avewright/proofwiki-math huggingface dataset:
Theorem: Quotient Epimorphism from Integers by Principal Ideal

Let $m$ be a strictly positive integer.
Let $\ideal m$ be the principal ideal of $\Z$ generated by $m$.
The restriction to $\N_m$ of the quotient epimorphism $q_m$ from the ring $\struct {\Z, +, \times}$ onto $\struct {\Z, +, \times} / \ideal m$ is an isomorphism from the ring $\struct {\N_m, +_m, \times_m}$ of integers modulo $m$ onto the quotient ring $\struct {\Z, +, \times} / \ideal m$.
In particular, $\struct {\Z, +, \times} / \ideal m$ has $m$ elements.

Proof:
Let $x, y \in \N_m$.
By the Division Theorem:
{{begin-eqn}}
{{eqn | q = \exists q, r \in \Z
      | l = x + y
      | r = m q + r
      | c = for $0 \le r < m$
}}
{{eqn | q = \exists p, s \in \Z
      | l = x y
      | r = m p + s
      | c = for $0 \le s < m$
}}
{{end-eqn}}
Then $x +_m y = r$ and $x \times_m y = s$, so:
{{begin-eqn}}
{{eqn | l = \map {q_m} {x +_m y}
      | r = \map {q_m} r
      | c = 
}}
{{eqn | r = \map {q_m} {m q} + \map {q_m} r
      | c = 
}}
{{eqn | r = \map {q_m} {m q + r}
      | c = 
}}
{{eqn | r = \map {q_m} {x + y}
      | c = 
}}
{{eqn | r = \map {q_m} x + \map {q_m} y
      | c = 
}}
{{end-eqn}}
and similarly:
:$\map {q_m} {x \times_m y} = \map {q_m} {x y} = \map {q_m} x \map {q_m} y$
So the restriction of $q_m$ to $\N_m$ is a homomorphism from $\struct {\N_m, +_m, \times_m}$ into $\struct {\Z / \ideal m, +_{\ideal m}, \times_{\ideal m} }$.
Let $a \in \Z$.
Then:
:$\exists q, r \in \Z: a = q m + r: 0 \le r < m$
so:
:$\map {q_m} a = \map {q_m} r \in q_m \sqbrk {\N_m}$
Therefore:
:$\Z / \ideal m = q_m \sqbrk \Z = q_m \sqbrk {\N_m}$
Therefore the restriction of $q_m$ to $\N_m$ is surjective.
If $0 < r < m$, then $r \notin \ideal m$ and thus $\map {q_m} r \ne 0$.
Thus the kernel of the restriction of $q_m$ to $\N_m$ contains only zero.
Therefore by the Quotient Theorem for Group Epimorphisms, the restriction of $q_m$ to $\N_m$ is an isomorphism from $\N_m$ to $\Z / \ideal m$.
{{wtd|Reference is made throughout to $\struct {\N_m, +_m, \times_m}$. It needs to be shown that this is the same (at least up to isomorphism) as the ring of integers modulo $m$ $\struct {\Z, +_m, \times_m}$.)}}
Theorem: Integer Divisor is Equivalent to Subset of Ideal

Let $\Z$ be the set of all integers.
Let $\Z_{>0}$ be the set of strictly positive integers.
Let $m \in \Z_{>0}$ and let $n \in \Z$.
Let $\ideal m$ be the principal ideal of $\Z$ generated by $m$.
Then:
:$m \divides n \iff \ideal n \subseteq \ideal m$

Proof:
The ring of integers is a principal ideal domain.
The result follows directly from Principal Ideals in Integral Domain.
{{qed}}
Theorem: Principal Ideal Domain is Unique Factorization Domain

Every principal ideal domain is a unique factorization domain.

Proof:
Let $R$ be a principal ideal domain whose zero is $0_R$.
Let $a \in R$ be an arbitrary element of $R$ which is neither $0_R$ nor a unit of $R$.
By Element of Principal Ideal Domain is Finite Product of Irreducible Elements, $a$ has a factorization into finitely many irreducible elements.
Note that this uses the Axiom of Dependent Choice.
{{explain|Why does it? That needs a proof}}
It remains to be demonstrated that this factorization is unique.
{{AimForCont}} there exist two or more different factorizations.
{{WLOG}}, let:
:$a = u p_1 \dotsm p_n = q_1 \dotsm q_m$
where:
:each $p_i$ and $q_j$ is irreducible
:$u$ is a unit of $R$.
:$\set {p_1, p_2, \ldots, p_n} \ne \set {q_1, q_2, \ldots, q_m}$
:$m \ge n$.
By:
:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal
:Maximal Ideal of Commutative and Unitary Ring is Prime Ideal
$\ideal {p_i}$ and $\ideal {q_j}$ are prime ideals for all $i$ and $j$.
Hence each $p_i$ and $q_j$ is a prime element of $R$.
Using primality of the factors, $p_1 \divides q_j$ for some $j \in \set {1, \ldots, m}$.
That is:
:$\exists u_1 \in R: p_1 u_1 = q_j$
Because:
:$q_j$ is irreducible
:$p_1$ is not a unit of $R$
$u_1$ is a unit.
By applying permutations to $\set {1, \ldots, m}$ we can take $p_1 u_1 = q_j$, say, so that:
{{begin-eqn}}
{{eqn | l = u_1^{-1} q_1 p_2 \dotsm p_n u
      | r = q_1 \dotsm q_m
      | c = 
}}
{{eqn | ll= \leadsto
      | l = q_1 \paren {u_1^{-1} p_2 \dotsm p_n u - q_2 \dotsm q_m}
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = u_1^{-1} p_2 \dotsm p_n u
      | r = u_1 q_2 \dotsm q_m
      | c = 
}}
{{end-eqn}}
by Cancellation Law for Ring Product of Integral Domain.
Repeating this $n - 1$ times establishes that, up to permutations of the indices:
:$u = u_1 \cdots u_n q_{m - n + 1} \cdots q_m$
where each $u_i$ is a unit of $R$.
Let $m > n$.
Then:
:$q_m \divides u$
and so $q_m$ is a unit of $R$ by definition.
This contradicts its irreducibility.
Hence, $m = n$.
Let us define $\pi: \set {1, \ldots, n} \to \set {1, \ldots, n}$ by:
:$\map \pi i = j$
where $p_i u_i = q_j$ as previously constructed.
We have shown that $\pi$ is a well-defined bijection.
For each $i \in \set {1, \ldots, n}$:
:$p_i = u_i^{-1} q_{\map \pi i}$
so:
:$p_i \divides q_{\map \pi i}$
and:
:$q_{\map \pi i} \divides p_i$
which makes $p_i$ and $q_{\map \pi i}$ associates for each $i \in \set {1, \ldots, n}$.
The bijection $\pi: \set {1, \dots, n} \to \set {1, \dots, n}$ establishes that $a = p_1 \cdots p_n$ and $a = q_1 \cdots q_n$ are equivalent factorizations.
This contradicts our assertion that there exist two or more factorizations of $a$ in $R$.
We have proved that an arbitrary $a \in R$ has a unique factorization in $R$
Hence, by definition, $R$ is a unique factorization domain. 
{{qed}}
Theorem: Maximal Ideal iff Quotient Ring is Field

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:
=== Maximal Ideal implies Quotient Ring is Field ===
{{:Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field}}{{qed|lemma}}
=== Quotient Ring is Field implies Ideal is Maximal ===
{{:Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal}}{{qed}}
Theorem: Maximal Ideal iff Quotient Ring is Field

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:
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 $R$ {{iff}} $\mathbb L_J = \set {J, R}$ and $J$ is a proper ideal of $R$.
By Ideals of Field, $R / J$ is a field {{iff}}:
:$\map {\mathbb L} {R / J} = \set {\map {q_J} J, \map {q_J} R}$
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 {{iff}} $J$ is a field.
{{qed}}
Theorem: Maximal Ideal iff Quotient Ring is Field

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:
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.
{{qed|lemma}}
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.
{{qed}}
Theorem: Prime Number iff Generates Principal Maximal Ideal

Let $\Z_{>0}$ be the set of strictly positive integers.
Let $p \in \Z_{>0}$.
Let $\ideal p$ be the principal ideal of $\Z$ generated by $p$. 
Then $p$ is prime {{iff}} $\ideal p$ is a maximal ideal of $\Z$.

Proof:
First, note that $\Z$ is a principal ideal domain, so all ideals are principal.
Theorem: Integral Domain of Prime Order is Field

Let $\struct {\Z_p, +_p, \times_p}$ be the ring of integers modulo $p$.
The following statements are equivalent:
:$(1): \quad p$ is a prime.
:$(2): \quad \struct {\Z_p, +_p, \times_p}$ is an integral domain.
:$(3): \quad \struct {\Z_p, +_p, \times_p}$ is a field.

Proof:
By Prime Number iff Generates Principal Maximal Ideal and Maximal Ideal iff Quotient Ring is Field, $(1)$ implies $(3)$, and from Field is Integral Domain, $(3)$ implies $(2)$.
By definition of integral domain, $\Z_p$ is an integral domain {{iff}} it has no proper zero divisors.
That is, {{iff}} $\struct {\Z_p^*, \times_p}$ is a semigroup.
Let $\ideal p$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $p$.
From Quotient Epimorphism from Integers by Principal Ideal:
:$\struct {\Z_p, +_p, \times_p}$ is isomorphic to $\struct {\Z, +, \times} / \ideal p$.
So, we can let $\map {q_p} m: \Z \to \Z_p$ be the quotient mapping from $\struct {\Z, +, \times}$ to $\struct {\Z_p, +_p, \times_p}$.
Let $0_p$ denote the zero of $\Z_p$.
{{AimForCont}} $p = m n$ where $1 < m < p, 1 < n < p$.
Then in the ring $\Z_p$ we have $\map {q_p} m \ne 0_p, \map {q_p} n \ne 0_p$.
But as $q_p$ is an epimorphism and therefore obeys the morphism property, $\map {q_p} m \times_p \map {q_p} n = \map {q_p} {m n} = \map {q_p} p = 0_p$.
But by definition, $0_p \notin \Z_p^*$.
Thus if $p = m n$, then $\struct {\Z_p^*, \times_p}$ is not a semigroup.
{{Qed}}
Theorem: Quotient Ring of Integers and Zero

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:
{{begin-eqn}}
{{eqn | l = \ideal 0
      | r = \set {\sum^n_{i \mathop = 1} r_i \times 0 \times s_i: n \in \N; r_i, s_i \in \Z}
      | c = {{Defof|Principal Ideal of Ring}}
}}
{{eqn | r = \set {\sum^n_{i \mathop = 1} 0: n \in \N}
      | c = $0$ is the zero under integer multiplication
}}
{{eqn | r = \set 0
      | c = Integer Addition Identity is Zero
}}
{{end-eqn}}
{{finish}}
Category:Quotient Rings
Category:Integers
0pghfflaz3nxk2mqcj13qe7xh1y0fpw
Theorem: Quotient Ring of Integers and Principal Ideal from Unity

Let $\struct {\Z, +, \times}$ be the integral domain of integers.
Let $\ideal 1$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $1$.
The quotient ring $\struct {\Z, +, \times} / \ideal 1$ is isomorphic to the null ring.

Proof:
{{ProofWanted|Integers form Integral Domain and Element in Integral Domain is Unit iff Principal Ideal is Whole Domain}}
Category:Ideal Theory
Category:Quotient Rings
Category:Integers
my5ivmidti5dm6u697af8prw2n7jmhk
Theorem: Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal

Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $\ideal p$ be the principal ideal of $D$ generated by $p$.
Then $p$ is irreducible {{iff}} $\ideal p$ is a maximal ideal of $D$.

Proof:
=== Necessary Condition ===
{{:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication}}
Theorem: Subring Generated by Unity of Ring with Unity

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let the mapping $g: \Z \to R$ be defined as $\forall n \in \Z: \map g n = n 1_R$, where $n 1_R$ the $n$th power of $1_R$.
Let $\ideal x$ be the principal ideal of $\struct {R, +, \circ}$ generated by $x$.
Then $g$ is an epimorphism from $\Z$ onto the subring $S$ of $R$ generated by $1_R$.
If $R$ has no proper zero divisors, then $g$ is the only nonzero homomorphism from $\Z$ into $R$.
The kernel of $g$ is either:
:$(1): \quad \ideal {0_R}$, in which case $g$ is an isomorphism from $\Z$ onto $S$
or:
:$(2): \quad \ideal p$ for some prime $p$, in which case $S$ is isomorphic to the field $\Z_p$.

Proof:
By the Index Law for Sum of Indices and Integral Multiple of Ring Element, we have:
:$\paren {n 1_R} \paren {m 1_R} = n \paren {m 1_R} = \paren {n m} 1_R$
Thus $g$ is an epimorphism from $\Z$ onto $S$.
{{AimForCont}} $R$ has no proper zero divisors.
By Kernel of Ring Epimorphism is Ideal, the kernel of $g$ is an ideal of $\Z$.
By Ring of Integers is Principal Ideal Domain, the kernel of $g$ is $\ideal p$ for some $p \in \Z_{>0}$.
By Kernel of Ring Epimorphism is Ideal (don't think this is the correct reference - check it), $S$ is isomorphic to $\Z_p$ and also has no proper zero divisors.
So from Integral Domain of Prime Order is Field either $p = 0$ or $p$ is prime.
Now we need to show that $g$ is unique.
Let $h$ be  a non-zero (ring) homomorphism from $\Z$ into $R$.
As $\map h 1 = \map h {1^2} = \paren {\map h 1}^2$, either $\map h 1 = 1_R$ or $\map h 1 = 0_R$ by Idempotent Elements of Ring with No Proper Zero Divisors.
But, by Homomorphism of Powers: Integers, $\forall n \in \Z: \map h n = \map h {n 1} = n \map h 1$
So if $\map h 1 = 0_R$, then $\forall n \in \Z: \map h n = n 0_R = 0_R$.
Hence $h$ would be a zero homomorphism, which contradicts our stipulation that it is not.
So $\map h 1 = 1_R$, and thus $\forall n \in \Z: \map h n = n 1 = \map g n$.
{{qed}}
{{Proofread}}
Theorem: Null Ring iff Characteristic is One

The only ring whose characteristic is $1$ is the null ring.

Proof:
From Null Ring iff Zero and Unity Coincide, $1_R \ne 0_R$ except when $R = \left\{{0_R}\right\}$.
{{qed}}
Category:Ring Theory
6oqldbeeezzek7hrj6fbojap6vj42gf
Theorem: Characteristic of Finite Ring with No Zero Divisors

Let $\struct {R, +, \circ}$ be a finite ring with unity with no proper zero divisors whose zero is $0_R$ and whose unity is $1_R$.
Let $n \ne 0$ be the characteristic of $R$.
Then:
:$(1): \quad n$ must be a prime number
:$(2): \quad n$ is the order of all non-zero elements in $\struct {R, +}$.
It follows that $\struct {R, +} \cong C_n$, where $C_n$ is the cyclic group of order $n$.

Proof:
Follows directly from Subring Generated by Unity of Ring with Unity.
{{qed}}
Theorem: Characteristic of Finite Ring with No Zero Divisors

Let $\struct {R, +, \circ}$ be a finite ring with unity with no proper zero divisors whose zero is $0_R$ and whose unity is $1_R$.
Let $n \ne 0$ be the characteristic of $R$.
Then:
:$(1): \quad n$ must be a prime number
:$(2): \quad n$ is the order of all non-zero elements in $\struct {R, +}$.
It follows that $\struct {R, +} \cong C_n$, where $C_n$ is the cyclic group of order $n$.

Proof:
Suppose $\Char R = n$ where $n$ is composite.
Let $n = r s$, where $r, s \in \Z, r > 1, s > 1$.
First note that:
{{begin-eqn}}
{{eqn | l = \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}
      | r = \paren {r s} \paren {1_R \circ 1_R}
      | c = Integral Multiple of Ring Element
}}
{{eqn | r = \paren {r s} 1_R
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}
      | r = n \cdot 1_R
      | c = 
}}
{{eqn | r = 0_R
      | c = 
}}
{{eqn | ll= \leadsto
      | l = r \cdot 1_R = 0_R
      | o = \lor
      | r = s \cdot 1_R = 0_R
      | c = 
}}
{{end-eqn}}
as $R$ has no proper zero divisors.
But both $r$ and $s$ are less than $n$ which contradicting the minimality of $n$.
So if $\Char R = n$ it follows that $n$ must be prime.
Now let $x \in R^*$.
Then by Characteristic times Ring Element is Ring Zero, $n \cdot x = 0_R$.
It follows from Element to Power of Multiple of Order is Identity that:
:$\order x \divides n$
Since $n$ is prime, either $\order x = 1$ or $\order x = n$.
It cannot be $1$, from Null Ring iff Characteristic is One, so the result follows.
{{qed}}
Theorem: Integral Domain with Characteristic Zero

In an integral domain with characteristic zero, every non-zero element has infinite order under ring addition.

Proof:
Let $\struct {D, +, \circ}$ be an integral domain, whose zero is $0_D$ and whose unity is $1_D$, such that $\Char D = 0$.
Let $x \in D, x \ne 0_D$.
Then:
{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{>0}
      | l = n \cdot x
      | r = n \cdot \paren {x \circ 1_D}
      | c = 
}}
{{eqn | r = \paren {n \circ 1_D} \cdot x
      | c = Integral Multiple of Ring Element
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = x
      | o = \ne
      | r = 0_D
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n \cdot 1_D
      | o = \ne
      | r = 0_D
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n \cdot x
      | o = \ne
      | r = 0_D
      | c = {{Defof|Integral Domain}}
}}
{{end-eqn}}
That is, $x$ has infinite order in $\struct {D, +}$.
{{qed}}
Theorem: Characteristic of Field is Zero or Prime

Let $F$ be a field.
Then the characteristic of $F$ is either zero or a prime number.

Proof:
From the definition, a field is a ring with no zero divisors.
So by Characteristic of Finite Ring with No Zero Divisors, if $\Char F \ne 0$ then it is prime.
{{Qed}}
Theorem: Characteristic of Field is Zero or Prime

Let $F$ be a field.
Then the characteristic of $F$ is either zero or a prime number.

Proof:
From the definition, a field is an instance of an integral domain.
The result follows from Characteristic of Integral Domain is Zero or Prime.
{{Qed}}
Theorem: Field of Characteristic Zero has Unique Prime Subfield

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:
Follows directly from:
:Subring Generated by Unity of Ring with Unity
:Quotient Theorem for Monomorphisms
{{qed}}
Theorem: Field of Prime Characteristic has Unique Prime Subfield

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:
{{begin-eqn}}
{{eqn | l = \map \phi a
      | r = \map \phi {n + k_1 p}
      | c = 
}}
{{eqn | r = \paren {n + k_1 p} \cdot 1_F
      | c = 
}}
{{eqn | r = n \cdot 1_F + k_1 p \cdot 1_F
      | c = 
}}
{{eqn | r = n \cdot 1_F
      | c = 
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \map \phi {C_a} + \map \phi {C_b}
      | r = \map \phi {a' + k_a p} + \map \phi {b' + k_b p}
      | c = 
}}
{{eqn | r = \paren {a' + k_a p} \cdot 1_F + \paren {b' + k_b p} \cdot 1_F
      | c = 
}}
{{eqn | r = \paren {a' + k_a p + b' + k_b p} \cdot 1_F
      | c = Integral Multiple Distributes over Ring Addition
}}
{{eqn | r = \paren {a' + b'} \cdot 1_F + \paren {\paren {k_a p + k_b p} \cdot 1_F}
      | c = 
}}
{{eqn | r = \paren {a' + b'} \cdot 1_F
      | c = 
}}
{{eqn | r = \map \phi {C_a +_p C_b}
      | c = 
}}
{{end-eqn}}
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 $P_1 = P_2$.
{{qed}}
Theorem: Intersection of All Division Subrings is Prime Subfield

Let $\struct {K, +, \circ}$ be a division ring.
Let $P$ be the intersection of the set of all division subrings of $K$.
Then $P$ is the prime subfield of $K$.

Proof:
By Intersection of Division Subrings is Division Subring, the intersection $P$ of the set of all division subrings of $K$ is a division ring.
Let $\map Z K$ be the center of $K$.
From Center of Ring is Commutative Subring, $\map Z K$ is a commutative subring of $K$.
Therefore $\map Z K$ is a commutative division ring
Thus $\map Z K$ is a subfield of $K$.
But as $P$ is contained in $\map Z K$, it is itself commutative.
By its definition, $P$ contains no proper subfield and hence is a prime field.
Also, $P$ is contained in every other subfield of $K$.
Therefore $P$ is the ''only'' prime subfield of $K$.
{{qed}}
Category:Subfields
Category:Division Subrings
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Theorem: Characteristic of Ordered Integral Domain is Zero

Let $\struct {D, +, \circ}$ be an ordered integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\Char D$ be the characteristic of $D$.
Then $\Char D = 0$.
Let $g: \Z \to D$ be the mapping defined as:
:$\forall n \in \Z: \map g n = n \cdot 1_D$
where $n \cdot 1_D$ is defined as the $n$th power of $1_D$.
Then $g$ is the only monomorphism from the ordered ring $\Z$ onto the ordered ring $D$.

Proof:
By Properties of Ordered Ring $(5)$:
:$\forall n \in \Z_{>0}: n \cdot 1_D > 0$
Thus:
:$\forall p > 0: \Char D \ne p$
Hence:
:$\Char D = 0$
and so $g$ is a monomorphism from $\Z$ into $D$.
Also, if $m < p$, then $p - m \in \Z_+$, so:
:$p \cdot 1_D - m \cdot 1_D > 0_D$
Hence:
:$\map g m < \map g p$
Thus by Monomorphism from Total Ordering, $g$ is a monomorphism from $\Z$ into $D$.
{{qed}}
{{Proofread}}
{{Improve|We have a new and more axiomatic definition of an ordered integral domain which we may want to bring this into line with. Ultimately it boils down to the fact that an ordering on an integral domain is a lot more rigid than for a ring, such that totality is a consequence of the trichotomy law.}}