avewright/proofwiki-math huggingface dataset:
Theorem: Ring of Square Matrices over Field is Ring with Unity

Let $F$ be a field.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {\map {\MM_F} n, +, \times}$ denote the ring of square matrices of order $n$ over $F$.
Then $\struct {\map {\MM_F} n, +, \times}$ is a ring with unity, but is not a commutative ring.

Proof:
We have by definition that a field is a division ring which is also commutative.
Hence $F$ is a commutative ring with unity.
So, from Ring of Square Matrices over Commutative Ring with Unity we have that $\struct {\map {\MM_F} n, +, \times}$ is a ring with unity.
From Matrix Multiplication is not Commutative, we have that $\struct {\map {\MM_F} n, +, \times}$ is not a commutative ring.
Hence the result.
{{qed}}
Theorem: Ring of Square Matrices over Real Numbers

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {\map {\MM_\R} n, +, \times}$ denote the ring of square matrices of order $n$ over $\R$.
Then $\struct {\map {\MM_\R} n, +, \times}$ is a ring with unity, but is not a commutative ring.

Proof:
Recall that Real Numbers form Field.
The result follows directly from Ring of Square Matrices over Field is Ring with Unity.
{{qed}}
Theorem: Matrix Multiplication is not Commutative/Order 2 Square Matrices

Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\map {\MM_R} 2$ denote the $2 \times 2$ matrix space over $R$.
The operation of (conventional) matrix multiplication is not commutative over $\map {\MM_R} 2$.

Proof:
As $R$ is a ring with unity, we have that:
{{begin-eqn}}
{{eqn | l = 0_R
      | o = \ne
      | r = 1_R
}}
{{eqn | l = 0_R \times 0_R
      | r = 0_R
      | c = 
}}
{{eqn | l = 0_R \times 1_R
      | r = 0_R = 1_R \times 0_R
      | c = 
}}
{{eqn | l = 1_R \times 1_R
      | r = 1_R
      | c = 
}}
{{end-eqn}}
Now let:
{{begin-eqn}}
{{eqn | l = \mathbf A
      | r = \begin {pmatrix} 0_R & 1_R \\ 0_R & 0_R \end {pmatrix}
}}
{{eqn | l = \mathbf B
      | r = \begin {pmatrix} 0_R & 0_R \\ 1_R & 0_R \end {pmatrix}
}}
{{end-eqn}}
By definition, both $\mathbf A$ and $\mathbf B$ are elements of $\map {\MM_R} 2$.
It will be demonstrated that $\mathbf A$ and $\mathbf B$ do not commute.
We have:
{{begin-eqn}}
{{eqn | l = \mathbf A \mathbf B
      | r = \begin {pmatrix} 0_R & 1_R \\ 0_R & 0_R \end {pmatrix} \begin {pmatrix} 0_R & 0_R \\ 1_R & 0_R \end {pmatrix}
      | c = 
}}
{{eqn | r = \begin {pmatrix} 0_R \times 0_R + 1_R \times 1_R & 0_R \times 0_R + 1_R \times 0_R \\ 0_R \times 0_R + 0_R \times 1_R & 0_R \times 0_R + 0_R \times 0_R \end{pmatrix}
      | c = 
}}
{{eqn | r = \begin {pmatrix} 1_R & 0_R \\ 0_R & 0_R \end{pmatrix}
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \mathbf B \mathbf A
      | r = \begin {pmatrix} 0_R & 0_R \\ 1_R & 0_R \end {pmatrix} \begin {pmatrix} 0_R & 1_R \\ 0_R & 0_R \end {pmatrix}
      | c = 
}}
{{eqn | r = \begin {pmatrix} 0_R \times 0_R + 0_R \times 0_R & 0_R \times 1_R + 0_R \times 0_R \\ 1_R \times 0_R + 0_R \times 0_R & 1_R \times 1_R + 0_R \times 0_R \end{pmatrix}
      | c = 
}}
{{eqn | r = \begin {pmatrix} 0_R & 0_R \\ 0_R & 1_R \end{pmatrix}
      | c = 
}}
{{end-eqn}}
and it is seen that:
:$\mathbf A \mathbf B \ne \mathbf B \mathbf A$
Thus, whatever the nature of the ring with unity $R$, it is never the case that matrix multiplication is commutative over $\map {\MM_R} 2$.
{{qed}}
Theorem: Ring Zero is Unique

Let $\struct {R, +, \circ}$ be a ring.
Then the ring zero of $R$ is unique.

Proof:
The ring zero is, by definition of a ring, the identity element of the additive group $\struct {R, +}$.
The result then follows from Identity of Group is Unique.
{{qed}}
Theorem: Ring Zero is Unique

Let $\struct {R, +, \circ}$ be a ring.
Then the ring zero of $R$ is unique.

Proof:
From Ring Product with Zero we have that the ring zero of $R$ is indeed a zero element, as suggested by its name.
The result then follows from Zero Element is Unique.
{{qed}}
Theorem: Ring Zero is Unique

Let $\struct {R, +, \circ}$ be a ring.
Then the ring zero of $R$ is unique.

Proof:
Suppose $0$ and $0'$ are both ring zeroes of $\struct {R, +, \circ}$.
Then by Ring Product with Zero:
:$0' \circ 0 = 0$ by dint of $0$ being a ring zero
:$0' \circ 0 = 0'$ by dint of $0'$ being a ring zero.
So $0 = 0' \circ 0 = 0'$.
So $0 = 0'$ and there is only one ring zero of $\struct {R, +, \circ}$ after all.
{{qed}}
Theorem: Ring Negative is Unique

Let $\struct {R, +, \circ}$ be a ring.
Let $a \in R$.
Then the ring negative $-a$ of $a$ is unique.

Proof:
The ring negative is, by definition of a ring, the inverse element of $a$ in the additive group $\struct {R, +}$.
The result then follows from Inverse in Group is Unique.
{{qed}}
Theorem: Negative of Ring Negative

Let $\struct {R, +, \circ}$ be a ring.
Let $a \in R$ and let $-a$ be the ring negative of $a$.
Then:
:$-\paren {-a} = a$

Proof:
The ring negative is, by definition of a ring, the inverse element of $a$ in the additive group $\struct {R, +}$.
The result then follows from Inverse of Group Inverse.
{{qed}}
Theorem: Even Perfect Number is Triangular

All perfect numbers which are even are triangular.

Proof:
Follows from:
: Even Perfect Number is Hexagonal
: Hexagonal Number is Triangular Number
{{qed}}
Theorem: Field has no Proper Zero Divisors

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:
{{begin-eqn}}
{{eqn | l = a
      | o = \ne
      | r = 0_F
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^{-1} \times \paren {a \times b}
      | r = a^{-1} \times 0_F
      | c = $a^{-1}$ exists because $a \ne 0_F$
}}
{{eqn | ll= \leadsto
      | l = \paren {a^{-1} \times a} \times b
      | r = a^{-1} \times 0_F
      | c = {{Field-axiom|M1}}
}}
{{eqn | ll= \leadsto
      | l = 1_F \times b
      | r = a^{-1} \times 0_F
      | c = {{Field-axiom|M4}}
}}
{{eqn | ll= \leadsto
      | l = b
      | r = a^{-1} \times 0_F
      | c = {{Field-axiom|M3}}
}}
{{eqn | ll= \leadsto
      | l = b
      | r = 0_F
      | c = Field Product with Zero
}}
{{end-eqn}}
{{qed}}
Theorem: Multiplicative Identity is Unique

Let $\struct {F, +, \times}$ be a field.
Then the multiplicative identity $1_F$ of $F$ is unique.

Proof:
From the definition of multiplicative identity, $1_F$ is the identity element of the multiplicative group $\struct {F^*, \times}$.
The result follows from Identity of Group is Unique.
{{qed}}
Theorem: Multiplicative Inverse in Field is Unique

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Then the multiplicative inverse $a^{-1}$ of $a$ is unique.

Proof:
From the definition of multiplicative inverse, $a^{-1}$ is the inverse element of the multiplicative group $\struct {F^*, \times}$.
The result follows from Inverse in Group is Unique.
{{qed}}
Theorem: Multiplicative Inverse in Field is Unique

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Then the multiplicative inverse $a^{-1}$ of $a$ is unique.

Proof:
From the definition of a field as a division ring, every element of $F^*$ is a unit.
The result follows from Product Inverse in Ring is Unique.
{{qed}}
Theorem: Inverse of Multiplicative Inverse

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.
Let $a \in F$ such that $a \ne 0_F$.
Let $a^{-1}$ be the multiplicative inverse of $a$.
Then $\paren {a^{-1} }^{-1} = a$.

Proof:
{{begin-eqn}}
{{eqn | l = \paren {a^{-1} } \times a
      | r = a \times \paren {a^{-1} }
      | c = {{Field-axiom|M2}}
}}
{{eqn | r = 1_F
      | c = {{Field-axiom|M4}}
}}
{{eqn | ll= \leadsto
      | l = a
      | r = \paren {a^{-1} }^{-1}
      | c = {{Defof|Multiplicative Inverse in Field}}
}}
{{end-eqn}}
{{qed}}
Theorem: Integral Multiple Distributes over Ring Addition

Let $\struct {R, +, \times}$ be a ring, or a field.
Let $a, b \in R$ and $m, n \in \Z$.
Then:
:$(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$
:$(2): \quad m \cdot \paren {a + b} = \paren {m \cdot a} + \paren {m \cdot b}$
where $m \cdot a$ is as defined in integral multiple.

Proof:
We have that the additive group $\struct {R, +}$ is an abelian group.
$(1): \quad \paren {m + n} \cdot a = \paren {m \cdot a} + \paren {n \cdot a}$:
This is an instance of Powers of Group Elements: Sum of Indices when expressed in additive notation:
:$\forall n, m \in \Z: \forall a \in R: m a + n a = \paren {m + n} a$
{{qed|lemma}}
$(2): \quad m \cdot \paren {a + b} = \paren {m \cdot a} + \paren {m \cdot b}$:
This is an instance of Power of Product in Abelian Group when expressed in additive notation:
:$\forall n \in \Z: \forall a, b \in R: n \paren {a + b} = n a + n b$
{{qed}}
Theorem: Integral Multiple of Integral Multiple

Let $\struct {F, +, \times}$ be a field.
Let $a \in F$ and $m, n \in \Z$.
Then:
:$\paren {m n} \cdot a = m \cdot \paren {n \cdot a}$
where $n \cdot a$ is as defined in integral multiple.

Proof:
We have that $\struct {F^*, \times}$ is the multiplicative group of $\struct {F, +, \times}$.
Let $a \in F^*$, that is, $a \in F: a \ne 0_F$, where $0_F$ is the zero of $F$.
This is an instance of Powers of Group Elements when expressed in additive notation:
:$\forall m, n \in \Z: \paren {m n} a = m \paren {n a}$
{{qed|lemma}}
Now suppose $a = 0_F$.
Then by definition of the zero element of $F$, we have that:
:$\paren {m n} \cdot a = 0_F = m \cdot \paren {n \cdot a}$
{{qed}}
Theorem: Product of Integral Multiples

Let $\struct {F, +, \times}$ be a field.
Let $a, b \in F$ and $m, n \in \Z$.
Then:
:$\paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$
where $m \cdot a$ is as defined in Integral Multiple.

Proof:
First let $m = 0$ or $n = 0$.
{{WLOG}}, let $m = 0$.
The case where $n = 0$ follows the same lines.
We have:
{{begin-eqn}}
{{eqn | q = \forall a, b \in R: \forall n \in \Z_{>0}
      | l = \paren {m \cdot a} \times \paren {n \cdot b}
      | r = \paren {0 \cdot a} \times \paren {n \cdot b}
      | c = Definition of $m$
}}
{{eqn | r = 0_F \times \paren {n \cdot b}
      | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}
}}
{{eqn | r = 0_F
      | c = {{Defof|Field Zero}}
}}
{{eqn | r = 0_F \times \paren {a \times b}
      | c = {{Defof|Field Zero}}
}}
{{eqn | r = 0 \cdot \paren {a \times b}
      | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}
}}
{{eqn | r = \paren {0 \times n} \cdot \paren {a \times b}
      | c = $0$ is zero of $\Z$
}}
{{eqn | r = \paren {m \times n} \cdot \paren {a \times b}
      | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}
}}
{{end-eqn}}
{{qed|lemma}}
Next let $m > 0$ and $n > 0$.
{{begin-eqn}}
{{eqn | q = \forall a, b \in R: \forall m, n \in \Z_{>0}
      | l = \paren {m \cdot a} \times \paren {n \cdot b}
      | r = \paren {\sum_{i \mathop = 1}^m a} \times \paren {\sum_{i \mathop = 1}^n b}
      | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}
}}
{{eqn | r = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {a \times b}
      | c = General Distributivity Theorem
}}
{{eqn | r = \sum_{i \mathop = 1}^m \paren {\sum_{j \mathop = 1}^n \paren {a \times b} }
      | c = {{Defof|Summation}}
}}
{{eqn | r = m \cdot \paren {n \cdot \paren {a \times b} }
      | c = {{Defof|Integral Multiple|subdef = Rings and Fields}}
}}
{{eqn | r = \paren {m n} \cdot \paren {a \times b}
      | c = Powers of Group Elements: Additive Notation
}}
{{end-eqn}}
{{qed|lemma}}
The results for $m < 0$ and $n < 0$ follow directly from Powers of Group Elements.
{{qed}}
Theorem: Annihilator of Ring Always Contains Zero

Let $\struct {R, +, \times}$ be a ring.
Let $\map {\mathrm {Ann} } R$ be the annihilator of $R$.
Then $0 \in \map {\mathrm {Ann} } R$.

Proof:
We have by definition of integral multiple that:
:$\forall r \in R: 0 \cdot r = 0_R$
where $0_R$ is the zero of $R$.
Hence the result by definition of annihilator.
{{qed}}
Category:Ring Theory
sky2zptmfq5vwwjsru78klub6slkzlg
Theorem: Non-Trivial Annihilator Contains Positive Integer

Let $\left({R, +, \times}\right)$ be a ring with unity.
Let $A = \operatorname{Ann} \left({R}\right)$ be the annihilator of $R$.
Let $a \in A$ such that $a \ne 0$.
Then $A$ contains at least one strictly positive integer.

Proof:
Let the zero of $R$ be $0_R$ and the unity of $R$ be $1_R$.
First we note that:
{{begin-eqn}}
{{eqn | l = a
      | o = \in
      | r = \operatorname{Ann} \left({R}\right)
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a \cdot 1_R
      | r = 0_R
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \left({-a}\right) \cdot 1_R
      | r = 0_R
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\left({a \cdot 1_R}\right)
      | r = 0_R
      | c = 
}}
{{end-eqn}}
So:
: $a \in A \implies -a \in A$
But if $a \ne 0$ then either $a > 0$ or $-a > 0$.
That is, either $a$ or $-a$ is positive.
So, if $\operatorname{Ann} \left({F}\right)$ contains at least one non-zero element, it contains at least one strictly positive integer.
{{qed}}
Category:Rings with Unity
bg2avar95sakmqjanxs4npzpr0q0vfa
Theorem: Area of Sector

Let $\CC = ABC$ be a circle whose center is $A$ and with radii $AB$ and $AC$.
Let $BAC$ be the sector of $\CC$ whose angle between $AB$ and $AC$ is $\theta$.
:300px
Then the area $\AA$ of sector $BAC$ is given by:
:$\AA = \dfrac {r^2 \theta} 2$
where:
:$r = AB$ is the length of the radius of the circle
:$\theta$ is measured in radians.

Proof:
{{questionable|To be replaced with something rigorous, based on calculus.}}
From Area of Circle, the area of $\CC$ is $\pi r^2$.
From Measurement of Full Angle, the angle within $\CC$ is $2 \pi$.
The fraction of the area of $\CC$ within the sector $BAC$ is therefore $\pi r^2 \times \dfrac \theta {2 \pi}$.
Hence the result.
{{qed}}
Theorem: Rational Numbers form Prime Field

The field of rational numbers $\struct {\Q, +, \times}$ is a prime field.
That is, the only subset of $\Q$ which sustains both addition and multiplication are $\Q$ and $\set 0$, and vacuously $\O$.

Proof:
Let $F$ be a subfield of $\struct {\Q, +, \times}$.
Then $F$ is by definition a field.
Thus by definition:
:$\struct {F, +}$ is an abelian group
and:
:$\struct {F^*, \times}$ is an abelian group, where $F^* = F \setminus \set 0$.
and $\times$ is distributive over $+$:
:$\forall a, b, c \in F: a \times \paren {b + c} = a \times b + a \times c$
Let $a \in F^*$.
Then:
{{begin-eqn}}
{{eqn | l = a
      | o = \in
      | r = F^*
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^{-1}
      | o = \in
      | r = F^*
      | c = {{Group-axiom|3}} in $\struct {F^*, \times}$
}}
{{eqn | ll= \leadsto
      | l = a^{-1} \times a = 1
      | o = \in
      | r = F^*
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1
      | o = \in
      | r = F
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -1
      | o = \in
      | r = F
      | c = {{Group-axiom|3}} in $\struct {F, +}$
}}
{{eqn | ll= \leadsto
      | l = 1 + \paren {-1} = 0
      | o = \in
      | r = F
      | c = 
}}
{{end-eqn}}
We have $1 \in F$.
Suppose $n \in \Z$ such that $n \in F$.
Then:
:$n + 1 \in F$
So by the Principle of Mathematical Induction, all positive integers are in $F$.
By {{Group-axiom|3}} in $\struct {F, +}$:
:$m \in F \implies -m \in F$
So all negative integers are also in $F$.
That is:
:$\forall m \in \Z: m \in \F$
Recall {{Group-axiom|3}} in $\struct {F^*, \times}$:
:$m \in F^* \implies m^{-1} \in F^*$
So:
:$\forall p \in \Z, q \in \Z_{\ne 0}: p \times q^{-1} \in F$
and it follows by the Subfield Test that $F = \Q$.
{{qed}}
Theorem: Field of Integers Modulo Prime is Prime Field

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.
{{qed}}
Theorem: Field has Prime Subfield

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 $F$.
{{qed}}
Theorem: Constant Mapping to Identity is Homomorphism/Rings

Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be rings with zeroes $0_1$ and $0_2$ respectively.
Let $\zeta$ be the zero homomorphism from $R_1$ to $R_2$, that is:
:$\forall x \in R_1: \map \zeta x = 0_2$
Then $\zeta$ is a ring homomorphism whose image is $\set {0_2}$ and whose kernel is $R_1$.

Proof:
The additive groups of $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ are $\struct {R_1, +_1}$ and $\struct {R_2, +_2}$ respectively.
Their identities are $0_1$ and $0_2$ respectively.
Thus from the Constant Mapping to Group Identity is Homomorphism we have that $\zeta: \struct {R_1, +_1} \to \struct {R_2, +_2}$ is a group homomorphism:
:$\map \zeta {x +_1 y} = \map \zeta x +_2 \map \zeta y$
Then we have:
{{begin-eqn}}
{{eqn | l = \map \zeta {x \circ_1 y}
      | r = 0_2
      | c = as $x \circ_1 y \in R_1$
}}
{{eqn | r = \map \zeta x \circ_2 \map \zeta y
      | c = as $\map \zeta x = 0_2$ and $\map \zeta y = 0_2$
}}
{{end-eqn}}
The results about image and kernel follow directly by definition.
{{qed}}
Theorem: Constant Mapping to Identity is Homomorphism/Groups

Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identities $e_1$ and $e_2$ respectively.
Let $\phi_e: \struct {G_1, \circ_1} \to \struct {G_2, \circ_2}$ be the constant mapping defined as:
:$\forall x \in G_1: \map {\phi_e} x = e_2$
Then $\phi_e$ is a group homomorphism whose image is $\set {e_2}$ and whose kernel is $G_1$.

Proof:
Let $x, y \in G_1$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\phi_e} {x \circ_1 y}
      | r = e_2
      | c = as $x \circ_1 y \in G_1$
}}
{{eqn | r = \map {\phi_e} x \circ_2 \map {\phi_e} y
      | c = as $\map {\phi_e} x = e_2$ and $\map {\phi_e} y = e_2$
}}
{{end-eqn}}
So $\phi_e$ is a group homomorphism.
The results about image and kernel follow directly by definition.
{{qed}}
Theorem: Identity Mapping is Automorphism/Groups

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then $I_G: \struct {G, \circ} \to \struct {G, \circ}$ is a group automorphism.
Its kernel is $\set e$.

Proof:
The main result Identity Mapping is Automorphism holds directly.
As $I_G$ is a bijection, the only element that maps to $e$ is $e$ itself.
Thus the kernel is $\set e$.
{{qed}}
Theorem: Identity Mapping is Automorphism/Rings

Let $\struct {R, +, \circ}$ be a ring whose zero is $0$.
Then $I_R: \struct {R, +, \circ} \to \struct {R, +, \circ}$ is a ring automorphism.
Its kernel is $\set 0$.

Proof:
The result Identity Mapping is Automorphism holds directly, for both $+$ and $\circ$.
As $I_R$ is a bijection, the only element that maps to $0$ is $0$ itself.
Thus the kernel is $\set 0$.
{{qed}}
Theorem: Ring Monomorphism from Integers to Rationals

Let $\phi: \Z \to \Q$ be the mapping from the integers $\Z$ to the rational numbers $\Q$ defined as:
:$\forall x \in \Z: \map \phi x = \dfrac x 1$
Then $\phi$ is a (ring) monomorphism, but specifically not an epimorphism.

Proof:
First note that:
:$\forall a, b \in \Z:  a \ne b \implies \dfrac a 1 = \map \phi a \ne \map \phi b = \dfrac b 1$
and so clearly $\phi$ is an injection.
However, take for example $\dfrac 1 2$:
:$\not \exists a \in \Z: \map \phi a = \dfrac 1 2$
as $\dfrac 1 2 \notin \Img \phi$.
So $\phi$ is not a surjection.
Next let $a, b \in \Z$.
{{begin-eqn}}
{{eqn | l = \map \phi {a + b}
      | r = \frac {a + b} 1
      | c = 
}}
{{eqn | r = \frac a 1 + \frac b 1
      | c = 
}}
{{eqn | r = \map \phi a + \map \phi b
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \map \phi {a \times b}
      | r = \frac {a \times b} 1
      | c = 
}}
{{eqn | r = \frac {a \times b} {1 \times 1}
      | c = 
}}
{{eqn | r = \frac a 1 \times \frac b 1
      | c = 
}}
{{eqn | r = \map \phi a \times \map \phi b
      | c = 
}}
{{end-eqn}}
So $\phi$ can be seen to be an injective, but not surjective, ring homomorphism.
Hence the result by definition of ring monomorphism and ring epimorphism.
{{qed}}
Theorem: Field Homomorphism Preserves Unity

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.
{{Qed}}
Theorem: Field Homomorphism Preserves Product Inverses

Let $\phi: \struct {F_1, +_1, \times_1} \to \struct {F_2, +_2, \times_2}$ be a field homomorphism.
Then:
:$\forall x \in F_1^*: \map \phi {x^{-1} } = \map \phi x^{-1}$

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 product inverse of $x$ in $\struct {F_1, +_1, \times_1}$ for $\times_1$ is the product inverse of $x$ in $\struct {F_1^*, \times_1}$
:the product inverse of $x$ in $\struct {F_2, +_2, \times_2}$ for $\times_2$ is the product inverse of $x$ in $\struct {F_2^*, \times_2}$
The result follows from Group Homomorphism Preserves Inverses.
{{qed}}
Theorem: Ring Homomorphism from Field is Monomorphism or Zero Homomorphism

Let $\struct {F, +_F, \circ}$ be a field whose zero is $0_F$.
Let $\struct {S, +_S, *}$ be a ring whose zero is $0_S$.
Let $\phi: F \to S$ be a ring homomorphism.
Then either:
:$(1): \quad \phi$ is a monomorphism (that is, $\phi$ is injective)
or
:$(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in F: \map \phi a = 0_S$).

Proof:
We have by definition that a field is a division ring.
The result can be seen to be an application of Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism.
{{qed}}
Theorem: Ring Homomorphism from Field is Monomorphism or Zero Homomorphism

Let $\struct {F, +_F, \circ}$ be a field whose zero is $0_F$.
Let $\struct {S, +_S, *}$ be a ring whose zero is $0_S$.
Let $\phi: F \to S$ be a ring homomorphism.
Then either:
:$(1): \quad \phi$ is a monomorphism (that is, $\phi$ is injective)
or
:$(2): \quad \phi$ is the zero homomorphism (that is, $\forall a \in F: \map \phi a = 0_S$).

Proof:
Let $\phi: F \to S$ be a ring homomorphism.
Suppose $\phi$ is not a monomorphism.
By definition, $\phi$ is not an injection.
So there must exist $a, b \in F: \map \phi a = \map \phi b$.
Let $k = a +_F \paren {-b}$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi k
      | r = \map \phi {a +_F \paren {-b} }
      | c = 
}}
{{eqn | r = \map \phi a +_S \map \phi {-b}
      | c = 
}}
{{eqn | r = \map \phi a +_S \paren {-\map \phi b}
      | c = 
}}
{{eqn | r = 0_S
      | c = as $\map \phi a = \map \phi b$
}}
{{end-eqn}}
As $a \ne b$ then $k \ne 0$ and so has a product inverse $\exists k^{-1} \in F$.
So for any $x \in F$ we can write $x = k \circ \paren {k^{-1} \circ x}$ and so:
{{begin-eqn}}
{{eqn | l = \map \phi x
      | r = \map \phi {k \circ \paren {k^{-1} \circ x} }
      | c = 
}}
{{eqn | r = \map \phi k * \map \phi {k^{-1} \circ x}
      | c = 
}}
{{eqn | r = 0_S * \map \phi {k^{-1} \circ x}
      | c = 
}}
{{eqn | r = 0_S
      | c = 
}}
{{end-eqn}}
So if $\phi$ is not a monomorphism, it is the zero homomorphism.
{{qed}}