nyuuzyou/wikis huggingface dataset:
Unit of Ring of Mappings iff Image is Subset of Ring Units

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.
Let $U_R$ be the set of units in $R$.
Let $S$ be a set.
Let $\struct {R^S, +', \circ'}$ be the ring of mappings on the set of mappings $R^S$.
Then:
:$f \in R^S$ is a unit of $R^S$ $\Img f \subseteq U_R$
where $\Img f$ is the image of $f$.
In this case, the inverse of $f$ is the mapping $f^{-1} : S \to U_R$ defined by:
:$\forall x \in S : \map {f^{-1} } x = \map f x^{-1}$
== Proof ==
From Structure Induced by Ring with Unity Operations is Ring with Unity, $\struct {R^S, +', \circ'}$ has a unity $f_{1_R}$ defined by:
:$\forall x \in S: \map {f_{1_R}} x = 1_R$
=== Unit of Ring of Mappings implies Image is Subset of Ring Units ===
=== Image is Subset of Ring Units implies Unit of Ring of Mappings ===
Category:Rings of Mappings
Category:Rings with Unity
Category:Unit of Ring of Mappings iff Image is Subset of Ring Units
Unit of Ring of Mappings iff Image is Subset of Ring Units/Image is Subset of Ring Units implies Unit of Ring of Mappings

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.
Let $U_R$ be the set of units in $R$.
Let $S$ be a set.
Let $\struct {R^S, +', \circ'}$ be the ring of mappings on the set of mappings $R^S$.
Let $\Img f \subseteq U_R$ where $\Img f$ is the image of $f$.
Then:
:$f \in R^S$ is a unit of $R^S$
and the inverse of $f$ is the mapping $f^{-1} : S \to U_R$ defined by:
:$\forall x \in S : \map {f^{-1}} {x} = \map f x^{-1}$
== Proof ==
From Structure Induced by Ring with Unity Operations is Ring with Unity, $\struct {R^S, +', \circ'}$ has a unity $f_{1_R}$ defined by:
:$\forall x \in S: \map {f_{1_R}} x = 1_R$
By assumption: 
:$\forall x \in S: \exists \map f x^{-1} : \map f x \circ \map f x^{-1} = \map f x^{-1} \circ \map f x = 1_R$
Let $f^{-1} : S \to U_R$ be defined by:
:$\forall x \in S : \map {f^{-1}} {x} = \map f x^{-1}$
Consider the mapping $f \circ’ f^{-1}$.
For all $x \in S$: 
Hence $f \circ’ f^{-1} = f_{1_R}$.
Similarly, $f^{-1} \circ’ f = f_{1_R}$.
Hence $f$ is a unit of $\struct {R^S, +', \circ'}$ and the inverse of $f$ is the mapping $f^{-1}$.
Category:Unit of Ring of Mappings iff Image is Subset of Ring Units
Unit of Ring of Mappings iff Image is Subset of Ring Units/Unit of Ring of Mappings implies Image is Subset of Ring Units

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.
Let $U_R$ be the set of units in $R$.
Let $S$ be a set.
Let $\struct {R^S, +', \circ'}$ be the ring of mappings on the set of mappings $R^S$.
Let $f \in R^S$ be a unit of $R^S$.
Then:
:$\Img f \subseteq U_R$
where $\Img f$ is the image of $f$.
In which case, the inverse of $f$ is the mapping $f^{-1} : S \to U_R$ defined by:
:$\forall x \in S : \map {f^{-1} } x = \map f x^{-1}$
== Proof ==
From Structure Induced by Ring with Unity Operations is Ring with Unity, $\struct {R^S, +', \circ'}$ has a unity $f_{1_R}$ defined by:
:$\forall x \in S: \map {f_{1_R} } x = 1_R$
Let $f^{-1}$ be the product inverse of $f$.
Let $x \in R$.
Then:
Similarly, $1_R = \map {f^{-1} } x \circ \map f x$
Hence $\map {f^{-1}} x$ is the product inverse of $\map f x$.
So $\map f x$ is a unit in $R$ and $\map f x^{-1} = \map {f^{-1} } x$.
The result follows.
Category:Unit of Ring of Mappings iff Image is Subset of Ring Units
Unit of System of Sets is Unique

== Theorem ==
The unit of a system of sets, if it exists, is unique.
If $U$ is the unit of a system of sets $\SS$, then $\forall A \in \SS: A \subseteq U$.
== Proof ==
Let $\SS$ be a system of sets.
Suppose $U$ and $U'$ are both units of $\SS$.
Then, by definition:
:$\forall A \in \SS: A \cap U = A$
:$\forall A \in \SS: A \cap U' = A$
This applies to both $U$ and $U'$, of course.
So $U \cap U' = U$ and $U' \cap U = U'$.
From Intersection with Subset is Subset‎ it follows that $U \subseteq U'$ and $U' \subseteq U$.
By definition of set equality:
:$U = U'$
We also see that from Intersection with Subset is Subset, $A \cap U = A \iff A \subseteq U$, which shows that:
:$\forall A \in \SS: A \subseteq U$.
Category:Set Systems
Unital C*-Algebra is Unital Banach Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra that is unital as an algebra, with identity element ${\mathbf 1}_A \ne {\mathbf 0}_A$.
Then $A$ is unital as a Banach algebra.
== Proof ==
We have:
From , we have $\norm { {\mathbf 1}_A} \ne 0$. 
Hence $\norm { {\mathbf 1}_A} = 1$.
== Sources ==
* : $2.1$: $\text C^\ast$-Algebras
Category:Unital Banach Algebras
Category:C*-Algebras
Unital Ring Homomorphism by Idempotent

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $e \in A$ be an idempotent element.
Let $\ideal e$ be the ideal of $A$ generated by $e$.
Then the mapping:
:$f: A \to \ideal e: a \mapsto e a$ 
is a surjective unital ring homomorphism from $\struct {A, +, \circ}$ to $\struct {\ideal e, +, \circ}$ with kernel the ideal $\ideal {1 - e}$ generated by $1 - e$.
== Proof ==
By Ring Homomorphism by Idempotent $f$ is a surjective ring homomorphism with kernel $\ideal {1 - e}$.
By Ring by Idempotent $A$ is a commutative ring with unity whose unity is $e$.
Then
Thus $f$ is a unital ring homomorphism.
Category:Ring Theory
Category:Commutative Rings
Category:Ring Homomorphisms
Unitary Matrix/Examples/Arbitrary Example 1

== Example of Unitary Matrix ==
The square matrix:
:$\begin {pmatrix} \dfrac {\sqrt 2} 2 & \dfrac {\sqrt 2} 2 \\ \dfrac {\sqrt 2} 2 i & -\dfrac {\sqrt 2} 2 i \end {pmatrix}$
is a unitary matrix.
== Proof ==
By definition of inverse matrix:
By definition of Hermitian conjugate:
:$\begin {pmatrix} \dfrac {\sqrt 2} 2 & \dfrac {\sqrt 2} 2 \\ \dfrac {\sqrt 2} 2 i & -\dfrac {\sqrt 2} 2 i \end {pmatrix}^\dagger = \begin {pmatrix} \dfrac {\sqrt 2} 2 & \dfrac {\sqrt 2} 2 i \\ \dfrac {\sqrt 2} 2 & -\dfrac {\sqrt 2} 2 i \end {pmatrix}$
Hence the result.
Category:Examples of Unitary Matrices
Unitary Module of All Mappings is Unitary Module

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring with unity whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Let $S$ be a set.
Let $\struct {G^S, +_G', \circ}_R$ be the module of all mappings from $S$ to $G$.
Then $\struct {G^S, +_G', \circ}_R$ is a unitary module.
== Proof ==
From Module of All Mappings is Module, we have that $\struct {G^S, +_G', \circ}_R$ is an $R$-module.
To show that $\struct {G^S, +_G', \circ}_R$ is a unitary $R$-module, we verify the following:
:$\forall f \in G^S: 1_R \circ f = f$
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module.
Then:
Thus:
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.4$
Category:Examples of Unitary Modules
Unitary R-Modules with n-Element Bases are Isomorphic

== Theorem ==
Let $G$ and $H$ be unitary $R$-modules for some ring with unity $R$.
Let $G$ and $G$ both have bases with $n$ elements.
Then $G$ and $H$ are isomorphic.
== Proof ==
From Isomorphism from $R^n$ via $n$-Term Sequence, both $G$ and $H$ are isomorphic to the $R$-module $R^n$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.5$: Passim
Category:Unitary Modules
Unitization of *-Algebra is *-Algebra

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $A_+$ be the unitization of $A$.
Define:
:$\tuple {a, \lambda}^\dagger = \tuple {a^\ast, \overline \lambda}$
for each $\tuple {a, \lambda} \in A_+$.
Then $\struct {A_+, \dagger}$ is a $\ast$-algebra.
== Proof ==
=== Proof of $(\text C^\ast 1)$ ===
Let $\tuple {a, \lambda} \in A_+$.
We have:
=== Proof of $(\text C^\ast 2)$ ===
Let $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$.
We have:
=== Proof of $(\text C^\ast 3)$ ===
Let $\tuple {a, \lambda}, \tuple {b, \mu} \in A_+$.
We have:
=== Proof of $(\text C^\ast 4)$ ===
Let $\tuple {a, \lambda} \in A_+$ and $c \in \C$. 
We have:
Category:*-Algebras
Category:Unitizations of Algebras over Fields
Unitization of Algebra over Field is Unital Algebra over Field

== Theorem ==
Let $K$ be a field.
Let $A$ be an algebra over $K$. 
Let $\struct {A_+, +_{A_+}, \cdot_{A_+}, \circ_{A_+} }_K$ be the unitization of $A$.
Then $\struct {A_+, +_{A_+}, \cdot_{A_+}, \circ_{A_+} }_K$ is a unital algebra over $K$.
== Proof ==
From Direct Product of Vector Spaces is Vector Space, $\struct {A_+, +_{A_+}, \cdot_{A_+} }_K = \struct {A \times K, +_{A \times K}, \cdot_{A \times K} }_K$ is a vector space over $K$.
We show that $\circ_{A_+} : A_+ \times A_+ \to A_+$ is a bilinear mapping.
Let $\tuple {u, \alpha} \in A_+$, $\tuple {v, \beta} \in A_+$ and $\tuple {w, \gamma} \in A_+$.
Let $\lambda \in K$.
We have:
and:
So $\circ_{A_+}$ is bilinear.
So $\struct {A_+, +_{A_+}, \cdot_{A_+}, \circ_{A_+} }_K$ is an algebra. 
We show that $\struct {A_+, +_{A_+}, \cdot_{A_+}, \circ_{A_+} }_K$ is unital.
Let $\tuple {x, \lambda} \in A_+$. 
Then, we have:
and:
So $\struct {A_+, +_{A_+}, \cdot_{A_+}, \circ_{A_+} }_K$ is a unital algebra over $K$.
Category:Algebras over Fields
Category:Unitizations of Algebras over Fields
Unitization of Algebra over Field preserves Subalgebra Relation

== Theorem ==
Let $K$ be a field.
Let $A$ be a non-unital commutative algebra over $K$.
Let $B$ be a subalgebra of $A$. 
Let $A_+$ and $B_+$ be the unitizations of $A$ and $B$ respectively.
Then $B_+$ is a unital subalgebra of $A_+$. 
== Proof ==
Let $\struct {x, s}, \tuple {y, t} \in B_+$.
Let $\lambda \in K$.
Then, we have:
:$\tuple {x, s} + \lambda \tuple {y, t} = \tuple {x + \lambda y, s + \lambda t}$
Since $\tuple {x, s}, \tuple {y, t} \in B_+$, we have $x, y \in B$. 
Since $B$ is a subalgebra of $A$, we have that:
:$B$ is a vector subspace of $A$
and hence:
:$x + \lambda y \in B$
Hence, we have:
:$\tuple {x + \lambda y, s + \lambda t} \in B_+$
So, we have:
:$\tuple {x, s} + \lambda \tuple {y, t} \in B_+$
From One-Step Vector Subspace Test, we therefore have that $B_+$ is a vector subspace of $A_+$.
To show that $B_+$ is a subalgebra of $A_+$, it remains to verify that:
:$\tuple {x, s} \tuple {y, t} \in B_+$ for all $\tuple {x, s}, \tuple {y, t} \in B_+$.
Let $\tuple {x, s}, \tuple {y, t} \in B_+$.
Then $x, y \in B$. 
From the definition of the unitization, we have:
:$\tuple {x, s} \tuple {y, t} = \tuple {x y + s y + t x, s t}$
Since $B$ is a subalgebra of $A$, we have:
:$x y \in B$
Since $B$ is a vector subspace of $A$, we have:
:$x y + s y + t x \in B$
Hence, we have:
:$\tuple {x y + s y + t x, s t} \in B_+$
and so:
:$\tuple {x, s} \tuple {y, t} \in B_+$
We therefore have that $B_+$ is a subalgebra of $A_+$.
It remains to show that $B_+$ is unital.
Since $B$ is a vector subspace of $A$, we have that:
:${\mathbf 0}_A \in B$
Hence, we have:
:$\tuple { {\mathbf 0}_A, 1_K} \in B_+$
From Unitization of Algebra over Field is Unital Algebra over Field, we conclude:
$B_+$ is a unital subalgebra of $A_+$.
Category:Unitizations of Algebras over Fields
Unitization of Commutative Algebra over Field is Commutative

== Theorem ==
Let $K$ be a field.
Let $A$ be a commutative algebra over $K$.
Let $A_+$ be the unitization of $A$.
Then $A_+$ is commutative.
== Proof ==
Let $\tuple {x, \lambda}, \tuple {y, \mu} \in A_+$. 
Then, we have:
So $A_+$ is commutative.
Category:Unitizations of Algebras over Fields
Unitization of Non-Unital Subalgebra of Unital Algebra is Isomorphic to Unital Subalgebra

== Theorem ==
Let $A$ be a unital algebra over $\C$.
Let $B$ be a subalgebra of $A$ that is not unital.
Let $B_+$ be the unitization of $B$.
Let $B \oplus \C {\mathbf 1}_A$ be the (internal) direct sum of $B$ and $\C {\mathbf 1}_A$. 
Define $f : B_+ \to B \oplus \C {\mathbf 1}_A$ by:
:$\map f {b, \lambda} = b + \lambda {\mathbf 1}_A$
for each $\tuple {b, \lambda} \in B_+$. 
Then $f$ is a unital algebra isomorphism.
== Proof ==
Since:
:$B \oplus {\mathbf 1}_A = \set {b + \lambda {\mathbf 1}_A : b \in B, \, \lambda \in \C} = \set {\map f {b, \lambda} : \tuple {b, \lambda} \in B_+}$
we have that $f$ is a surjection.
We show that $f$ is linear.
Let $\tuple {b_1, \lambda_1}, \tuple {b_2, \lambda_2} \in B_+$ and $t \in \C$.
We then have:
So $f$ is indeed linear.
Further we have:
So $f$ is an algebra homomorphism.
From Linear Transformation is Injective iff Kernel Contains Only Zero, to finish we only need to show that $\ker f = \set { \tuple { {\mathbf 0}_A, 0} }$.
Suppose that:
:$\map f {b, \lambda} = 0$
for $\tuple {b, \lambda} \in B_+$.
Suppose first that $\lambda \ne 0$. 
From:
:$b = -\lambda {\mathbf 1}_A$
we then have:
:${\mathbf 1}_A = -\lambda^{-1} b \in B$ 
This contradicts that $B$ is not unital.
Hence we have $\lambda = 0$.
Then we have $b = {\mathbf 0}_A$.
Hence $\ker f = \set { \tuple { {\mathbf 0}_A, 0} }$.
So $f$ is an algebra isomorphism.
Finally, we have:
:$\map f { {\mathbf 0}_A, 1} = {\mathbf 1}_A$
so $f$ is a unital algebra isomorphism.
Category:Algebras
Category:Unitizations of Algebras over Fields
Unitization of Normed Algebra is Banach Algebra iff Original Algebra is Banach Algebra

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$ that is not unital as an algebra. 
Let $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ be the normed unitization of $\struct {A, \norm {\, \cdot \,} }$.
Then $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a Banach algebra $\struct {A, \norm {\, \cdot \,} }$ is a Banach algebra.
== Proof ==
=== Necessary Condition ===
Suppose that $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a Banach algebra.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {A, \norm {\, \cdot \,} }$.
We then have, for each $n, m \in \N$: 
:$\norm {\tuple {x_n, 0} - \tuple {x_m, 0} }_{A_+} = \norm {\tuple {x_n - x_m, 0} }_{A_+} = \norm {x_n - x_m}$
Let $\epsilon > 0$.
Since $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence, there exists $N \in \N$ such that:
:$\norm {x_n - x_m} < \epsilon$ for $n, m \ge N$:
so that:
:$\norm {\tuple {x_n, 0} - \tuple {x_m, 0} }_{A_+} < \epsilon$ for $n, m \ge N$.
Since $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a Banach algebra, there exists $\tuple {x, \lambda} \in A_+$ such that $\tuple {x_n, 0} \to \tuple {x, \lambda}$.
From Norm on Unitization of Normed Algebra is Equivalent to Direct Product Norm, $\norm {\, \cdot \,}_{A_+}$ is equivalent to the direct product norm $\norm {\, \cdot \,}_{A \times \GF}$.
From Convergence in Direct Product Norm and Convergent Sequences in Vector Spaces with Equivalent Norms Coincide, we have $x_n \to x$ and $\lambda = 0$.
Then, we have: 
:$\norm {\tuple {x_n, 0} - \tuple {x, 0} }_{A_+} \to 0$
while:
:$\norm {\tuple {x_n, 0} - \tuple {x, 0} }_{A_+} = \norm {\tuple {x_n - x, 0} }_{A_+} = \norm {x_n - x}$
so that:
:$\norm {x_n - x} \to 0$
From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have $x_n \to x$ in $A$. 
So every Cauchy sequence in $\struct {A, \norm {\, \cdot \,} }$ converges.
=== Sufficient Condition ===
Suppose that $\struct {A, \norm {\, \cdot \,} }$ is a Banach algebra.
In particular, $\struct {A, \norm {\, \cdot \,} }$ is a Banach space.
From:
:Real Number Line is Banach Space if $\GF = \R$
:Complex Plane is Banach Space if $\GF = \C$
we have that $\GF = \C$ is also a Banach space. 
From Direct Product of Banach Spaces is Banach Space, the direct product $A \times \GF = A_+$ equipped with the direct product norm $\norm {\, \cdot \,}_{A \times \GF}$ is a Banach space.
From Norm on Unitization of Normed Algebra is Equivalent to Direct Product Norm, $\norm {\, \cdot \,}_{A_+}$ and $\norm {\, \cdot \,}_{A \times \GF}$ are equivalent.
From Norm Equivalence Preserves Completeness, $\struct {A, \norm {\, \cdot \,}_{A_+} }$ is therefore a Banach space, hence a Banach algebra.
Category:Unitizations of Normed Algebras
Category:Banach Algebras
Unitization of Normed Algebra is Unital Normed Algebra

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra that is not unital as an algebra. 
Let $A_+$ be the unitization of $A$.
Define $\norm {\, \cdot \,}_{A_+} : A_+ \to \hointr 0 \infty$ by: 
:$\norm {\tuple {x, \lambda} }_{A_+} = \norm x + \cmod \lambda$
for each $\tuple {x, \lambda} \in A_+$.
Then $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a unital normed algebra.
== Proof ==
From Unitization of Algebra over Field is Unital Algebra over Field, $A_+$ is a unital algebra.
We show that $\norm {\, \cdot \,}_{A_+}$ is an algebra norm and that $\norm { {\mathbf 1}_{A_+} } = 1$.
=== Proof of ===
Suppose that $\tuple {x, \lambda} \in A_+$ is such that:
:$\norm {\tuple {x, \lambda} }_{A_+} = 0$
Then, we have:
:$\norm x + \cmod \lambda = 0$
So $\norm x = 0$ and $\cmod \lambda = 0$.
From , we obtain that $x = {\mathbf 0}_X$.
Since $\cmod \lambda = 0$, we have $\lambda = 0$.
So we have $\tuple {x, \lambda} = {\mathbf 0}_{A_+}$. 
Hence we have proved .
=== Proof of ===
Let $\tuple {x, \lambda} \in A_+$.
Let $\mu \in \GF$.
Then, we have:
So we have proved .
=== Proof of ===
Let $\struct {x, \lambda}, \struct {y, \mu} \in A_+$.
Then, we have:
So we have proved . 
=== Proof of Submultiplicativity ===
Let $\struct {x, \lambda}, \struct {y, \mu} \in A_+$.
Suppose that $\norm {\tuple {x, \lambda} }_{A_+} \norm {\tuple {y, \mu} }_{A_+} = 0$.
Then we have $\norm {\tuple {x, \lambda} }_{A_+} = 0$ or $\norm {\tuple {y, \mu} }_{A_+} = 0$.
From , we have $\tuple {x, \lambda} = {\mathbf 0}_{A_+}$ or $\tuple {y, \mu} = {\mathbf 0}_{A_+}$.
In this case, we have $\tuple {x, \lambda} \tuple {y, \mu} = {\mathbf 0}_{A_+}$.
Now take $\tuple {x, \lambda} \ne {\mathbf 0}_{A_+}$ and $\tuple {y, \mu} \ne {\mathbf 0}_{A_+}$.
We have:
=== Proof of $\norm { {\mathbf 1}_{A_+} }_{A_+} = 1$ ===
We have:
So $\norm {\, \cdot \,}_{A_+}$ is an algebra norm.
So $\struct {A_+, \norm {\, \cdot \,}_{A_+} }$ is a unital normed algebra.
Category:Unitizations of Normed Algebras
Units of 5th Cyclotomic Ring

== Theorem ==
Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
The units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are $1$ and $-1$.
== Proof ==
Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$.
Let $z_1 \in \Z \sqbrk {i \sqrt 5}$ be a unit of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
Thus by definition:
:$\exists z_2 \in \Z \sqbrk {i \sqrt 5}: z_1 \times z_2 = 1$
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
Then:
The result follows.
== Sources ==
* : Chapter $9$: Rings: Exercise $19 \ \text {(i)}$
Category:Cyclotomic Rings
Units of Field of Complex Numbers form Group

== Theorem ==
:$\C^\times = \C \setminus \set 0$
where $\C^\times$ denotes the group of units of $\C$.
== Proof ==
By Complex Numbers form Field, $\C$ is a field.
From Group of Units of Field it follows that:
: $\C^\times = \C \setminus \set 0$
Category:Examples of Groups
Category:Complex Numbers
Units of Gaussian Integers

== Theorem ==
Let $\struct {\Z \sqbrk i, +, \times}$ be the ring of Gaussian integers.
The set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {1, i, -1, -i}$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $9$: Rings: Exercise $3$
Category:Gaussian Integers
Category:Units of Gaussian Integers
Units of Gaussian Integers/Proof 1

== Theorem ==
== Proof ==
Let $a + b i$ be a unit of $\struct {\Z \sqbrk i, +, \times}$.
Then $a$ and $b$ are not both $0$ as then $a + b i$ would be the zero of $\struct {\Z \sqbrk i, +, \times}$.
Then:
:$\exists c, d \in \Z: \paren {a + b i} \paren {c + d i} = 1 + 0 i$
This leads (after algebra) to:
:$c = \dfrac a {a^2 + b^2}, d = \dfrac {-b} {a^2 + b^2}$
Let $a^2 + b^2 = n$.
We have that $n \in \Z, n > 0$.
If $c$ and $d$ are integers, then $a$ and $b$ must both be divisible by $n$.
Let $a = n x, b = n y$.
Then:
:$n^2 x^2 + n^2 y^2 = n$
and so:
:$n \paren {x^2 + y^2} = 1$
Thus $n = a^2 + b^2 = 1$ and so as $a, b \in \Z$ we have:
:$a^2 = 1, b^2 = 0$
or:
:$a^2 = 0, b^2 = 1$
from which the result follows.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Example $50$
Category:Units of Gaussian Integers
Units of Gaussian Integers/Proof 2

== Theorem ==
== Proof ==
Let $\alpha = a + b i$ be a unit of $\struct {\Z \sqbrk i, +, \times}$.
Then by definition of unit:
:$\exists\beta = c + d i \in \Z \sqbrk i: \alpha \beta = 1$
Let $\cmod \alpha$ denote the modulus of $\alpha$.
Then:
By Divisors of One:
:$\cmod a^2 = 1$ or $-1$
Since $\cmod \alpha$ and $\cmod \beta$ are positive integers:
:$\cmod \alpha^2 = a^2 + b^2 = 1$
and so either:
:$\cmod a = 1$ and $\cmod b = 0$
or:
:$\cmod b = 1$ and $\cmod a = 0$.
Hence the set of units of $\struct {\Z \sqbrk i, +, \times}$ is $\set {\pm 1, \pm i}$.
Category:Units of Gaussian Integers
Units of Gaussian Integers form Group

== Theorem ==
Let $U_\C$ be the set of units of the Gaussian integers:
:$U_\C = \set {1, i, -1, -i}$
where $i$ is the imaginary unit: $i = \sqrt {-1}$.
Let $\struct {U_\C, \times}$ be the algebraic structure formed by $U_\C$ under the operation of complex multiplication.
Then $\struct {U_\C, \times}$ forms a cyclic group under complex multiplication.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 6$: Examples of Finite Groups: $\text{(i)}$
* : $\S 34$. Examples of groups: $(6) \ \text{(i)}$
* : $\S 3.2$: Groups; the axioms: Exercise $(1)$
Category:Units of Gaussian Integers form Group
Category:Group of Gaussian Integer Units
Category:Complex Roots of Unity
Units of Gaussian Integers form Group/Proof 2

== Theorem ==
== Proof ==
From Gaussian Integer Units are 4th Roots of Unity:
: $\left\{{1, i, -1, -i}\right\}$ constitutes the set of the $4$th roots of unity.
The result follows from Roots of Unity under Multiplication form Cyclic Group.
Category:Units of Gaussian Integers form Group
Units of Gaussian Integers form Group/Proof 3

== Theorem ==
== Proof ==
From Units of Gaussian Integers, $U_\C$ is the set of units of the ring of Gaussian integers.
From Group of Units is Group, $\left({U_\C, \times}\right)$ forms a group.
It remains to note that:
thus demonstrating that $U_\C$ is cyclic.
Category:Units of Gaussian Integers form Group
Units of Quadratic Integers over 2

== Theorem ==
Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
:$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Let $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$ be the integral domain where $+$ and $\times$ are conventional addition and multiplication on real numbers.
Then numbers of the form $a + b \sqrt 2$ such that $a^2 - 2 b^2 = \pm 1$ are all units of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$.
== Proof ==
For $a + b \sqrt 2$ to be a unit of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$, we require that:
:$\exists c, d \in \Z: \paren {a + b \sqrt 2} \paren {c + d \sqrt 2} = 1$
In Quadratic Integers over 2 are Not a Field it is shown that the product inverse of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2 b^2} + \dfrac {b \sqrt 2} {a^2 - 2 b^2}$.
So if $a^2 - 2 b^2 = \pm 1$ it follows that $c$ and $d$ are integers.
Hence the result.
== Note ==
This is not to say that the only units of $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$.
This simple analysis does not rule out the possibility of there being other $a, b \in \Z$ such that $\dfrac a {a^2 - 2 b^2} + \dfrac {b \sqrt 2} {a^2 - 2 b^2} \in \Z \sqbrk {\sqrt 2}$.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 26$. Divisibility: Example $49$
Category:Examples of Integral Domains
Category:Units of Rings
Category:Quadratic Integers
Units of Ring Direct Product are Ring Direct Product of Units

== Theorem ==
Let $R_1, R_2, \ldots, R_m$ be rings with identity.
Then the following group isomorphism holds:
:$\map U {R_1 \oplus R_2 \oplus \cdots \oplus R_n} \simeq \map U {R_1} \oplus \map U {R_2} \oplus \cdots \oplus \map U {R_n}$.
== Proof ==
It suffices to prove the case $m = 2$.
By the finite direct sum 
is isomorphic to the finite direct product.
Category:Ring Direct Products