nyuuzyou/wikis huggingface dataset:
Quotient Norm is Norm

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\Bbb F$. 
Let $N$ be a closed linear subspace of $X$.
Let $X/N$ be the quotient vector space of $X$ modulo $N$.
Let $\norm {\, \cdot \,}_{X/N}$ be the quotient norm on $X/N$.
Then $\norm {\, \cdot \,}_{X/N}$ is indeed a norm. 
== Proof ==
=== Norm is Well-Defined and Finite ===
Let $\pi$ be the quotient map associated with $X/N$.
We show that if $x, x' \in X$ have $\map \pi x = \map \pi {x'}$, then: 
:$\ds \inf_{z \mathop \in N} \norm {x - z} = \inf_{z \mathop \in N} \norm {x' - z}$
From Quotient Mapping is Linear Transformation:
:$\map \pi {x' - x} = 0$
From Kernel of Quotient Mapping:
:$x' - x \in N$
Since $N$ is a linear subspace of $X$:
:$\paren {x' - x} + N = N$
So, we may manipulate: 
and so the quotient norm is well-defined.
Since: 
:$\norm {x - z} \ge 0$
for all $z \in N$, we also have: 
:$\ds \inf_{z \mathop \in N} \norm {x - z} \ge 0$ 
=== Proof of ===
First, we can calculate: 
Conversely, suppose that: 
:$\ds \inf_{z \mathop \in N} \norm {x - z} = 0$ 
Then by the definition of infimum, for each $n \in \N$ there exists $z_n \in N$ such that: 
:$\ds \norm {x - z_n} < \frac 1 n$
By the Squeeze Theorem, we then have: 
:$\ds \lim_{n \mathop \to \infty} \norm {x - z_n} = 0$ 
From Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we then have: 
:$z_n \to x$ 
Since $N$ is closed:
:$x \in N$
So from Kernel of Quotient Mapping:
:$\map \pi x = 0_{X/N}$
and hence:
:$\norm {\map \pi x}_{X/N} = 0$
So we have: 
:$\norm {\map \pi x}_{X/N} = 0$ $x = 0_{X/N}$
 
=== Proof of ===
Let $t \in \Bbb F$. 
Clearly if $t = 0$, we have: 
:$\norm {t \map \pi x}_{X/N} = 0$ 
So take $t \ne 0$. 
We have: 
Since $N$ is a linear subspace of $X$: 
:$\ds \paren {\frac 1 t} N = N$ 
So: 
=== Proof of ===
We first argue that: 
:$\ds \inf_{z \mathop \in N} \norm {x - z} = \inf_{z_1, z_2 \mathop \in N} \norm {x - \paren {z_1 + z_2} }$
First, for each $z_1, z_2 \in N$ we have:
:$z_1 + z_2 \in N$
so:
:$\ds \inf_{z \mathop \in N} \norm {x - z} \le \inf_{z_1, z_2 \mathop \in N} \norm {x - \paren {z_1 + z_2} }$
Conversely, we have $0 \in N$ and so $z, 0 \in N$ have $z + 0 = z$, and so we also get:
:$\ds \inf_{z_1, z_2 \mathop \in N} \norm {x - \paren {z_1 + z_2} } \le \inf_{z \mathop \in N} \norm {x - z}$ 
and hence: 
:$\ds \inf_{z \mathop \in N} \norm {x - z} = \inf_{z_1, z_2 \mathop \in N} \norm {x - \paren {z_1 + z_2} }$
We now have for $x, y \in X$: 
Category:Examples of Norms
Category:Quotient Norms
Quotient Norm on Quotient of C*-Algebra by Closed Ideal in terms of Positive Elements of Closed Unit Ball

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $I$ be a closed ideal of $A$.
Let $\struct {A/I, \norm {\, \cdot \,}_{A/I} }$ be the quotient normed algebra of $A$ modulo $I$.
Let $\pi : A \to A/I$ be the quotient mapping.
Let $B_I$ be the closed unit ball of $I$.
Let $B_I^+ \subseteq B_I$ be the set of positive elements of $A$ contained in $B_I$.
Then:
:$\norm {\map \pi a}_{A/I} = \inf \set {\norm {a - a x} : x \in B_I^+}$
for each $a \in A$.
== Proof ==
Let $a \in A$ and:
:$\alpha = \inf \set {\norm {a - a x} : x \in B_I^+}$
Since $I$ is an ideal, we have $a x \in I$ for each $x \in B_I^+$.
Hence we have:
:$\ds \alpha \ge \inf \set {\norm {a - i} : i \in I}$
From the definition of the quotient norm, we have:
:$\norm {\map \pi a}_{A/I} \le \alpha$ 
We now want to show that $\norm {\map \pi a}_{A/I} \ge \alpha$.
Let $y \in I$. 
We show that $\norm {a + y} \ge \epsilon$.
From Closed Ideal of C*-Algebra is Self-Adjoint, $I$ is a closed $\ast$-subalgebra of $A$.
Hence from Closed *-Subalgebra of C*-Algebra is C*-Algebra, $I$ is a $\text C^\ast$-algebra.
From Existence of Approximate Identity of C*-Algebra arising from Dense Ideal, there exists a directed set $\struct {\Lambda, \preceq}$ and approximate identity $\family {e_\lambda}_{\lambda \mathop \in \Lambda}$ for $I$.
Let $\le_A$ be the canonical preordering of $A$.
From the definition of an approximate identity we have:
:$\map {\sigma_A} {e_\lambda} \subseteq \closedint 0 1$ for each $\lambda \in \Lambda$.
Hence from Spectral Radius of Normal Element of C*-Algebra Equal to Norm, we have $\norm {e_\lambda} \le 1$ and hence $e_\lambda \in B_I^+$ for each $\lambda \in \Lambda$. 
From the Spectral Mapping Theorem for Polynomials, we have:
:$\map {\sigma_A} { {\mathbf 1}_A - e_\lambda} \subseteq \closedint 0 1$ for each $\lambda \in \Lambda$.
From Spectral Radius of Normal Element of C*-Algebra Equal to Norm, we have:
:$\norm { {\mathbf 1}_A - e_\lambda} \le 1$ for each $\lambda \in \Lambda$. 
We then have:
:$\norm {\paren {a + y} \paren { {\mathbf 1}_A - e_\lambda} } \le \norm {a + y} \norm { {\mathbf 1}_A - e_\lambda} \le \norm {a + y}$
from the definition of an algebra norm.
Hence we have:
Let $\epsilon > 0$. 
Since $\family {y e_\lambda}_{\lambda \mathop \in \Lambda}$ converges to $y$, there exists $\lambda \in \Lambda$ such that:
:$\norm {y - y e_\lambda} < \epsilon$
Then we have:
:$\norm {a + y} \ge \alpha - \norm {y - y e_\lambda} \ge \alpha - \epsilon$
for each $\epsilon > 0$.
Hence we have:
:$\norm {a + y} \ge \alpha$
Hence:
:$\norm {\map \pi a}_{A/I} = \sup \set {\norm {a + y} : y \in I} \ge \alpha$
We conclude $\norm {\map \pi a}_{A/I} = \alpha$.
== Sources ==
* : Lemma $5.5$
Category:Quotient C*-Algebras
Category:Positive Elements of C*-Algebras
Quotient Normed Algebra is Normed Algebra

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,}_R}$ be a normed division ring. 
Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra.
Let $I$ be a closed ideal of $A$.
Let $A/I$ be the quotient algebra of $A$ modulo $I$.
Let $\norm {\, \cdot \,}_{A/I}$ be the quotient norm.
Then $\struct {A/I, \norm {\, \cdot \,}_{A/I} }$ is a normed algebra.
== Proof ==
Since $I$ is a closed linear subspace of $A$, Quotient Norm is Norm shows that $\struct {A/I, \norm {\, \cdot \,}_{A/I} }$ is a normed vector space.
It remains to show that for $a, b \in A$ we have:
:$\norm {a b + I}_{A/I} \le \norm {a + I}_{A/I} \norm {b + I}_{A/I}$
Let $\epsilon > 0$.
By the definition of the quotient norm, there exists $a', b' \in I$ such that:
:$\norm {a + a'}_{A/I} < \epsilon + \norm {a + I}_{A/I}$
and:
:$\norm {b + b'}_{A/I} < \epsilon + \norm {b + I}_{A/I}$
We then have:
Since $I$ is an ideal, we have:
:$a' b + a b' + a' b' \in I$
Hence we have:
:$\ds \norm {a b + I}_{A/I} = \inf_{i \mathop \in I} \norm {a b + i} < \paren {\epsilon + \norm {a + I}_{A/I} } \paren {\epsilon + \norm {b + I}_{A/I} }$
Taking $\epsilon \to 0^+$, we obtain:
:$\norm {a b + I}_{A/I} \le \norm {a + I}_{A/I} \norm {b + I}_{A/I}$
So $\norm {\, \cdot \,}_{A/I}$ is an algebra norm on $A/I$. 
So $\struct {A/I, \norm {\, \cdot \,}_{A/I} }$ is a normed algebra.
== Sources ==
* : $1.1$: Banach Algebras: Theorem $1.1.1$
Category:Normed Algebras
Quotient Normed Algebra of Unital Normed Algebra is Unital Normed Algebra

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,}_R}$ be a normed division ring.
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital normed algebra with identity element ${\mathbf 1}_A$. 
Let $I$ be a closed proper ideal of $A$. 
Let $\struct {A/I, \norm {\, \cdot \,}_{A/I} }$ be the quotient normed algebra of $A$ modulo $I$.
Then:
:$\norm { {\mathbf 1}_A + I}_{A/I} = 1$ 
== Proof ==
From Norm of Identity Element in Normed Algebra Lower Bounded by One, we have $1 \le \norm { {\mathbf 1}_A + I}_{A/I}$.
Since $I$ is a linear subspace of $A$, we have ${\mathbf 0}_A \in I$. 
Hence we have:
:$\ds \inf_{i \mathop \in I} \norm { {\mathbf 1}_A + i} \le \norm { {\mathbf 1}_A + {\mathbf 0}_A} = \norm { {\mathbf 1}_A}$
From the definition of a unital normed algebra, we have:
:$\norm { {\mathbf 1}_A} = 1$
Hence we obtain:
:$\norm { {\mathbf 1}_A + I}_{A/I} \le 1$
So:
:$\norm { {\mathbf 1}_A + I}_{A/I} = 1$ 
Category:Quotient Normed Algebras
Category:Unital Normed Algebras
Quotient Ring/Examples/Principal Ideal Generated by 3

== Examples of Quotient Ring ==
Let $\struct {\Z, + \times}$ denote the ring of integers.
Let $\struct {3 \Z, +, \times}$ denote the principal ideal of $\Z$ generated by $3$.
Let $\Z / 3 \Z$ denote the quotient ring of $\Z$ by $3 \Z$.
Then $\Z / 3 \Z$ has $3$ elements:
== Proof ==
== Sources ==
* 
* 
Category:Examples of Quotient Rings
Quotient Ring Defined by Ring Itself is Null Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {R / R, +, \circ}$ be the quotient ring defined by $R$.
Then $\struct {R / R, +, \circ}$ is a null ring.
== Proof ==
From Ring is Ideal of Itself, it is clear we can form the quotient ring $\struct {R / R, +, \circ}$.
But $R / R = 0_R$ and so is the null ring.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
* : Chapter $5$: Rings: $\S 22$. Quotient Rings: Example $40$
Category:Ideal Theory
Category:Quotient Rings
Quotient Ring by Null Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {\set {0_R}, +, \circ}$ be the null ideal of $\struct {R, +, \circ}$.
Let $\struct {R / \set {0_R}, +, \circ}$ be the quotient ring of $R$ defined by $\set {0_R}$.
Then $\struct {R / \set {0_R}, +, \circ}$ is isomorphic to $\struct {R, +, \circ}$.
== Proof ==
Consider the additive group $\struct {R, +}$.
From Trivial Quotient Group is Quotient Group:
:$\struct {R, +} / \set {0_R} \cong \struct {R, +}$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old
* : Chapter $5$: Rings: $\S 22$. Quotient Rings: Example $40$
Category:Ideal Theory
Category:Quotient Rings
Quotient Ring is Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$.
Then $R / J$ is also a ring.
== Proof ==
First, it is to be shown that $+$ and $\circ$ are in fact well-defined operations on $R / J$.
=== Well-definition of $+$ ===
=== Well-definition of $\circ$ ===
Now to prove that $\struct {R / J, +, \circ}$ is a ring, proceed by verifying the ring axioms in turn:
=== ===
From:
:Ideal is Additive Normal Subgroup
:The definition of a quotient group
:Quotient Group is Group
it follows that $\struct {R / J, +}$ is a group.
=== ===
By definition of $\circ$ in $R / J$, it follows that $\struct {R / J, \circ}$ is closed.
=== ===
Associativity can be deduced from the fact that $\circ$ is associative on $R$:
=== ===
Distributivity can be deduced from the fact that $\circ$ is distributive on $R$:
Having verified all of the ring axioms, it follows that $\struct {R / J, +, \circ}$ is a ring.
== Sources ==
* : Chapter $0$: Algebraic Concepts
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.2$
* : $\S 60.2$ Factor rings: $\text{(i)}$
Category:Quotient Rings
Category:Ideal Theory
Quotient Ring is Ring/Quotient Ring Addition is Well-Defined

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$.
Then $+$ is well-defined on $R / J$, that is:
:$x_1 + J = x_2 + J, y_1 + J = y_2 + J \implies \paren {x_1 + y_1} + J = \paren {x_2 + y_2} + J$
== Proof ==
From Ideal is Additive Normal Subgroup that $J$ is a normal subgroup of $R$ under $+$.
Thus, the quotient group $\struct {R / J, +}$ is defined, and as a Quotient Group is Group, $+$ is well-defined.
== Sources ==
* : Chapter $5$: Rings: $\S 22$. Quotient Rings: Theorem $41$
* : $\S 2.2$: Homomorphisms
Category:Quotient Rings
Quotient Ring is Ring/Quotient Ring Product is Well-Defined

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$.
Then $\circ$ is well-defined on $R / J$, that is:
:$x_1 + J = x_2 + J, y_1 + J = y_2 + J \implies x_1 \circ y_1 + J = x_2 \circ y_2 + J$
== Proof ==
From Left Cosets are Equal iff Product with Inverse in Subgroup, we have:
and:
Hence from the definition of ideal:
Thus:
== Also see ==
* Ideal induces Congruence Relation on Ring
== Sources ==
* : Chapter $5$: Rings: $\S 22$. Quotient Rings: Theorem $41$
* : $\S 2.2$: Homomorphisms
* : $\S 60.1$ Factor rings
Category:Quotient Rings
Quotient Ring of Cauchy Sequences is Division Ring

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$.
Let $\NN$ be the set of null sequences.
Then the quotient ring $\CC / \NN$ is a division ring.
=== Corollary ===
Let $\struct {R, \norm {\, \cdot \,} }$ be a valued field.
== Proof ==
By Null Sequences form Maximal Left and Right Ideal then $\NN$ is an ideal of the ring $\CC$ that is also a maximal left ideal.
By Maximal Left and Right Ideal iff Quotient Ring is Division Ring then the quotient ring $\CC / \NN$ is a division ring
== Sources ==
* : $\S 3.2$: Completions
* : $\S 1.3$: Construction of the completion of a normed field: Theorem $1.19$
Category:Cauchy Sequences
Category:Normed Division Rings
Category:Completion of Normed Division Ring
Category:Rings of Sequences
Quotient Ring of Cauchy Sequences is Division Ring/Corollary 1

== Theorem ==
Let $\struct {F, \norm {\, \cdot \,} }$ be a valued field.
Let $\CC$ be the ring of Cauchy sequences over $F$.
Let $\NN$ be the set of null sequences.
Then the quotient ring $\CC \,\big / \NN$ is a field.
== Proof ==
By Quotient Ring of Cauchy Sequences is Division Ring then $\CC \,\big / \NN$ is a division ring.
By Corollary to Cauchy Sequences form Ring with Unity then $\CC$ is a commutative ring with unity.
By Quotient Ring of Commutative Ring is Commutative then $\CC \,\big / \NN$ is a commutative division ring, that is, a field.
== Sources ==
* : $\S 3.2$: Completions
* : $\S 1.3$: Construction of the completion of a normed field, Theorem $1.19$
Category:Completion of Normed Division Ring
Quotient Ring of Cauchy Sequences is Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
For all $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the left coset $\sequence {x_n} + \NN$
Let $\norm {\, \cdot \,}_1: \CC \,\big / \NN \to \R_{\ge 0}$ be defined by:
:$\ds \forall \eqclass {x_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} }_1 = \lim_{n \mathop \to \infty} \norm {x_n}$
Then:
:$\struct {\CC \,\big / \NN, \norm {\, \cdot \,}_1 }$ is a normed division ring.
=== Corollary ===
Let $\struct {R, \norm {\, \cdot \,} }$ be a valued field.
== Proof ==
By Quotient Ring of Cauchy Sequences is Division Ring then $\CC \,\big / \NN$ is a division ring.
It remains to be proved that:
:$\norm {\, \cdot \,}_1$ is well-defined
:$\norm {\, \cdot \,}_1$ satisfies the norm axioms.
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
=== Lemma 4 ===
The result follows.
== Sources ==
* : $\S 3.2$: Completions: Definition $3.2.11$, Problems $81$, $82$
* : $\S 1.3$: Construction of the completion of a normed field: Theorem $1.21$
Category:Cauchy Sequences
Category:Normed Division Rings
Category:Completion of Normed Division Ring
Quotient Ring of Cauchy Sequences is Normed Division Ring/Corollary 1

== Theorem ==
Let $\struct {F, \norm {\, \cdot \,} }$ be a valued field.
Let $\CC$ be the ring of Cauchy sequences over $F$
Let $\NN$ be the set of null sequences.
For all $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the left coset $\sequence {x_n} + \NN$
Let $\norm {\, \cdot \,}_1: \CC \,\big / \NN \to \R_{\ge 0}$ be defined by:
:$\ds \forall \eqclass {x_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} }_1 = \lim_{n \mathop \to \infty} \norm {x_n}$
Then $\struct {\CC \,\big / \NN, \norm {\, \cdot \,}_1 }$ is a valued field.
== Proof ==
By Quotient Ring of Cauchy Sequences is Normed Division Ring then $\CC \,\big / \NN$ is a normed division ring.
By Corollary to Quotient Ring of Cauchy Sequences is Normed Division Ring then $\CC \,\big / \NN$ is a field.
The result follows.
== Sources ==
* : $\S 3.2$: Completions
* : $\S 1.3$: Construction of the completion of a normed field, Theorem $1.21$
Category:Completion of Normed Division Ring
Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 1

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
For all $\sequence {x_n} \in \CC$, let $\eqclass {x_n}{}$ denote the left coset $\sequence {x_n} + \NN$
Let $\norm {\, \cdot \,}_1: \CC \, \big / \NN \to \R_{\ge 0}$ be defined by:
:$\ds \forall \eqclass {x_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n}{} }_1 = \lim_{n \mathop \to \infty} \norm {x_n}$
Then:
:$\norm {\, \cdot \,}_1$ is well-defined.
That is, 
:$(1): \quad \ds \forall \eqclass {x_n}{}: \lim_{n \mathop \to \infty} \norm{x_n}$ exists
:$(2): \quad \ds \forall \eqclass {x_n}{}, \eqclass {y_n}{} \in \CC \,\big / \NN: \eqclass {x_n}{} = \eqclass {y_n}{} \implies \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n}$
== Proof ==
By Norm Sequence of Cauchy Sequence has Limit then:
:for each $\eqclass {x_n}{}$ the $\ds \lim_{n \mathop \to \infty} \norm{x_n}$ exists.
Suppose $\eqclass {x_n}{} = \eqclass {y_n}{}$.
By Left Cosets are Equal iff Difference in Subgroup then:
:$\sequence {x_n} - \sequence {y_n} = \sequence {x_n - y_n} \in \NN$
By Equivalent Cauchy Sequences have Equal Limits of Norm Sequences then:
:$\ds \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n}$
Hence:
:$\ds \norm {\eqclass {x_n}{} }_1 = \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n} = \norm {\eqclass {x_n}{} }_1$
The result follows.
== Sources ==
* : $\S 3.2$: Completions
Category:Completion of Normed Division Ring
Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 2

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$.
Let $\NN$ be the set of null sequences.
For all $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the left coset $\sequence {x_n} + \NN$.
Let $\norm {\, \cdot \,}_1: \CC \,\big / \NN \to \R_{\ge 0}$ be defined by:
:$\ds \forall \eqclass {x_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n}{} }_1 = \lim_{n \mathop \to \infty} \norm{x_n}$
Then:
:$\norm {\, \cdot \,}_1$ satisfies 
That is:
:$\forall \eqclass {x_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} }_1 = 0 \iff \eqclass {x_n} {} = \eqclass {0_R} {} $
== Proof ==
By Quotient Ring of Cauchy Sequences is Division Ring the zero of $\CC \,\big / \NN$ is $\eqclass {0_R} {}$.
The result follows.
== Sources ==
* : $\S 3.2$: Completions
Category:Completion of Normed Division Ring
Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 3

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$.
Let $\NN$ be the set of null sequences.
For all $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the left coset $\sequence {x_n} + \NN$.
Let $\norm {\, \cdot \,}_1: \CC \,\big / \NN \to \R_{\ge 0}$ be defined by:
:$\ds \forall \eqclass {x_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n}{} }_1 = \lim_{n \mathop \to \infty} \norm {x_n}$
Then:
:$\norm {\, \cdot \,}_1$ satisfies the .
That is:
:$\forall \eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} \eqclass {y_n} {} }_1 = \norm {\eqclass {x_n} {} }_1 \times \norm {\eqclass {y_n} {} }_1$
== Proof ==
Let $\eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN$
== Sources ==
* : $\S 3.2$: Completions
Category:Completion of Normed Division Ring
Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 4

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
For all $\sequence {x_n} \in \CC$, let $\eqclass {x_n} {}$ denote the left coset $\sequence {x_n} + \NN$
Let $\norm {\, \cdot \,}_1: \CC \,\big / \NN \to \R_{\ge 0}$ be defined by:
:$\ds \forall \eqclass {x_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} }_1 = \lim_{n \mathop \to \infty} \norm {x_n}$
Then:
:$\norm {\, \cdot \,}_1$ satisfies the .
That is:
:$\forall \eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} + \eqclass {y_n} {} }_1 \le \norm {\eqclass {x_n} {} }_1 + \norm {\eqclass {y_n} {} }_1$
== Proof ==
Let $\eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN$
By :
:$\forall n: \norm {x_n + y_n} \le \norm {x_n} + \norm {y_n}$
So:
== Sources ==
* : $\S 3.2$: Completions
Category:Completion of Normed Division Ring
Quotient Ring of Commutative Ring is Commutative

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$.
If $\struct {R, +, \circ}$ is a commutative ring, then so is $\struct {R / J, +, \circ}$.
== Proof ==
Let $\struct {R, +, \circ}$ be a commutative ring
That means $\circ$ is commutative on $R$.
Thus:
== Sources ==
* : $\S 60.2$ Factor rings: $\text{(iii)}$
Category:Quotient Rings
Quotient Ring of Integers and Principal Ideal from Unity

== Theorem ==
Let $\left({\Z, +, \times}\right)$ be the integral domain of integers.
Let $\left({1}\right)$ be the principal ideal of $\left({\Z, +, \times}\right)$ generated by $1$.
The quotient ring $\left({\Z, +, \times}\right) / \left({1}\right)$ is isomorphic to the null ring.
== Proof ==
Category:Ideal Theory
Category:Quotient Rings
Category:Integers
Quotient Ring of Integers and Zero

== Theorem ==
Let $\struct {\Z, +, \times}$ be the integral domain of integers.
Let $\ideal 0$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $0$.
The quotient ring $\struct {\Z / \ideal 0, +, \times}$ is isomorphic to $\struct {\Z, +, \times}$.
== Proof ==
Category:Quotient Rings
Category:Integers
Quotient Ring of Integers with Principal Ideal

== Theorem ==
Let $\struct {\Z, +, \times}$ be the integral domain of integers.
Let $n \in \Z$.
Let $\ideal n$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $n$.
The quotient ring $\struct {\Z, +, \times} / \ideal n$ is isomorphic to $\struct {\Z_n, +_n, \times_n}$, the ring of integers modulo $n$.
Note the special cases where $n = 0$ or $1$:
:Quotient Ring of Integers and Zero
:Quotient Ring of Integers and Principal Ideal from Unity
== Proof ==
== Sources ==
* : Chapter $5$: Rings: $\S 22$. Quotient Rings: Example $39$
* : $\S 60$. Factor rings: Illustration
Category:Ideal Theory
Category:Quotient Rings
Category:Integers
Quotient Ring of Kernel of Ring Epimorphism

== Theorem ==
Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism.
Let $K = \map \ker \phi$ be the kernel of $\phi$.
There exists a unique ring isomorphism $g: R_1 / K \to R_2$ such that:
:$g \circ q_K = \phi$
== Proof ==
From the Quotient Theorem for Epimorphisms, there is one and only one isomorphism that satisfies the conditions for each of the operations on $R_1$.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $22$. New Rings from Old: Theorem $22.6: \ 1^\circ$
Category:Ring Epimorphisms
Category:Ring Isomorphisms
Category:Quotient Rings
Quotient Ring of Noetherian Ring is Noetherian

== Theorem ==
Let $A$ be a Noetherian ring.
Let $\mathfrak a \subseteq A$ be an ideal.
Let $A / \mathfrak a$ be the quotient ring of $A$ by $\mathfrak a$.
Then $A / \mathfrak a$ is a Noetherian ring.
== Proof ==
Observe that:
:$0 \longrightarrow \mathfrak a \longrightarrow A \longrightarrow A / \mathfrak a \longrightarrow 0$
is a short exact sequence of 
$A$-modules.
By Short Exact Sequence Condition of Noetherian Modules, $A / \mathfrak a$ is a Noetherian $A$-module.
As $A / \mathfrak a$ is an $A / \mathfrak a$-module, $A / \mathfrak a$ is also a Noetherian $A / \mathfrak a$-module.
By , $A / \mathfrak a$ is a Noetherian ring.
Category:Noetherian Rings