avewright/proofwiki-math huggingface dataset:
Theorem: Power of Product of Commuting Elements in Semigroup equals Product of Powers

:$\forall n \in \N_{>0}: \map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$

Proof:
The proof proceeds by the Principle of Mathematical Induction:
Let $\map P n$ be the proposition:
:$\map {\circ^n} {a \circ b} = \paren {\circ^n a} \circ \paren {\circ^n b}$
Theorem: Number of Compositions

A $k$-composition of a positive integer $n$ is an ordered $k$-tuple: $c = \tuple {c_1, c_2, \ldots, c_k}$ such that $c_1 + c_2 + \cdots + c_k = n$ and $c_i $ are strictly positive integers.
The number of $k$-composition of $n$ is $\dbinom {n - 1} {k - 1}$ and the total number of compositions of $n$ is $2^{n - 1}$ (that is for $k = 1, 2, 3, \ldots, n$).

Proof:
Consider the following array consisting of $n$ ones and $n - 1$ blanks:
:$\begin{bmatrix} 1 \ \_ \ 1 \ \_ \ \cdots \ \_ \ 1 \ \_ \ 1 \end{bmatrix}$
In each blank we can either put a comma or a plus sign.
Each way of choosing $,$ or $+$ will give a composition of $n$ with the commas separating the individual $c_i$'s.
It follows easily that there are $2^{n-1}$ ways of doing this, since there are two choices for each of $n-1$ blanks.
The result follows from the Product Rule for Counting.
Similarly if we want specifically $k$ different $c_i$'s then we are left with choosing $k - 1$ out of $n - 1$ blanks to place the $k - 1$ commas.
The number of ways of doing so is $\dbinom {n - 1} {k - 1}$ from the Binomial Theorem.
{{qed}}
Category:Combinatorics
8iu4de40dv0u51hbtzi3vkc7k08zpiv
Theorem: Sum of Sequences of Fifth and Seventh Powers

:$\ds \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$

Proof:
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^5 + \sum_{i \mathop = 1}^n i^7 = 2 \paren {\sum_{i \mathop = 1}^n i}^4$
Theorem: Cardinality of Set of Strictly Increasing Mappings

Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be tosets.
Let the cardinality of $S$ and $T$ be:
:$\card S = m, \card T = n$
Then the number of strictly increasing mappings from $S$ to $T$ is:
:$\dbinom n m = \dfrac {n!} {m! \ \paren {n - m}!}$.
where $\dbinom n m$ is a binomial coefficient.

Proof:
From:
:Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing
and:
:Strictly Monotone Mapping with Totally Ordered Domain is Injective
a strictly increasing mapping $\phi$ from $S$ to $T$ is an order isomorphism from $S$ to $\map \phi S$.
Let $\mathbb F$ be the set of all strictly increasing mappings from $S$ to $T$.
Let $\mathbb G$ be the set of all subsets of $T$ with $m$ elements.
By Unique Isomorphism between Equivalent Finite Totally Ordered Sets, the mapping $\Phi: \mathbb F \to \mathbb G$ defined as:
:$\forall \phi \in \mathbb F: \Phi: \phi \to \map \phi S$
is a bijection.
The result follows from Cardinality of Set of Subsets.
{{qed}}
Theorem: Construction of Inverse Completion/Cartesian Product with Cancellable Elements

Let $\struct {S \times C, \oplus}$ be the external direct product of $\struct {S, \circ}$ and $\struct {C, \circ {\restriction_C} }$, where $\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ {\restriction_C}$ on $C$.
That is:
:$\forall \tuple {x, y}, \tuple {u, v} \in S \times C: \tuple {x, y} \oplus \tuple {u, v} = \tuple {x \circ u, y \mathop {\circ {\restriction_C} } v}$
Then $\struct {S \times C, \oplus}$ is a commutative semigroup.

Proof:
By Cancellable Elements of Semigroup form Subsemigroup, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {S, \circ}$, where $\circ {\restriction_C}$ is the restriction of $\circ$ to $C$.
By Restriction of Commutative Operation is Commutative, as $\struct {C, \circ {\restriction_C} }$ is a substructure of a commutative structure, it is also commutative.
From:
:the external direct product preserves the nature of semigroups
:the external direct product preserves commutativity
we see that $\struct ({S \times C, \oplus}$ is a commutative semigroup.
{{Qed}}
Theorem: Construction of Inverse Completion/Equivalence Relation

The cross-relation $\boxtimes$ is an equivalence relation on $\struct {S \times C, \oplus}$.

Proof:
From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Also from Semigroup is Subsemigroup of Itself, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {C, \circ {\restriction_C} }$.
The result follows from Cross-Relation is Equivalence Relation.
{{qed}}
Theorem: Construction of Inverse Completion/Quotient Structure

Let the quotient structure defined by $\boxtimes$ be:
:$\struct {T', \oplus'} := \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$
where $\oplus_\boxtimes$ is the operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$.
=== Quotient Structure is Commutative Semigroup ===
{{:Construction of Inverse Completion/Quotient Structure is Commutative Semigroup}}
=== Quotient Mapping is Injective ===
{{:Construction of Inverse Completion/Quotient Mapping is Injective}}
=== Quotient Mapping is Monomorphism ===
{{:Construction of Inverse Completion/Quotient Mapping is Monomorphism}}
=== Image of Quotient Mapping is Subsemigroup ===
{{:Construction of Inverse Completion/Image of Quotient Mapping is Subsemigroup}}
=== Quotient Mapping to Image is Isomorphism ===
{{:Construction of Inverse Completion/Quotient Mapping to Image is Isomorphism}}
=== Image of Cancellable Elements in Quotient Mapping ===
{{:Construction of Inverse Completion/Quotient Mapping/Image of Cancellable Elements}}

Proof:
From the defined equivalence relation, we have that:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
is a congruence relation on $\struct {S \times C, \oplus}$.
From the definition of the members of the equivalence classes:
:$(1) \quad \forall x, y \in S, a, b \in C: \tuple {x \circ a, a} \boxtimes \tuple {y \circ b, b} \iff x = y$
:$(2) \quad \forall x, y \in S, a, b \in C: \eqclass {\tuple {x \circ a, y \circ a} } \boxtimes = \eqclass {\tuple {x, y} } \boxtimes$
From the definition of the equivalence class of equal elements:
:$(3) \quad \forall c, d \in C: \tuple {c, c} \boxtimes \tuple {d, d}$
where $\eqclass {\tuple {x, y} } \boxtimes$ is the equivalence class of $\tuple {x, y}$ under $\boxtimes$.
Hence we are justified in asserting the existence of the quotient structure:
:$\struct {T', \oplus'} = \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$
{{qed}}
Theorem: Integers form Commutative Ring

The set of integers $\Z$ forms a commutative ring under addition and multiplication.

Proof:
We have that:
:From Integers under Addition form Abelian Group, the algebraic structure $\struct {\Z, +}$ is an abelian group.
:From Integers under Multiplication form Countably Infinite Commutative Monoid, the algebraic structure $\struct {\Z, \times}$ is a commutative monoid and therefore a commutative semigroup.
:Integer Multiplication Distributes over Addition.
Thus all the ring axioms are fulfilled, and $\struct {\Z, +, \times}$ is a commutative ring.
By Integer Multiplication has Zero, the zero is $0$.
{{Qed}}
Theorem: Symmetric Group is Group

Let $S$ be a set.
Let $\map \Gamma S$ denote the set of all permutations on $S$.
Then $\struct {\map \Gamma S, \circ}$, the symmetric group on $S$, forms a group.

Proof:
Taking the group axioms in turn:
=== {{Group-axiom|0|nolink}} ===
By Composite of Permutations is Permutation, $S$ is itself a permutation on $S$.
Thus $\struct {\map \Gamma S, \circ}$ is closed.
{{qed|lemma}}
=== {{Group-axiom|1|nolink}} ===
From Set of all Self-Maps under Composition forms Monoid, we have that $\struct {\map \Gamma S, \circ}$ is associative.
{{qed|lemma}}
=== {{Group-axiom|2|nolink}} ===
From Set of all Self-Maps under Composition forms Monoid, we have that $\struct {\map \Gamma S, \circ}$ has an identity, that is, the identity mapping.
{{qed|lemma}}
=== {{Group-axiom|3|nolink}} ===
By Inverse of Permutation is Permutation, if $f$ is a permutation of $S$, then so is its inverse $f^{-1}$.
{{qed|lemma}}
Thus all the group axioms have been fulfilled, and the result follows.
{{qed}}
Theorem: Symmetric Group is Group

Let $S$ be a set.
Let $\map \Gamma S$ denote the set of all permutations on $S$.
Then $\struct {\map \Gamma S, \circ}$, the symmetric group on $S$, forms a group.

Proof:
A direct application of Set of Invertible Mappings forms Symmetric Group.
{{qed}}
Theorem: Quotient Ring Defined by Ring Itself is Null Ring

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {R / R, +, \circ}$ be the quotient ring defined by $R$.
Then $\struct {R / R, +, \circ}$ is a null ring.

Proof:
From Ring is Ideal of Itself, it is clear we can form the quotient ring $\struct {R / R, +, \circ}$.
But $R / R = 0_R$ and so is the null ring.
{{qed}}
Theorem: Set of Subrings forms Complete Lattice

Let $\struct {K, +, \circ}$ be a ring.
Let $\mathbb K$ be the set of all subrings of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a complete lattice.

Proof:
Let $\O \subset \mathbb S \subseteq \mathbb K$.
By Intersection of Subrings is Largest Subring Contained in all Subrings:
:$\bigcap \mathbb S$ is the largest subring of $K$ contained in each of the elements of $\mathbb S$.
By Intersection of Subrings Containing Subset is Smallest:
:The intersection of the set of all subrings of $K$ containing $\bigcup \mathbb S$ is the smallest subring of $K$ containing $\bigcup \mathbb S$.
Thus:
:Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum.
:The supremum of $\mathbb S$ is the intersection of the set of all subrings of $K$.
Therefore $\struct {\mathbb K, \subseteq}$ is a complete lattice.
{{qed}}
Theorem: Set of Ideals forms Complete Lattice

Let $\struct {K, +, \circ}$ be a ring.
Let $\mathbb K$ be the set of all ideals of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a complete lattice.

Proof:
Let $\O \subset \mathbb S \subseteq \mathbb K$.
By Intersection of Ring Ideals is Largest Ideal Contained in all Ideals:
:$\bigcap \mathbb S$ is the largest ideal of $K$ contained in each of the elements of $\mathbb S$.
By Intersection of Ring Ideals Containing Subset is Smallest:
:The intersection of the set of all ideals of $K$ containing $\bigcup \mathbb S$ is the smallest ideal of $K$ containing $\bigcup \mathbb S$.
Thus, not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum.
The supremum of $\mathbb S$ is the join of the set of all ideals of $\mathbb S$.
From Sum of Ideals is Ideal: General Result, this supremum is:
:$\ds \sum_{S \mathop \in \mathbb S} \struct {S, +, \circ}$
That is:
:$\struct {S_1, +, \circ} + \struct {S_2, +, \circ} + \dotsb$
where addition of ideals is as defined in subset product.
Therefore $\struct {\mathbb K, \subseteq}$ is a complete lattice.
{{qed}}
Theorem: Subset Product of Abelian Subgroups

Let $\left({G, \circ}\right)$ be an abelian group.
Let $H_1$ and $H_2$ be subgroups of $G$.
Then $H_1 \circ H_2$ is a subgroup of $G$.

Proof:
From Subgroup of Abelian Group is Normal, $H_1$ and $H_2$ are normal.
The result follows from Subset Product with Normal Subgroup is Subgroup.
{{qed}}
Category:Abelian Groups
Category:Subset Products
42ddr5761fjsy89srlm2c6znwzueywi
Theorem: Sum of Ideals is Ideal

Let $J_1$ and $J_2$ be ideals of a ring $\struct{R, +, \circ}$.
Then:
: $J = J_1 + J_2$ is an ideal of $R$
where $J_1 + J_2$ is as defined in subset product with respect to $\struct{R, +}$.

Proof:
By definition, $\struct {R, +}$ is an abelian group.
So from Subset Product of Abelian Subgroups, we have that:
:$\struct{J, +} = \struct{J_1, +} + \struct{J_2, +}$
is itself a subgroup of $R$.
Now consider $a \circ b$ where $a \in J, b \in R$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ b
      | r = \paren {a_1 + a_2} \circ b
      | c = by definition of $a$
}}
{{eqn | r = \paren {a_1 \circ b} + \paren {a_2 \circ b}
      | c = as $\circ$ is distributive over $+$
}}
{{eqn | o = \in
      | r = J_1 + J_2
      | c = as $\paren {a_i \circ b} \in J_i$ because $J_i$ is an ideal of $R$
}}
{{end-eqn}}
Similarly, $b \circ a \in J_1 + J_2$
So by definition $J_1 + J_2$ is an ideal of $R$.
{{qed}}
Theorem: Principal Ideal from Element in Center of Ring

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $b \in R$ be in the center of $R$.
Then:
:$\ideal b = R \circ b = \set {x \circ b: x \in R}$
where $\ideal b$ is the principal ideal generated by $b$.

Proof:
Let $J = R \circ b$
The center of $R$ is defined as:
:$\map Z R = \set {x \in R: \forall s \in R: s \circ x = x \circ s}$
Therefore:
:$R \circ b = b \circ R = \set {x \circ b: x \in R} = \set {b \circ x: x \in R}$
and so:
:$x \in J \implies x \circ b \in J \land b \circ x \in J$
Thus $J$ is an ideal of $R$ and so $J = \ideal b$.
{{qed}}
Theorem: External Direct Sum of Rings is Ring

Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be rings.
Then their (external) direct product:
:$\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$
is a ring.

Proof:
Consider the structures $\struct {R_1, +_1}, \struct {R_2, +_2}, \ldots, \struct {R_n, +_n}$.
By the definition of a ring, these are all groups.
From External Direct Product of Groups is Group we have that the their external direct product:
:$\ds \struct {R, +} = \prod_{k \mathop = 1}^n \struct {R_k, +_k}$
is a group.
Similarly, consider the structures $\struct {R_1, \circ_1}, \struct {R_2, \circ_2}, \ldots, \struct {R_n, \circ_n}$.
By the definition of a ring, these are all semigroups.
From External Direct Product of Semigroups we have that the their external direct product:
:$\ds \struct {R, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, \circ_k}$
is a semigroup.
Finally we note that from External Direct Product of Ringoids is Ringoid, $\circ$ as defined here is distributive over $+$.
Hence the result, by definition of ring.
{{qed}}
Theorem: External Direct Product of Ringoids is Ringoid

Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be ringoids.
Let $\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$ be their external direct product.
Then $\struct {R, +, \circ}$ is a ringoid.

Proof:
By definition of ringoid, $\circ_k$ distributes over $+_k$ for all $k = 1, 2, \ldots, n$.
Let $x, y, z \in R$.
Then:
{{begin-eqn}}
{{eqn | l = x \circ \paren {y + z}
      | r = \tuple {x_1, x_2, \ldots, x_n} \circ \paren {\tuple {y_1, y_2, \ldots, y_n} + \tuple {z_1, z_2, \ldots, z_n} }
}}
{{eqn | r = \tuple {x_1, x_2, \ldots, x_n} \circ \tuple {y_1 +_1 z_1, y_2 +_2 z_2, \ldots, y_n +_n z_n}
}}
{{eqn | r = \tuple {x_1 \circ_1 \paren {y_1 +_1 z_1}, x_2 \circ_2 \paren {y_2 +_2 z_2}, \ldots, x_n \circ_n \paren {y_n +_n z_n} }
}}
{{eqn | r = \tuple {\paren {x_1 \circ_1 y_1} +_1 \paren {x_1 \circ_1 z_1}, \paren {x_2 \circ_2 y_2} +_2 \paren {x_2 \circ_2 z_2}, \ldots, \paren {x_n \circ_n y_n} +_n \paren {x_n \circ_n z_n} }
}}
{{eqn | r = \tuple {\paren {x_1 \circ_1 y_1}, \paren {x_2 \circ_2 y_2}, \ldots, \paren {x_n \circ_n y_n} } + \tuple {\left({x_1 \circ_1 z_1}\right), \paren {x_2 \circ_2 z_2}, \ldots, \paren {x_n \circ_n z_n} }
}}
{{eqn | r = \paren {\tuple {x_1, x_2, \ldots, x_n} \circ \paren {y_1, y_2, \ldots, y_n} } + \paren {\tuple {x_1, x_2, \ldots, x_n} \circ \tuple {z_1, z_2, \ldots, z_n} }
}}
{{eqn | r = \paren {x \circ y} + \paren {x \circ z}
}}
{{end-eqn}}
In the same way:
:$\paren {y + z} \circ x = \paren {y \circ x} + \paren {z \circ x}$
Thus $\circ$ is distributive over $+$.
By definition of ringoid, $\struct {R, +, \circ}$ is a ringoid.
{{qed}}
Category:External Direct Products
Category:Distributive Operations
3j6ha5v597f4ayrj1lrlx88q6k6x3az
Theorem: Ideal of External Direct Sum of Rings

Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be rings.
Let 
:$\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$
be their direct product.
For each $k \in \closedint 1 n$, let:
:${R_k}' = \set {\tuple {x_1, \ldots, x_n} \in R: \forall j \ne k: x_j = 0}$
Then:
:$\forall k \in \closedint 1 n: {R_k}'$ is an ideal of $R$.

Proof:
Let $y = \tuple {y_1, y_2, \ldots, y_n} \in R$.
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in R_k$
By definition of direct product, we have:
:$x \circ y = \tuple {x_1 \circ y_1, x_2 \circ y_2, \ldots, x_n \circ y_n}$
:$y \circ x = \tuple {y_1 \circ x_1, y_2 \circ x_2, \ldots, y_n \circ x_n}$
But we have:
:$\forall j \ne k: x_j \circ y_j = 0 = y_j \circ x_j$
because $x_j = 0$.
So it follows that $x \circ y \in R_k$ and $y \circ x \in R_k$.
Hence ${R_k}'$ is an ideal of $R$.
{{qed}}
Theorem: Canonical Injection from Ideal of External Direct Sum of Rings

Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \dotsc, \struct {R_n, +_n, \circ_n}$ be rings.
Let $\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$ be their external direct product.
For each $k \in \closedint 1 n$, let:
:$R'_k = \set {\tuple {x_1, \dotsc, x_n} \in R: \forall j \ne k: x_j = 0}$
Let $\inj_k: R_k \to R$ be the canonical injection on the $k$th coordinate from $R_k$ into $\struct {R, +, \circ}$.
Let $\pr_k: R \to R'_k$ be the projection on the $k$th coordinate of $\struct {R, +, \circ}$ onto $R'_k$.
Then:
:$\inj_k: R_k \to R'_k$ is an isomorphism
:Its inverse is the restriction of $\pr_k$ to $R'_k$.

Proof:
From Ideal of External Direct Sum of Rings we have that $R'_k$ is an ideal of $R$, and thus a subring of $R$.
The result follows by application of Canonical Injection is Monomorphism.
{{qed}}
Theorem: Projection onto Ideal of External Direct Sum of Rings

Let $\struct {R_1, +_1, \circ_1}, \struct {R_2, +_2, \circ_2}, \ldots, \struct {R_n, +_n, \circ_n}$ be rings.
Let $\ds \struct {R, +, \circ} = \prod_{k \mathop = 1}^n \struct {R_k, +_k, \circ_k}$ be their direct product.
For each $k \in \closedint 1 n$, let:
:${R_k}' = \set {\tuple {x_1, \ldots, x_n} \in R: \forall j \ne k: x_j = 0}$
Let $\pr_k: R \to {R_k}'$ be the projection on the $k$th coordinate of $\struct {R, +, \circ}$ onto ${R_k}'$.
Then $\pr_k$ is an epimorphism.

Proof:
From Ideal of External Direct Sum of Rings we already have that ${R_k}'$ is an ideal of $R$.
The result follows by application of Projection is Epimorphism.
{{qed}}
Theorem: Conditions for Internal Ring Direct Sum

Let $\struct {R, +, \circ}$ be a ring.
Let $\sequence {\struct {S_k, +, \circ} }$ be a sequence of subrings of $R$.
Then $R$ is the ring direct sum of $\sequence {S_k}_{1 \mathop \le k \mathop \le n}$ {{iff}}:
:$(1): \quad R = S_1 + S_2 + \cdots + S_n$
:$(2): \quad \sequence {\struct {S_k, +} }_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups of $\struct {R, +}$
:$(3): \quad \forall k \in \closedint 1 n: S_k$ is an ideal of $R$.

Proof:
Let $S$ be the cartesian product of $\sequence {\struct {S_k, +} }$
First note that $\phi$ is a group homomorphism from $\struct {S, +}$ to $\struct {R, +}$, as:
:$\ds \sum_{j \mathop = 1}^n \paren {x_j + y_j} = \sum_{j \mathop = 1}^n x_j + \sum_{j \mathop = 1}^n y_j$ from Associativity on Indexing Set.
So $\phi$ is a ring homomorphism {{iff}}:
:$\ds \paren {\sum_{j \mathop = 1}^n x_j} \circ \paren {\sum_{j \mathop = 1}^n y_j} = \sum_{j \mathop = 1}^n \paren {x_j \circ y_j}$
Now, let:
:$\struct {S, +, \circ}$ be the cartesian product of $\sequence {\struct {S_k, +, \circ} }$
:$\phi: S \to R$ be the mapping defined as:
::$\map \phi {x_1, x_2, \ldots, x_n} = x_1 + x_2 + \cdots x_n$
Clearly $\phi$ is a surjection {{iff}} $(1)$ holds.
By Internal Direct Product Generated by Subgroups, $\phi$ is a ring isomorphism {{iff}} $(1)$ and $(2)$ hold.
Let $\phi: S \to R$ be a ring isomorphism.
By Canonical Injection from Ideal of External Direct Sum of Rings, $\map {\inj_k} {S_k}$ is an ideal of $S$.
So $\map \phi {\map {\inj_k} {S_k} }$ is an ideal of $R$.
But $\phi$ and $\pr_k$ coincide on $\map {\inj_k} {S_k}$.
So:
:$\map \phi {\map {\inj_k} {S_k} } = \map {\pr_k} {\map {\inj_k} {S_k} } = S_k$
and so $(3)$ holds.
Now suppose $(3)$ holds, and the $S_k$ are all ideals of $R$.
By $(2)$ and by definition of independent subgroups:
:$i \ne j \implies S_i \cap S_j = \set {0_R}$
So for all $\tuple {x_1, x_2, \ldots, x_n}, \tuple {y_1, y_2, \ldots, y_n} \in S$:
{{begin-eqn}}
{{eqn | l = \map \phi {x_1, x_2, \ldots, x_n} \circ \map \phi {y_1, y_2, \ldots, y_n}
      | r = \paren {\sum_{i \mathop = 1}^n x_i} \circ \paren {\sum_{j \mathop = 1}^n y_j}
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {\sum_{j \mathop = 1}^n x_i \circ y_j}
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n x_i \circ y_i
      | c = 
}}
{{eqn | r = \map \phi {x_1, x_2, \ldots, x_n} \circ \map \phi {y_1, y_2, \ldots, y_n}
      | c = 
}}
{{end-eqn}}
because as $S_i, S_j$ are ideals, we have:
:$x_i \circ j_j = S_i \cap S_j = \set {0_R}$
Hence the three conditions are sufficient for $\phi$ to be a ring isomorphism.
{{qed}}
Theorem: Structure Induced by Ring Operations is Ring

Let $\struct {R, +, \circ}$ be a ring.
Let $S$ be a set.
Then $\struct {R^S, +', \circ'}$ is a ring, where $+'$ and $\circ'$ are the pointwise operations induced on $R^S$ by $+$ and $\circ$.

Proof:
As $R$ is a ring, both $+$ and $\circ$ are closed on $R$ by definition.
From Closure of Pointwise Operation on Algebraic Structure, it follows that both $+'$ and $\circ'$ are closed on $R^S$:
:$\forall f, g \in R^S: f +' g \in R^S$
:$\forall f, g \in R^S: f \circ' g \in R^S$
By Structure Induced by Abelian Group Operation is Abelian Group, $\struct {R^S, +'}$ is an abelian group.
By Structure Induced by Associative Operation is Associative, $\struct {R^S, \circ'}$ is a semigroup.
From Pointwise Operation on Distributive Structure is Distributive, $\circ'$ is distributive over $+'$.
The result follows by definition of ring.
{{qed}}
Theorem: Division Subring Test

Let $\struct {K, +, \circ}$ be a division ring, and let $L$ be a subset of $K$.
Then $\struct {L, +, \circ}$ is a division subring of $\struct {K, +, \circ}$ {{iff}} these all hold:
:$(1) \quad L^* \ne \O$
:$(2) \quad \forall x, y \in L: x + \paren {-y} \in L$
:$(3) \quad \forall x, y \in L: x \circ y \in L$
:$(4) \quad x \in L^* \implies x^{-1} \in L^*$

Proof:
=== Necessary Condition ===
Suppose $\struct {L, +, \circ}$ is a division subring of $\struct {K, +, \circ}$.
The conditions $(1)$ to $(3)$ hold by virtue of the Subring Test.
Then $(4)$ also holds by the definition of a division ring:
:$\forall x \in L^*: \exists ! x^{-1} \in L^*: x^{-1} \circ x = x \circ x^{-1} = 1_L$
{{qed|lemma}}
Theorem: Set of Division Subrings forms Complete Lattice

Let $\struct {D, +, \circ}$ be a division ring.
Let $\mathbb K$ be the set of all division subrings of $K$.
Then $\struct {\mathbb K, \subseteq}$ is a complete lattice.

Proof:
Let $\O \subset \mathbb S \subseteq \mathbb K$.
By Intersection of Division Subrings is Largest Division Subring Contained in all Division Subrings:
:$\bigcap \mathbb S$ is the largest division subring of $K$ contained in each of the elements of $\mathbb S$.
By Intersection of Division Subrings Containing Subset is Smallest:
:The intersection of the set of all division subrings of $K$ containing $\bigcup \mathbb S$ is the smallest division subring of $K$ containing $\bigcup \mathbb S$.
Thus:
:Not only is $\bigcap \mathbb S$ a lower bound of $\mathbb S$, but also the largest, and therefore an infimum.
:The supremum of $\mathbb S$ is the intersection of the set of all division subrings of $K$.
Therefore $\struct {\mathbb K, \subseteq}$ is a complete lattice.
{{qed}}