nyuuzyou/wikis huggingface dataset:
Maximal Element need not be Greatest Element

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $M \in S$ be a maximal element of $S$.
Then $M$ is not necessarily the greatest element of $S$.
== Proof ==
Proof by Counterexample:
Let $S = \set {a, b, c}$.
Let $\preccurlyeq$ be defined as:
:$x \preccurlyeq y \iff \tuple {x, y} \in \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {a, c} }$
A straightforward but laborious process determines that $\preccurlyeq$ is a partial ordering on $S$.
We have that:
:$c \preccurlyeq x \implies c = x$
and:
:$b \preccurlyeq x \implies b = x$
and so by definition, both $b$ and $c$ are maximal elements of $S$.
Suppose $b$ is the greatest element of $S$.
Then from Greatest Element is Unique it follows that $c$ can not be the greatest element of $S$.
Hence the result.
In fact, from the definition of the greatest element of $S$:
:$x$ is the greatest element of $S$ $\forall y \in S: y \preccurlyeq x$
it can be seen directly that neither $b$ nor $c$ is the greatest element of $S$.
== Also see ==
* Maximal Element in Toset is Unique and Greatest
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
* : Chapter $1$: Sets and mappings: $\S 1.5$: Ordered Sets
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Maximal Elements
Category:Greatest Elements
Maximal Element need not be Unique

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
It is possible for $S$ to have more than one maximal element.
== Proof ==
Proof by Counterexample:
Consider the set $T$ defined as:
:$T = \set {0, 1}$
Let $S$ be defined as:
:$S := \powerset T \setminus T$
where $\powerset T$ denotes the power set of $T$.
That is:
:$S = \set {\O, \set 0, \set 1}$
Let $\preccurlyeq$ be the relation defined on $S$ as:
:$\forall a, b \in S: a \preccurlyeq b \iff a \subseteq b$
That is, $\preccurlyeq$ is the subset relation on $S$.
From Subset Relation is Ordering, $\struct {S, \preccurlyeq}$ is an ordered set.
Let $a \in S$ such that $\set 1 \preccurlyeq a$.
Then by inspection it is apparent that:
:$a = \set 1$
That is, $\set 1$ is a maximal element of $\struct {S, \preccurlyeq}$.
Similarly, let $a \in S$ such that $\set 0 \preccurlyeq a$.
Then by inspection it is apparent that:
:$a = \set 0$
That is, $\set 0$ is also a maximal element of $\struct {S, \preccurlyeq}$.
Hence $S$ has more than one maximal element.
== Examples ==
== Also see ==
* Greatest Element is Unique
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
* 
* 
Category:Maximal Element need not be Unique
Category:Maximal Elements
Maximal Element need not be Unique/Examples/Arbitrary Example 1

== Example of Maximal Element need not be Unique ==
Consider the set $S = \set {a, b, c, d, e}$ with the partial ordering $\preccurlyeq$ defined as:
:${\preccurlyeq} := \set {\tuple {c, a}, \tuple {d, a}, \tuple {e, a}, \tuple {d, b}, \tuple {e, b}, \tuple {c, b}, \tuple {c, e} }$
This can be illustrated using the following Hasse diagram:
:
It can be seen by inspection that both $a$ and $b$ are maximal elements of the partially ordered set $\struct {S, \preccurlyeq}$.
== Sources ==
* 
* 
Category:Maximal Element need not be Unique
Maximal Element need not be Unique/Examples/Arbitrary Example 2

== Example of Maximal Element need not be Unique ==
Consider the set $S = \set {1, 3, 5, 7, 9}$ under the subset ordering.
Let $T \subseteq \powerset S$ be the set of subsets of $S$ that do not contain both $3$ and $5$.
Then the subsets $\set {1, 3, 7, 9}$ and $\set {1, 5, 7, 9}$ of $S$ are maximal elements of the partially ordered set $\struct {T, \subseteq}$.
== Sources ==
* 
Category:Maximal Element need not be Unique
Maximal Element of Chain is Greatest Element

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $C$ be a chain in $S$.
Let $m$ be a maximal element of $C$.
Then $m$ is the greatest element of $C$.
== Proof ==
Let $x \in C$.
Since $m$ is maximal in $C$, $m \nprec x$.
Since $C$ is a chain, $x = m$ or $x \prec m$.
Thus for each $x \in C$, $x \preceq m$.
Therefore $m$ is the greatest element of $C$.
== Also see ==
* Minimal Element of Chain is Smallest Element
== Sources ==
* : $\S\text I.3$
Category:Maximal Elements
Category:Greatest Elements
Category:Chains (Order Theory)
Maximal Element of Complement of Filter is Meet Irreducible

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $F$ be a filter in $L$.
Let $p \in S$.
Let $p = \map \max {\relcomp S F}$.
Then $p$ is meet irreducible.
== Proof ==
Let $x, y \in S$.
:$p = x \wedge y$ and $p \ne x$ and $p \ne y$
By Meet Precedes Operands:
:$p \preceq x$ and $p \preceq y$
By definition of $\prec$:
:$p \prec x$ and $p \prec y$
We will prove that
:$x \notin F$ or $y \notin F$
:$x \in F \land y \in F$
By definition of filtered:
:$\exists z \in F: z \preceq x \land z \preceq y$
By definition of infimum:
:$z \preceq p$
By definition of upper section:
:$p \in F$
Thus this contradicts $p \in \relcomp S F$ by definition of greatest element.
By definition of relative complement:
:$x \in \relcomp S F$ or $y \in \relcomp S F$
Thus by definition of greatest element:
this contradicts $p \prec x$ and $p \prec y$
== Sources ==
* 
* 
Category:Order Theory
Category:Meet Irreducible Elements
Maximal Element of Nest is Greatest Element

== Theorem ==
Let $A$ be a nest.
Let $x$ be a maximal element of $A$.
Then $x$ is the greatest element of $A$.
== Proof ==
Let $x$ be a maximal element of $A$.
Let $y \in A$ be arbitrary such that $x \ne y$.
We have by definition of maximal element that:
:$x \not \subset y$
and as $x \ne y$:
:$x \nsubseteq y$
But because $A$ is a nest:
:$x \subseteq y$ or $y \subseteq x$
from which it follows that it must be the case that:
:$y \subseteq x$
As $y$ is arbitrary, the result follows by definition of greatest element.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Maximal Elements
Category:Nests
Maximal Element under Subset Relation need not be Greatest Element

== Theorem ==
Let $A$ be a class.
Let $M \in A$ be a maximal element of $A$ under the subset relation
Then $M$ is not necessarily the greatest element of $A$.
== Proof ==
Let $A = \set {x, y}$ such that:
:$x = \set \O$
:$y = \set {\set \O}$
Then:
:$x$ and $y$ are both maximal elements of $A$ by definition.
However:
:$x \not \subseteq y$
and:
:$y \not \subseteq x$
and so neither $x$ nor $y$ are the greatest element of $A$.
== Also see ==
* Maximal Element of Nest is Greatest Element
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Maximal Elements
Category:Greatest Elements
Maximal Ergodic Theorem

== Theorem ==
Let $\struct {X, \BB, \mu, T}$ be a measure-preserving dynamical system.
Let $g : X \to \overline \R$ be a $\mu$-integrable function.
Let $\alpha \in \R$.
Let:
:$\ds E_\alpha := \set {x \in X : \sup_{n \ge 1} \frac 1 n \sum_{i \mathop = 0}^{n-1} \map g {T^i x} > \alpha }$
Then:
:$\ds \alpha \map \mu {E_\alpha} \le \int_{E_\alpha} g \rd \mu$
== Proof ==
Let $\map {L^1} \mu$ be a real-valued $L^1$ space with respect to $\mu$.
Consider $U_T : \map {L^1} \mu \to \map {L^1} \mu$ defined by:
:$f \mapsto f \circ T$
Observe:
where:
:$\ds F_N := \set {\max_{1 \le n \le N} \sum_{i \mathop = 0}^{n-1} \map {U_T^i} {g - \alpha} > 0 }$
Moreover, $F_N \subseteq F_{N+1}$ for all $N$.
Thus:
== Sources ==
* $2.6$ Pointwise Ergodic Theorem
Category:Ergodic Theory
Category:Named Theorems
Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice

== Theorem ==
=== Assertions ===
Then $M$ is a prime ideal.
== Proof ==
 $M$ is not a prime ideal.
Then by Prime Ideal in Lattice, there are elements $a$ and $b$ of $L$ such that
:$a \wedge b \in M$
:$a \notin M$
:$b \notin M$
=== Lemma 1 ===
, we can thus suppose that:
:$\forall m \in M: m \vee a \notin F$
Let $N = \set {x \in L: \exists m \in M: x \le m \vee a}$.
=== Lemma 2 ===
=== Lemma 3 ===
=== Lemma 4 ===
By assuming that $M$ is not a prime ideal, we have constructed an ideal $N$ properly containing $M$ that is disjoint from $F$.
This contradicts the maximality of $M$.
Thus $M$ is a prime ideal.
Category:Order Theory
Category:Lattice Theory
Category:Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice
Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice/Assertions

== Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice Assertions ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a distributive lattice.
Let $F$ be a filter in $L$.
Let $M$ be an ideal in $L$ which is disjoint from $F$ such that:
:no ideal in $L$ larger than $M$ is disjoint from $F$.
Category:Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice
Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice/Lemma 1

== Lemma for Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice ==
There do not exist $m$ and $n$ in $M$ such that $m \vee a \in F$ and $n \vee b \in F$.
== Proof ==
 such exist.
Since:
:$m \vee \paren {n \vee b} \ge n \vee b$
:$n \vee b \in F$
:$F$ is a filter
it follows that:
:$m \vee \paren {n \vee b} \in F$
Applying associativity yields:
:$\paren {m \vee n} \vee b \in F$
By the same argument:
:$\paren {m \vee n} \vee a \in F$
By the definition of a filter:
:$\paren {\paren {m \vee n} \vee b} \wedge \paren {\paren {n \vee m} \vee a} \in F$
Distributing:
:$\paren {m \vee n} \wedge \paren {b \vee a} \in F$
But by assumption:
:$b \vee a \in M$
and by the definition of an ideal:
:$m \vee n \in M$
so again by the definition of an ideal:
:$\paren {m \vee n} \wedge \paren {b \vee a} \in M$
contradicting the supposition that $M$ is disjoint from $F$.
Category:Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice
Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice/Lemma 2

== Lemma for Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice ==
Let $N = \set {x \in L: \exists m \in M: x \le m \vee a}$.
$N$ is an ideal in $L$.
== Proof ==
Let:
:$x \in N$
:$y \in L$
:$y \le x$
Then by the definition of $N$ there exists an $m \in M$ such that:
:$x \le m \vee a$
Since $y \le x$, it follows that:
:$y \le m \vee a$
so $y \in N$.
Let:
:$x \in N$
:$y \in N$
Then there exist $m_x$ and $m_y$ in $M$ such that:
:$x \le m_x \vee a$
:$y \le m_y \vee a$
Then:
:$x \vee y \le \paren {m_x \vee a} \vee \paren {m_y \vee a} = \paren {m_x \vee m_y} \vee a$
But $m_x \vee m_y \in M$, so:
:$x \vee y \in N$
Category:Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice
Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice/Lemma 3

== Lemma for Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice ==
Let $N = \set {x \in L: \exists m \in M: x \le m \vee a}$.
$M \subsetneq N$
== Proof ==
Let $m \in M$.
Then:
:$m \le \paren {m \vee a}$
so $m \in N$.
Thus $M \subseteq N$.
We have:
:$a \le \paren {m \vee a}$
so:
:$a \in N$
but:
:$a \notin M$
Thus:
:$M \subsetneq N$
Category:Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice
Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice/Lemma 4

== Lemma for Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice ==
Let $N = \set {x \in L: \exists m \in M: x \le m \vee a}$.
Then we have that:
:$N \cap F = \O$
== Proof ==
 $x \in N \cap F$.
Then:
:$x \in N$
so for some $m \in M$:
:$x \le m \vee a$
Furthermore, $x \in F$.
So by the definition of a filter:
:$m \vee a \in F$
But this contradicts our assumption that $\forall m \in M: m \vee a \notin F$.
Category:Maximal Ideal WRT Filter Complement is Prime in Distributive Lattice
Maximal Ideal iff Quotient Ring is Field

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
The following are equivalent:
:$(1): \quad$ $J$ is a maximal ideal.
:$(2): \quad$ The quotient ring $R / J$ is a field.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Maximal Left and Right Ideal iff Quotient Ring is Division Ring
== Sources ==
* : $\S 2$: Exercise $12$
* : $\S 1.8$ Algebraic properties of $p$-adic integers
Category:Quotient Rings
Category:Maximal Ideal iff Quotient Ring is Field
Category:Maximal Ideals of Rings
Category:Field Theory
Maximal Ideal iff Quotient Ring is Field/Proof 1

== Theorem ==
== Proof Outline ==
The hard part is proving the existence of inverses.
Take an element $x$ of the Ring to invert.
Define a set $K$, that contains the Ideal and is contained by the ring.
It is the set of all members of the Ideal, each added to a multiple of $x$.
Prove that this set is an ideal and contains our original maximal ideal.
== Proof ==
=== Maximal Ideal implies Quotient Ring is Field ===
=== Quotient Ring is Field implies Ideal is Maximal ===
Category:Maximal Ideal iff Quotient Ring is Field
Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let $J$ be a maximal ideal.
Then the quotient ring $R / J$ is a field.
== Proof ==
Let $J$ be a maximal ideal.
Because $J \subset R$, it follows from Quotient Ring of Commutative Ring is Commutative and Quotient Ring of Ring with Unity is Ring with Unity that $R / J$ is a commutative ring with unity.
We now need to prove that every non-zero element of $\struct {R / J, +, \circ}$ has a product inverse in $R / J$.
Let $x \in R$ such that $x + J \ne J$, that is: $x \notin J$.
Thus $x + J \in R / J$ is not the zero element of $R / J$.
Take $K \subseteq R$ such that:
:$K = \set {j + r \circ x: j \in J, r \in R}$
that is, the subset of $R$ which can be expressed as a sum of an element of $J$ and a product in $R$ of $x$.
Now $0_R \in K$ as $0_R \in J$ and $0_R \in R$, giving $0_R + 0_R \circ x = 0_R$.
So:
:$(1): \quad K \ne \O$
Now let $g, h \in K$.
That is:
:$g = j_1 + r_1 \circ x, h = j_2 + r_2 \circ x$
Then:
:$-h = -j_2 + \paren {-r_2} \circ x$
But $j_1 - j_2 \in J$ from Test for Ideal.
Similarly $-r_2 \in R$.
So $-h \in K$ and we have:
:$(2) \quad g + \paren {-h} = \paren {j_1 - j_2} + \paren {r_1 - r_2} \circ x$
Now consider $g \in J, y \in R$.
Then:
:$g \circ y = \paren {j_1 + r_1 \circ x} \circ y = \paren {j_1 \circ y} + \paren {r_1 \circ y} \circ x$
which is valid by the fact that $R$ is a commutative ring.
But as $J$ is an ideal:
:$\paren {j_1 \circ y} \in J$
while $r_1 \circ y \in R$.
Thus:
:$(3) \quad g \circ y \in K$
and similarly:
:$(3) \quad y \circ g \in K$
So Test for Ideal can be applied to statements $(1)$ to $(3)$, and it is seen that $K$ is an ideal of $R$.
Now:
and because $x = 0_R + 1_R \circ x$ (remember $0_R \in J$), then $x \in K$ too.
So, because $x \notin J$, $K$ is an ideal such that $J \subset K \subseteq R$.
Because $J$ is a maximal ideal, then $K = R$.
Thus $1_R \in K$ and thus:
:$\exists j_0 \in J, s \in R: 1_R = j_0 + s \circ x$
So:
:$1_R + \paren {-s \circ x} = j_0 \in J$
Hence:
:$1_R + J = s \circ x + J = \paren {s + J} \circ \paren {x + J}$
So in the commutative ring $\struct {R / J, +, \circ}$, the product inverse of $x + J$ is $s + J$.
The result follows.
== Sources ==
* : $\S 63.1$ Construction of fields as factor rings
Category:Maximal Ideal iff Quotient Ring is Field
Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $J$ be an ideal of $R$.
Let the quotient ring $R / J$ be a field.
Then $J$ is a maximal ideal.
== Proof ==
Let $R / J$ be a field.
Let $K$ be a left ideal of $R$ such that $J \subsetneq K \subseteq R$.
We have that $J$ is the zero of $R / J$.
Let $x \in K \setminus J$.
Because $x \notin J$ then $x + J \ne J$.
Because $R / J$ is a field then $x + J \in R / J$ has a product inverse, say $s + J$.
Hence:
:$1_R + J = \paren {s + J} \circ \paren {x + J} = \paren {s \circ x } + J$
By Left Cosets are Equal iff Product with Inverse in Subgroup:
:$1_R - s \circ x \in J \subsetneq K$
By the definition of an ideal:
:$x \in K, s \in R \implies s \circ x \in K$
:$1_R - s \circ x \in K, s \circ x \in K \implies \paren {1_R - s \circ x} + \paren {s \circ x} = 1_R \in K$
:$1_R \in K \implies \forall y \in R: y \circ 1_R = y \in K$
Hence $K = R$.
The result follows.
== Sources ==
* : Chapter $9$: Rings: Exercise $18$
Category:Maximal Ideal iff Quotient Ring is Field
Maximal Ideal iff Quotient Ring is Field/Proof 2

== Theorem ==
== Proof ==
Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$.
Let the ordered set $\struct {\map {\mathbb L} {R / J}, \subseteq}$ be the set of all ideals of $R / J$.
Let the mapping $\Phi_J: \struct {\mathbb L_J, \subseteq} \to \struct {\map {\mathbb L} {R / J}, \subseteq}$ be defined as:
:$\forall a \in \mathbb L_J: \map {\Phi_J} a = \map {q_J} a$
where $q_J: a \to a / J$ is the quotient epimorphism from $a$ to $a / J$.
Then from Ideals Containing Ideal Isomorphic to Quotient Ring, $\Phi_J$ is an isomorphism.
Now from Quotient Ring Defined by Ring Itself is Null Ring, $\map {q_J} J$ is the null ideal of $R / J$.
At the same time, $\map {q_J} R$ is the entire ring $R / J$.
Let $R / J$ not be the null ring.
Then by Epimorphism Preserves Rings and Epimorphism Preserves Commutativity, $R / J$ is a commutative ring with unity.
By definition, $J$ is a maximal ideal of $R$ $\mathbb L_J = \set {J, R}$ and $J$ is a proper ideal of $R$.
By Ideals of Field, $R / J$ is a field :
:$\map {\mathbb L} {R / J} = \set {\map {q_J} J, \map {q_J} R}$
and the null ideal $\map {q_J} J$ is a proper ideal of $R / J$.
As $\Phi_J: \mathbb L_J \to \map {\mathbb L} {R / J}$ is an isomorphism, $J$ is a maximal ideal $J$ is a field.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.6$
Category:Maximal Ideal iff Quotient Ring is Field
Maximal Ideal iff Quotient Ring is Field/Proof 3

== Theorem ==
== Proof ==
Let $J$ be a maximal ideal.
We have by definition of quotient ring that $J$ is the zero element of $R / J$.
Let $A \in R / J$ be a non-zero element of $R / J$.
Let $x \in A$.
Since $A \ne J$, we have that $x \notin J$.
Let the ideal $K = J + A$ of $R$ be formed.
This contains all the elements of the form $j + r a$, with $j \in J$ and $r \in R$.
As $J$ is maximal and $J \subsetneq K$, it follows that:
:$K = R$
and so:
:$1_R \in K$
That is:
:$\exists j \in J, r \in R: j + r a = 1_R$
Thus:
:$\paren {r + J} \paren {a + J} = \paren {1 - u} J = 1_R + J$
and so $\paren {r + J}$ is the product inverse of $\paren {a + J}$.
So every non-zero element of $R / J$ has a product inverse.
That is, $R / J$ is a field.
Let $R / J$ be a field.
Let $K$ be an ideal of $R$ such that:
:$J \subsetneq K \subseteq R$
Let $a \in K$ such that $a \notin J$.
Then:
:$J + \ideal a \subsetneq K$
But as $a \notin J$, we have that $a + J$ is a non-zero element of $R / J$.
Thus as $R / J$ is a field, $a + J$ has a product inverse $r + J$:
:$\paren {r + J} \paren {a + J} = 1_R + J$
So:
:$\exists r \in R, j \in J: r a + \paren {-1_R} = j$
That is:
:$r a + \paren {-1_R} \in J$
So:
:$j + r a = 1_R$
and from Ideal of Unit is Whole Ring: Corollary this implies:
:$J + \ideal a = R$
So:
:$K = R$
and it follows by definition that $J$ is maximal.
== Sources ==
* : Chapter $5$: Rings: $\S 23$. Maximal Ideals: Theorem $43$
Category:Maximal Ideal iff Quotient Ring is Field
Maximal Ideal in Unital Banach Algebra is Closed

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.
Let $I$ be a maximal ideal of $A$.
Then $I$ is closed.
== Proof ==
From Closure of Proper Ideal in Unital Banach Algebra is Proper Ideal, the closure $I^-$ is a proper ideal of $A$ with $I \subseteq I^-$.
Since $I$ is a maximal ideal, we have $I = I^-$.
From Set is Closed iff Equals Topological Closure, we conclude that $I$ is closed.
Category:Banach Algebras
Maximal Ideal of Commutative and Unitary Ring is Prime Ideal

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $M$ be a maximal ideal of $R$.
Then $M$ is a prime ideal of $R$.
== Proof ==
From Maximal Ideal iff Quotient Ring is Field:
:the quotient ring $R / M$ is a field.
It follows from Field is Integral Domain that $R / M$ is an integral domain.
Finally it follows from Prime Ideal iff Quotient Ring is Integral Domain that $M$ is a prime ideal.
== Sources ==
* : Chapter $9$: Rings: Exercise $13$
Category:Maximal Ideals of Rings
Category:Prime Ideals of Commutative and Unitary Rings
Maximal Ideal of Division Ring

== Theorem ==
Let $\struct {D, +, \circ}$ be a Division Ring whose zero is $0$.
Let $\struct {J, +, \circ}$ be a maximal ideal of $D$.
Then:
:$J = \set 0$
== Proof ==
From Ideals of Division Ring, the only ideals of a Division Ring $\struct {D, +, \circ}$ are $\struct {D, +, \circ}$ and $\struct {\set 0, +, \circ}$.
Hence the result by definition of maximal ideal.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
Category:Ideal Theory
Category:Division Rings
Maximal Ideal of Unital Commutative Banach Algebra is Kernel of Character

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a commutative unital Banach algebra over $\C$. 
Let $M$ be a maximal ideal of $A$. 
Then there exists a character $\phi$ such that:
:$\ker \phi = M$ 
== Proof ==
From Maximal Ideal in Unital Banach Algebra is Closed, $M$ is closed.
Let $A/M$ be the quotient algebra of $A$ modulo $M$.
From Quotient Normed Algebra of Unital Normed Algebra is Unital Normed Algebra, $A/M$ is a unital normed algebra. 
From Quotient Algebra of Commutative Algebra is Commutative, $A/M$ is a commutative algebra.
Hence from the Gelfand-Mazur Theorem, there exists an isometric algebra isomorphism $q_2 : A/M \to \C$.
From Quotient Mapping on Quotient Algebra is Algebra Homomorphism, $q_1 : A \to A/M$ defined by:
:$\map {q_1} x = x + M$
for each $x \in A$ is a algebra homomorphism.
From Composition of Algebra Homomorphisms is Algebra Homomorphism, $q_2 \circ q_1 : A \to \C$ is an algebra homomorphism.
Since $q_2$ is an algebra isomorphism, for any $x + M \in A/M \setminus \set { {\mathbf 0}_{A/M} }$, we have $\map {q_2} {x + M} = \map {q_2} {\map {q_1} x} \ne 0$.
Hence $q_2 \circ q_1 : A \to \C$ is a non-zero algebra homomorphism.
Hence $\phi = q_2 \circ q_1$ is a character on $A$.
We just need to assure ourselves that $\ker \phi = M$. 
Since $q_2$ is an algebra isomorphism we have $\ker {q_2} = \set { {\mathbf 0}_A + M}$.
Hence if $\map {q_2} {\map {q_1} x} = 0$ for $x \in A$ we have $\map {q_1} x = {\mathbf 0}_{A/M}$. 
From Kernel of Quotient Mapping, we then have $x \in M$.
Conversely, if $x \in M$ then $\map {q_1} x = 0$ so $\map \phi x = 0$. 
So $\ker \phi = M$ as desired. 
Category:Unital Banach Algebras
Category:Characters (Banach Algebras)
Category:Commutative Banach Algebras
Maximal Inequality for Positive Operators

== Theorem ==
Let $\struct {X, \BB, \mu}$ be a probability space.
Let $\map {L^1} \mu$ be a real-valued $L^1$ space with respect to $\mu$.
Let $U : \map {L^1} \mu \to \map {L^1} \mu$ be a positive linear operator, that is:
:$\forall f \in \map {L^1} \mu : f \ge 0 \implies U f \ge 0$
Suppose:
:$\norm U \le 1$
where $\norm \cdot$ denotes the operator norm.
Then, for all $N \in \N_{>0}$:
:$\ds \int_{\set {F_N > 0} } f \rd \mu \ge 0$
where:
:$F_N := \max \set {f_n : 1 \le n \le N}$
and:
:$f_n := f + U f + \cdots + U^{n-1} f$
== Proof ==
Let:
:$\map {G_N} x := \max \set {\map {f_n} x : 0 \le n \le N}$
where $f_0 := 0$.
Let $0 \le n \le N$.
:$U G_N + f \ge U f_n + f = f_{n+1}$
In particular:
:$\paren 1 :\quad U G_N + f \ge F_N$
Let $x \in X$.
If $\map {F_N} x > 0$, then:
Further, from $G_N \ge 0$ follows:
:$\paren 3 :\quad U G_N \ge 0$ 
Therefore:
== Sources ==
* $2.6$ Pointwise Ergodic Theorem
Category:Ergodic Theory
Category:Operator Theory