_Unity_Coincide | https://proofwiki.org/wiki/Null_Ring_iff_Zero_and_Unity_Coincide | [
"Null Ring"
] | [
"Definition:Null Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring Zero"
] | [
"Definition:Element",
"Definition:Null Ring",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Ring (Abstract Algebra)",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Non-Null Ring",
"Definition:Ring Zero",
"Definition... |
proofwiki-649 | Unity is Unit | The unity in a ring is a unit. | This is a special case of Direct Product of Unitary Modules is Unitary Module.
{{qed}} | The [[Definition:Unity of Ring|unity]] in a [[Definition:Ring (Abstract Algebra)|ring]] is a [[Definition:Unit of Ring|unit]]. | This is a special case of [[Direct Product of Unitary Modules is Unitary Module]].
{{qed}} | Module on Cartesian Product of Ring with Unity is Unitary Module/Proof 1 | https://proofwiki.org/wiki/Unity_is_Unit | https://proofwiki.org/wiki/Module_on_Cartesian_Product_of_Ring_with_Unity_is_Unitary_Module/Proof_1 | [
"Rings with Unity"
] | [
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Unit of Ring"
] | [
"Direct Product of Unitary Modules is Unitary Module"
] |
proofwiki-650 | Unity is Unit | The unity in a ring is a unit. | This is a special case of the Unitary Module of All Mappings is Unitary Module, where $S$ is the set $\closedint 1 n \subset \N_{>0}$.
{{qed}} | The [[Definition:Unity of Ring|unity]] in a [[Definition:Ring (Abstract Algebra)|ring]] is a [[Definition:Unit of Ring|unit]]. | This is a special case of the [[Unitary Module of All Mappings is Unitary Module]], where $S$ is the [[Definition:Set|set]] $\closedint 1 n \subset \N_{>0}$.
{{qed}} | Module on Cartesian Product of Ring with Unity is Unitary Module/Proof 2 | https://proofwiki.org/wiki/Unity_is_Unit | https://proofwiki.org/wiki/Module_on_Cartesian_Product_of_Ring_with_Unity_is_Unitary_Module/Proof_2 | [
"Rings with Unity"
] | [
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Unit of Ring"
] | [
"Unitary Module of All Mappings is Unitary Module",
"Definition:Set"
] |
proofwiki-651 | Unity is Unit | The unity in a ring is a unit. | This is a special case of a Finite Direct Product of Unitary Modules is Unitary Module where each of the $G_k$ is the $R$-module $R$.
{{qed}} | The [[Definition:Unity of Ring|unity]] in a [[Definition:Ring (Abstract Algebra)|ring]] is a [[Definition:Unit of Ring|unit]]. | This is a special case of a [[Finite Direct Product of Unitary Modules is Unitary Module]] where each of the $G_k$ is the [[Definition:Module over Ring|$R$-module]] $R$.
{{qed}} | Module on Cartesian Product of Ring with Unity is Unitary Module/Proof 3 | https://proofwiki.org/wiki/Unity_is_Unit | https://proofwiki.org/wiki/Module_on_Cartesian_Product_of_Ring_with_Unity_is_Unitary_Module/Proof_3 | [
"Rings with Unity"
] | [
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Unit of Ring"
] | [
"Finite Direct Product of Unitary Modules is Unitary Module",
"Definition:Module over Ring"
] |
proofwiki-652 | Unity is Unit | The unity in a ring is a unit. | Let $\left({R, +, \circ}\right)$ be a ring with unity $1_R$.
From Identity is Self-Inverse:
:$1_R^{-1} = 1_R \in R \implies 1_R \in U_R$
{{qed}}
Category:Rings with Unity
1kgq0vcsz8xqaswu08wx5qsyj7zb6a3 | The [[Definition:Unity of Ring|unity]] in a [[Definition:Ring (Abstract Algebra)|ring]] is a [[Definition:Unit of Ring|unit]]. | Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring with Unity|ring with unity]] $1_R$.
From [[Identity is Self-Inverse]]:
:$1_R^{-1} = 1_R \in R \implies 1_R \in U_R$
{{qed}}
[[Category:Rings with Unity]]
1kgq0vcsz8xqaswu08wx5qsyj7zb6a3 | Unity is Unit | https://proofwiki.org/wiki/Unity_is_Unit | https://proofwiki.org/wiki/Unity_is_Unit | [
"Rings with Unity"
] | [
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Unit of Ring"
] | [
"Definition:Ring with Unity",
"Inverse of Identity Element is Itself",
"Category:Rings with Unity"
] |
proofwiki-653 | Unity and Negative form Subgroup of Units | Let $\struct {R, +, \circ}$ be a ring with unity.
Then:
:$\struct {\set {1_R, -1_R}, \circ} \le U_R$
That is, the set consisting of the unity and its negative forms a subgroup of the group of units. | From Unity is Unit:
:$1_R \in U_R$
It remains to be shown that $-1_R \in U_R$.
From {{Ring-axiom|A}}, $\struct {R, +}$ is an abelian group.
Therefore:
:$1_R \in R \implies -1_R \in R$.
From Product of Ring Negatives:
:$-1_R \circ -1_R = 1_R \circ 1_R = 1_R$
Thus $-1_R$ has a ring product inverse (itself) and therefore ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
Then:
:$\struct {\set {1_R, -1_R}, \circ} \le U_R$
That is, the [[Definition:Set|set]] consisting of the [[Definition:Unity of Ring|unity]] and its [[Definition:Ring Negative|negative]] forms a [[Definition:Subgroup|subgroup]] of the [[D... | From [[Unity is Unit]]:
:$1_R \in U_R$
It remains to be shown that $-1_R \in U_R$.
From {{Ring-axiom|A}}, $\struct {R, +}$ is an [[Definition:Abelian Group|abelian group]].
Therefore:
:$1_R \in R \implies -1_R \in R$.
From [[Product of Ring Negatives]]:
:$-1_R \circ -1_R = 1_R \circ 1_R = 1_R$
Thus $-1_R$ has a [[... | Unity and Negative form Subgroup of Units | https://proofwiki.org/wiki/Unity_and_Negative_form_Subgroup_of_Units | https://proofwiki.org/wiki/Unity_and_Negative_form_Subgroup_of_Units | [
"Rings with Unity",
"Subgroups"
] | [
"Definition:Ring with Unity",
"Definition:Set",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring Negative",
"Definition:Subgroup",
"Definition:Group of Units/Ring"
] | [
"Unity is Unit",
"Definition:Abelian Group",
"Product of Ring Negatives",
"Definition:Product Inverse",
"Definition:Ring (Abstract Algebra)/Product",
"Two-Step Subgroup Test",
"Category:Rings with Unity",
"Category:Subgroups"
] |
proofwiki-654 | Negative of Product Inverse | Let $\struct {R, +, \circ}$ be a ring with unity.
Let $z \in U_R$, where $U_R$ is the set of units.
Then:
:$\paren {-z}^{-1} = -\paren {z^{-1} }$
where $z^{-1}$ is the ring product inverse of $z$. | Let the unity of $\struct {R, +, \circ}$ be $1_R$.
{{begin-eqn}}
{{eqn | l = \paren {-\paren {z^{-1} } } \circ \paren {-z}
| r = z^{-1} \circ z
| c = Product of Ring Negatives
}}
{{eqn | r = 1_R
| c = Inverse under $\circ$
}}
{{eqn | r = z \circ z^{-1}
| c = Inverse under $\circ$
}}
{{eqn | r = ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
Let $z \in U_R$, where $U_R$ is the [[Definition:Set|set]] of [[Definition:Unit of Ring|units]].
Then:
:$\paren {-z}^{-1} = -\paren {z^{-1} }$
where $z^{-1}$ is the [[Definition:Ring Product Inverse|ring product inverse]] of $z$. | Let the [[Definition:Unity of Ring|unity]] of $\struct {R, +, \circ}$ be $1_R$.
{{begin-eqn}}
{{eqn | l = \paren {-\paren {z^{-1} } } \circ \paren {-z}
| r = z^{-1} \circ z
| c = [[Product of Ring Negatives]]
}}
{{eqn | r = 1_R
| c = [[Definition:Inverse Element|Inverse under $\circ$]]
}}
{{eqn | r =... | Negative of Product Inverse | https://proofwiki.org/wiki/Negative_of_Product_Inverse | https://proofwiki.org/wiki/Negative_of_Product_Inverse | [
"Rings with Unity"
] | [
"Definition:Ring with Unity",
"Definition:Set",
"Definition:Unit of Ring",
"Definition:Product Inverse"
] | [
"Definition:Unity (Abstract Algebra)/Ring",
"Product of Ring Negatives",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Product of Ring Negatives"
] |
proofwiki-655 | Product of Negative with Product Inverse | Let $\struct {R, +, \circ}$ be a ring with unity.
Let $z \in U_R$, where $U_R$ is the set of units.
Then:
:$(1): \quad \forall x \in R: -\paren {x \circ z^{-1} } = \paren {-x} \circ z^{-1} = x \circ \paren {\paren {-z}^{-1} }$
:$(2): \quad \forall x \in R: -\paren {z^{-1} \circ x} = z^{-1} \circ \paren {-x} = \paren {\... | {{begin-eqn}}
{{eqn | n = 1
| l = -\paren {x \circ z^{-1} }
| r = \paren {-x} \circ z^{-1}
| c = Product with Ring Negative
}}
{{eqn | r = x \circ \paren {-\paren {z^{-1} } }
| c = Product with Ring Negative
}}
{{eqn | r = x \circ \paren {\paren {-z}^{-1} }
| c = Negative of Product Invers... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
Let $z \in U_R$, where $U_R$ is the [[Definition:Set|set]] of [[Definition:Unit of Ring|units]].
Then:
:$(1): \quad \forall x \in R: -\paren {x \circ z^{-1} } = \paren {-x} \circ z^{-1} = x \circ \paren {\paren {-z}^{-1} }$
:$(2): \qua... | {{begin-eqn}}
{{eqn | n = 1
| l = -\paren {x \circ z^{-1} }
| r = \paren {-x} \circ z^{-1}
| c = [[Product with Ring Negative]]
}}
{{eqn | r = x \circ \paren {-\paren {z^{-1} } }
| c = [[Product with Ring Negative]]
}}
{{eqn | r = x \circ \paren {\paren {-z}^{-1} }
| c = [[Negative of Prod... | Product of Negative with Product Inverse | https://proofwiki.org/wiki/Product_of_Negative_with_Product_Inverse | https://proofwiki.org/wiki/Product_of_Negative_with_Product_Inverse | [
"Rings with Unity"
] | [
"Definition:Ring with Unity",
"Definition:Set",
"Definition:Unit of Ring"
] | [
"Product with Ring Negative",
"Product with Ring Negative",
"Negative of Product Inverse",
"Product with Ring Negative",
"Product with Ring Negative",
"Negative of Product Inverse"
] |
proofwiki-656 | Negative of Division Product | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Then:
:$\forall x \in R: -\dfrac x z = \dfrac {-x} z = \dfrac x {-z}$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is the division product of $x$ by $z$. | Follows directly from Product of Negative with Product Inverse and the definition of division product.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\struct {U_R, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {R, +, \circ}$.
Then:
:$\forall x \in R: -\dfrac x z = \dfrac {-x} z = \dfrac x {-z}$
where $\dfrac x z$ is defi... | Follows directly from [[Product of Negative with Product Inverse]] and the definition of [[Definition:Division Product|division product]].
{{qed}} | Negative of Division Product | https://proofwiki.org/wiki/Negative_of_Division_Product | https://proofwiki.org/wiki/Negative_of_Division_Product | [
"Commutative Algebra"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Group of Units/Ring",
"Definition:Division Product"
] | [
"Product of Negative with Product Inverse",
"Definition:Division Product"
] |
proofwiki-657 | Addition of Division Products | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, c \in R, b, d \in U_R$.
Then:
:$\dfrac a b + \dfrac c d = \dfrac {a \circ d + b \circ c} {b \circ d}$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $... | First we demonstrate the operation has the specified property:
{{begin-eqn}}
{{eqn | l = \frac a b + \frac c d
| r = a \circ b^{-1} + c \circ d^{-1}
| c = {{Defof|Division Product}}
}}
{{eqn | r = a \circ b^{-1} \circ d \circ d^{-1} + c \circ d^{-1} \circ b \circ b^{-1}
| c = {{Defof|Inverse Element}}... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\struct {U_R, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {R, +, \circ}$.
Let $a, c \in R, b, d \in U_R$.
Then:
:$\dfrac a b + \dfrac c d = \dfrac {a \circ d + b \circ ... | First we demonstrate the operation has the specified property:
{{begin-eqn}}
{{eqn | l = \frac a b + \frac c d
| r = a \circ b^{-1} + c \circ d^{-1}
| c = {{Defof|Division Product}}
}}
{{eqn | r = a \circ b^{-1} \circ d \circ d^{-1} + c \circ d^{-1} \circ b \circ b^{-1}
| c = {{Defof|Inverse Element}... | Addition of Division Products | https://proofwiki.org/wiki/Addition_of_Division_Products | https://proofwiki.org/wiki/Addition_of_Division_Products | [
"Commutative Algebra"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Group of Units/Ring",
"Definition:Division Product"
] | [
"Definition:Commutative Ring"
] |
proofwiki-658 | Equality of Division Products | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, b \in R, c, d \in U_R$.
Then:
:$\dfrac a c = \dfrac b d \iff a \circ d = b \circ c$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$. | {{begin-eqn}}
{{eqn | l = \frac a c
| r = \frac b d
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = a \circ c^{-1}
| r = b \circ d^{-1}
| c = {{Defof|Division Product}}
}}
{{eqn | ll= \leadstoandfrom
| l = a \circ c^{-1} \circ c \circ d
| r = b \circ d^{-1} \circ c \circ d
| ... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\struct {U_R, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {R, +, \circ}$.
Let $a, b \in R, c, d \in U_R$.
Then:
:$\dfrac a c = \dfrac b d \iff a \circ d = b \circ c$
w... | {{begin-eqn}}
{{eqn | l = \frac a c
| r = \frac b d
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = a \circ c^{-1}
| r = b \circ d^{-1}
| c = {{Defof|Division Product}}
}}
{{eqn | ll= \leadstoandfrom
| l = a \circ c^{-1} \circ c \circ d
| r = b \circ d^{-1} \circ c \circ d
| ... | Equality of Division Products | https://proofwiki.org/wiki/Equality_of_Division_Products | https://proofwiki.org/wiki/Equality_of_Division_Products | [
"Commutative Algebra"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Group of Units/Ring",
"Definition:Division Product"
] | [] |
proofwiki-659 | Product of Division Products | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, b \in R, c, d \in U_R$.
Then:
:$\dfrac a c \circ \dfrac b d = \dfrac {a \circ b} {c \circ d}$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divid... | {{begin-eqn}}
{{eqn | l = \frac a c \circ \frac b d
| r = a \circ c^{-1} \circ b \circ d^{-1}
| c = {{Defof|Division Product}}
}}
{{eqn | r = \paren {a \circ b} \circ \paren {d^{-1} \circ c^{-1} }
| c = {{Defof|Commutative Operation}}
}}
{{eqn | r = \paren {a \circ b} \circ \paren {c \circ d}^{-1}
... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\struct {U_R, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {R, +, \circ}$.
Let $a, b \in R, c, d \in U_R$.
Then:
:$\dfrac a c \circ \dfrac b d = \dfrac {a \circ b} {c \c... | {{begin-eqn}}
{{eqn | l = \frac a c \circ \frac b d
| r = a \circ c^{-1} \circ b \circ d^{-1}
| c = {{Defof|Division Product}}
}}
{{eqn | r = \paren {a \circ b} \circ \paren {d^{-1} \circ c^{-1} }
| c = {{Defof|Commutative Operation}}
}}
{{eqn | r = \paren {a \circ b} \circ \paren {c \circ d}^{-1}
... | Product of Division Products | https://proofwiki.org/wiki/Product_of_Division_Products | https://proofwiki.org/wiki/Product_of_Division_Products | [
"Commutative Algebra"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Group of Units/Ring",
"Definition:Division Product"
] | [
"Inverse of Product"
] |
proofwiki-660 | Inverse of Division Product | Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, b \in U_R$.
Then:
:$\paren {\dfrac a b}^{-1} = \dfrac {1_R} {\paren {a / b}} = \dfrac b a$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided ... | {{begin-eqn}}
{{eqn | l = \frac {1_R} {\paren {a / b} }
| r = 1_R / \paren {a \circ b^{-1} }
| c = {{Defof|Division Product}}
}}
{{eqn | r = 1_R \circ \paren {a \circ b^{-1} }^{-1}
| c = {{Defof|Division Product}}
}}
{{eqn | r = \paren {a \circ b^{-1} }^{-1}
| c = {{Defof|Identity Element}} of $... | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\struct {U_R, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {R, +, \circ}$.
Let $a, b \in U_R$.
Then:
:$\paren {\dfrac a b}^{-1} = \dfrac {1_R} {\paren {a / b}} = \dfrac ... | {{begin-eqn}}
{{eqn | l = \frac {1_R} {\paren {a / b} }
| r = 1_R / \paren {a \circ b^{-1} }
| c = {{Defof|Division Product}}
}}
{{eqn | r = 1_R \circ \paren {a \circ b^{-1} }^{-1}
| c = {{Defof|Division Product}}
}}
{{eqn | r = \paren {a \circ b^{-1} }^{-1}
| c = {{Defof|Identity Element}} of $... | Inverse of Division Product | https://proofwiki.org/wiki/Inverse_of_Division_Product | https://proofwiki.org/wiki/Inverse_of_Division_Product | [
"Commutative Algebra"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Group of Units/Ring",
"Definition:Division Product"
] | [
"Inverse of Group Product",
"Inverse of Group Inverse"
] |
proofwiki-661 | Zero Product with Proper Zero Divisor is with Zero Divisor | Let $\struct {R, +, \circ}$ be a ring.
Let $x \in R$ be a proper zero divisor of $R$.
Then:
:$\paren {x \divides 0_R} \land \paren {x \circ y = 0_R} \land \paren {y \ne 0_R} \implies y \divides 0_R$
That is, if $x$ is a proper zero divisor, then whatever non-zero element you form the product with it by to get zero must... | Follows directly from the definition of proper zero divisor.
If $y \ne 0_R$ and $x \circ y = 0_R$ and $x \in R^*$ (which is has to be if it's a proper zero divisor), then all the criteria of being a zero divisor are fulfilled by $y$.
{{qed}}
Category:Zero Divisors
ggtpz8ldh9x4o8hpodh433socagnn6f | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $x \in R$ be a [[Definition:Proper Zero Divisor|proper zero divisor]] of $R$.
Then:
:$\paren {x \divides 0_R} \land \paren {x \circ y = 0_R} \land \paren {y \ne 0_R} \implies y \divides 0_R$
That is, if $x$ is a [[Definition:Proper Z... | Follows directly from the definition of [[Definition:Proper Zero Divisor|proper zero divisor]].
If $y \ne 0_R$ and $x \circ y = 0_R$ and $x \in R^*$ (which is has to be if it's a proper zero divisor), then all the criteria of being a [[Definition:Zero Divisor of Ring|zero divisor]] are fulfilled by $y$.
{{qed}}
[[Cat... | Zero Product with Proper Zero Divisor is with Zero Divisor | https://proofwiki.org/wiki/Zero_Product_with_Proper_Zero_Divisor_is_with_Zero_Divisor | https://proofwiki.org/wiki/Zero_Product_with_Proper_Zero_Divisor_is_with_Zero_Divisor | [
"Zero Divisors"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Proper Zero Divisor",
"Definition:Proper Zero Divisor",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Ring Zero",
"Definition:Zero Divisor/Ring"
] | [
"Definition:Proper Zero Divisor",
"Definition:Zero Divisor/Ring",
"Category:Zero Divisors"
] |
proofwiki-662 | Unit of Ring is not Zero Divisor | Let $\struct {R, +, \circ}$ be a non-null ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $x$ be a unit of $\struct {R, +, \circ}$.
Then $x$ is neither a left zero divisor nor a right zero divisor of $\struct {R, +, \circ}$. | {{AimForCont}} $x$ is either a left zero divisor or a right zero divisor of $\struct {R, +, \circ}$.
{{WLOG}}, suppose $x$ is a left zero divisor of $\struct {R, +, \circ}$.
That is:
:$x \circ y = 0_R$
for some $y \in R \setminus \set {0_R}$.
Then:
{{begin-eqn}}
{{eqn | l = y
| r = 1_R \circ y
| c = {{Defof... | Let $\struct {R, +, \circ}$ be a [[Definition:Non-Null Ring|non-null]] [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Let $x$ be a [[Definition:Unit of Ring|unit]] of $\struct {R, +, \circ}$.
Then $x$ is neither a [[De... | {{AimForCont}} $x$ is either a [[Definition:Left Zero Divisor|left zero divisor]] or a [[Definition:Right Zero Divisor|right zero divisor]] of $\struct {R, +, \circ}$.
{{WLOG}}, suppose $x$ is a [[Definition:Left Zero Divisor|left zero divisor]] of $\struct {R, +, \circ}$.
That is:
:$x \circ y = 0_R$
for some $y \... | Unit of Ring is not Zero Divisor | https://proofwiki.org/wiki/Unit_of_Ring_is_not_Zero_Divisor | https://proofwiki.org/wiki/Unit_of_Ring_is_not_Zero_Divisor | [
"Rings with Unity",
"Units of Rings",
"Zero Divisors"
] | [
"Definition:Non-Null Ring",
"Definition:Ring with Unity",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Unit of Ring",
"Definition:Left Zero Divisor",
"Definition:Right Zero Divisor"
] | [
"Definition:Left Zero Divisor",
"Definition:Right Zero Divisor",
"Definition:Left Zero Divisor",
"Ring Product with Zero",
"Definition:Contradiction",
"Proof by Contradiction",
"Definition:Left Zero Divisor",
"Definition:Right Zero Divisor",
"Category:Rings with Unity",
"Category:Units of Rings",
... |
proofwiki-663 | Zero Divisor Product is Zero Divisor | The ring product of a zero divisor with any ring element is a zero divisor. | Let $\struct {R, +, \circ}$ be a ring.
Let $x \divides 0_R$ in $R$.
Then:
{{begin-eqn}}
{{eqn | q = \exists y \in R, y \ne 0_R
| l = x \circ y
| r = 0_R
| c = {{Defof|Zero Divisor of Ring}}
}}
{{eqn | ll= \leadsto
| q = \forall z \in R
| l = z \circ \paren {x \circ y}
| r = z \circ 0... | The [[Definition:Ring Product|ring product]] of a [[Definition:Zero Divisor of Ring|zero divisor]] with any ring element is a [[Definition:Zero Divisor of Ring|zero divisor]]. | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Let $x \divides 0_R$ in $R$.
Then:
{{begin-eqn}}
{{eqn | q = \exists y \in R, y \ne 0_R
| l = x \circ y
| r = 0_R
| c = {{Defof|Zero Divisor of Ring}}
}}
{{eqn | ll= \leadsto
| q = \forall z \in R
| l = z \cir... | Zero Divisor Product is Zero Divisor | https://proofwiki.org/wiki/Zero_Divisor_Product_is_Zero_Divisor | https://proofwiki.org/wiki/Zero_Divisor_Product_is_Zero_Divisor | [
"Zero Divisors"
] | [
"Definition:Ring (Abstract Algebra)/Product",
"Definition:Zero Divisor/Ring",
"Definition:Zero Divisor/Ring"
] | [
"Definition:Ring (Abstract Algebra)",
"Ring Product with Zero",
"Definition:Associative Operation",
"Category:Zero Divisors"
] |
proofwiki-664 | Product is Zero Divisor means Zero Divisor | If the ring product of two elements of a ring is a zero divisor, then one of the two elements must be a zero divisor. | {{begin-eqn}}
{{eqn | l = \paren {x \circ y}
| o = \divides
| r = 0_R
}}
{{eqn | ll= \leadsto
| q = \exists z \divides 0_R \in R
| l = \paren {x \circ y} \circ z
| r = 0_R, x \ne 0_R, y \ne 0_R
| c = {{Defof|Zero Divisor of Ring}}
}}
{{eqn | ll= \leadsto
| l = x \circ \paren {y... | If the [[Definition:Ring Product|ring product]] of two [[Definition:Element|elements]] of a [[Definition:Ring (Abstract Algebra)|ring]] is a [[Definition:Zero Divisor of Ring|zero divisor]], then one of the two elements must be a [[Definition:Zero Divisor of Ring|zero divisor]]. | {{begin-eqn}}
{{eqn | l = \paren {x \circ y}
| o = \divides
| r = 0_R
}}
{{eqn | ll= \leadsto
| q = \exists z \divides 0_R \in R
| l = \paren {x \circ y} \circ z
| r = 0_R, x \ne 0_R, y \ne 0_R
| c = {{Defof|Zero Divisor of Ring}}
}}
{{eqn | ll= \leadsto
| l = x \circ \paren {y... | Product is Zero Divisor means Zero Divisor | https://proofwiki.org/wiki/Product_is_Zero_Divisor_means_Zero_Divisor | https://proofwiki.org/wiki/Product_is_Zero_Divisor_means_Zero_Divisor | [
"Ring Theory"
] | [
"Definition:Ring (Abstract Algebra)/Product",
"Definition:Element",
"Definition:Ring (Abstract Algebra)",
"Definition:Zero Divisor/Ring",
"Definition:Zero Divisor/Ring"
] | [
"Zero Product with Proper Zero Divisor is with Zero Divisor",
"Category:Ring Theory"
] |
proofwiki-665 | Ring Element is Zero Divisor iff not Cancellable | Let $\struct {R, +, \circ}$ be a ring which is not null.
Let $z \in R^*$.
Then $z$ is a zero divisor {{iff}} $z$ is not cancellable for $\circ$. | === Sufficient Condition ===
Let $z$ be a zero divisor.
Then either $z \circ x = 0_R$ or $x \circ z = 0_R$ for some $x \in R^*$.
{{Recall|Cancellable Element}}
{{:Definition:Cancellable Element}}
{{WLOG}}, suppose $z \circ x = 0_R$ for some $x \in R^*$.
Then:
{{begin-eqn}}
{{eqn | l = z \circ x
| r = 0_R
| ... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] which is [[Definition:Non-Null Ring|not null]].
Let $z \in R^*$.
Then $z$ is a [[Definition:Zero Divisor of Ring|zero divisor]] {{iff}} $z$ is not [[Definition:Cancellable Element|cancellable]] for $\circ$. | === Sufficient Condition ===
Let $z$ be a [[Definition:Zero Divisor of Ring|zero divisor]].
Then either $z \circ x = 0_R$ or $x \circ z = 0_R$ for some $x \in R^*$.
{{Recall|Cancellable Element}}
{{:Definition:Cancellable Element}}
{{WLOG}}, suppose $z \circ x = 0_R$ for some $x \in R^*$.
Then:
{{begin-eqn}}
{{eq... | Ring Element is Zero Divisor iff not Cancellable | https://proofwiki.org/wiki/Ring_Element_is_Zero_Divisor_iff_not_Cancellable | https://proofwiki.org/wiki/Ring_Element_is_Zero_Divisor_iff_not_Cancellable | [
"Zero Divisors"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Non-Null Ring",
"Definition:Zero Divisor/Ring",
"Definition:Cancellable Element"
] | [
"Definition:Zero Divisor/Ring",
"Definition:Cancellable Element",
"Definition:Cancellable Element",
"Definition:Zero Divisor/Ring"
] |
proofwiki-666 | Ring Less Zero is Semigroup for Product iff No Proper Zero Divisors | Let $\struct {R, +, \circ}$ be a non-null ring.
Then $R$ has no zero divisors {{iff}} $\struct {R^*, \circ}$ is a semigroup. | === 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... | Let $\struct {R, +, \circ}$ be a [[Definition:Non-Null Ring|non-null]] [[Definition:Ring (Abstract Algebra)|ring]].
Then $R$ has no [[Definition:Zero Divisor of Ring|zero divisors]] {{iff}} $\struct {R^*, \circ}$ is a [[Definition:Semigroup|semigroup]]. | === Necessary Condition ===
Let $\struct {R, +, \circ}$ be a [[Definition:Non-Null Ring|non-null]] [[Definition:Ring (Abstract Algebra)|ring]] with no [[Definition:Zero Divisor of Ring|zero divisors]].
The set $R^* = R \setminus \set {0_R} \ne \O$ as $\struct {R, +, \circ}$ is [[Definition:Non-Null Ring|non-null]].
... | Ring Less Zero is Semigroup for Product iff No Proper Zero Divisors | https://proofwiki.org/wiki/Ring_Less_Zero_is_Semigroup_for_Product_iff_No_Proper_Zero_Divisors | https://proofwiki.org/wiki/Ring_Less_Zero_is_Semigroup_for_Product_iff_No_Proper_Zero_Divisors | [
"Ring Theory",
"Semigroups",
"Zero Divisors"
] | [
"Definition:Non-Null Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Zero Divisor/Ring",
"Definition:Semigroup"
] | [
"Definition:Non-Null Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Zero Divisor/Ring",
"Definition:Non-Null Ring",
"Definition:Zero Divisor/Ring",
"Definition:Cancellable Element",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Zero Divisor/Ring",
"Restriction of... |
proofwiki-667 | Idempotent Elements of Ring with No Proper Zero Divisors | Let $\struct {R, +, \circ}$ be a non-null ring with no (proper) zero divisors.
Let $x \in R$.
Then:
:$x \circ x = x \iff x \in \set {0_R, 1_R}$
That is, the only elements of $\struct {R, \circ}$ that are idempotent are zero and unity. | We have $0_R \circ 0_R = 0_R$, so that sorts out one element.
Let $R^*$ be the ring $R$ without the zero: $R^* = R \setminus \set {0_R}$.
By Ring Element is Zero Divisor iff not Cancellable, all elements of $R^*$ that are not zero divisors are cancellable.
Therefore all elements of $R^*$ are cancellable.
Suppose $x \ci... | Let $\struct {R, +, \circ}$ be a [[Definition:Non-Null Ring|non-null]] [[Definition:Ring (Abstract Algebra)|ring]] with no [[Definition:Proper Zero Divisor|(proper) zero divisors]].
Let $x \in R$.
Then:
:$x \circ x = x \iff x \in \set {0_R, 1_R}$
That is, the only [[Definition:Element|elements]] of $\struct {R, \... | We have $0_R \circ 0_R = 0_R$, so that sorts out one [[Definition:Element|element]].
Let $R^*$ be the [[Definition:Ring Less Zero|ring $R$ without the zero]]: $R^* = R \setminus \set {0_R}$.
By [[Ring Element is Zero Divisor iff not Cancellable]], all elements of $R^*$ that are not [[Definition:Zero Divisor|zero div... | Idempotent Elements of Ring with No Proper Zero Divisors | https://proofwiki.org/wiki/Idempotent_Elements_of_Ring_with_No_Proper_Zero_Divisors | https://proofwiki.org/wiki/Idempotent_Elements_of_Ring_with_No_Proper_Zero_Divisors | [
"Ring Theory"
] | [
"Definition:Non-Null Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Proper Zero Divisor",
"Definition:Element",
"Definition:Idempotence/Element",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring"
] | [
"Definition:Element",
"Definition:Ring (Abstract Algebra)/Ring Less Zero",
"Ring Element is Zero Divisor iff not Cancellable",
"Definition:Zero Divisor",
"Definition:Cancellable Element",
"Definition:Cancellable Element",
"Definition:Cancellable Element"
] |
proofwiki-668 | Non-Zero Elements of Division Ring form Group | Let $\struct {R, +, \circ}$ be a division ring.
Then $\struct {R^*, \circ}$ is a group. | A division ring by definition is a ring with unity, and therefore not null.
A division ring by definition has no zero divisors, so $\struct {R^*, \circ}$ is a semigroup.
$1_R \in \struct {R^*, \circ}$ and so the identity of $\circ$ is in $\struct {R^*, \circ}$.
By the definition of a division ring, each element of $\s... | Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]].
Then $\struct {R^*, \circ}$ is a [[Definition:Group|group]]. | A [[Definition:Division Ring|division ring]] by definition is a [[Definition:Ring with Unity|ring with unity]], and therefore not [[Definition:Null Ring|null]].
A [[Definition:Division Ring|division ring]] by definition has no [[Definition:Zero Divisor of Ring|zero divisors]], so $\struct {R^*, \circ}$ is a [[Ring Le... | Non-Zero Elements of Division Ring form Group | https://proofwiki.org/wiki/Non-Zero_Elements_of_Division_Ring_form_Group | https://proofwiki.org/wiki/Non-Zero_Elements_of_Division_Ring_form_Group | [
"Division Rings",
"Group Theory"
] | [
"Definition:Division Ring",
"Definition:Group"
] | [
"Definition:Division Ring",
"Definition:Ring with Unity",
"Definition:Null Ring",
"Definition:Division Ring",
"Definition:Zero Divisor/Ring",
"Ring Less Zero is Semigroup for Product iff No Proper Zero Divisors",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Division Ring",
... |
proofwiki-669 | Subring Test | Let $S$ be a subset of a ring $\struct {R, +, \circ}$.
Then $\struct {S, +, \circ}$ is a subring of $\struct {R, +, \circ}$ {{iff}} these all hold:
:$(1): \quad S \ne \O$
:$(2): \quad \forall x, y \in S: x + \paren {-y} \in S$
:$(3): \quad \forall x, y \in S: x \circ y \in S$ | === Necessary Condition ===
If $S$ is a subring of $\struct {R, +, \circ}$, the conditions hold by virtue of the ring axioms as applied to $S$. | Let $S$ be a [[Definition:Subset|subset]] of a [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$.
Then $\struct {S, +, \circ}$ is a [[Definition:Subring|subring]] of $\struct {R, +, \circ}$ {{iff}} these all hold:
:$(1): \quad S \ne \O$
:$(2): \quad \forall x, y \in S: x + \paren {-y} \in S$
:$(3):... | === Necessary Condition ===
If $S$ is a [[Definition:Subring|subring]] of $\struct {R, +, \circ}$, the conditions hold by virtue of the [[Axiom:Ring Axioms|ring axioms]] as applied to $S$. | Subring Test | https://proofwiki.org/wiki/Subring_Test | https://proofwiki.org/wiki/Subring_Test | [
"Subrings"
] | [
"Definition:Subset",
"Definition:Ring (Abstract Algebra)",
"Definition:Subring"
] | [
"Definition:Subring",
"Axiom:Ring Axioms",
"Axiom:Ring Axioms",
"Definition:Subring"
] |
proofwiki-670 | Subdomain Test | Let $S$ be a subset of an integral domain $\struct {R, +, \circ}$.
Then $\struct {S, + {\restriction_S}, \circ {\restriction_S} }$ is a subdomain of $\struct {R, +, \circ}$ {{iff}} these conditions hold:
:$(1): \quad \struct {S, + {\restriction_S}, \circ {\restriction_S} }$ is a subring of $\struct {R, +, \circ}$
:$(2)... | By Idempotent Elements of Ring with No Proper Zero Divisors, it follows that the unity of a subdomain is the unity of the integral domain it's a subdomain of.
{{qed}} | Let $S$ be a [[Definition:Subset|subset]] of an [[Definition:Integral Domain|integral domain]] $\struct {R, +, \circ}$.
Then $\struct {S, + {\restriction_S}, \circ {\restriction_S} }$ is a [[Definition:Subdomain|subdomain]] of $\struct {R, +, \circ}$ {{iff}} these conditions hold:
:$(1): \quad \struct {S, + {\restri... | By [[Idempotent Elements of Ring with No Proper Zero Divisors]], it follows that the [[Definition:Unity of Ring|unity]] of a [[Definition:Subdomain|subdomain]] is the [[Definition:Unity of Ring|unity]] of the [[Definition:Integral Domain|integral domain]] it's a [[Definition:Subdomain|subdomain]] of.
{{qed}} | Subdomain Test | https://proofwiki.org/wiki/Subdomain_Test | https://proofwiki.org/wiki/Subdomain_Test | [
"Integral Domains",
"Subrings"
] | [
"Definition:Subset",
"Definition:Integral Domain",
"Definition:Subdomain",
"Definition:Subring",
"Definition:Unity (Abstract Algebra)/Ring"
] | [
"Idempotent Elements of Ring with No Proper Zero Divisors",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Subdomain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Integral Domain",
"Definition:Subdomain"
] |
proofwiki-671 | Centralizer of Ring Subset is Subring | Let $S$ be a subset of a ring $\struct {R, +, \circ}$
Then $\map {C_R} S$, the centralizer of $S$ in $R$, is a subring of $R$. | Certainly $0_R \in \map {C_R} S$ as $0_R$ commutes (trivially) with all elements of $R$.
Suppose $x, y \in \map {C_R} S$.
Then:
{{begin-eqn}}
{{eqn | q = \forall s \in S
| l = s \circ \paren {x + \paren {-y} }
| r = s \circ x + s \circ \paren {-y}
| c = Distributivity of $\circ$ over $+$
}}
{{eqn | r ... | Let $S$ be a [[Definition:Subset|subset]] of a [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$
Then $\map {C_R} S$, the [[Definition:Centralizer of Ring Subset|centralizer]] of $S$ in $R$, is a [[Definition:Subring|subring]] of $R$. | Certainly $0_R \in \map {C_R} S$ as $0_R$ [[Definition:Commute|commutes]] (trivially) with all [[Definition:Element|elements]] of $R$.
Suppose $x, y \in \map {C_R} S$.
Then:
{{begin-eqn}}
{{eqn | q = \forall s \in S
| l = s \circ \paren {x + \paren {-y} }
| r = s \circ x + s \circ \paren {-y}
| c ... | Centralizer of Ring Subset is Subring | https://proofwiki.org/wiki/Centralizer_of_Ring_Subset_is_Subring | https://proofwiki.org/wiki/Centralizer_of_Ring_Subset_is_Subring | [
"Ring Theory",
"Subrings"
] | [
"Definition:Subset",
"Definition:Ring (Abstract Algebra)",
"Definition:Centralizer/Ring Subset",
"Definition:Subring"
] | [
"Definition:Commutative/Elements",
"Definition:Element",
"Definition:Distributive Operation",
"Definition:Centralizer/Ring Subset",
"Definition:Distributive Operation",
"Element Commutes with Product of Commuting Elements",
"Subring Test",
"Definition:Subring"
] |
proofwiki-672 | Center of Ring is Commutative Subring | The center $\map Z R$ of a ring $R$ is a commutative subring of $R$. | Follows directly from the definition of center and Centralizer of Ring Subset is Subring.
{{qed}} | The [[Definition:Center of Ring|center]] $\map Z R$ of a [[Definition:Ring (Abstract Algebra)|ring]] $R$ is a [[Definition:Commutative Ring|commutative]] [[Definition:Subring|subring]] of $R$. | Follows directly from the definition of [[Definition:Center of Ring|center]] and [[Centralizer of Ring Subset is Subring]].
{{qed}} | Center of Ring is Commutative Subring | https://proofwiki.org/wiki/Center_of_Ring_is_Commutative_Subring | https://proofwiki.org/wiki/Center_of_Ring_is_Commutative_Subring | [
"Ring Theory",
"Subrings"
] | [
"Definition:Center (Abstract Algebra)/Ring",
"Definition:Ring (Abstract Algebra)",
"Definition:Commutative Ring",
"Definition:Subring"
] | [
"Definition:Center (Abstract Algebra)/Ring",
"Centralizer of Ring Subset is Subring"
] |
proofwiki-673 | Ideal is Subring | Let $\struct {R, +, \circ}$ be a ring, and let $J$ be an ideal of $R$.
Then $J$ is a subring of $R$. | This follows directly from the definition of an ideal and Subring Test.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]], and let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
Then $J$ is a [[Definition:Subring|subring]] of $R$. | This follows directly from the definition of an [[Definition:Ideal of Ring|ideal]] and [[Subring Test]].
{{qed}} | Ideal is Subring | https://proofwiki.org/wiki/Ideal_is_Subring | https://proofwiki.org/wiki/Ideal_is_Subring | [
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring",
"Definition:Subring"
] | [
"Definition:Ideal of Ring",
"Subring Test"
] |
proofwiki-674 | Ring is Ideal of Itself | Let $\struct {R, +, \circ}$ be a ring.
Then $R$ is an ideal of $R$. | 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 $R$.
{{qed}} | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
Then $R$ is an [[Definition:Ideal of Ring|ideal]] of $R$. | From [[Ring is Subring of Itself]], $\struct {R, +, \circ}$ is a [[Definition:Subring|subring]] of $\struct {R, +, \circ}$.
Also:
:$\forall x, y \in \struct {R, +, \circ}: x \circ y \in R$
thus fulfilling the [[Definition:Ideal of Ring|condition]] for $R$ to be an [[Definition:Ideal of Ring|ideal]] of $R$.
{{qed}} | Ring is Ideal of Itself | https://proofwiki.org/wiki/Ring_is_Ideal_of_Itself | https://proofwiki.org/wiki/Ring_is_Ideal_of_Itself | [
"Ideal Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring"
] | [
"Ring is Subring of Itself",
"Definition:Subring",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring"
] |
proofwiki-675 | Ideal of Unit is Whole Ring | Let $\struct {R, +, \circ}$ be a ring with unity.
Let $J$ be an ideal of $R$.
If $J$ contains a unit of $R$, then $J = R$. | Let $u \in J$, where $u \in U_R$.
Also by definition, we have $u^{-1} \in U_R$.
Let $x \in R$.
{{begin-eqn}}
{{eqn | o =
| r = x \in R
| c =
}}
{{eqn | o = \leadsto
| r = x \circ u^{-1} \in R
| c = as $R$ is closed
}}
{{eqn | o = \leadsto
| r = \paren {x \circ u^{-1} } \circ u \in J
... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
If $J$ contains a [[Definition:Unit of Ring|unit]] of $R$, then $J = R$. | Let $u \in J$, where $u \in U_R$.
Also by definition, we have $u^{-1} \in U_R$.
Let $x \in R$.
{{begin-eqn}}
{{eqn | o =
| r = x \in R
| c =
}}
{{eqn | o = \leadsto
| r = x \circ u^{-1} \in R
| c = as $R$ is [[Definition:Closed Algebraic Structure|closed]]
}}
{{eqn | o = \leadsto
| r ... | Ideal of Unit is Whole Ring | https://proofwiki.org/wiki/Ideal_of_Unit_is_Whole_Ring | https://proofwiki.org/wiki/Ideal_of_Unit_is_Whole_Ring | [
"Ideal Theory"
] | [
"Definition:Ring with Unity",
"Definition:Ideal of Ring",
"Definition:Unit of Ring"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-676 | Test for Ideal | Let $J$ be a subset of a ring $\struct {R, +, \circ}$.
Then $J$ is an ideal of $\struct {R, +, \circ}$ {{iff}} these all hold:
:$(1): \quad J \ne \O$
:$(2): \quad \forall x, y \in J: x + \paren {-y} \in J$
:$(3): \quad \forall j \in J, r \in R: r \circ j \in J, j \circ r \in J$ | === Necessary Condition ===
Let $J$ be an ideal of $\struct {R, +, \circ}$.
Then conditions $(1)$ to $(3)$ hold by virtue of the ring axioms and $J$ being an ideal.
{{qed|lemma}} | Let $J$ be a [[Definition:Subset|subset]] of a [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$.
Then $J$ is an [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$ {{iff}} these all hold:
:$(1): \quad J \ne \O$
:$(2): \quad \forall x, y \in J: x + \paren {-y} \in J$
:$(3): \quad \forall... | === Necessary Condition ===
Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$.
Then conditions $(1)$ to $(3)$ hold by virtue of the [[Axiom:Ring Axioms|ring axioms]] and $J$ being an [[Definition:Ideal of Ring|ideal]].
{{qed|lemma}} | Test for Ideal | https://proofwiki.org/wiki/Test_for_Ideal | https://proofwiki.org/wiki/Test_for_Ideal | [
"Ideal Theory"
] | [
"Definition:Subset",
"Definition:Ring (Abstract Algebra)",
"Definition:Ideal of Ring"
] | [
"Definition:Ideal of Ring",
"Axiom:Ring Axioms",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring",
"Definition:Ideal of Ring"
] |
proofwiki-677 | Epimorphism Preserves Rings | Let $\struct {R_1, +_1, \circ_1}$ be a ring, and $\struct {R_2, +_2, \circ_2}$ be a closed algebraic structure.
Let $\phi: R_1 \to R_2$ be an epimorphism.
Then $\struct {R_2, +_2, \circ_2}$ is a ring. | From Epimorphism Preserves Groups, we have that if $\struct {R_1, +_1}$ is a group then so is $\struct {R_2, +_2}$.
From Epimorphism Preserves Semigroups, we have that if $\struct {R_1, \circ_1}$ is a semigroup then so is $\struct {R_2, \circ_2}$.
From Epimorphism Preserves Distributivity, we have that if $\circ_1$ dis... | Let $\struct {R_1, +_1, \circ_1}$ be a [[Definition:Ring (Abstract Algebra)|ring]], and $\struct {R_2, +_2, \circ_2}$ be a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
Let $\phi: R_1 \to R_2$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
Then $\struct {R_2, +_2, \circ_2}$ ... | From [[Epimorphism Preserves Groups]], we have that if $\struct {R_1, +_1}$ is a [[Definition:Group|group]] then so is $\struct {R_2, +_2}$.
From [[Epimorphism Preserves Semigroups]], we have that if $\struct {R_1, \circ_1}$ is a [[Definition:Semigroup|semigroup]] then so is $\struct {R_2, \circ_2}$.
From [[Epimorphi... | Epimorphism Preserves Rings | https://proofwiki.org/wiki/Epimorphism_Preserves_Rings | https://proofwiki.org/wiki/Epimorphism_Preserves_Rings | [
"Ring Theory",
"Ring Epimorphisms"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Ring (Abstract Algebra)"
] | [
"Epimorphism Preserves Groups",
"Definition:Group",
"Epimorphism Preserves Semigroups",
"Definition:Semigroup",
"Epimorphism Preserves Distributivity",
"Definition:Ring (Abstract Algebra)",
"Definition:Ring (Abstract Algebra)"
] |
proofwiki-678 | Ring Homomorphism of Addition is Group Homomorphism | 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. | 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.
{{Qed}}
Category:Ring Homomorphisms
Category:Group Homomorphisms
njds2yzpn0mudnikuageex15uclu56s | Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
Then $\phi: \struct {R_1, +_1} \to \struct {R_2, +_2}$ is a [[Definition:Group Homomorphism|group homomorphism]]. | From the definition of a [[Definition:Ring (Abstract Algebra)|ring]], both $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ are [[Definition:Abelian Group|abelian groups]].
The result follows from the definitions of [[Definition:Ring Homomorphism|ring homomorphism]] and [[Definition:Group Homomorphism|group homomorphism]... | Ring Homomorphism of Addition is Group Homomorphism | https://proofwiki.org/wiki/Ring_Homomorphism_of_Addition_is_Group_Homomorphism | https://proofwiki.org/wiki/Ring_Homomorphism_of_Addition_is_Group_Homomorphism | [
"Ring Homomorphisms",
"Group Homomorphisms"
] | [
"Definition:Ring Homomorphism",
"Definition:Group Homomorphism"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Abelian Group",
"Definition:Ring Homomorphism",
"Definition:Group Homomorphism",
"Category:Ring Homomorphisms",
"Category:Group Homomorphisms"
] |
proofwiki-679 | Element of Integral Domain Divides Zero | Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Then every element of $D$ divides $0_D$:
:$\forall x \in D: x \divides 0_D$ | By definition, an integral domain is a ring.
So, from Ring Product with Zero:
:$\forall x \in D: 0_D = x \circ 0_D$
The result follows from the definition of divisor.
{{qed}} | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
Then every [[Definition:Element|element]] of $D$ [[Definition:Divisor of Ring Element|divides]] $0_D$:
:$\forall x \in D: x \divides 0_D$ | By definition, an [[Definition:Integral Domain|integral domain]] is a [[Definition:Ring (Abstract Algebra)|ring]].
So, from [[Ring Product with Zero]]:
:$\forall x \in D: 0_D = x \circ 0_D$
The result follows from the definition of [[Definition:Divisor of Ring Element|divisor]].
{{qed}} | Element of Integral Domain Divides Zero | https://proofwiki.org/wiki/Element_of_Integral_Domain_Divides_Zero | https://proofwiki.org/wiki/Element_of_Integral_Domain_Divides_Zero | [
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Element",
"Definition:Divisor (Algebra)/Ring with Unity"
] | [
"Definition:Integral Domain",
"Definition:Ring (Abstract Algebra)",
"Ring Product with Zero",
"Definition:Divisor (Algebra)/Ring with Unity"
] |
proofwiki-680 | Unity Divides All Elements | Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Then unity is a divisor of every element of $D$:
:$\forall x \in D: 1_D \divides x$
Also:
:$\forall x \in D: -1_D \divides x$ | The element $1_D$ is the unity of $\struct {D, +, \circ}$, and so:
:$1_D \in D: x = 1_D \circ x$
Similarly, from Product of Ring Negatives:
:$-1_D \in D: x = \paren {-1_D} \circ \paren {-x}$
The result follows from the definition of divisor.
{{qed}} | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$.
Then [[Definition:Unity of Ring|unity]] is a [[Definition:Divisor of Ring Element|divisor]] of every [[Definition:Element|element]] of $D$:
:$\forall x \in D: 1_D \divides x$
Also:
:$\... | The element $1_D$ is the [[Definition:Unity of Ring|unity]] of $\struct {D, +, \circ}$, and so:
:$1_D \in D: x = 1_D \circ x$
Similarly, from [[Product of Ring Negatives]]:
:$-1_D \in D: x = \paren {-1_D} \circ \paren {-x}$
The result follows from the definition of [[Definition:Divisor of Ring Element|divisor]].
{{... | Unity Divides All Elements/Proof 1 | https://proofwiki.org/wiki/Unity_Divides_All_Elements | https://proofwiki.org/wiki/Unity_Divides_All_Elements/Proof_1 | [
"Integral Domains",
"Unity Divides All Elements",
"Divisibility"
] | [
"Definition:Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Element"
] | [
"Definition:Unity (Abstract Algebra)/Ring",
"Product of Ring Negatives",
"Definition:Divisor (Algebra)/Ring with Unity"
] |
proofwiki-681 | Unity Divides All Elements | Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Then unity is a divisor of every element of $D$:
:$\forall x \in D: 1_D \divides x$
Also:
:$\forall x \in D: -1_D \divides x$ | This is a special case of Unit of Integral Domain divides all Elements, as Unity is Unit.
Furthermore, from Unity and Negative form Subgroup of Units we also have that $-1_D$ is a unit of $D$.
Hence the result.
{{qed}} | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$.
Then [[Definition:Unity of Ring|unity]] is a [[Definition:Divisor of Ring Element|divisor]] of every [[Definition:Element|element]] of $D$:
:$\forall x \in D: 1_D \divides x$
Also:
:$\... | This is a special case of [[Unit of Integral Domain divides all Elements]], as [[Unity is Unit]].
Furthermore, from [[Unity and Negative form Subgroup of Units]] we also have that $-1_D$ is a [[Definition:Unit of Ring|unit of $D$]].
Hence the result.
{{qed}} | Unity Divides All Elements/Proof 2 | https://proofwiki.org/wiki/Unity_Divides_All_Elements | https://proofwiki.org/wiki/Unity_Divides_All_Elements/Proof_2 | [
"Integral Domains",
"Unity Divides All Elements",
"Divisibility"
] | [
"Definition:Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Element"
] | [
"Unit of Integral Domain divides all Elements",
"Unity is Unit",
"Unity and Negative form Subgroup of Units",
"Definition:Unit of Ring"
] |
proofwiki-682 | Element of Integral Domain is Divisor of Itself | Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Then every element of $D$ is a divisor of itself:
:$\forall x \in D: x \divides x$ | Follows directly from the definition of divisor:
:$\forall x \in D: \exists 1_D \in D: x = 1_D \circ x$
{{qed}}
Category:Integral Domains
8xrney68i9349jnuparynsqc0h08i4k | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$.
Then every [[Definition:Element|element]] of $D$ is a [[Definition:Divisor of Ring Element|divisor]] of itself:
:$\forall x \in D: x \divides x$ | Follows directly from the definition of [[Definition:Divisor of Ring Element|divisor]]:
:$\forall x \in D: \exists 1_D \in D: x = 1_D \circ x$
{{qed}}
[[Category:Integral Domains]]
8xrney68i9349jnuparynsqc0h08i4k | Element of Integral Domain is Divisor of Itself | https://proofwiki.org/wiki/Element_of_Integral_Domain_is_Divisor_of_Itself | https://proofwiki.org/wiki/Element_of_Integral_Domain_is_Divisor_of_Itself | [
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Element",
"Definition:Divisor (Algebra)/Ring with Unity"
] | [
"Definition:Divisor (Algebra)/Ring with Unity",
"Category:Integral Domains"
] |
proofwiki-683 | Unit of Integral Domain divides all Elements | Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Let $\struct {U_D, \circ}$ be the group of units of $\struct {D, +, \circ}$.
Then:
:$\forall x \in D: \forall u \in U_D: u \divides x$
That is, every unit of $D$ is a divisor of every element of $D$. | {{begin-eqn}}
{{eqn | q = \forall x \in D, u \in U_D
| l = x
| r = u \circ \paren {u^{-1} \circ x}
| c = {{Defof|Unit of Ring}}
}}
{{eqn | ll= \leadsto
| l = u
| o = \divides
| r = x
| c = {{Defof|Divisor of Ring Element}}
}}
{{end-eqn}}
{{qed}} | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$.
Let $\struct {U_D, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {D, +, \circ}$.
Then:
:$\forall x \in D: \forall u \in U_D: u \divides x$
That is, e... | {{begin-eqn}}
{{eqn | q = \forall x \in D, u \in U_D
| l = x
| r = u \circ \paren {u^{-1} \circ x}
| c = {{Defof|Unit of Ring}}
}}
{{eqn | ll= \leadsto
| l = u
| o = \divides
| r = x
| c = {{Defof|Divisor of Ring Element}}
}}
{{end-eqn}}
{{qed}} | Unit of Integral Domain divides all Elements | https://proofwiki.org/wiki/Unit_of_Integral_Domain_divides_all_Elements | https://proofwiki.org/wiki/Unit_of_Integral_Domain_divides_all_Elements | [
"Integral Domains"
] | [
"Definition:Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Group of Units/Ring",
"Definition:Unit of Ring",
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Element"
] | [] |
proofwiki-684 | Divisor of Unit is Unit | Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Let $\struct {U_D, \circ}$ be the group of units of $\struct {D, +, \circ}$.
Then:
:$x \in D, u \in U_D: x \divides u \implies x \in U_D$
That is, if $x$ is a divisor of a unit, $x$ must itself be a unit. | Let $x \in D, u \in U_D$ such that $x \divides u$.
By definition:
:$\exists t \in D: u = t \circ x$
Thus:
:$1_D = u^{-1} \circ u = u^{-1} \circ t \circ x$
Also, as $D$ is an integral domain and hence a commutative ring, we have:
:$u^{-1} \circ t \circ x = 1_D = x \circ u^{-1} \circ t$
It follows by definition that $x$ ... | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$.
Let $\struct {U_D, \circ}$ be the [[Definition:Group of Units of Ring|group of units]] of $\struct {D, +, \circ}$.
Then:
:$x \in D, u \in U_D: x \divides u \implies x \in U_D$
That i... | Let $x \in D, u \in U_D$ such that $x \divides u$.
By [[Definition:Divisor of Ring Element|definition]]:
:$\exists t \in D: u = t \circ x$
Thus:
:$1_D = u^{-1} \circ u = u^{-1} \circ t \circ x$
Also, as $D$ is an [[Definition:Integral Domain|integral domain]] and hence a [[Definition:Commutative Ring|commutative ri... | Divisor of Unit is Unit | https://proofwiki.org/wiki/Divisor_of_Unit_is_Unit | https://proofwiki.org/wiki/Divisor_of_Unit_is_Unit | [
"Integral Domains",
"Units of Rings"
] | [
"Definition:Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Group of Units/Ring",
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Unit of Ring",
"Definition:Unit of Ring"
] | [
"Definition:Divisor (Algebra)/Ring with Unity",
"Definition:Integral Domain",
"Definition:Commutative Ring",
"Definition:Unit of Ring",
"Definition:Inverse (Abstract Algebra)/Inverse"
] |
proofwiki-685 | Associatehood is Equivalence Relation | Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\sim$ be the relation defined on $D$ as:
:$\forall x, y \in D: x \sim y$ {{iff}} $x$ is an associate of $y$
Then $\sim$ is an equivalence relation. | Checking in turn each of the criteria for equivalence: | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
Let $\sim$ be the [[Definition:Relation|relation]] defined on $D$ as:
:$\forall x, y \in D: x \sim y$ {{iff}} $x$ is an [[Definition:Asso... | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Associatehood is Equivalence Relation | https://proofwiki.org/wiki/Associatehood_is_Equivalence_Relation | https://proofwiki.org/wiki/Associatehood_is_Equivalence_Relation | [
"Integral Domains",
"Examples of Equivalence Relations",
"Associates"
] | [
"Definition:Integral Domain",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Relation",
"Definition:Associate/Integral Domain",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-686 | Divisor Relation in Integral Domain is Transitive | Let $\struct {D, +, \circ}$ be an integral domain.
Let $x, y, z \in D$.
Then:
:$x \divides y \land y \divides z \implies x \divides z$ | Let $x \divides y \land y \divides z$.
Then from the definition of divisor, we have:
:$x \divides y \iff \exists s \in D: y = s \circ x$
:$y \divides z \iff \exists t \in D: z = t \circ y$
Then:
:$z = t \circ \paren {s \circ x} = \paren {t \circ s} \circ x$
Thus:
:$\exists \paren {t \circ s} \in D: z = \paren {t \circ ... | Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
Let $x, y, z \in D$.
Then:
:$x \divides y \land y \divides z \implies x \divides z$ | Let $x \divides y \land y \divides z$.
Then from the definition of [[Definition:Divisor of Ring Element|divisor]], we have:
:$x \divides y \iff \exists s \in D: y = s \circ x$
:$y \divides z \iff \exists t \in D: z = t \circ y$
Then:
:$z = t \circ \paren {s \circ x} = \paren {t \circ s} \circ x$
Thus:
:$\exists \p... | Divisor Relation in Integral Domain is Transitive | https://proofwiki.org/wiki/Divisor_Relation_in_Integral_Domain_is_Transitive | https://proofwiki.org/wiki/Divisor_Relation_in_Integral_Domain_is_Transitive | [
"Integral Domains"
] | [
"Definition:Integral Domain"
] | [
"Definition:Divisor (Algebra)/Ring with Unity"
] |
proofwiki-687 | Integers form Unique Factorization Domain | The integers $\struct {\Z, +, \times}$ form a unique factorization domain. | Follows directly from:
:$(1) \quad$ the fundamental theorem of arithmetic
:$(2) \quad$ the fact that $\struct {\Z, +, \times}$ is an integral domain
:$(3) \quad$ the definitions complete factorization and equivalent factorizations.
{{qed}}
Category:Integral Domains
Category:Integers
Category:Factorization
Category:Uniq... | The [[Integers form Integral Domain|integers]] $\struct {\Z, +, \times}$ form a [[Definition:Unique Factorization Domain|unique factorization domain]]. | Follows directly from:
:$(1) \quad$ the [[Fundamental Theorem of Arithmetic|fundamental theorem of arithmetic]]
:$(2) \quad$ the fact that [[Integers form Integral Domain|$\struct {\Z, +, \times}$ is an integral domain]]
:$(3) \quad$ the definitions [[Definition:Complete Factorization|complete factorization]] and [[De... | Integers form Unique Factorization Domain | https://proofwiki.org/wiki/Integers_form_Unique_Factorization_Domain | https://proofwiki.org/wiki/Integers_form_Unique_Factorization_Domain | [
"Integral Domains",
"Integers",
"Factorization",
"Unique Factorization Domains"
] | [
"Integers form Integral Domain",
"Definition:Unique Factorization Domain"
] | [
"Fundamental Theorem of Arithmetic",
"Integers form Integral Domain",
"Definition:Complete Factorization",
"Definition:Equivalent Factorizations",
"Category:Integral Domains",
"Category:Integers",
"Category:Factorization",
"Category:Unique Factorization Domains"
] |
proofwiki-688 | Trivial Ordering Compatibility in Boolean Ring | Let $\struct {S, +, \circ}$ be a Boolean ring.
Then the trivial ordering is the only ordering on $S$ compatible with both its operations. | That the trivial ordering is compatible with $\circ$ and $*$ follows from Trivial Ordering is Universally Compatible.
Conversely, suppose that $\preceq$ is an ordering compatible with $\circ$ and $*$.
We recall the definition of the trivial ordering:
{{:Definition:Trivial Ordering}}
Let $a, b \in S$ such that $a \prece... | Let $\struct {S, +, \circ}$ be a [[Definition:Boolean Ring|Boolean ring]].
Then the [[Definition:Trivial Ordering|trivial ordering]] is the only [[Definition:Ordering|ordering]] on $S$ [[Definition:Relation Compatible with Operation|compatible]] with both its [[Definition:Binary Operation|operations]]. | That the [[Definition:Trivial Ordering|trivial ordering]] is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$ and $*$ follows from [[Trivial Ordering is Universally Compatible]].
Conversely, suppose that $\preceq$ is an [[Definition:Ordering|ordering]] [[Definition:Relation Compatible with Op... | Trivial Ordering Compatibility in Boolean Ring | https://proofwiki.org/wiki/Trivial_Ordering_Compatibility_in_Boolean_Ring | https://proofwiki.org/wiki/Trivial_Ordering_Compatibility_in_Boolean_Ring | [
"Boolean Rings",
"Order Theory"
] | [
"Definition:Boolean Ring",
"Definition:Trivial Ordering",
"Definition:Ordering",
"Definition:Relation Compatible with Operation",
"Definition:Operation/Binary Operation"
] | [
"Definition:Trivial Ordering",
"Definition:Relation Compatible with Operation",
"Trivial Ordering is Universally Compatible",
"Definition:Ordering",
"Definition:Relation Compatible with Operation",
"Definition:Trivial Ordering",
"Definition:Relation Compatible with Operation",
"Definition:Boolean Ring... |
proofwiki-689 | Smallest Field is Field | The ring $\struct {\set {0_R, 1_R}, +, \circ}$ is the smallest algebraic structure which is a field. | From Field Contains at least 2 Elements, a field must contain at least two elements.
Hence the null ring, which contains one element, is ''not'' a field.
For $\struct {\set {0_R, 1_R}, +, \circ}$ to be a field:
:$\struct {\set {0_R, 1_R}, +}$ must be an abelian group.
This is fulfilled as this is the parity group.
:$\s... | The [[Definition:Ring (Abstract Algebra)|ring]] $\struct {\set {0_R, 1_R}, +, \circ}$ is the smallest [[Definition:Algebraic Structure|algebraic structure]] which is a [[Definition:Field (Abstract Algebra)|field]]. | From [[Field Contains at least 2 Elements]], a [[Definition:Field (Abstract Algebra)|field]] must contain at least two [[Definition:Element|elements]].
Hence the [[Definition:Null Ring|null ring]], which contains one [[Definition:Element|element]], is ''not'' a [[Definition:Field (Abstract Algebra)|field]].
For $\st... | Smallest Field is Field | https://proofwiki.org/wiki/Smallest_Field_is_Field | https://proofwiki.org/wiki/Smallest_Field_is_Field | [
"Field Theory"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Algebraic Structure",
"Definition:Field (Abstract Algebra)"
] | [
"Field Contains at least 2 Elements",
"Definition:Field (Abstract Algebra)",
"Definition:Element",
"Definition:Null Ring",
"Definition:Element",
"Definition:Field (Abstract Algebra)",
"Definition:Field (Abstract Algebra)",
"Definition:Abelian Group",
"Definition:Parity Group",
"Definition:Commutat... |
proofwiki-690 | Field is Integral Domain | Every field is an integral domain. | Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Suppose $\exists x, y \in F: x \circ y = 0_F$.
Suppose $x \ne 0_F$.
Then, by the definition of a field, $x^{-1}$ exists in $F$ and:
:$y = 1_F \circ y = x^{-1} \circ x \circ y = x^{-1} \circ 0_F = 0_F$.
Otherwise $x = 0_F$.
So if $x \ci... | Every [[Definition:Field (Abstract Algebra)|field]] is an [[Definition:Integral Domain|integral domain]]. | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
Suppose $\exists x, y \in F: x \circ y = 0_F$.
Suppose $x \ne 0_F$.
Then, by the definition of a [[Definition:Field (Abstract Algebra)|f... | Field is Integral Domain/Proof 1 | https://proofwiki.org/wiki/Field_is_Integral_Domain | https://proofwiki.org/wiki/Field_is_Integral_Domain/Proof_1 | [
"Field is Integral Domain",
"Field Theory",
"Integral Domains"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Integral Domain"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Field Zero",
"Definition:Multiplicative Identity",
"Definition:Field (Abstract Algebra)"
] |
proofwiki-691 | Field is Integral Domain | Every field is an integral domain. | This result follows directly from:
:Field has no Proper Zero Divisors
:By definition, that a field is a commutative ring.
{{qed}} | Every [[Definition:Field (Abstract Algebra)|field]] is an [[Definition:Integral Domain|integral domain]]. | This result follows directly from:
:[[Field has no Proper Zero Divisors]]
:By definition, that a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Commutative Ring|commutative ring]].
{{qed}} | Field is Integral Domain/Proof 2 | https://proofwiki.org/wiki/Field_is_Integral_Domain | https://proofwiki.org/wiki/Field_is_Integral_Domain/Proof_2 | [
"Field is Integral Domain",
"Field Theory",
"Integral Domains"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Integral Domain"
] | [
"Field has no Proper Zero Divisors",
"Definition:Field (Abstract Algebra)",
"Definition:Commutative Ring"
] |
proofwiki-692 | Finite Integral Domain is Galois Field | A finite integral domain is a Galois field. | Let $\struct {R, +, \circ}$ be a finite integral domain whose unity is $1$ and whose zero is $0$.
Let $a \in R$ be arbitrary such that $a \ne 0$.
We wish to show that $a$ has a product inverse in $R$.
Consider the map $f: R \to R$ defined by $f: x \mapsto a \circ x$.
We first show that the kernel of $f$ is the null rin... | A [[Definition:Finite Integral Domain|finite integral domain]] is a [[Definition:Galois Field|Galois field]]. | Let $\struct {R, +, \circ}$ be a [[Definition:Finite Integral Domain|finite integral domain]] whose [[Definition:Unity of Ring|unity]] is $1$ and whose [[Definition:Ring Zero|zero]] is $0$.
Let $a \in R$ be [[Definition:Arbitrary|arbitrary]] such that $a \ne 0$.
We wish to show that $a$ has a [[Definition:Ring Produ... | Finite Integral Domain is Galois Field/Proof 1 | https://proofwiki.org/wiki/Finite_Integral_Domain_is_Galois_Field | https://proofwiki.org/wiki/Finite_Integral_Domain_is_Galois_Field/Proof_1 | [
"Galois Fields",
"Integral Domains",
"Finite Integral Domain is Galois Field"
] | [
"Definition:Finite Integral Domain",
"Definition:Galois Field"
] | [
"Definition:Finite Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring Zero",
"Definition:Arbitrary",
"Definition:Product Inverse",
"Definition:Mapping",
"Definition:Kernel of Ring Homomorphism",
"Definition:Null Ring",
"Definition:Integral Domain",
"Definition:Proper Ze... |
proofwiki-693 | Finite Integral Domain is Galois Field | A finite integral domain is a Galois field. | Let $R$ be a finite integral domain with unity $1$ and zero $0$.
Let $R^*$ denote $R \setminus \set 0$, the set $R$ without the zero.
As $R$ is finite, we may enumerate the elements of $R$ as:
:$x_0 = 0, x_1 = 1, x_2, x_3, \ldots, x_n$
Let $R^*$.
Consider the elements:
:$x_k x_1, x_k x_2, \ldots, x_k x_n$
which are me... | A [[Definition:Finite Integral Domain|finite integral domain]] is a [[Definition:Galois Field|Galois field]]. | Let $R$ be a [[Definition:Finite Integral Domain|finite integral domain]] with [[Definition:Unity of Ring|unity]] $1$ and [[Definition:Ring Zero|zero]] $0$.
Let $R^*$ denote $R \setminus \set 0$, the set $R$ without the [[Definition:Ring Zero|zero]].
As $R$ is [[Definition:Finite Set|finite]], we may enumerate the el... | Finite Integral Domain is Galois Field/Proof 2 | https://proofwiki.org/wiki/Finite_Integral_Domain_is_Galois_Field | https://proofwiki.org/wiki/Finite_Integral_Domain_is_Galois_Field/Proof_2 | [
"Galois Fields",
"Integral Domains",
"Finite Integral Domain is Galois Field"
] | [
"Definition:Finite Integral Domain",
"Definition:Galois Field"
] | [
"Definition:Finite Integral Domain",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Ring Zero",
"Definition:Ring Zero",
"Definition:Finite Set",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Ring (Abstract Algebra)/Product",
"Definition:Integral Domain",
"Definiti... |
proofwiki-694 | Finite Integral Domain is Galois Field | A finite integral domain is a Galois field. | An integral domain is by definition a ring which has no proper zero divisors.
By definition, a Galois field is a field whose underlying set is finite.
The result follows from Finite Ring with No Proper Zero Divisors is Field.
{{qed}} | A [[Definition:Finite Integral Domain|finite integral domain]] is a [[Definition:Galois Field|Galois field]]. | An [[Definition:Integral Domain|integral domain]] is by definition a [[Definition:Ring (Abstract Algebra)|ring]] which has no [[Definition:Proper Zero Divisor|proper zero divisors]].
By definition, a [[Definition:Galois Field|Galois field]] is a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Underlyi... | Finite Integral Domain is Galois Field/Proof 3 | https://proofwiki.org/wiki/Finite_Integral_Domain_is_Galois_Field | https://proofwiki.org/wiki/Finite_Integral_Domain_is_Galois_Field/Proof_3 | [
"Galois Fields",
"Integral Domains",
"Finite Integral Domain is Galois Field"
] | [
"Definition:Finite Integral Domain",
"Definition:Galois Field"
] | [
"Definition:Integral Domain",
"Definition:Ring (Abstract Algebra)",
"Definition:Proper Zero Divisor",
"Definition:Galois Field",
"Definition:Field (Abstract Algebra)",
"Definition:Underlying Set/Abstract Algebra",
"Definition:Finite Set",
"Finite Ring with No Proper Zero Divisors is Field"
] |
proofwiki-695 | Finite Integral Domain is Galois Field | A finite integral domain is a Galois field. | {{AimForCont}} $\struct {D, +, \circ}$ is a finite integral domain which is not a field.
From Non-Field Integral Domain has Infinite Number of Ideals, $\struct {D, +, \circ}$ has an infinite number of distinct ideals.
But this contradicts the assertion that $\struct {D, +, \circ}$ is finite.
Hence the result by Proof b... | A [[Definition:Finite Integral Domain|finite integral domain]] is a [[Definition:Galois Field|Galois field]]. | {{AimForCont}} $\struct {D, +, \circ}$ is a [[Definition:Finite Set|finite]] [[Definition:Integral Domain|integral domain]] which is not a [[Definition:Field (Abstract Algebra)|field]].
From [[Non-Field Integral Domain has Infinite Number of Ideals]], $\struct {D, +, \circ}$ has an [[Definition:Infinite Set|infinite n... | Finite Integral Domain is Galois Field/Proof 4 | https://proofwiki.org/wiki/Finite_Integral_Domain_is_Galois_Field | https://proofwiki.org/wiki/Finite_Integral_Domain_is_Galois_Field/Proof_4 | [
"Galois Fields",
"Integral Domains",
"Finite Integral Domain is Galois Field"
] | [
"Definition:Finite Integral Domain",
"Definition:Galois Field"
] | [
"Definition:Finite Set",
"Definition:Integral Domain",
"Definition:Field (Abstract Algebra)",
"Non-Field Integral Domain has Infinite Number of Ideals",
"Definition:Infinite Set",
"Definition:Distinct/Plural",
"Definition:Ideal of Ring",
"Definition:Contradiction",
"Definition:Finite Set",
"Proof ... |
proofwiki-696 | Field is Subfield of Itself | Let $\struct {F, +, \circ}$ be a field.
Then $\struct {F, +, \circ}$ is a subfield of $\struct {F, +, \circ}$. | $F$ is a field and $F \subseteq F$ from Set is Subset of Itself.
{{Qed}} | Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]].
Then $\struct {F, +, \circ}$ is a [[Definition:Subfield|subfield]] of $\struct {F, +, \circ}$. | $F$ is a [[Definition:Field (Abstract Algebra)|field]] and $F \subseteq F$ from [[Set is Subset of Itself]].
{{Qed}} | Field is Subfield of Itself | https://proofwiki.org/wiki/Field_is_Subfield_of_Itself | https://proofwiki.org/wiki/Field_is_Subfield_of_Itself | [
"Subfields"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Subfield"
] | [
"Definition:Field (Abstract Algebra)",
"Set is Subset of Itself"
] |
proofwiki-697 | Center of Division Ring is Subfield | Let $\struct {K, +, \circ}$ be an division ring.
Let $\map Z K$ be the center of $K$.
Then $\map Z K$ is a subfield of $K$. | For $\map Z K$ to be a subfield of $K$, it needs to be a division ring that is commutative.
Thus the result follows directly from Center of Ring is Commutative Subring.
{{qed}} | Let $\struct {K, +, \circ}$ be an [[Definition:Division Ring|division ring]].
Let $\map Z K$ be the [[Definition:Center of Ring|center]] of $K$.
Then $\map Z K$ is a [[Definition:Subfield|subfield]] of $K$. | For $\map Z K$ to be a [[Definition:Subfield|subfield]] of $K$, it needs to be a [[Definition:Division Ring|division ring]] that is [[Definition:Commutative Ring|commutative]].
Thus the result follows directly from [[Center of Ring is Commutative Subring]].
{{qed}} | Center of Division Ring is Subfield | https://proofwiki.org/wiki/Center_of_Division_Ring_is_Subfield | https://proofwiki.org/wiki/Center_of_Division_Ring_is_Subfield | [
"Division Rings",
"Subfields"
] | [
"Definition:Division Ring",
"Definition:Center (Abstract Algebra)/Ring",
"Definition:Subfield"
] | [
"Definition:Subfield",
"Definition:Division Ring",
"Definition:Commutative Ring",
"Center of Ring is Commutative Subring"
] |
proofwiki-698 | Ideals of Field | Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Then $\struct {R, +, \circ}$ is a field {{iff}} the only ideals of $\struct {R, +, \circ}$ are $\struct {R, +, \circ}$ and $\set {0_R}$. | === Necessary Condition ===
Let $\struct {R, +, \circ}$ be a field.
The result follows from Field has 2 Ideals.
{{qed|lemma}} | Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
Then $\struct {R, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]] {{iff}} the only [[Definition:Ide... | === Necessary Condition ===
Let $\struct {R, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]].
The result follows from [[Field has 2 Ideals]].
{{qed|lemma}} | Ideals of Field | https://proofwiki.org/wiki/Ideals_of_Field | https://proofwiki.org/wiki/Ideals_of_Field | [
"Field Theory",
"Ideal Theory",
"Commutative Algebra"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Ring Zero",
"Definition:Unity (Abstract Algebra)/Ring",
"Definition:Field (Abstract Algebra)",
"Definition:Ideal of Ring"
] | [
"Definition:Field (Abstract Algebra)",
"Field has 2 Ideals"
] |
proofwiki-699 | Epimorphism from Division Ring to Ring | Let $\struct {K, +, \circ}$ be a division ring whose zero is $0_K$.
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\phi: K \to R$ be a ring epimorphism.
Then one of the following applies:
:$(1): \quad R$ is a null ring
:$(2): \quad R$ is a division ring and $\phi$ is a ring isomorphism. | We have that the kernel of $K$ is an ideal.
From Ideals of Division Ring, $\map \ker K$ must therefore either be $0_K$ or $K$.
Let $\map \ker K = 0_K$.
Then by Ring Epimorphism with Trivial Kernel is Isomorphism $\phi$ is an ring isomorphism.
Thus $R$ a division ring like $K$.
Let $\map \ker K = K$.
Then:
:$\forall x \... | Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]] whose [[Definition:Ring Zero|zero]] is $0_K$.
Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $\phi: K \to R$ be a [[Definition:Ring Epimorphism|ring epimorphism]]... | We have that the [[Kernel of Ring Epimorphism is Ideal|kernel of $K$ is an ideal]].
From [[Ideals of Division Ring]], $\map \ker K$ must therefore either be $0_K$ or $K$.
Let $\map \ker K = 0_K$.
Then by [[Ring Epimorphism with Trivial Kernel is Isomorphism]] $\phi$ is an [[Definition:Ring Isomorphism|ring isomorph... | Epimorphism from Division Ring to Ring | https://proofwiki.org/wiki/Epimorphism_from_Division_Ring_to_Ring | https://proofwiki.org/wiki/Epimorphism_from_Division_Ring_to_Ring | [
"Ring Theory",
"Division Rings",
"Ring Epimorphisms"
] | [
"Definition:Division Ring",
"Definition:Ring Zero",
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Ring Epimorphism",
"Definition:Null Ring",
"Definition:Division Ring",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism"
] | [
"Kernel of Ring Epimorphism is Ideal",
"Ideals of Division Ring",
"Ring Epimorphism with Trivial Kernel is Isomorphism",
"Definition:Isomorphism (Abstract Algebra)/Ring Isomorphism",
"Definition:Division Ring",
"Definition:Ring Epimorphism",
"Definition:Surjection"
] |
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