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

Let $R$ be a commutative ring with unity.
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.
However, for $n \ge 2$, $\struct {\map {\MM_R} n, +, \times}$ is not a commutative ring.

Proof:
From Ring of Square Matrices over Ring with Unity we have that $\struct {\map {\MM_R} n, +, \times}$ is a ring with unity.
However, Matrix Multiplication is not Commutative.
Hence $\struct {\map {\MM_R} n, +, \times}$ is not a commutative ring for $n \ge 2$.
For $n = 1$ we have that:
{{begin-eqn}}
{{eqn | q = \forall \mathbf A, \mathbf B \in \map {\MM_R} 1
      | l = \mathbf A \mathbf B
      | r = a_{11} b_{11}
      | c = where $\mathbf A = \begin {pmatrix} a_11 \end {pmatrix}$ and $\mathbf B = \begin {pmatrix} b_11 \end {pmatrix}$
}}
{{eqn | r = b_{11} a_{11}
      | c = as $R$ is a commutative ring
}}
{{eqn | r = \mathbf {B A}
      | c = 
}}
{{end-eqn}}
Thus, for $n = 1$, $\struct {\map {\MM_R} n, +, \times}$ ''is'' a commutative ring.
{{qed}}
Theorem: Change of Basis Matrix is Nonsingular

Let $R$ be a ring with unity.
Let $M$ be a free $R$-module of finite dimension $n > 0$.
Let $\AA$ and $\BB$ be ordered bases of $M$.
Let $\mathbf P$ be the change of basis matrix from $\AA$ to $\BB$.
Then $\mathbf P$ is nonsingular, and its inverse $\mathbf P^{-1}$ is the change of basis matrix from $\BB$ to $\AA$.

Proof:
From Product of Change of Basis Matrices and Change of Basis Matrix Between Equal Bases:
:$\sqbrk {I_M; \AA, \BB} \sqbrk {I_M; \BB, \AA} = \sqbrk {I_M; \AA, \AA} = I_n$
:$\sqbrk {I_M; \BB, \AA} \sqbrk {I_M; \AA, \BB} = \sqbrk {I_M; \BB, \BB} = I_n$
Hence the result.
{{qed}}
Theorem: Invertible Matrix Corresponds with Change of Basis

Let $R$ be a commutative ring with unity.
Let $G$ be an $n$-dimensional unitary $R$-module.
Let $\sequence {a_n}$ be an ordered basis of $G$.
Let $\mathbf P = \sqbrk \alpha_n$ be a square matrix of order $n$ over $R$.
Let $\ds \forall j \in \closedint 1 n: b_j = \sum_{i \mathop = 1}^n \alpha_{i j} a_i$.
Then $\sequence {b_n}$ is an ordered basis of $G$ {{iff}} $\mathbf P$ is nonsingular.

Proof:
From Change of Basis Matrix is Nonsingular, if $\sequence {b_n}$ is an ordered basis of $G$ then $\mathbf P$ is nonsingular.
Now let $\mathbf P$ be nonsingular.
Then by {{Corollary|Set of Linear Transformations is Isomorphic to Matrix Space}}, there is an automorphism $u$ of $G$ which satisfies $\mathbf P = \sqbrk {u; \sequence {a_n} }$.
Therefore, as $\forall j \in \closedint 1 n: b_j = \map u {a_j}$, it follows that $\sequence {b_n}$ is also an ordered basis of $G$.
{{Qed}}
Theorem: Matrix Equivalence is Equivalence Relation

Matrix equivalence is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Matrix Similarity is Equivalence Relation

Matrix similarity is an equivalence relation.

Proof:
Follows directly from Matrix Equivalence is Equivalence Relation.
{{qed}}
Theorem: Matrix Similarity is Equivalence Relation

Matrix similarity is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
=== Reflexive ===
$\mathbf A = \mathbf{I_n}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all order $n$ square matrices $\mathbf A$.
So matrix similarity is reflexive.
{{qed|lemma}}
=== Symmetric ===
Let $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.
As $\mathbf P$ is {{apriori}} nonsingular, we have:
{{begin-eqn}}
{{eqn | l = \mathbf P \mathbf B \mathbf P^{-1}
      | r = \mathbf P \mathbf P^{-1} \mathbf A \mathbf P \mathbf P^{-1}
      | c = 
}}
{{eqn | r = \mathbf{I_n} \mathbf A \mathbf{I_n}
      | c = 
}}
{{eqn | r = \mathbf A
      | c = 
}}
{{end-eqn}}
So matrix similarity is symmetric.
{{qed|lemma}}
=== Transitive ===
Let $\mathbf B = \mathbf P_1^{-1} \mathbf A \mathbf P_1$ and $\mathbf C = \mathbf P_2^{-1} \mathbf B \mathbf P_2$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf C
      | r = \mathbf P_2^{-1} \mathbf P_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2
}}
{{eqn | r = \paren {\mathbf P_1 \mathbf P_2}^{-1} \mathbf A \paren {\mathbf P_1 \mathbf P_2}
      | c = Inverse of Matrix Product
}}
{{end-eqn}}
By Product of Matrices is Nonsingular iff Matrices are Nonsingular, $\paren {\mathbf P_1 \mathbf P_2}$ is nonsingular.
So matrix similarity is transitive.
{{qed|lemma}}
So, by definition, matrix similarity is an equivalence relation.
{{qed}}
Theorem: Similar Matrices are Equivalent

If two square matrices over a ring with unity $R$ are similar, then they are equivalent.
That is:
:every equivalence class for the similarity relation on $\map {\MM_R} n$ is contained in an equivalence class for the relation of matrix equivalence.
where $\map {\MM_R} n$ denotes the $n \times n$ matrix space over $R$.

Proof:
If $\mathbf A \sim \mathbf B$ then $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.
Let $\mathbf Q = \mathbf P$.
Then $\mathbf A$ are equivalent to $\mathbf B$, as:
:$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
{{qed}}
Theorem: Equivalent Matrices have Equal Rank

Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over a field $K$.
Let $\map \phi {\mathbf A}$ denote the rank of $\mathbf A$.
Let $\mathbf A \equiv \mathbf B$ denote that $\mathbf A$ and $\mathbf B$ are matrix equivalent.
Then:
:$\mathbf A \equiv \mathbf B$
{{iff}}:
:$\map \phi {\mathbf A} = \map \phi {\mathbf B}$

Proof:
Let $\mathbf A$ and $\mathbf B$ be $m \times n$ matrices over a field $K$ such that $\mathbf A \equiv \mathbf B$.
Let $S$ and $T$ be vector spaces of dimensions $n$ and $m$ over $K$.
Let $\mathbf A$ be the matrix of a linear transformation $u: S \to T$ relative to the ordered bases $\sequence {a_n}$ of $S$ and $\sequence {b_m}$ of $T$.
Let $\psi: K^m \to T$ be the isomorphism defined as:
:$\ds \map \psi {\sequence {\lambda_m} } = \sum_{k \mathop = 1}^m \lambda_k b_k$
Then $\psi$ takes the $j$th column of $\mathbf A$ into $\map u {a_j}$.
Hence it takes the subspace of $K^m$ generated by the columns of $\mathbf A$ onto the codomain of $u$.
Thus $\map \rho {\mathbf A} = \map \rho u$, and equivalent matrices over a field have the same rank.
Now let $\map \LL {K^n, K^m}$ be the set of all linear transformations from $K^n$ to $K^m$.
Let $u, v \in \map \LL {K^n, K^m}$ such that $\mathbf A$ and $\mathbf B$ are respectively the matrices of $u$ and $v$ relative to the standard ordered bases of $K^n$ and $K^m$.
Let $r = \map \phi {\mathbf A}$.
By Linear Transformation from Ordered Basis less Kernel, there exist ordered bases $\sequence {a_n}, \sequence {a'_n}$ of $K^n$ such that:
:$\sequence {\map u {a_r} }$ and $\sequence {\map v {a'_r} }$ are ordered bases of $\map u {K^n}$ and $\map v {K^n}$ respectively
and such that:
:$\set {a_k: k \in \closedint {r + 1} n}$ and $\set {a'_k: k \in \closedint {r + 1} n}$ are respectively bases of the kernels of $u$ and $v$.
Thus, by Results concerning Generators and Bases of Vector Spaces there exist ordered bases $\sequence {b_m}$ and $\sequence {b'_m}$ of $K^m$ such that $\forall k \in \closedint 1 r$:
{{begin-eqn}}
{{eqn | l = b_k
      | r = \map u {a_k}
      | c = 
}}
{{eqn | l = b'_k
      | r = \map v {a'_k}
      | c = 
}}
{{end-eqn}}
Let $z$ be the automorphism of $K^n$ which satisfies $\forall k \in \closedint 1 n: \map z {a'_k} = a_k$.
Let $w$ be the automorphism of $K^m$ which satisfies $\forall k \in \closedint 1 m: \map w {b'_k} = b_k$.
Then:
:<nowiki>$\map {\paren {w^{-1} \circ u \circ z} } {a'_k} = \begin {cases}
\map {w^{-1} } {b_k} = b'_k = \map v {a'_k} & : k \in \closedint 1 r \\
0 = \map v {a'_k} & : k \in \closedint {r + 1} n
\end {cases}$</nowiki>
So:
:$w^{-1} \circ u \circ z = v$
Now let $\mathbf P$ be the matrix of $z$ relative to the standard ordered bases of $K^n$, and let $\mathbf Q$ be the matrix of $w$ relative to the standard ordered bases of $K^m$.
Then $\mathbf P$ and $\mathbf Q$ are nonsingular and:
:$\mathbf Q^{-1} \mathbf A \mathbf P = \mathbf B$
and thus:
:$\mathbf A \equiv \mathbf B$
{{Qed}}
{{Proofread}}
Theorem: Number of Matrix Equivalence Classes

Let $K$ be a field.
Let $\map {\MM_K} {m, n}$ be the $m \times n$ matrix space over $K$.
Let $\mathbf A$ be an $m \times n$ matrix of rank $r$ over $K$.
Then:
:<nowiki>$\mathbf A \equiv \begin{cases}
\sqbrk {0_K}_{m n} & : r = 0 \\
& \\
\begin{bmatrix}
  \mathbf I_r & \bszero \\
  \bszero & \bszero 
\end{bmatrix} & : 0 < r < \min \set {n, m} \\
& \\
\begin{bmatrix}
  \mathbf I_r & \bszero 
\end{bmatrix} & : r = m < n \\
& \\
\begin{bmatrix}
  \mathbf I_r \\
  \bszero 
\end{bmatrix} & : r = n < m \\
& \\
\mathbf I_r & : r = m = n
\end{cases}$</nowiki>
Thus there are exactly $\min \set {m, n} + 1$ equivalence classes for the relation of equivalence on $\map {\MM_K} {m, n}$, one of which contains only the zero matrix.

Proof:
Follows from Equivalent Matrices have Equal Rank.
{{qed}}
Theorem: Transpose of Matrix Product

Let $\mathbf A$ and $\mathbf B$ be matrices over a commutative ring such that $\mathbf A \mathbf B$ is defined.
Then $\mathbf B^\intercal \mathbf A^\intercal$ is defined, and:
:$\paren {\mathbf A \mathbf B}^\intercal = \mathbf B^\intercal \mathbf A^\intercal$
where $\mathbf X^\intercal$ is the transpose of $\mathbf X$.

Proof:
Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{n p}$
Let $\mathbf A \mathbf B = \sqbrk c_{m p}$.
Then from the definition of matrix product:
:$\ds \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
So, let $\paren {\mathbf A \mathbf B}^\intercal = \sqbrk r_{p m}$.
The dimensions are correct, because $\mathbf A \mathbf B$ is an $m \times p$ matrix, thus making $\paren {\mathbf A \mathbf B}^\intercal$ a $p \times m$ matrix.
Thus:
:$\ds \forall j \in \closedint 1 p, i \in \closedint 1 m: r_{j i} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
Now, let $\mathbf B^\intercal \mathbf A^\intercal = \sqbrk s_{p m}$
Again, the dimensions are correct because $\mathbf B^\intercal$ is a $p \times n$ matrix and $\mathbf A^\intercal$ is an $n \times m$ matrix.
Thus:
:$\ds \forall j \in \closedint 1 p, i \in \closedint 1 m: s_{j i} = \sum_{k \mathop = 1}^n b_{k j} \circ a_{i k}$
As the underlying structure of $\mathbf A$ and $\mathbf B$ is a commutative ring, then $a_{i k} \circ b_{k j} = b_{k j} \circ a_{i k}$.
Note the order of the indices in the term in the summation sign on the {{RHS}} of the above.
They are reverse what they would normally be because we are multiplying the transposes together.
Thus it can be seen that $r_{j i} = s_{j i}$ and the result follows.
{{qed}}
Theorem: Rank is Dimension of Subspace

Let $K$ be a field.
Let $\mathbf A$ be an $m \times n$ matrix over $K$.
Then the rank of $\mathbf A$ is the dimension of the subspace of $K^n$ generated by the rows of $\mathbf A$.

Proof:
Let $u: K^n \to K^m$ be the linear transformation such that $\mathbf A$ is the matrix of $u$ relative to the standard ordered bases of $K^n$ and $K^m$.
Let $\map \rho {\mathbf A}$ be the rank of $\mathbf A$.
Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$.
Similar notations on $u$ denote the rank and transpose of $u$.
We have:
:$\map \rho {\mathbf A} = \map \rho u$
and:
:$\map \rho {\mathbf A^\intercal} = \map \rho {u^\intercal}$
but from Rank and Nullity of Transpose:
:$\map \rho {u^\intercal} = \map \rho u$
{{qed}}
Theorem: General Linear Group is Group

Let $K$ be a field.
Let $\GL {n, K}$ be the general linear group of order $n$ over $K$.
Then $\GL {n, K}$ is a group.

Proof:
Taking the group axioms in turn:
Theorem: Transpose of Row Matrix is Column Matrix

Let $\mathbf x = \sqbrk x_{1 n} = \begin {bmatrix} x_1 & x_2 & \cdots & x_n \end {bmatrix}$ be a row matrix.
Then $\mathbf x^\intercal$, the transpose of $\mathbf x$, is a column matrix:
:$\begin {bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}^\intercal = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$

Proof:
Self-evident.
{{Qed}}
Theorem: Transpose of Transpose of Matrix

Let $\mathbf A$ be a matrix.
Let $\mathbf A^\intercal$ be the transpose of $\mathbf A$.
Then:
:$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$

Proof:
Follows directly from the definition of the transpose of a matrix.
{{Qed}}
Theorem: Solution to Simultaneous Linear Equations

Let $\ds \forall i \in \closedint 1 m: \sum _{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$ be a system of simultaneous linear equations
where all of $\alpha_1, \ldots, a_n, x_1, \ldots x_n, \beta_i, \ldots, \beta_m$ are elements of a field $K$.
Then $x = \tuple {x_1, x_2, \ldots, x_n}$ is a solution of this system {{iff}}:
:$\sqbrk \alpha_{m n} \sqbrk x_{n 1} = \sqbrk \beta_{m 1}$
where $\sqbrk a_{m n}$ is an $m \times n$ matrix.

Proof:
We can see the truth of this by writing them out in full.
:$\ds \sum_{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$
can be written as:
{{begin-eqn}}
{{eqn | l = \alpha_{1 1} x_1 + \alpha_{1 2} x_2 + \ldots + \alpha_{1 n} x_n
      | r = \beta_1
      | c = 
}}
{{eqn | l = \alpha_{2 1} x_1 + \alpha_{2 2} x_2 + \ldots + \alpha_{2 n} x_n
      | r = \beta_2
      | c =
}}
{{eqn | o = \vdots
}}
{{eqn | l = \alpha_{m 1} x_1 + \alpha_{m 2} x_2 + \ldots + \alpha_{m n} x_n
      | r = \beta_m
      | c = 
}}
{{end-eqn}}
while $\sqbrk \alpha_{m n} \sqbrk x_{n 1} = \sqbrk \beta_{m 1}$ can be written as:
:<nowiki>$\begin {bmatrix}
\alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\
\alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\
      \vdots &       \vdots & \ddots &       \vdots \\
\alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n}
\end {bmatrix}
\begin {bmatrix}
x_1 \\ x_2 \\ \vdots \\ x_n
\end {bmatrix}
= \begin {bmatrix}
\beta_1 \\ \beta_2 \\ \vdots \\ \beta_m
\end {bmatrix}$</nowiki>
So the question:
:Find a solution to the following system of $m$ simultaneous linear equations in $n$ variables
is equivalent to:
:Given the following element $\mathbf A \in \map {\MM_K} {m, n}$ and $\mathbf b \in \map {\MM_K} {m, 1}$, find the set of all $\mathbf x \in \map {\MM_K} {n, 1}$ such that $\mathbf A \mathbf x = \mathbf b$
where $\map {\MM_K} {m, n}$ is the $m \times n$ matrix space over $S$.
{{qed}}
Theorem: Infinite Cyclic Group is Isomorphic to Integers

Let $G$ be an infinite cyclic group.
Then $G$ is isomorphic to the additive group of integers: $G \cong \struct {\Z, +}$.

Proof:
From the definition of an infinite cyclic group, we have:
:$G = \gen a = \set {a^k: k \in \Z}$
Let us define the mapping:
:$\phi: \Z \to G: \map \phi k = a^k$.
We now show that $\phi$ is an isomorphism.
From Mapping from Additive Group of Integers to Powers of Group Element is Homomorphism, $\phi$ is a homomorphism.
Now we show that $\phi$ is a surjection.
As $G$ is cyclic, every element of $G$ is apower of $a$ for some $a \in G$ such that $G = \gen a$.
Thus:
:$\forall x \in G: \exists k \in \Z: x = a^k$
By the definition of $\phi$:
:$\map \phi k = a^k = x$
Thus $\phi$ is surjective.
Now we show that $\phi$ is an injection.
This follows directly from Powers of Infinite Order Element, where:
:$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$
Thus $\phi$ is an injective, surjective homomorphism, thus:
: $G \cong \struct {\Z, +}$
as required.
{{Qed}}
Theorem: Subgroup of Infinite Cyclic Group is Infinite Cyclic Group

Let $G = \gen a$ be an infinite cyclic group generated by $a$, whose identity is $e$.
Let $g \in G, g \ne e: \exists k \in \Z, k \ne 0: g = a^k$.
Let $H = \gen g$.
Then $H \le G$ and $H \cong G$.
Thus, all non-trivial subgroups of an infinite cyclic group are themselves infinite cyclic groups.
A subgroup of $G = \gen a$ is denoted as follows:
:$n G := \gen {a^n}$
This notation is usually used in the context of $\struct {\Z, +}$, where $n \Z$ is (informally) understood as '''the set of integer multiples of $n$'''.

Proof:
The fact that $H \le G$ follows from the definition of subgroup generator.
By Infinite Cyclic Group is Isomorphic to Integers:
:$G \cong \struct {\Z, +}$
Now we show that $H$ is of infinite order.
Suppose $\exists h \in H, h \ne e: \exists r \in \Z, r > 0: h^r = e$.
But:
:$h \in H \implies \exists s \in \Z, s > 0: h = g^s$
where $g = a^k$.
Thus:
:$e = h^r = \paren {g^s}^r = \paren {\paren {a^k}^s}^r = a^{k s r}$
and thus $a$ is of finite order.
This would mean that $G$ was also of finite order.
So $H$ must be of infinite order.
From Subgroup of Cyclic Group is Cyclic, as $G$ is cyclic, then $H$ must also be cyclic.
From Infinite Cyclic Group is Isomorphic to Integers:
:$H \cong \struct {\Z, +}$
Therefore, as $G \cong \struct {\Z, +}$:
:$H \cong G$
{{qed}}
Theorem: Quotient Group of Infinite Cyclic Group by Subgroup

Let $C_n$ be the cyclic group of order $n$.
Then:
:$C_n \cong \dfrac {\struct {\Z, +} } {\struct {n \Z, +} } = \dfrac \Z {n \Z}$
where:
:$\Z$ is the additive group of integers
:$n \Z$ is the additive group of integer multiples
:$\Z / n \Z$ is the quotient group of $\Z$ by $n \Z$.
Thus, every cyclic group is isomorphic to one of:
:$\Z, \dfrac \Z \Z, \dfrac \Z {2 \Z}, \dfrac \Z {3 \Z}, \dfrac \Z {4 \Z}, \ldots$

Proof:
Let $C_n = \gen {a: a^n = e_{C_n} }$, that is, let $a$ be a generator of $C_n$.
Let us define $\phi: \struct {\Z, +} \to C_n$ such that:
:$\forall k \in \Z: \map \phi k = a^k$
Then from the First Isomorphism Theorem:
:$\Img \phi = C_n = \struct {\Z, +} / \map \ker \phi$
We now need to show that $\map \ker \phi = n \Z$.
We have:
:$\map \ker \phi = \set {k \in \Z: a^k = e_{C_n} }$
Let $x \in \map \ker \phi$.
Then $a^x = e_{C_n}$ and thus:
:$n \divides x$
The result follows.
{{qed}}
Theorem: Ideals Containing Ideal Form Lattice

Let $J$ be an ideal of a ring $R$.
Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$.
Then the ordered set $\struct {\mathbb L_J, \subseteq}$ is a lattice.

Proof:
Let $b_1, b_2 \in \mathbb L_J$.
Then from Set of Ideals forms Complete Lattice:
:$(1): \quad b_1 + b_2 \in \mathbb L_J$ and is the supremum of $\set {b_1, b_2}$
:$(2): \quad b_1 \cap b_2 \in \mathbb L_J$ and is the infimum of $\set {b_1, b_2}$
Thus $\struct {\mathbb L_J, \subseteq}$ is a lattice.
{{qed}}
Theorem: Ideals Containing Ideal Isomorphic to Quotient Ring

Let $J$ be an ideal of a ring $R$.
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 $a / J$ from the definition of quotient ring.
Then $\Phi_J$ is an isomorphism.

Proof:
Let $b \in \mathbb L_J$.
From the way $\mathbb L_J$ is defined:
:$J \subseteq b$
Thus by Preimage of Image of Subring under Ring Homomorphism:
:$\map {q_J^{-1} } {\map {q_J} b} = b + J = b$
Let $c$ be an ideal of $R / J$.
Then, by Image of Preimage of Subring under Ring Epimorphism:
:$\map {q_J} {\map {q_J^{-1} } c} = c$
Thus by Bijection iff Left and Right Inverse, $\Phi_J$ is a bijection.
Hence:
:$\forall c \in \map {\mathbb L} {R / J}: \map {q_J^{-1} } {\Phi_J} c = \map {q_J^{-1} } c$
Now to show that $\Phi_J$ is an isomorphism.
Let $b_1, b_2 \in \mathbb L_J$.
Let $b_1 \subseteq b_2$.
Then from Subset Maps to Subset:
:$\map {q_J} {b_1} \subseteq \map {q_J} {b_2}$
Conversely, suppose $\map {q_J} {b_1} \subseteq \map {q_J} {b_2}$.
By what we have just proved:
:$b_1 = \map {q_J^{-1} } {\map {q_J} {b_1} } \subseteq \map {q_J^{-1} } {\map {q_J} {b_2} } = b_2$
Thus $\Phi_J$ is an isomorphism.
{{Qed}}
Theorem: Ring of Integers is Principal Ideal Domain

The integers $\Z$ form a principal ideal domain.

Proof:
Let $J$ be an ideal of $\Z$.
Then $J$ is a subring of $\Z$, and so $\left({J, +}\right)$ is a subgroup of $\left({\Z, +}\right)$.
But by Integers under Addition form Infinite Cyclic Group, the group $\left({\Z, +}\right)$ is cyclic, generated by $1$.
Thus by Subgroup of Cyclic Group is Cyclic, $\left({J, +}\right)$ is cyclic, generated by some $m \in \Z$.
Therefore from the definition of principal ideal, $J = \left\{{k m: k \in \Z}\right\} = \left({m}\right)$, and is thus a principal ideal.
{{qed}}
Theorem: Ring of Integers is Principal Ideal Domain

The integers $\Z$ form a principal ideal domain.

Proof:
We have that Integers are Euclidean Domain.
Then we have that Euclidean Domain is Principal Ideal Domain.
Hence the result.
{{qed}}
Theorem: Ring of Integers is Principal Ideal Domain

The integers $\Z$ form a principal ideal domain.

Proof:
Let $U$ be an arbitrary ideal of $\Z$.
Let $c$ be a non-zero element of $U$.
Then both $c$ and $-c$ belong to $\ideal a$ and one of them is positive.
Thus $U$ contains strictly positive elements.
Let $b$ be the smallest strictly positive element of $U$.
By the Set of Integers Bounded Below by Integer has Smallest Element, $b$ is guaranteed to exist.
If $\ideal b$ denotes the ideal generated by $b$, then $\ideal b \subseteq U$ because $b\in U$ and $U$ is an ideal.
Let $a \in U$.
By the Division Theorem:
:$\exists q, r \in \Z, 0 \le r < b: a = b q + r$
As $a, b \in U$ it follows that so does $r = a - b q$.
By definition of $b$ it follows that $r = 0$.
Thus:
:$a = b q \in \ideal b$
and so:
:$U \subseteq \ideal b$
From the above:
:$U = \ideal b$
It follows by definition that $U$ is a principal ideal of $\Z$.
Recall that $U$ was an arbitrary ideal of $\Z$.
Hence by definition $\Z$ is a principal ideal domain.
{{qed}}
Theorem: Principal Ideals of Integers

Let $J$ be a non-zero ideal of $\Z$.
Then $J = \ideal b$ where $b$ is the smallest strictly positive integer belonging to $J$.

Proof:
It follows from Ring of Integers is Principal Ideal Domain that $J$ is a principal ideal.
Let $c \in J, c \ne 0$.
Then $-c \in J$ and by Natural Numbers are Non-Negative Integers, exactly one of them is strictly positive.
Thus $J$ ''does'' actually contain strictly positive elements, so that's a start.
Let $b$ be the smallest strictly positive element of $J$.
This exists because Natural Numbers are Non-Negative Integers and the Well-Ordering Principle.
By definition of a principal ideal, we have $\ideal b \subseteq J$ as $b \in J$.
We need to show that $J \subseteq \ideal b$.
So, let $a \in J$.
By the Division Theorem, $\exists q, r: a = b q + r, 0 \le r < b$.
As $a, b \in J$, then so does $r = a - b q$.
So, by the definition of $b$, it follows that $r = 0$.
Thus $a = b q \in \ideal b$.
{{qed}}
Theorem: Natural Numbers Set Equivalent to Ideals of Integers

Let $S$ be the set of all ideals of $\Z$.
Let the mapping $\psi: \N \to S$ be defined as:
:$\forall b \in \N: \map \psi b = \ideal b$
where $\ideal b$ is the principal ideal of $\Z$ generated by $b$.
Then $\psi$ is a bijection.

Proof:
First we show that $\psi$ is injective.
Suppose $0 < b < c$.
Then $b \in \ideal b$, but $b \notin \ideal c$, because from Principal Ideals of Integers, $c$ is the smallest positive integer in $\ideal c$.
Thus $\ideal b \ne \ideal c$.
It is also apparent that $b > 0 \implies \ideal b \ne \ideal 0$ as $\ideal 0 = \set 0$.
Thus $\psi$ is injective.
Surjectivity follows from Principal Ideals of Integers: every integer is the smallest strictly positive element of a principal ideal of $\Z$.
{{qed}}