nyuuzyou/wikis huggingface dataset:
Ring of Polynomial Forms is Integral Domain

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Let $X \in R$ be transcendental over $D$.
Let $D \sqbrk X$ be the ring of polynomials in $X$ over $D$.
Then $D \sqbrk X$ is an integral domain.
== Proof ==
By Ring of Polynomial Forms is Commutative Ring with Unity we know that $D \sqbrk X$ is a commutative ring with unity.
Let neither $\ds \map f X = \sum_{k \mathop = 0}^n a_k x^k$ nor $\ds \map g X = \sum_{k \mathop = 0}^m b_k X^k$ be the null polynomial.
Then their leading coefficients $a_n$ and $b_m$ are non-zero.
Therefore, as $D$ is an integral domain and $a_n, b_m \in D$, so is their product $a_n b_m$.
By the definition of polynomial multiplication, it follows that $f g$ is not the null polynomial.
It follows that $D \sqbrk X$ has no proper zero divisors.
Hence $D \sqbrk X$ is an integral domain.
== Also see ==
* Ring of Polynomial Forms over Integral Domain is Integral Domain
== Sources ==
* : Chapter $1$: Integral Domains: $\S 5$. Further Examples of Integral Domains: Example $5$
* : $\S 64.3$ Polynomial rings over an integral domain
Category:Polynomial Rings
Category:Integral Domains
Ring of Polynomial Forms is not necessarily Isomorphic to Ring of Polynomial Functions

== Theorem ==
Let $D$ be an integral domain.
Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.
Let $\map P D$ be the ring of polynomial functions over $D$.
Then it is not necessarily the case that $D \sqbrk X$ is isomorphic with $\map P D$.
== Proof ==
Proof by Counterexample:
Consider the integral domain $\struct {\Z_2, +, \times}$.
From Ring of Integers Modulo Prime is Integral Domain, it is seen that $\struct {\Z_2, +, \times}$ is indeed an integral domain.
Consider the ring of polynomial forms $\Z_2 \sqbrk X$.
This is an infinite ring, as it can be seen that $S \subseteq \Z_2 \sqbrk X$ where:
:$S = \set {1, X, X^2, X^3, \dotsc}$
But the ring of polynomial functions $\map P D$ is finite, as:
:$\map P D \subseteq \Z_2^{Z_2}$
where $\Z_2^{Z_2}$ is the set of all mappings from $\Z_2$ to $\Z_2$, and has $4$ elements.
== Also see ==
* Epimorphism from Polynomial Forms to Polynomial Functions, where it is shown that there exists an epimorphism from $D \sqbrk X$ to $\map P D$.
== Sources ==
* : $\S 64$ Polynomial rings over an integral domain
Category:Polynomial Theory
Ring of Polynomial Forms over Field is Vector Space

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose unity is $1_F$.
Let $F \sqbrk X$ be the ring of polynomials over $F$.
Then $F \sqbrk X$ is an vector space over $F$.
=== Corollary ===
== Proof ==
Let the operation $\times': F \to F \sqbrk X$ be defined as follows.
Let $x \in F$.
Let $\mathbf y \in F \sqbrk X$ be defined as:
:$\mathbf y = \ds \sum_{k \mathop = 0}^n y_k X^k$
where $n = \map \deg {\mathbf y}$ denotes the degree of $\mathbf y$
Thus:
:$x \times' \mathbf y := \ds x \sum_{k \mathop = 0}^n y_k X^k = \sum_{k \mathop = 0}^n \paren {x \times y_k} X^k$
We have that $\times': F \to F \sqbrk X$ is an instance of polynomial multiplication where the multiplier $x$ is a polynomial of degree $0$.
Hence, let the supposed vector space over $F$ in question be denoted in full as:
:$\mathbf V = \struct {F \sqbrk X, +', \times'}_F$
where:
:$+': F \sqbrk X \to F \sqbrk X$ denotes polynomial addition
:$\times': F \to F \sqbrk X$ denotes the operation as defined above. 
We already have that $F \sqbrk X$ is an integral domain.
Thus vector space axioms $\text V 0$ to $\text V 4$ are fulfilled.
By definition of $\times'$, it is seen that the remaining vector space axioms are fulfilled as follows:
Let $\lambda, \mu \in F$.
Let $\mathbf x, \mathbf y \in F \sqbrk X$ such that $\map \deg {\mathbf x} = m$ and $\map \deg {\mathbf y} = n$.
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All vector space axioms are hence seen to be fulfilled.
Hence the result.
== Sources ==
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Example $63$
Category:Examples of Vector Spaces
Ring of Polynomial Forms over Field is Vector Space/Corollary

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose unity is $1_F$.
Let $F \sqbrk X$ be the ring of polynomials over $F$.
Let $S \subseteq F \sqbrk X$ denote the subset of $F \sqbrk X$ defined as:
:$S = \set {\mathbf x \in F \sqbrk X: \map \deg {\mathbf x} < d}$
for some $d \in \Z_{>0}$.
Then $S$ is an vector space over $F$.
== Proof ==
From Ring of Polynomial Forms over Field is Vector Space we note that $\struct {F, +, \times}$ is a vector space over $F$.
The remaining question is that $S$ remains closed under polynomial addition and scalar multiplication.
Let $\mathbf x, \mathbf y \in S$ such that $\map \deg {\mathbf x} = m$ and $\map \deg {\mathbf y} = n$.
We have:
As both $m < d$ and $n < d$ by dint of them belonging to $S$, it follows that $\max \set {m, n} < d$.
Hence $\mathbf x + \mathbf y \in S$ and so $S$ is closed under polynomial addition.
Let the operation $\times': F \to F \sqbrk X$ be defined as follows.
Let $\lambda \in F$.
Let $\mathbf x \in F \sqbrk X$ be defined as:
:$\mathbf x = \ds \sum_{k \mathop = 0}^n x_k X^k$
where $n = \map \deg {\mathbf x}$ denotes the degree of $\mathbf x$.
Thus:
:$\lambda \times' \mathbf x := \ds \lambda \times' \sum_{k \mathop = 0}^n x_k X^k = \sum_{k \mathop = 0}^n \paren {\lambda \times x_k} X^k$
We have that $\times': F \to F \sqbrk X$ is an instance of polynomial multiplication where the multiplier $\lambda$ is a polynomial of degree $0$.
By definition of polynomial multiplication:
The result follows.
== Sources ==
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Example $63$
Category:Examples of Vector Spaces
Ring of Polynomial Forms over Integral Domain is Integral Domain

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Let $\struct {D \sqbrk X, \oplus, \odot}$ be the ring of polynomial forms over $D$ in the indeterminate $X$.
Then $\struct {D \sqbrk X, \oplus, \odot}$ is an integral domain.
== Proof ==
By definition an integral domain is a commutative ring with unity. 
From Ring of Polynomial Forms is Commutative Ring with Unity it follows that $\struct {D \sqbrk X, +, \circ}$ is a commutative ring with unity.
Suppose $f, g \in D \sqbrk X$ such that neither $f$ nor $g$ are the null polynomial.
Let $\map \deg f = n$ and $\map \deg g = m$.
From Degree of Product of Polynomials over Integral Domain the degree of $f \odot g$ is $n + m$.
Thus by definition $f \odot g$ is not the null polynomial of $D \sqbrk X$.
Thus neither $f$ nor $g$ is a proper zero divisor of $D \sqbrk X$.
This holds for any two arbitrary non-null polynomials elements of $D \sqbrk X$.
Hence $\struct {D \sqbrk X, \oplus, \odot}$ is a commutative ring with unity with no proper zero divisors.
That is, $\struct {D \sqbrk X, \oplus, \odot}$ is an integral domain.
== Sources ==
* : $\S 3.2$: Polynomial rings: Lemma $3.7 \ \text{(iii)}$ 
Category:Polynomial Rings
Category:Integral Domains
Ring of Polynomial Functions is Commutative Ring with Unity

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $R \sqbrk {\set {X_j: j \in J} }$ be the ring of polynomial forms over $R$ in the indeterminates $\set {X_j: j \in J}$.
Let $R^J$ be the free module on $J$.
Let $A$ be the set of all polynomial functions $R^J \to R$.
Let $\struct {A, +, \circ}$ be the ring of polynomial functions on $R$.
Then $\struct {A, +, \circ}$ is a commutative ring with unity.
== Proof ==
First we check that the operations of ring product and ring addition are closed in $A$.
Let $Z$ be the set of all multiindices indexed by $J$.
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k, \ g = \sum_{k \mathop \in Z} b_k \mathbf X^k \in R \sqbrk {\set {X_j: j \in J} }$.
Under the evaluation homomorphism, $f$ and $g$ map to:
:$\ds A \owns \hat f: \forall x \in R^J: \map {\hat f} x = \sum_{k \mathop \in Z} a_k x^k$
:$\ds A \owns \hat g: \forall x \in R^J: \map {\hat g} x = \sum_{k \mathop \in Z} b_k x^k$
Then the induced pointwise sum of $\hat f$ and $\hat g$ is:
Thus polynomial functions are closed under ring addition.
The [[induced pointwise product of $\hat f$ and $\hat g$ is:
Thus polynomial functions are closed under ring product.
Finally, we invoke Structure Induced by Ring Operations is Ring, which shows that $\struct {A, +, \circ}$ is a commutative ring with unity.
Category:Polynomial Theory
Ring of Polynomials over Reals is not Field

== Theorem ==
Let $\R \sqbrk X$ be the ring of polynomials in an indeterminate $X$ over $\R$.
Then $\R \sqbrk X$ is not a field.
== Proof ==
Consider the polynomial $x + 1$ in $\R \sqbrk X$.
There exists no polynomial $\map f x$ such that:
:$\paren {x + 1} \map f x = 1$
This is because the has degree $1$, and the has degree $0$.
== Sources ==
* : Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $19$
Category:Polynomial Rings
Category:Field Theory
Ring of Quaternions is Ring

== Theorem ==
The set $\mathbb H$ of quaternions forms a ring under the operations of addition and multiplication.
== Proof ==
From Quaternion Addition forms Abelian Group‎, $\mathbb H$ forms an abelian group under quaternion addition.
From the definition it is clear that quaternion multiplication is closed.
We have from Matrix Form of Quaternion that quaternions can be expressed in matrix form.
From Quaternion Multiplication‎ we have that quaternion multiplication can be expressed in terms of matrix multiplication, which is associative.
So $\mathbb H$ forms a semigroup under quaternion multiplication.
Finally, we have that Matrix Multiplication Distributes over Matrix Addition.
So all the conditions are fulfilled for $\mathbb H$ to be a ring.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $9$
Category:Quaternions
Category:Examples of Rings
Ring of Sets Generated by Semiring

== Theorem ==
Let $\SS$ be a semiring of sets.
Let $\map \RR \SS$ be the minimal ring generated by $\SS$.
Let $\LL$ be the system of sets $A$ with the finite expansions:
:$\ds A = \bigcup_{k \mathop = 1}^n A_k$
with respect to the sets $A_k \in \SS$.
Then $\LL = \map \RR \SS$.
== Proof ==
First we need to show that $\LL$ is a ring of sets.
Let $A, B \in \LL$.
Then by definition of $\LL$, they have expansions:
Since $\SS$ is a semiring of sets, we have:
:$C_{ij} = A_i \cap B_j \in \SS$
By Pairwise Disjoint Subsets in Semiring Part of Partition, there exist finite expansions:
From these, it follows that $A \cap B$ and $A \ast B$ have the finite expansions:
Hence both $A \cap B \in \LL$ and $A \ast B \in \LL$.
So by definition, $\LL$ is a ring of sets.
From the details of the above construction, the fact that $\LL$ is the minimal ring generated by $\SS$ follows immediately.
Category:Rings of Sets
Category:Semirings of Sets
Ring of Sets is Closed under Finite Intersection

== Theorem ==
Let $\RR$ be a ring of sets.
Let $A_1, A_2, \ldots, A_n \in \RR$.
Then:
:$\ds \bigcap_{j \mathop = 1}^n A_j \in \RR$
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \bigcap_{j \mathop = 1}^n A_j \in \RR$
$\map P 1$ is true, as this just says $A_1 \in \RR$.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$A_1 \cap A_2 \in \RR$
which is immediate, from definition 1 of ring of sets.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \bigcap_{j \mathop = 1}^k A_j \in \RR$
Then we need to show:
:$\ds \bigcap_{j \mathop = 1}^{k + 1} A_j \in \RR$
=== Induction Step ===
This is our induction step:
We have that:
:$\ds \bigcap_{j \mathop = 1}^{k + 1} A_j = \bigcap_{j \mathop = 1}^k A_j \cap A_{k + 1}$
But from the induction hypothesis we have that:
:$\ds \bigcap_{j \mathop = 1}^k A_j \in \RR$
Hence from the basis for the induction, it follows that:
:$\ds \bigcap_{j \mathop = 1}^{k + 1} A_j \in \RR$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \bigcap_{j \mathop = 1}^n A_j \in \RR$
== Sources ==
* : $\S 1.1$: Rings and Algebras
Category:Rings of Sets
Ring of Sets is Closed under Finite Union

== Theorem ==
Let $\RR$ be a ring of sets.
Let $A_1, A_2, \ldots, A_n \in \RR$.
Then:
:$\ds \bigcup_{j \mathop = 1}^n A_j \in \RR$
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \bigcup_{j \mathop = 1}^n A_j \in \RR$
$\map P 1$ is true, as this just says $A_1 \in \RR$.
=== Basis for the Induction ===
$\map P 2$ is the case:
:$A_1 \cup A_2 \in \RR$
which is immediate, from definition 2 of ring of sets.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\ds \bigcup_{j \mathop = 1}^k A_j \in \RR$
Then we need to show:
:$\ds \bigcup_{j \mathop = 1}^{k + 1} A_j \in \RR$
=== Induction Step ===
This is our induction step:
We have that:
:$\ds \bigcup_{j \mathop = 1}^{k + 1} A_j = \bigcup_{j \mathop = 1}^k A_j \cup A_{k+1}$
But from the induction hypothesis we have that:
:$\ds \bigcup_{j \mathop = 1}^k A_j \in \RR$
Hence from the basis for the induction, it follows that:
:$\ds \bigcup_{j \mathop = 1}^{k + 1} A_j \in \RR$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \bigcup_{j \mathop = 1}^n A_j \in \RR$
== Sources ==
* : $\S 1.1$: Rings and Algebras
Category:Rings of Sets
Ring of Sets is Commutative Ring

== Theorem ==
A ring of sets $\struct {\RR, \symdif, \cap}$ is a commutative ring whose zero is $\O$.
== Proof ==
By definition, the operations $\cap$ and $\symdif$ are closed in $\RR$.
Hence we can apply the following results:
* Set System Closed under Symmetric Difference is Abelian Group: $\struct {\RR, \symdif}$ is an abelian group.
* Set System Closed under Intersection is Commutative Semigroup: $\struct {\RR, \cap}$ is a commutative semigroup.
* Intersection Distributes over Symmetric Difference.
* The identity of $\struct {\RR, \symdif}$ is $\O$, and this, by definition, is the zero.
So $\struct {\RR, \symdif, \cap}$ is a commutative ring whose zero is $\O$.
Category:Commutative Rings
Category:Rings of Sets
Ring of Sets is Semiring of Sets

== Theorem ==
Let $\RR$ be a ring of sets.
Then $\RR$ is also a semiring of sets.
== Proof ==
Let $A \in \RR$.
Suppose $A_1 \subseteq A$.
Let $A_2 = A \setminus A_1$, where $A \setminus A_1$ denotes set difference.
By definition, $A_2$ is then the relative complement of $A_1$ with respect to $A$.
From Union with Relative Complement it then follows that $A_1 \cup A_2 = A$.
By Ring of Sets Closed under Various Operations:
:$A \setminus A_1 = A_2 \in \RR$
But by Set Difference Intersection with Second Set is Empty Set:
:$\paren {A \setminus A_1} \cap A_1 = \O$
Hence for any given $A$ for which we have $A_1 \subseteq A$, we can represent $A$ as the finite expansion $A_1 \cup A_2$ such that $A_1 \cap A_2 = \O$ and $A_1, A_2 \in \RR$
Thus, by definition, $\RR$ is a semiring of sets.
Category:Rings of Sets
Category:Semirings of Sets
Ring of Square Matrices over Commutative Ring with Unity

== Theorem ==
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:
Thus, for $n = 1$, $\struct {\map {\MM_R} n, +, \times}$ is a commutative ring.
== Notation ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.2$
Category:Commutative Algebra
Category:Rings of Square Matrices
Category:Examples of Rings with Unity
Ring of Square Matrices over Field is Ring with Unity

== Theorem ==
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.
== Notation ==
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $7$
Category:Matrix Algebra
Category:Examples of Rings with Unity
Ring of Square Matrices over Real Numbers

== Theorem ==
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.
== Notation ==
== Examples ==
== Sources ==
* : $\S 54$. The definition of a ring and its elementary consequences
Category:Matrix Algebra
Category:Rings with Unity
Category:Ring of Square Matrices over Real Numbers
Ring of Square Matrices over Ring is Ring

== Theorem ==
Let $R$ be a ring.
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.
== Proof ==
From Matrix Entrywise Addition forms Abelian Group we have that $\struct {\map {\MM_R} n, +}$ is an abelian group, because $\struct {R, +}$ is itself an abelian group.
Similarly, it is clear that $\struct {\map {\MM_R} n, \times}$ is a semigroup, as Matrix Multiplication over Order n Square Matrices is Closed and Matrix Multiplication is Associative.
Finally, we note that Matrix Multiplication Distributes over Matrix Addition.
== Notation ==
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $3$
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Rings of Square Matrices
Ring of Square Matrices over Ring with Unity

== Theorem ==
Let $R$ be a 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.
== Proof ==
From Ring of Square Matrices over Ring is Ring we have that $\struct {\map {\MM_R} n, +, \times}$ is a ring.
As $R$ has a unity, the unit matrix can be formed.
The unity of $\struct {\map {\MM_R} n, +, \times}$ is this unit matrix.
== Notation ==
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $3$
Category:Rings of Square Matrices
Category:Rings with Unity
Ring with Multiplicative Norm has No Proper Zero Divisors

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let its zero be denoted by $0_R$.
Let $\norm {\,\cdot\,}$ be a multiplicative norm on $R$.
Then $R$ has no proper zero divisors. 
That is:
:$\forall x, y \in R^*: x \circ y \ne 0_R$
where $R^*$ is defined as $R \setminus \set {0_R}$.
== Proof ==
:
:$\exists x, y \in {R^*} : x \circ y = 0_R$
By positive definiteness:
:$x, y \ne 0_R \iff \norm x, \norm y \ne 0$
Thus:
:$\norm x \norm y \ne 0$
But we also have:
a contradiction.
Category:Norm Theory
Category:Zero Divisors
Ring with Unity has Prime Ideal

== Theorem ==
Let $A$ be a non-trivial commutative ring with unity.
Then $A$ has a prime ideal.
== Proof ==
By Krull's Theorem, $A$ has a maximal ideal.
By Maximal Ideal of Commutative and Unitary Ring is Prime Ideal, $A$ has a prime ideal.
== Also see ==
* Spectrum of Ring is Nonempty
Category:Commutative Algebra
Ring with Unity is Module over Itself

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity $1_R$.
Then $\struct {R, +, \circ}_R$ is a unitary $R$-module. 
== Proof ==
From Ring is Module over Itself we have that $\struct {R, +, \circ}_R$ is an $R$-module. 
We have that $\struct {R, +, \circ}$ has a unity $1_R$. 
For $\struct {R, +, \circ}_R$ to be unitary, it must satisfy the additional axiom:
The axiom follows from the definition of a unity.
Category:Module Theory
Category:Ring is Module over Itself
Rings of Polynomials in Ring Elements are Isomorphic

== Theorem ==
Let $R_1, R_2$ be commutative rings with unity.
Let $D$ be an integral subdomain of both $R_1$ and $R_2$.
Let $X_1, X_2 \in R$ be transcendental over $D$.
Let $D \sqbrk {X_1}, D \sqbrk {X_2}$ be the rings of polynomials in $X_1$ and $X_2$ over $D$.
Then $D \sqbrk {X_1}$ is isomorphic to $D \sqbrk {X_2}$.
== Proof ==
First it is shown that the mapping $\phi: D \sqbrk {X_1} \to D \sqbrk {X_2}$ given by:
:$\ds \map \phi {\sum_{k \mathop = 0}^n a_k \circ X_1^k} = \sum_{k \mathop = 0}^n a_k \circ X_2^k$
is a bijection.
Let $p, q \in \phi: D \sqbrk {X_1}$.
Suppose $\map \phi p = \map \phi q$.
Then the coefficients of $\map \phi p$ and $\map \phi q$ are equal, and $p_1 = q_1$.
Thus, by definition, $\phi$ is an injection.
By the same argument, the mapping $\psi: D \sqbrk {X_2} \to D \sqbrk {X_1}$ defined as:
:$\ds \map \psi {\sum_{k \mathop = 0}^n a_k \circ X_2^k} = \sum_{k \mathop = 0}^n a_k \circ X_1^k$
is similarly an injection.
Thus by Injection is Bijection iff Inverse is Injection, $\phi$ is a bijection.
It remains to show that $\phi$ is a ring homomorphism. 
Let $\ds p = \sum_{k \mathop = 0}^n a_k \circ X_1^k, q = \sum_{k \mathop = 0}^n b_k \circ X_2^k \in D \sqbrk {X_1}$. 
For convenience we set $a_k = 0$, $k > n$ and $b_k = 0$, $k > m$.
We have:
Similarly for multiplication:
This completes the proof.
== Motivation ==
Thus we see that the ring in which an integral domain is embedded is (to a certain extent) irrelevant -- it is only the integral domain itself which is important.
== Sources ==
* : $\S 64.4$ Polynomial rings over an integral domain
Category:Polynomial Theory
Category:Ring Isomorphisms
Rising Factorial as Factorial by Binomial Coefficient

== Theorem ==
Let $x \in \R$ be a real number.
Let $n \in \Z_{\ge 0}$ be a positive integer.
:$x^{\overline n} = n! \dbinom {x + n - 1} n$
where:
:$x^{\overline n}$ denotes $x$ to the $n$ rising
:$n!$ denotes the factorial of $n$
:$\dbinom {x + n - 1} n$ denotes a binomial coefficient.
== Proof ==
By definition of $x$ to the $n$ rising:
:$x^{\overline n} = x \paren {x + 1} \cdots \paren {x + n - 1}$
By definition of the binomial coefficient of a real number:
:$\dbinom {x + n - 1} n = \dfrac {\paren {x + n - 1} \paren {x + n - 2} \cdots x} {n!}$
Hence the result.
Category:Binomial Coefficients
Category:Factorials
Category:Rising Factorials
Rising Factorial as Quotient of Factorials

== Theorem ==
Let $x \in \Z_{\ge 0}$ be a positive integer.
Then:
:$x^{\overline n} = \dfrac {\paren {x + n - 1}!} {\paren {x - 1}!} = \dfrac {\map \Gamma {x + n} } {\map \Gamma x}$
where:
:$x^{\overline n}$ denotes the $n$th rising factorial power of $x$
:$\map \Gamma x$ denotes the Gamma function of $x$.
== Proof ==
== Also see ==
* Falling Factorial as Quotient of Factorials
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: $(21)$
Category:Rising Factorials
Category:Factorials
Rising Factorial in terms of Falling Factorial

== Theorem ==
:$x^{\overline n} = \paren {x + n - 1}^{\underline n}$
where:
:$x^{\underline n}$ is the $n$th falling factorial power of $x$
:$x^{\overline n}$ is the $n$th rising factorial power of $x$.
== Proof ==
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: $(20)$
Category:Falling Factorials
Category:Rising Factorials