nyuuzyou/wikis huggingface dataset:
Field Homomorphism Preserves Subfields

== Theorem ==
Let $\struct {F_1, +_1, \circ_1}$ and $\struct {F_2, +_2, \circ_2}$ be fields.
Let $\phi: F_1 \to F_2$ be a field homomorphism such that $\phi$ is not the trivial homomorphism.
If $K$ is a subfield of $F_1$, then $\phi \sqbrk K$ is a subfield of $F_2$.
=== Corollary ===
== Proof ==
First note that if $\phi$ is the trivial homomorphism then $\phi \sqbrk K = 0_{F_2}$ and so is not a field.
Since $K$ is a field, we have that:
:$0_{F_1} \in K$
:$1_{F_1} \in K$
and so by Ring Homomorphism Preserves Zero and Field Homomorphism Preserves Unity:
:$\map \phi {0_{F_1} } = 0_{F_2} \in \phi \sqbrk K$
:$\map \phi {1_{F_1} } = 1_{F_2} \in \phi \sqbrk K$
So $\phi \sqbrk K$ contains at least the zero and unity.
From Group Homomorphism Preserves Subgroups:
:$\struct {\phi \sqbrk K, +_2}$ is a subgroup of $\struct {F_2, +_2}$
:$\struct {\phi \sqbrk K, \circ_2}$ is a subgroup of $\struct {F_2, \circ_2}$.
Thus, as $\struct {F_2, +_2}$ and $\struct {F_2, \circ_2}$ are both groups, the result follows.
Category:Field Homomorphisms
Category:Subfields
Field Homomorphism Preserves Subfields/Corollary

== Corollary to Field Homomorphism Preserves Subfields ==
Let $\struct {F_1, +_1, \circ_1}$ and $\struct {F_2, +_2, \circ_2}$ be fields.
The image of a field homomorphism $\phi: F_1 \to F_2$ is a subfield of $F_2$.
== Proof ==
From Field is Subfield of Itself, $F_1$ is a subfield of $F_1$.
The result then follows directly from Field Homomorphism Preserves Subfields.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms
Category:Field Homomorphisms
Category:Subfields
Field Homomorphism Preserves Unity

== Theorem ==
Let $\phi: \struct {F_1, +_1, \times_1} \to \struct {F_2, +_2, \times_2}$ be a field homomorphism.
Let:
:$1_{F_1}$ be the unity of $F_1$
:$1_{F_2}$ be the unity of $F_2$.
Then:
:$\map \phi {1_{F_1} } = 1_{F_2}$
== Proof ==
By definition, if $\struct {F_1, +_1, \times_1}$ and $\struct {F_2, +_2, \times_2}$ are fields then $\struct {F_1^*, \times_1}$ and $\struct {F_2^*, \times_2}$ are groups.
Again by definition:
:the unity of $\struct {F_1, +_1, \times_1}$ is the identity of $\struct {F_1^*, \times_1}$
:the unity of $\struct {F_2, +_2, \times_2}$ is the identity of $\struct {F_2^*, \times_2}$.
The result follows from Group Homomorphism Preserves Identity.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms
Category:Field Homomorphisms
Field Homomorphism is either Trivial or Injection

== Theorem ==
Let $\struct {E, +_E, \times_E}$ and $\struct {F, +_F, \times_F}$ be fields.
Let $\phi: E \to F$ be a (field) homomorphism.
Then $\phi$ is either an injection or the trivial homomorphism.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 87 \eta$
Category:Field Homomorphism is either Trivial or Injection
Category:Field Homomorphisms
Field Homomorphism is either Trivial or Injection/Proof 2

== Theorem ==
== Proof ==
Let $\phi: E \to F$ be a field homomorphism.
Suppose $\phi$ is not an injection.
So there must exist $a, b \in F: \map \phi a = \map \phi b$.
Let $k = a +_E \paren {-b}$.
Then:
As $a \ne b$ then $k \ne 0_E$ and so has a product inverse $\exists k^{-1} \in E$.
So for any $x \in E$ we can write $x = k \circ \paren {k^{-1} \circ x}$ and so:
So if $\phi$ is not an injection, it is the trivial homomorphism.
Category:Field Homomorphism is either Trivial or Injection
Field Norm of Complex Number is Multiplicative Function

== Theorem ==
Let $\C$ denote the set of complex numbers.
Let $N: \C \to \R_{\ge 0}$ denote the field norm on complex numbers:
:$\forall z \in \C: \map N z = \cmod z^2$
where $\cmod z$ denotes the complex modulus of $z$.
Then $N$ is a multiplicative function on $\C$.
== Proof ==
So $N$ is a multiplicative function by definition.
== Sources ==
* : Chapter $9$: Rings: Exercise $19$
Category:Field Norm of Complex Number
Field Norm of Complex Number is Positive Definite

== Theorem ==
Let $\C$ denote the set of complex numbers.
Let $N: \C \to \R_{\ge 0}$ denote the field norm on complex numbers:
:$\forall z \in \C: \map N z = \cmod z^2$
where $\cmod z$ denotes the complex modulus of $z$.
Then $N$ is positive definite on $\C$.
== Proof ==
First it is shown that $\map N z = 0 \iff z = 0$.
Let $z = x + i y$.
Then we have:
Hence the result by definition of positive definite.
Category:Field Norm of Complex Number
Field Norm of Complex Number is not Norm

== Theorem ==
Let $\C$ denote the set of complex numbers.
Let $N: \C \to \R_{\ge 0}$ denote the field norm on complex numbers:
:$\forall z \in \C: \map N z = \cmod z^2$
where $\cmod z$ denotes the complex modulus of $z$.
Then $N$ is not a norm on $\C$.
== Proof ==
Proof by Counterexample:
Let $z_1 = z_2 = 1$.
Then:
But:
So we have that for these instances of $z_1$ and $z_2$:
:$\map N {z_1 + z_2} > \map N {z_1} + \map N {z_2}$
and so the triangle inequality is not satisfied.
Hence by definition $N$ is not a norm on $\C$.
Category:Field Norm of Complex Number
Field Norm of Quaternion is Multiplicative

== Theorem ==
Let $\mathbf x$ be a quaternion.
Let $\overline {\mathbf x}$ be the conjugate of $\mathbf x$.
The field norm of $\mathbf x$:
:$\map n {\mathbf x} := \size {\mathbf x \overline {\mathbf x} }$
is a multiplicative function.
== Proof ==
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Quaternions
Field Norm of Quaternion is Positive Definite

== Theorem ==
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion.
Let $\overline {\mathbf x}$ be the conjugate of $\mathbf x$.
The field norm of $\mathbf x$:
:$\map n {\mathbf x} := \cmod {\mathbf x \overline {\mathbf x} }$
is positive definite.
== Proof ==
Hence $n$ is positive definite by definition.
== Sources ==
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $3$. FIELD
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Quaternions
Field Norm of Quaternion is not Norm

== Theorem ==
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion.
Let $\overline {\mathbf x}$ be the conjugate of $\mathbf x$.
The field norm of $\mathbf x$:
:$\map n {\mathbf x} := \size {\mathbf x \overline {\mathbf x} }$
is not a norm in the abstract algebraic context of a division ring.
== Proof ==
Each of the norm axioms is examined in turn:
=== ===
This is proved in Field Norm of Quaternion is Positive Definite‎.
=== ===
This is proved in Field Norm of Quaternion is Multiplicative.
=== ===
For example:
:$\map n {1 + 1} = 4 > 2 = \map n 1 + \map n 1$
and so is not satisfied.
Not all the norm axioms are fulfilled.
Hence the result.
Category:Quaternions
Category:Norm Theory
Field Norm on 5th Cyclotomic Ring

== Theorem ==
Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
Let $\alpha = a + i b \sqrt 5$ be an arbitrary element of $\Z \sqbrk {i \sqrt 5}$.
The field norm of $\alpha$ is given by:
:$\map N \alpha = a^2 + 5 b^2$
== Proof ==
== Sources ==
* : Chapter $9$: Rings: Exercise $19$
Category:Cyclotomic Rings
Field Operations of P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\norm {\,\cdot\,}_p$ be the p-adic norm on the rationals $\Q$.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
That is, $\Q_p$ is the quotient ring $\CC \, \big / \NN$ where:
:$\CC$ denotes the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$
:$\NN$ denotes the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$.
Then $\Q_p$ is a field and the field operations are defined by:
:$+ : \quad \forall \sequence {x_n} + \NN, \sequence {y_n} + \NN \in \CC \, \big / \NN$:
:::$\quad \paren {\sequence {x_n} + \NN} + \paren {\sequence {y_n} + \NN} = \sequence {x_n + y_n} + \NN$
:$\circ : \quad \forall \sequence {x_n} + \NN, \sequence {y_n} + \NN \in \CC \, \big / \NN$: 
:::$\quad \paren {\sequence {x_n} + \NN} \paren {\sequence {y_n} + \NN} = \sequence {x_n y_n} + \NN$
== Proof ==
By Corollary to Cauchy Sequences form Ring with Unity, $\CC$ is a commutative ring of Cauchy sequences with the ring operations defined by:
:$+ : \quad \forall \sequence {x_n}, \sequence {y_n} \in \CC$:
:::$\quad \sequence {x_n} + \sequence {y_n} = \sequence {x_n + y_n}$
:$\circ : \quad \forall \sequence {x_n}, \sequence {y_n} \in \CC$:
:::$\quad \sequence {x_n} \circ \sequence {y_n} = \sequence {x_n y_n}$
By Corollary to Null Sequences form Maximal Left and Right Ideal, $\NN$ is a maximal ideal of $\CC$. 
By Corollary to Quotient Ring of Cauchy Sequences is Division Ring, the quotient ring $\Q_p = \CC \big / \NN$ is a field with field operations defined by:
:$+ :\quad \forall \sequence {x_n} + \NN, \sequence {y_n} + \NN \in \CC \, \big / \NN$:
:::$\quad \paren {\sequence {x_n} + \NN} + \paren {\sequence {y_n} + \NN} = \paren {\sequence {x_n} + \sequence {y_n} } + \NN$
:$\circ :\quad \forall \sequence {x_n} + \NN, \sequence {y_n} + \NN \in \CC \, \big / \NN$: 
:::$\quad \paren {\sequence {x_n} + \NN} \paren {\sequence {y_n} + \NN} = \paren {\sequence {x_n} \circ \sequence {y_n} } + \NN$
Putting the operations on $\CC$ with those on $\CC \, \big / \NN$ gives:
:$+ :\quad \forall \sequence {x_n} + \NN, \sequence{y_n} + \NN \in \CC \, \big / \NN$:
:::$\quad \paren {\sequence {x_n} + \NN} + \paren {\sequence {y_n} + \NN} = \sequence {x_n + y_n} + \NN$
:$\circ :\quad \forall \sequence {x_n} + \NN, \sequence {y_n} + \NN \in \CC \, \big / \NN$: 
:::$\quad \paren {\sequence {x_n} + \NN} \paren {\sequence {y_n} + \NN} = \sequence {x_n y_n} + \NN$
Category:P-adic Number Theory
Field Product with Non-Zero Element yields Unique Solution

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $a, b, x \in F$ such that $b \ne 0_F$.
Let:
:$b \times x = a$
Then:
:$x = a b^{-1}$
That is:
:$x = \dfrac a b$
where $\dfrac a b$ denotes division.
== Proof ==
== Sources ==
* : Chapter $4$: Fields: $\S 14$. Definition of a Field
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(iii)}$
Category:Field Theory
Field Product with Zero

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$.
Then:
:$a \times 0_F = 0_F$
That is, $0_F$ acts as a zero element of $F$, which justifies its name.
== Proof ==
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 87 \alpha$
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(iii)}$
Category:Field Theory
Field Unity Divided by Element equals Multiplicative Inverse

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $a \in F$.
Then:
:$\dfrac {1_F} a = a^{-1}$
where $\dfrac {1_F} a$ denotes division.
== Proof ==
== Sources ==
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(iv)}$
Category:Field Theory
Field has 2 Ideals

== Theorem ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Then the only ideals of $\struct {F, +, \circ}$ are $\struct {F, +}$ and $\set {0_F}$.
That is, $\struct {F, +, \circ}$ has no non-null proper ideals.
== Proof ==
By definition, a field is a division ring.
From Null Ring is Ideal and Ring is Ideal of Itself, it is always the case that $\set {0_F}$ and $\struct {F, +}$ are ideals of $\struct {F, +, \circ}$.
From Ideals of Division Ring, it follows that the only ideals of $\struct {F, +, \circ}$ are $\struct {F, +}$ and $\set {0_F}$.
== Sources ==
* : Chapter $5$: Rings: $\S 21$. Ideals: Theorem $36$
* : $\S 58.3$ Ideals
Category:Commutative Algebra
Category:Field Theory
Category:Ideal Theory
Field has Algebraic Closure

== Theorem ==
Every field has an algebraic closure. 
== Proof ==
Let $F$ be a field.
Let $\FF$ be the collection of all extensions of $F$. 
Define an ordering on $\FF$ thus:
:$\forall K, L \in \FF: K \preceq L \iff L$ is an extension of $K$.
	
Let $C$ be a chain in $\FF$.
Let $\ds E = \bigcup_{K \mathop \in C} K$.
$E$ satisfies all field axioms, so $E \in \FF$.
By Set is Subset of Union, $E$ is an upper bound for $C$. 
	
By Zorn's Lemma, $C$ has a maximal element $m$.
Then $m$ is an algebraic closure of $F$.
Category:Field Theory
Field has Characteristic of Zero iff exists Monomorphism from Rationals

== Theorem ==
Let $F$ be a field.
Then:
:there exists a field monomorphism $\phi: \Q \to F$ from the field of rational numbers $\Q$ and $F$.
:
:$\Char F = 0$
where $\Char F$ denotes the characteristic of $F$.
== Proof ==
=== Necessary Condition ===
Let $\Char F = 0$.
Then from Field of Characteristic Zero has Unique Prime Subfield, $F$ has a unique prime subfield $K$ such that:
:$K \cong \Q$
where $\cong$ denotes isomorphism.
Thus there exists an isomorphism from $\Q$ to a subfield of $F$.
From Injection to Image is Bijection, it follows that $\phi$ is an injection into $F$.
Hence the result by definition of field monomorphism.
=== Sufficient Condition ===
Let there exists a field monomorphism $\phi: \Q \to F$ from the field of rational numbers $\Q$ and $F$.
Let $K := \Img \phi$.
Then $K$ is isomorphic to $\Q$.
As $K$ is a subfield of $F$, both $0$ and $1$ are in $K$.
From Characteristic of Field is Zero or Prime, either $\Char F = 0$ or $\Char F = p$ for some prime number $p$.
 $\Char F = p$.
Then:
:$p \cdot 1 = 0$
But as $1 \in K$ it follows that $p \cdot 1 \in K$ as $K$ is closed under addition.
But as $K \cong \Q$ this cannot happen
Hence there is no prime number $p$ such that $\Char F = p$.
The result follows by Proof by Contradiction.
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 89 \beta$
Category:Characteristics of Fields
Category:Rational Numbers
Category:Field Monomorphisms
Field has Prime Characteristic p iff exists Monomorphism from Field of Integers Modulo p

== Theorem ==
Let $F$ be a field.
Then:
:there exists some prime number $p$ such that $\Char F = p$
:
:there exists a field monomorphism $\phi: \Z_p \to F$
where:
:$\Char F$ denotes the characteristic of $F$.
:$\Z_p$ denotes the field of integers modulo $p$.
== Proof ==
Let $\struct {F, +, \times}$ be a field whose zero is $0$ and whose unity is $1$.
=== Sufficient Condition ===
Let there exists some prime number $p$ such that $\Char F = p$.
Let us define the mapping $\phi: \Z_p \to F$ as:
:$\forall \eqclass n p \in \Z_p: \map \phi {\eqclass n p} = n \cdot 1$
where $n \cdot 1$ denotes the power of $1$ in the context of the additive group $\struct {F, +}$:
:$n \cdot 1 = \begin {cases}
0 & : n = 0 \\
\paren {\paren {n - 1} \cdot 1} + 1 & : n > 0
\end {cases}$
We show that $\phi$ is the field monomorphism required.
By definition, a field monomorphism is a field homomorphism which is also an injection.
We have:
Then by definition $\phi$ is a (field) homomorphism.
It remains to be shown that $\phi$ is an injection.
Let $\eqclass a p, \eqclass b p \in \Z_p$ such that:
:$\phi {\eqclass a p} = \phi {\eqclass b p}$
=== Necessary Condition ===
Let there exist a field monomorphism $\phi: \Z_p \to F$.
We require to show that $\Char F = p$.
It is sufficient to show that:
:$\forall a \in F: n \circ a = 0 \iff n = p$
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 89 \beta$
* : $\S 1.8$ Algebraic properties of $p$-adic integers
Category:Characteristics of Fields
Category:Field Monomorphisms
Category:Field of Integers Modulo Prime
Field has Prime Subfield

== Theorem ==
Let $\struct {F, +, \times}$ be a field.
Then $F$ has a subfield which is a prime field.
== Proof ==
By definition of field, $F$ is a division ring where $\times$ is commutative.
Therefore all division subrings of $F$ are in fact subfields of $F$.
By Intersection of All Division Subrings is Prime Subfield, the intersection of all subfields of $F$ is a prime field which is a subfield of $F$.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Subfields
Field has no Proper Zero Divisors

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Then $\struct {F, +, \times}$ has no proper zero divisors.
That is:
:$a \times b = 0_F \implies a = 0_F \lor b = 0_F$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 87 \gamma$
Category:Field Theory
Category:Field has no Proper Zero Divisors
Field has no Proper Zero Divisors/Proof 2

== Theorem ==
== Proof ==
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : Chapter $4$: Fields: $\S 14$. Definition of a Field
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $3 \ \text {(v)}$
Category:Field has no Proper Zero Divisors
Field is Galois over Fixed Field of Automorphism Group

== Theorem ==
Let $E/F$ be a finite field extension.
Let $K = \operatorname{Fix}_E(\operatorname{Aut}(E/F))$ be the fixed field of the automorphism group of $E/F$.
Then $E/K$ is Galois.
== Proof ==
Follows from Closed Fields in Galois Connection for Field Extension (and does not use Fundamental Theorem of Galois Theory).
Category:Galois Theory
Field is Integral Domain/Proof 1

== Theorem ==
== Proof ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Suppose $\exists x, y \in F: x \circ y = 0_F$.
Suppose $x \ne 0_F$.
Then, by the definition of a field, $x^{-1}$ exists in $F$ and:
:$y = 1_F \circ y = x^{-1} \circ x \circ y = x^{-1} \circ 0_F = 0_F$.
Otherwise $x = 0_F$.
So if $x \circ y = 0_F$, either $x = 0_F$ or $y = 0_F$ as we were to show.
Note that this is equivalent to the statement $\forall x, y \in F: x \ne 0_F \land y \ne 0_F \implies x \circ y \ne 0_F$, so the set $F^*$ is closed under product.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $3$: Some special classes of rings
* : $\S 55.1$ Special types of ring and ring elements
Category:Field is Integral Domain
Field is Integral Domain/Proof 2

== Theorem ==
== Proof ==
This result follows directly from:
:Field has no Proper Zero Divisors
:By definition, that a field is a commutative ring.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
* : Chapter $4$: Fields: $\S 14$. Definition of a Field
Category:Field is Integral Domain
Field is Principal Ideal Domain

== Theorem ==
Let $F$ be a field.
Then $F$ is a principal ideal domain.
== Proof ==
Let $F$ be a field.
Let $I \subset F$ be a non-null ideal of $F$.
Let $a \in I$ be non-zero.
Since $F$ is a field, $a^{-1}$ exists.
We have that $1 = a^{-1} \cdot a \in I$.
Since $1 \in I$, for every element $b \in F$:
:$b = b \cdot 1 \in I$
we have that $I = F = \ideal 1$ if $I \ne \set 0$.
Thus the only ideals of $F$ are $\ideal 0 = \set 0$ and $\ideal 1 = F$, which are both principal ideals.
Hence $F$ is a principal ideal domain.
Category:Principal Ideal Domains
Category:Field Theory
Field is Subfield of Itself

== Theorem ==
Let $\struct {F, +, \circ}$ be a field.
Then $\struct {F, +, \circ}$ is a subfield of $\struct {F, +, \circ}$.
== Proof ==
$F$ is a field and $F \subseteq F$ from Set is Subset of Itself.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
* : $\S 56.1$ Subrings and subfields
Category:Subfields
Field of Characteristic Zero has Unique Prime Subfield

== Theorem ==
Let $F$ be a field, whose zero is $0_F$ and whose unity is $1_F$, with characteristic zero.
Then there exists a unique $P \subseteq F$ such that:
:$(1): \quad P$ is a subfield of $F$
:$(2): \quad P$ is isomorphic to the field of rational numbers $\struct {\Q, +, \times}$.
That is, $P \cong \Q$ 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 1 ==
== Proof 2 ==
Category:Characteristics of Fields
Category:Subfields
Category:Field of Characteristic Zero has Unique Prime Subfield
Field of Characteristic Zero has Unique Prime Subfield/Proof 1

== Theorem ==
== Proof ==
Follows directly from:
:Subring Generated by Unity of Ring with Unity
:Quotient Theorem for Monomorphisms
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.10$
Category:Field of Characteristic Zero has Unique Prime Subfield
Field of Characteristic Zero has Unique Prime Subfield/Proof 2

== Theorem ==
== Proof ==
Let $\struct {F, +, \circ}$ be a field such that $\Char F = 0$.
Let $P$ be a prime subfield of $F$.
From Field has Prime Subfield, this has been shown to exist.
As $P$ is a subfield of $F$, we apply Zero and Unity of Subfield and see that the unity of $P$ is $1_F$.
As $P$ is closed:
:$\forall m \in \Z: m \cdot 1_F \in P$ of which $0 \cdot 1_F = 0_F$ the only one that is zero
:$\forall n \in \Z, n \ne 0: \paren {n \cdot 1_F}^{-1} \in P$
So $P$ contains all elements of $F$ of the form:
: $\paren {m \cdot 1_F} \circ \paren {n \cdot 1_F}\^{-1}$
where $m, n \in \Z, n \ne 0$.
This can be expressed more clearly in division notation as:
:$\dfrac {m \cdot 1_F} {n \cdot 1_F}$
Now let $P'$ consist of all the elements of $F$ of the form:
:$\dfrac {m \cdot 1_F} {n \cdot 1_F}$
Let $\dfrac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \in P'$ and $\dfrac {m_2 \cdot 1_F} {n_2 \cdot 1_F} \in P'$.
Then:
So $P'$ is closed under $+$.
Next:
So $P'$ is closed under $\circ$.
Next:
So $P'$ is closed under taking inverses of $+$.
Next, assuming that $m \ne 0$:
So $P' \setminus \set {0_F}$ is closed under taking inverses of $\circ$.
Thus by Subfield Test, $P'$ is a subfield of $F$.
It follows that $P = P'$, and so $P$ contains precisely the elements of the form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$.
We can consistently define a mapping $\phi: \Q \to P$ by:
:$\forall m, n \in \Z: n \ne 0: \map \phi {\dfrac m n} = \dfrac {m \cdot 1_F} {n \cdot 1_F}$
First we show that $\phi$ is well-defined.
Suppose $\dfrac {m_1} {n_1} = \dfrac {m_2} {n_2}$.
Then by multiplying both sides by $n_1 n_2$:
:$m_1 n_2 = m_2 n_1$
We need to show that:
:$\map \phi {\dfrac {m_1} {n_1} } = \map \phi {\dfrac {m_2} {n_2} }$
So:
That is:
:$\map \phi {\dfrac {m_1} {n_1} } = \map \phi {\dfrac {m_2} {n_2} }$
demonstrating that $\phi$ is well-defined.
Next, we need to show that $\phi$ is a ring isomorphism.
So:
and:
thus proving that $\phi$ is a ring homomorphism.
From Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi$ is a ring monomorphism.
It is also clear that $\phi$ is a surjection, as every element of $P$ is the image of some element of $\Q$.
It follows that $\phi$ is a ring isomorphism.
Now let $K$ be a subfield of $F$.
Let $P = \Img \phi$ as defined above.
We know that $1_F \in K$.
Thus $K$ contains a subfield $P$ such that $P$ is isomorphic to $\Q$.
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:Field of Characteristic Zero has Unique Prime Subfield
Field of Integers Modulo Prime is Prime Field

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Z_p, +, \times}$‎ be the field of integers modulo $p$.
Then $\struct {\Z_p, +, \times}$‎ is a prime field.
== Proof ==
If $\struct {F, +, \times}$ is a subfield of $\struct {\Z_p, +, \times}$, then $\struct {F, +}$ is a subgroup of $\struct {\Z_p, +}$.
But from Prime Group has no Proper Subgroups, $\struct {\Z_p, +}$ has no proper subgroup except the trivial group.
Hence $F = \Z_p$ and so follows the result.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties: Example $2$
Category:Modulo Arithmetic
Category:Prime Fields