nyuuzyou/wikis huggingface dataset:
Field of Prime Characteristic has Unique Prime Subfield

== Theorem ==
Let $F$ be a field whose characteristic is $p$.
Then there exists a unique $P \subseteq F$ such that:
:$(1): \quad P$ is a subfield of $F$
:$(2): \quad P \cong \Z_p$.
That is, $P \cong \Z_p$ is a unique minimal subfield of $F$, and all other subfields of $F$ contain $P$.
This field $P$ is called the prime subfield of $F$.
== Proof ==
Let $\struct {F, +, \times}$ be a field whose unity is $1_F$ such that $\Char F = p$.
Let $P$ be a prime subfield of $F$.
From Field has Prime Subfield, this has been shown to exist.
We can consistently define a mapping $\phi: \Z_p \to F$ by:
:$\forall n \in \Z_p: \map \phi {\eqclass n p} = n \cdot 1_F$
Suppose $a, b \in \eqclass n p$.
Then:
:$a = n + k_1 p, b = n + k_2 p$
So:
and similarly for $b$, showing that $\phi$ is well-defined.
Let $C_a, C_b \in \Z_p$.
Let $a \in C_a, b \in C_b$ such that $a = a' + k_a p, b = b' + k_b p$.
Then:
Similarly for $\map \phi {C_a} \times \map \phi {C_b}$.
So $\phi$ is a ring homomorphism.
From Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi$ is a ring monomorphism.
Thus it follows that $P = \Img \phi$ is a subfield of $F$ such that $P \cong \Z_p$.
Let $K$ be a subfield of $F$.
let $P = \Img \phi$ as defined above.
We know that $1_F \in K$.
It follows that $1_F \in K \implies P \subseteq K$.
Thus $K$ contains a subfield $P$ such that $P$ is isomorphic to $\Z_p$.
The uniqueness of $P$ follows from the fact that if $P_1$ and $P_2$ are both minimal subfields of $F$, then $P_1 \subseteq P_2$ and $P_2 \subseteq P_1$, thus $P_1 = P_2$.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms: Theorem $3.2$
* : $\S 61$. Characteristic of an integral domain or field
Category:Prime Fields
Field of Quotients is Divisible Module

== Theorem ==
Let $D$ be an integral domain.
Let $\map {\operatorname {Quot} } D$ be the field of quotients of $D$.
Then $\map {\operatorname {Quot} } D$ is a divisible $D$-module.
== Proof ==
Let $a \in D$ be a non zero divisor.
Let $x, y \in D$ such that $y \ne 0$.
Then $\dfrac x y \in \map {\operatorname {Quot} } D$.
By definition of integral domain:
:$a \ne 0$
Thus $\dfrac x {a y}$ is defined in $\map {\operatorname {Quot} } D$.
It follows that:
:$a \cdot \dfrac x {a y} = \dfrac x y$
Thus $\map {\operatorname {Quot} } D$ is a divisible $D$-module.
Category:Integral Domains
Category:Module Theory
Category:Homological Algebra
Field of Quotients is Unique

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain.
Let $K, L$ be field of quotients of $\struct {D, +, \circ}$.
Then there is one and only one (field) isomorphism $\phi: K \to L$ satisfying:
:$\forall x \in D: \map \phi x = x$
== Proof ==
Follows directly from the Quotient Theorem for Monomorphisms.
== Motivation ==
It follows from this result that when discussing an integral domain $\struct {D, +, \circ}$, all we need to do is select any particular field of quotients $K$ of $D$, and call $K$ the field of quotients of $D$.
If $D$ is already a subdomain of a specified field $L$, then the field of quotients selected will usually be the subfield of $L$ consisting of all the elements $x / y$ where $x \in D, y \in D^*$ (see Field of Quotients of Subdomain).
This is also clearly the subfield of $L$ generated by $D$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.10$: Corollary
Category:Fields of Quotients
Category:Field Isomorphisms
Category:Integral Domains
Field of Quotients of Ring of Polynomial Forms on Reals that yields Complex Numbers

== Theorem ==
Let $\struct {\R, +, \times}$ denote the field of real numbers.
Let $X$ be transcendental over $\R$.
Let $\R \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Consider the field of quotients:
:$\R \sqbrk X / \ideal p$
where:
:$p = X^2 + 1$
:$\ideal p$ denotes the ideal generated by $p$.
Then $\R \sqbrk X / \ideal p$ is the field of complex numbers.
== Proof ==
It is taken as read that $X^2 + 1$ is irreducible in $\R \sqbrk X$.
Hence by Polynomial Forms over Field form Principal Ideal Domain: Corollary 1, $\R \sqbrk X / \ideal p$ is indeed a field.
Let $\nu$ be the quotient epimorphism from $\R \sqbrk X$ onto $\R \sqbrk X / \ideal p$.
From Quotient Ring Epimorphism is Epimorphism:
:$\map \ker {\map \nu \R} = \R \cap \ideal p = \set 0$
So $\map \nu \R$ is a monomorphism from $\R \sqbrk X$ to $\R \sqbrk X / \ideal p$.
Thus $\map \nu \R$ is an isomorphic copy of $\R$ inside $\R \sqbrk X / \ideal p$.
We identify this isomorphic copy of $\R$ with $\R$ itself, ignoring the difference between $\map \nu x$ and $x$ when $x = \R$.
Hence:
:$(1): \quad \R \subseteq \R \sqbrk X / \ideal p$
Let $f \in \R \sqbrk X$ be arbitrary.
By Division Theorem for Polynomial Forms over Field:
:$\exists q, r \in \R \sqbrk X: f = q p + r$
where $r =a + b X$ for some $a, b \in \R$.
Hence:
As $\nu$ is an epimorphism, it is a fortiori surjection.
Hence:
:$(2): \quad$ Every element $w$ of $\R \sqbrk X / \ideal p$ can be expressed in the form:
::$w = a + b i$
:for some $a, b \in \R$.
Because $X^2 + 1 \in \ker \nu$, we have:
Hence we have that:
:$(3): \quad$ In $\R \sqbrk X / \ideal p$, $i^2 = -1$
Thus $(2)$ can be improved to:
:$(4): \quad$ Every element $w$ of $\R \sqbrk X / \ideal p$ can be expressed uniquely in the form:
::$w = a + b i$
:for some $a, b \in \R$.
From $(1)$, $(3)$ and $(4)$, the field $\R \sqbrk X / \ideal p$ is recognised as the field of complex numbers.
== Sources ==
* : $\S 65$. Some properties of $F \sqbrk X$, where $F$ is a field
Category:Polynomial Theory
Category:Fields of Quotients
Category:Real Numbers
Category:Complex Numbers
Field of Quotients of Subdomain

== Theorem ==
Let $\struct {F, +, \circ}$ be a field whose unity is $1_F$.
Let $\struct {D, +, \circ}$ be a subdomain of $\struct {F, +, \circ}$ whose unity is $1_D$.
Let:
:$K = \set {\dfrac x y: x \in D, y \in D^*}$
where $\dfrac x y$ is the division product of $x$ by $y$.
Then $\struct {K, +, \circ}$ is a field of quotients of $\struct {D, +, \circ}$.
== Proof ==
$1_D = 1_F$ by Subdomain Test.
The sum and product of two elements of $K$ are also in $K$ by Addition of Division Products and Product of Division Products.
The additive and product inverses of $K$ are also in $K$ by Negative of Division Product and Inverse of Division Product.
Thus by Subfield Test, $\struct {K, +, \circ}$ is a subfield of $\struct {F, +, \circ}$ which clearly contains $\struct {D, +, \circ}$.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers: Theorem $23.8$
Category:Fields of Quotients
Field of Rational Functions is Field

== Theorem ==
Let $K$ be a field.
Let $K \sqbrk x$ be the integral domain of polynomial forms on $K$.
Let $\map K x$ be the field of rational functions on $K$.
Then $\map K x$ forms a field.
If the characteristic of $K$ is $p$, then the characteristic of $\map K x$ is non-zero.
== Proof ==
== Comment ==
Thus we see that despite Characteristic of Finite Ring is Non-Zero (and by implication that of a finite field), it is not necessarily the case that the characteristic of an infinite field is zero.
== Sources ==
* : Chapter $4$: Fields: $\S 17$. The Characteristic of a Field: Example $26$
Category:Field Theory
Category:Polynomial Theory
Field of Uncountable Cardinality K has Transcendence Degree K

== Theorem ==
Let $F$ be a field of uncountable cardinality $\kappa$.
Then $F$ has transcendence degree $\kappa$ over its prime field.
== Proof ==
We prove the theorem for fields with characteristic $p = 0$.
In the case where $p$ is a prime, the proof is similar, but instead we view the fields as extensions of $\Z / \Z_p$.
The main idea is to exploit the lower cardinality of sets of polynomials in order to keep finding algebraically independent elements of $F$.
Since each characteristic $0$ field contains a copy of $\Q$ as its prime field, we can view $F$ as a field extension over $\Q$.
We will show that $F$ has a subset of cardinality $\kappa$ which is algebraically independent over $\Q$.
Since $\kappa$ is the largest possible cardinality for a subset of $F$, this will establish the theorem.
We build the claimed subset of $F$ by transfinite induction and implicit use of the axiom of choice.
For each ordinal $\alpha < \kappa$ we define a set $S_\alpha$.
We will build the sets so that each $S_\alpha$ has cardinality equal to that of $\alpha$ and is algebraically independent over $\Q$.
Let $S_0 = \O$.
Let $S_1$ be a singleton containing some element of $F$ which is not algebraic over $\Q$.
This is possible from Algebraic Numbers are Countable.
Define $S_\beta$ for successor ordinals $\beta = \alpha + 1 < \kappa$ to be $S_\alpha$ together with an element of $F$ which is not a root of any non-trivial polynomial with coefficients in $\Q \cup S_\alpha$.
From Set of Polynomials over Infinite Set has Same Cardinality there are only $\card {\Q \cup S_\alpha} = \aleph_0 + \card \alpha < \kappa$ many such polynomials.
hence the above construction is possible.
From Polynomial over Field has Finitely Many Roots and the Cardinality of Infinite Union of Infinite Sets, the union of $\kappa$ many finite sets is at most size $\kappa$.
Thus there are are this many roots of such polynomials
Define $S_\beta$ for limit ordinals by $\ds S_\beta = \bigcup_{\alpha \mathop < \beta} S_\alpha$.
We can then define $S_\kappa$ to be $\ds \bigcup_{\alpha \mathop < \kappa} S_\alpha$.
This is a set of size $\kappa$ since its elements can be paired with the ordinals less than $\kappa$. 
It is now demonstrated that it is algebraically independent over $\Q$.
Let $\map P {x_1, \ldots, x_n}$ be a non-trivial polynomial with coefficients in $\Q$ and elements $a_1, \ldots, a_n$ in $F$. 
, it is assumed that $a_n$ was added at an ordinal $\alpha + 1$ later than the other elements. 
Then $\map P {a_1, \ldots, a_{n - 1}, x_n}$ is a polynomial with coefficients in $\Q \cup S_\alpha$. 
$a_n$ was selected at stage $\alpha + 1$ so as not to be root of such a polynomial.
So $\map P {a_1, \ldots, a_{n - 1}, x_n} \ne 0$.
Thus it is algebraically independent over $\Q$.
Category:Field Theory
Category:Uncountable Sets
Field with 4 Elements has only Order 2 Elements

== Theorem ==
Let $\struct {\GF, +, \times}$ be a field which has exactly $4$ elements.
Then:
:$\forall a \in \GF: a + a = 0_\GF$
where $0_\GF$ is the zero of $\GF$.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Galois Field of Order $4$, where it is demonstrated that such an $\GF$:
::$\struct {\GF, +} \cong \Z_2 \times \Z_2$ and $\struct {\GF^*, \times} \cong \Z_3$
:is actually a field.
== Sources ==
* : $\S 2.4$: The rational numbers and some finite fields: Exercise $5$
Category:Galois Fields
Category:Field with 4 Elements has only Order 2 Elements
Field with 4 Elements has only Order 2 Elements/Proof 1

== Theorem ==
== Proof ==
By definition of field, both the algebraic structures $\struct {\GF, +}$ and $\struct {\GF^*, \times}$ are (abelian) groups, where $\GF^* := \GF \setminus \set 0$.
By definition:
:$\struct {\GF, +}$ is of order $4$
:$\struct {\GF^*, \times}$ is of order $3$.
From Classification of Groups of Order up to 15, there are only two possibilities:
:$(1): \quad \struct {\GF, +} \cong \Z_4$ and $\struct {\GF^*, \times} \cong \Z_3$
:$(2): \quad \struct {\GF, +} \cong \Z_2 \times \Z_2$ and $\struct {\GF^*, \times} \cong \Z_3$.
In case $(2)$:
:$\forall a \in \GF: a + a = 0_\GF$
but in case $(1)$:
:$\exists a, b \in \GF: a + a = b \ne 0_\GF$ where $b + b = 0_\GF$
as there exists an element of $\struct {\GF, +}$ of order $4$.
 $(1)$ describes a field.
Then:
But then:
As $\struct {\GF, +} \cong \Z_4$, both $a + a = b$ and $\paren {-a} + \paren {-a} = b$.
The only way for $\paren {b \times b} + \paren {b \times b} = 0_\GF$ is for $b \times b = b$ or $b \times b = 0_\GF$.
The second case is eliminated as $0_\GF \notin \struct {\GF^*, \times}$.
So it must be the case that $b \times b = b$.
So as $\struct {\GF^*, \times} \cong \Z_3$ it must follow that $a \times b = a, a \times \paren {-a} = b$ and then $a \times a = -a$.
It follows that the Cayley tables of $\struct {\GF, +}$ and $\struct {\GF^*, \times}$ must therefore be as follows:
:$\begin{array}{c|cccc}
+ & 0_\GF & a & b & -a \\
\hline
0_\GF & 0_\GF & a & b & -a \\
a & a & b & -a & 0_\GF \\
b & b & -a & 0_\GF & a \\
-a & -a & 0_\GF & a & b \\
\end{array}
\qquad
\begin{array}{c|cccc}
\times & 0_\GF & b & a & -a \\
\hline
0_\GF & 0_\GF & 0_\GF & 0_\GF & 0_\GF \\
b & 0_\GF & b & a & -a \\
a & 0_\GF & a & -a & b \\
-a & 0_\GF & -a & b & a \\
\end{array}$
But:
So:
:$\paren {a + b} \times a \ne \paren {a \times a} + \paren {b \times a}$
demonstrating that $\times$ is not distributive over $+$.
Thus $\GF$ as has been defined:
:$\struct {\GF, +} \cong \Z_4$ and $\struct {\GF^*, \times} \cong \Z_3$
is not a field.
It follows that our supposition that $\struct {\GF, +} \cong \Z_4$ was false.
Thus, if $\GF$ is a field, then $\struct {\GF, +} \cong \Z_2 \times \Z_2$.
That is:
:$\forall a \in \GF: a + a = 0_\GF$
as needed to be proved.
Category:Field with 4 Elements has only Order 2 Elements
Field with 4 Elements has only Order 2 Elements/Proof 2

== Theorem ==
== Proof ==
Let $\struct {\GF, +, \times}$ be a field of order $4$ whose zero is $0_\GF$.
By definition, $\GF$ is a Galois field.
The additive group $\struct {\GF, +}$ of $\GF$ can be one of two:
:$(1): \quad$ the cyclic group of order $4$, generated by the identity of $\struct {\GF, +}$ which is $0_\GF$
or:
:$(2): \quad$ the Klein $4$-group, whose elements are all of the form $a + a = 0_\GF$.
 $(1)$ is the additive group of a field.
Then the characteristic of $\GF$ is $4$.
But from Characteristic of Galois Field is Prime it follows that $\GF$ is not a field.
From that contradiction it follows that the additive group $\struct {\GF, +}$ of $F$ cannot be the cyclic group of order $4$.
Hence for $\GF$ to be a field at all, it is necessary for $\struct {\GF, +}$ to be the Klein $4$-group.
Category:Field with 4 Elements has only Order 2 Elements
Fifth Apotome/Example

== Example of Fifth Apotome==
Let $a, b \in \set {x \in \R_{>0} : x^2 \in \Q}$ be two rationally expressible numbers such that $a - b$ is an apotome.
By definition, $a - b$ is a fifth apotome :
:$(1): \quad b \in \Q$
:$(2): \quad \dfrac {\sqrt {a^2 - b^2}} a \notin \Q$
where $\Q$ denotes the set of rational numbers.
Let $a = \sqrt {13}$ and $b = 3$.
Then:
Therefore $\sqrt {13} - 3$ is a fifth apotome.
Category:Examples of Apotome
Fifth Power expressible as Sum of 4 Fifth Powers

== Theorem ==
$61 \, 917 \, 364 \, 224$ can be expressed as the sum of $4$ fifth powers:
:$61 \, 917 \, 364 \, 224 = 144^5 = 27^5 + 84^5 + 110^5 + 133^5$
== Proof ==
== Sources ==
* : $61,917,364,224$
* : $61,917,364,224$
Category:Fifth Powers
Category:61,917,364,224
Fifth Power of Complex Number

== Theorem ==
Let $z = a + i b$ be a complex number.
Then its fifth power is given by:
:$z^5 = a^5 - 10 a^3 b^2 + 5 a b^4 + i \paren {5 a^4 b - 10 a^2 b^3 + b^5}$
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.21$
Category:Complex Numbers
Category:Fifth Powers
Fifth Root/Examples/2

== Decimal Expansion ==
The decimal expansion of the $5$th root of $2$ starts:
:$\sqrt [5] 2 \approx 1 \cdotp 14869 \, 83554 \, 99703 \, 50067 \, 986 \ldots$
== Sources ==
* : $\S 1$: Special Constants: $1.8$
Category:Examples of Fifth Roots
Category:2
Fifth Root/Examples/3

== Decimal Expansion ==
The decimal expansion of the $5$th root of $3$ starts:
:$\sqrt [5] 3 \approx 1 \cdotp 24573 \, 09396 \, 15517 \, 32596 \, 668 \ldots$
== Sources ==
* : $\S 1$: Special Constants: $1.9$
Category:Examples of Fifth Roots
Category:3
Fifth Sylow Theorem

== Theorem ==
The number of Sylow $p$-subgroups of a finite group is a divisor of their common index.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
Some sources call this the fourth Sylow theorem and merge it with what we call the Fourth Sylow Theorem.
Others merge this result with what we call the Fourth Sylow Theorem and call it the third Sylow theorem.
Others merge this with what we call the Third Sylow Theorem and call it the third Sylow theorem.
== Also see ==
* Sylow Theorems
== Historical Note ==
Category:Sylow Theorems
Category:Fifth Sylow Theorem
Fifth Sylow Theorem/Proof 1

== Theorem ==
== Proof ==
By the Orbit-Stabilizer Theorem, the number of conjugates of $P$ is equal to the index of the normalizer $\map {N_G} P$.
Thus by Lagrange's Theorem, the number of Sylow $p$-subgroups divides $\order G$.
Let $m$ be the number of Sylow $p$-subgroups, and let $\order G = k p^n$.
 
From the Fourth Sylow Theorem, $m \equiv 1 \pmod p$.
So it follows that $m \nmid p \implies m \nmid p^n$.
Thus $m \divides k$ which is the index of the Sylow $p$-subgroups in $G$.
== Sources ==
* : Chapter $11$: The Sylow Theorems: Corollary $11.11$
Category:Fifth Sylow Theorem
Fifth Sylow Theorem/Proof 2

== Theorem ==
== Proof ==
Let $G$ be a finite group of order $p^n m$, where $p \nmid m$ and $n > 0$.
Let $r$ be the number of Sylow $p$-subgroups of $G$.
It is to be shown that $r \divides m$.
Let $H$ be a Sylow $p$-subgroup of $G$.
We have that:
:$\order H = p^n$
:$\index G H = m$
Let $S_1, S_2, \ldots, S_m$ denote the elements of the left coset space of $G / H$.
We have that $G$ acts on $G / H$ by the rule:
:$g * S_i = g S_i$
for $S_i \in G / H$.
There is only one orbit under this action, namely the whole of $G / H$.
Therefore the stabilizer of each $S_i$ is a subgroup of $G$ of index $m$ and order $p^n$.
In other words, each $S_i$ has a Sylow $p$-subgroup as a stabilizer.
Now it is shown that each Sylow $p$-subgroup is the stabilizer of one or more of the cosets $S_1, S_2, \ldots, S_m$.
We have that $H$ is the stabilizer of the coset $H$, which must be one of $S_1, S_2, \ldots, S_m$.
Let $S_1, S_2, \ldots, S_k$ be the elements of $G / H$ whose stabilizer is $H$.
By the Third Sylow Theorem, any other Sylow $p$-subgroup of $H$ is a conjugates $g H g^{-1}$ of $H$.
Thus it is seen that $g H g^{-1}$ is a stabilizer of the cosets $g S_1, g S_2, \ldots, g S_k$.
So each of the $r$ distinct Sylow $p$-subgroups of $G$ is the stabilizer of exactly $k$ elements of $G / H$.
Thus:
:$m = k r$
and so:
:$r \divides m$
as required.
== Sources ==
* : Chapter $2$: The Sylow Theorems: $\S 58$: Third Sylow Theorem
Category:Fifth Sylow Theorem
Figures in Perspective from Line are in Perspective from Point

== Theorem ==
Let $A$ and $B$ be plane geometric figures.
Let $A$ and $B$ be in perspective from a line.
Then $A$ and $B$ are also in perspective from a point.
== Proof ==
== Also see ==
* Figures in Perspective from Point are in Perspective from Line
== Sources ==
* 
* 
Category:Perspective
Figures in Perspective from Point are in Perspective from Line

== Theorem ==
Let $A$ and $B$ be plane geometric figures.
Let $A$ and $B$ be in perspective from a point.
Then $A$ and $B$ are also in perspective from a line.
== Proof ==
== Also see ==
* Figures in Perspective from Line are in Perspective from Point
== Sources ==
* 
* 
Category:Perspective
Filter Basis Generates Filter

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\BB \subset \powerset S$.
Then:
:$\FF = \set {V \subseteq S: \exists U \in \BB: U \subseteq V}$ is a filter on $S$
:
:$(1): \quad \forall V_1, V_2 \in \BB: \exists U \in \BB: U \subseteq V_1 \cap V_2$
:$(2): \quad \O \notin \BB, \BB \ne \O$
That is, $\BB$ is a filter basis of the filter $\FF$, which is generated by $\BB$.
== Proof ==
Assume first that $\FF$ is a filter on $S$.
Then $S \in \FF$ and thus $\BB \ne \O$.
Because $\O \notin \FF$ we have that $\O \notin \BB$, since $\BB \subseteq \FF$.
Let $V_1, V_2 \in \BB$.
Then:
:$V_1, V_2 \in \FF$
Because $\FF$ is a filter:
:$V := V_1 \cap V_2 \in \FF$
The definition of $\FF$ implies therefore that there is $U \in \BB$ such that $U \subseteq V = V_1 \cap V_2$.
Assume now that $\BB$ satisfies conditions $(1)$ and $(2)$.
To show that $\FF$ is a filter, note that because $\BB \ne \O$ there is a $B \in \BB$.
Because $\BB \subset \powerset S$:
:$B \subseteq S$
and thus by the definition of filter:
:$S \in \FF$
Because:
:the only subset of $\O$ is $\O$
and:
:since $\O \notin \BB$
it follows that:
:$\O \notin \FF$
Let $V_1, V_2 \in \FF$.
Then there exist $U_1, U_2 \in \BB$ such that $U_1 \subseteq V_1$ and $U_2 \subseteq V_2$.
Because $\BB$ satisfies condition $(1)$ there exists a set $U \in \BB$ for which $U \subseteq U_1 \cap U_2$ holds.
Because $U_1 \cap U_2 \subseteq V_1 \cap V_2$ this implies that:
:$U \subseteq V_1 \cap V_2$
and therefore:
:$V_1 \cap V_2 \in \FF$
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Filters
Category:Filter Theory
Filter Containing Complements is Not Proper

== Theorem ==
Let $L = \struct {S, \lor, \land, \preceq}$ be a bounded lattice.
Let $F \subseteq S$ be a filter on $L$.
Suppose there exist $a, b \in F$ such that:
:$b$ is a complement of $a$.
Then:
:$F = S$
== Proof ==
By filter axiom $\paren 2$:
:$\exists c \in F: c \preceq a \land c \preceq b$
By definition of complement:
:$b \land a = \bot$
Thus, by definition of meet:
:$c \preceq \bot$
Therefore:
Hence, by definition of set equality:
:$F = S$
Category:Bounded Lattices
Category:Filter Theory
Filter Contains Greatest Element

== Theorem ==
Let $\struct{S, \preceq}$ be an ordered set with greatest element $\top$.
Let $\FF$ be a filter in $\struct{S, \preceq}$.
Then:
:$\top \in \FF$
== Proof ==
By :
:$\exists x \in \FF$
By definition of greatest element:
$x \preceq \top$
By :
:$\top \in \FF$
Category:Ordered Sets
Category:Filter Theory
Category:Greatest Elements
Filter is Finer iff Sets of Basis are Subsets

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ denote the power set of $S$.
Let $\FF, \FF' \subset \powerset S$ be two filters on $S$.
Let $\FF$ have a basis $\BB$.
Let $\FF'$ have a basis $\BB'$.
$\FF$ is finer than $\FF'$ for every set of $\BB'$, there is a set of $\BB$ subset to it.
== Proof ==
=== Necessary Condition ===
Suppose for every set of $\BB'$, there is a set of $\BB$ subset to it.
Pick any $U \in \FF'$.
Then from definition of a basis:
:$\exists V' \in \BB': V' \subseteq U$
By our assumption:
:$\exists V \in \BB: V \subseteq V' \subseteq U \subseteq S$
By definition of a filter:
:$U \in \FF$.
Hence we have $\FF' \subseteq \FF$.
=== Sufficient Condition ===
We prove the contrapositive.
Suppose there is some set of $\BB'$ with no set of $\BB$ subset to it.
Let $U$ be such a set.
Then $U \in \BB' \subseteq \FF'$.
From our assumption:
:$\nexists V \in \BB: V \subseteq U$
From the definition of a basis:
:$U \notin \FF$
This gives $\FF' \nsubseteq \FF$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Filters
Category:Filter Theory
Filter is Ideal in Dual Ordered Set

== Theorem ==
Let $P = \struct {S, \preceq}$ be an ordered set.
Let $X$ be a subset of $S$.
Then
:$X$ is filter in $P$
:$X$ is ideal in $P^{-1}$
where $P^{-1} = \struct {S, \succeq}$ denotes the dual of $P$.
== Proof ==
By Dual of Dual Ordering:
:dual of $P^{-1}$ is $P$.
Hence by Ideal is Filter in Dual Ordered Set:
:the result fallows.
== Sources ==
* 
* 
Category:Order Theory
Category:Dual Orderings