nyuuzyou/wikis huggingface dataset:
Polygon not Equal to Triangle has Chord

== Theorem ==
Let $P$ be a polygon that is not a triangle.
Then $P$ has a chord that lies completely in the interior of $P$.
== Proof ==
From Polygon has Salient Angle, it follows that $P$ has a convex internal angle.
Let $A, B, C$ be vertices of $P$ such that $\angle BAC$ is a convex internal angle of the vertex $A$.
Let $BC$ subtend the angle $\angle BAC$.
Suppose $BC$ is equal to a side of $P$.
Then $P = \triangle ABC$, which would contradict our assumption that $P$ is not a triangle.
Suppose $BC$ is a chord of $P$ that lies completely in the interior of $P$.
Then the proof is done.
Suppose $BC$ does not lie completely in the interior of $P$.
Then either a side of $P$ intersects $BC$, or a vertex of $P$ lies on $BC$.
In either case, one or more vertices of $P$ lie inside $\triangle ABC$, or on $BC$.
Let $\mathcal V$ be the set of those vertices.
Define $D$ as the vertex in $\mathcal V$ that lies furthest away from $BC$.
If more than one vertex in $\mathcal V$ have the same maximal distance from $BC$, pick $D$ among them arbitrarily.
Now $AD$ is a chord of $P$ that lies completely in the interior.
As any side of $P$ that would intersect $AD$ would either intersect $AB$ or $AC$, or end in a vertex that lie further away from $BC$ than $D$ does.
In all of these cases, we have a contradiction.
== Sources ==
* : $\S 1.3.1$
Category:Chords
Category:Polygons
Polygon of Forces

== Theorem ==
Let $n$ forces act on a particle where $n \ge 3$.
Let the magnitude and direction of these forces be represented by vectors in space such that the magnitude of the forces is proportional to the length of the vectors.
Then the forces are in equilibrium the vectors form a polygon.
This is known as a polygon of forces.
== Proof ==
== Also see ==
* Definition:Force Polygon
* Triangle of Forces
== Sources ==
* 
* 
Category:Force Polygons
Category:Equilibrium (Mechanics)
Category:Force
Polynomial Addition is Associative

== Theorem ==
Addition of polynomials is an associative operation.
== Proof ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\set {X_j: j \in J}$ be a set of indeterminates.
Let $Z$ be the set of all multiindices indexed by $\set {X_j: j \in J}$.
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
:$\ds g = \sum_{k \mathop \in Z} b_k \mathbf X^k$ 
:$\ds h = \sum_{k \mathop \in Z} c_k \mathbf X^k$ 
be arbitrary polynomials in the indeterminates $\set {X_j: j \in J}$ over $R$.
Then:
 
 
Category:Polynomial Theory
Polynomial Addition is Commutative

== Theorem ==
Addition of polynomials is commutative.
== Proof ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity with zero $0_R$.
Let $\set {X_j: j \in J}$ be a set of indeterminates.
Let $Z$ be the set of all multiindices indexed by $\set {X_j: j \in J}$.
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
:$\ds g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
be arbitrary polynomials in the indeterminates $\set {X_j: j \in J}$ over $R$.
Then:
 
 
So $f + g = g + f$ for all polynomials $f$ and $g$.
Therefore polynomial addition is commutative.
Category:Polynomial Theory
Polynomial Divisor Modulo Integer/Examples/Arbitrary Example 1

== Examples of Polynomial Divisors Modulo $m$ ==
Let $\map f x$ be the polynomial:
:$\map f x = 2 x^4 - 4 x - 3$
Then $\map f x$ has the following (polynomial) divisors modulo $7$:
:$2 x^2 + 3 x + 3$
:$x - 2$
:$x + 4$
== Proof ==
== Sources ==
* 
* 
Category:Examples of Polynomial Divisors Modulo Integer
Polynomial Equation expressed in Homogeneous Coordinates becomes Homogeneous Equation

== Theorem ==
Let $\map P {x, y}$ be a polynomial equation expressing a locus in Cartesian coordinates on the Cartesian plane.
Let $\map P {x, y}$ be converted to homogeneous Cartesian coordinates $\map P {X, Y, Z}$.
Then $\map P {X, Y, Z}$ is a homogeneous equation.
== Proof ==
== Sources ==
* 
* 
Category:Homogeneous Cartesian Coordinates
Category:Homogeneous Equations
Polynomial Factor Theorem

== Theorem ==
Let $\map P x$ be a polynomial in $x$ over a field $K$ of degree $n$.
Then:
:$\xi \in K: \map P \xi = 0 \iff \map P x = \paren {x - \xi} \map Q x$
where $Q$ is a polynomial of degree $n - 1$.
Hence, if $\xi_1, \xi_2, \ldots, \xi_n \in K$ such that all are different, and $\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$, then:
:$\ds \map P x = k \prod_{j \mathop = 1}^n \paren {x - \xi_j}$
where $k \in K$.
=== Corollary ===
== Proof ==
Let $P = \paren {x - \xi} Q$.
Then:
:$\map P \xi = \map Q \xi \cdot 0 = 0$
Conversely, let $\map P \xi = 0$.
By the Division Theorem for Polynomial Forms over Field, there exist polynomials $Q$ and $R$ such that:
:$P = \map Q {x - \xi} + R$
and:
:$\map \deg R < \map \deg {x - \xi} = 1$
Evaluating at $\xi$ we have:
:$0 = \map P \xi = \map R \xi$
But:
:$\deg R = 0$
so:
:$R \in K$
In particular:
:$R = 0$
Thus:
:$P = \map Q {x - \xi}$
as required.
The fact that $\map \deg Q = n - 1$ follows from:
:Ring of Polynomial Forms is Integral Domain
and:
:Degree of Product of Polynomials over Integral Domain.
We can then apply this result to:
:$\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$
We can progressively work through:
:$\map P x = \paren {x - \xi_1} \map {Q_{n - 1} } x$ 
where $\map {Q_{n - 1} } x$ is a polynomial of order $n - 1$.
Then, substituting $\xi_2$ for $x$:
:$0 = \map P {\xi_2} = \paren {\xi_2 - \xi_1} \map {Q_{n - 1} } x$
Since $\xi_2 \ne \xi_1$:
:$\map {Q_{n - 1} } {\xi_2} = 0$
and we can apply the above result again:
:$\map {Q_{n - 1} } x = \paren {x - \xi_2} \map {Q_{n - 2} } x$
Thus:
:$\map P x = \paren {x - \xi_1} \paren {x - \xi_2} \map {Q_{n - 2} } x$
and we then move on to consider $\xi_3$.
Eventually we reach:
:$\map P x = \paren {x - \xi_1} \paren {x - \xi_2} \dotsm \paren {x - \xi_n} \map {Q_0} x$
$\map {Q_0} x$ is a polynomial of zero degree, that is a constant polynomial.
The result follows.
== Also known as ==
== Sources ==
* : Chapter $9$: Rings: Exercise $24$
Category:Polynomial Factor Theorem
Category:Polynomial Theory
Category:Named Theorems
Polynomial Factor Theorem/Also known as

== Polynomial Factor Theorem: Also known as == 
Some sources refer to the Polynomial Factor Theorem (and its variants) as the Factor Theorem, but there are multiple theorems with this name.
== Sources ==
* : Chapter $1$: Polynomials; The Remainder and Factor Theorems; Undetermined Coefficients; Partial Fractions: $1.2$. The remainder and factor theorems
* 
* 
Category:Polynomial Factor Theorem
Polynomial Factor Theorem/Corollary

== Corollary to Polynomial Factor Theorem ==
Let $\map P x$ be a polynomial in $x$ over the real numbers $\R$ of degree $n$.
Suppose there exists $\xi \in \R: \map P \xi = 0$.
Then $\map P x = \paren {x - \xi} \map Q x$, where $\map Q x$ is a polynomial of degree $n - 1$.
Hence, if $\xi_1, \xi_2, \dotsc, \xi_n \in \R$ such that all are different, and $\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$, then:
:$\ds \map P x = k \prod_{j \mathop = 1}^n \paren {x - \xi_j}$
where $k \in \R$.
=== Complex Number form ===
== Proof ==
Recall that Real Numbers form Field.
The result then follows from the Polynomial Factor Theorem.
== Also known as ==
== Sources ==
* : Chapter $1$: Polynomials; The Remainder and Factor Theorems; Undetermined Coefficients; Partial Fractions: $1.2$. The remainder and factor theorems
* : $\S 3$: Natural Numbers: Exercise $\S 3.11 \ (3)$
* 
* 
Category:Polynomial Factor Theorem
Category:Polynomial Theory
Polynomial Factor Theorem/Corollary/Complex Numbers

== Corollary to Polynomial Factor Theorem ==
Let $\map P z$ be a polynomial in $z$ over the complex numbers $\C$ of degree $n$.
Suppose there exists $\zeta \in \C: \map P \zeta = 0$.
Then:
:$\map P z = \paren {x - \zeta} \map Q z$
where $\map Q z$ is a polynomial of degree $n - 1$.
Hence, if $\zeta_1, \zeta_2, \ldots, \zeta_n \in \C$ such that all are different, and $\map P {\zeta_1} = \map P {\zeta_2} = \dotsb = \map P {\zeta_n} = 0$, then:
:$\ds \map P z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$
where $k \in \C$.
== Proof ==
Recall that Complex Numbers form Field.
The result then follows from the Polynomial Factor Theorem.
== Also known as ==
== Sources ==
* : $\S 3$. Roots of Unity
Category:Polynomial Factor Theorem
Category:Complex Polynomial Functions
Polynomial Forms is PID Implies Coefficient Ring is Field

== Theorem ==
Let $D$ be an integral domain.
Let $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.
Let $D \sqbrk X$ be a principal ideal domain;
Then $D$ is a field.
== Proof ==
Let $y \in D$ be non-zero.
Then, using the principal ideal property, for some $f \in D \sqbrk X$ we have:
:$\gen {y, X} = \gen f \subseteq D \sqbrk X$
Therefore:
:$\exists p, q \in D \sqbrk X: y = f p, X = f q$
By Properties of Degree we conclude that $f = a$ and $q = b + c X$ for some $a, b, c \in D$.
Substituting into the equation $X = f q$ we obtain:
:$X = a b + a c X$
which implies that:
:$a c = 1$
That is:
:$a \in D^\times$
where $D^\times$ denotes the group of units of $D$.
Therefore:
:$\gen f = \gen 1 = D \sqbrk X$
Therefore:
:$\exists r, s \in D \sqbrk X: r y + s X = 1$
If $d$ is the constant term of $r$, then we have $y d = 1$.
Therefore $y \in D^\times$.
Our choice of $y$ was arbitrary, so this shows that $D^\times \supseteq D \setminus \set 0$.
This says precisely that $D$ is a field.
== Also see ==
* Polynomial Forms over Field form Principal Ideal Domain
Category:Polynomial Theory
Category:Principal Ideal Domains
Polynomial Forms over Field form Integral Domain/Formulation 1

== Theorem ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental in $F$.
Let $F \sqbrk X$ be the ring of polynomial forms in $X$ over $F$.
Then $F \sqbrk X$ is an integral domain.
== Proof 1 ==
== Proof 2 ==
Category:Field Theory
Category:Polynomial Theory
Category:Ideal Theory
Category:Polynomial Forms over Field form Integral Domain
Polynomial Forms over Field form Integral Domain/Formulation 1/Proof 1

== Theorem ==
== Proof ==
We already have from Ring of Polynomial Forms is Commutative Ring with Unity that $F \sqbrk X$ is a ring.
Suppose $f$ and $g$ are polynomials in $F \sqbrk X$ such that $f \ne 0_F, g \ne 0_F$.
If $\map \deg f = \map \deg g = 0$ then $f$ and $g$ are elements of $F$.
As $F$ is a field and a field is an integral domain, $f g \ne 0_f$.
Otherwise from Degree of Product of Polynomials over Integral Domain:
:$\map \deg {f g} = \map \deg f + \map \deg g$
and so:
:$\map \deg {f g} > 0$
which means:
:$f g \ne 0_F$
Hence the result, by definition of integral domain.
Category:Polynomial Forms over Field form Integral Domain
Polynomial Forms over Field form Integral Domain/Formulation 2

== Theorem ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\GF$ be the set of all polynomials over $\struct {F, +, \circ}$ defined as sequences.
Let polynomial addition and polynomial multiplication be defined as:
:$\forall f = \sequence {a_k} = \tuple {a_0, a_1, a_2, \ldots}, g = \sequence {b_k} = \tuple {b_0, b_1, b_2, \ldots} \in \GF$:
::$f \oplus g := \tuple {a_0 + b_0, a_1 + b_1, a_2 + b_2, \ldots}$
::$f \otimes g := \tuple {c_0, c_1, c_2, \ldots}$ where $\ds c_i = \sum_{j \mathop + k \mathop = i} a_j \circ b_k$
Then $\struct {\GF, \oplus, \otimes}$ is an integral domain.
== Proof ==
As $\struct {F, +, \circ}$ is a field, it is also by definition a ring.
Thus from Polynomial Ring of Sequences is Ring we have that $\struct {\GF, \oplus, \otimes}$ is a ring.
From Field is Integral Domain, a field is also by definition an integral domain.
Let $f, g \in \GF$ such that neither $f$ nor $g$ are the null polynomial.
Let:
:$\deg f = m, \deg g = n$
where $\deg$ denotes the degree of $f$ and $g$ respectively.
By Degree of Product of Polynomials over Integral Domain, the degree of $f \times g$ is $m + n$.
Then by definition of polynomial multiplication, its leading coefficient is $a_m \circ b_n$.
As by definition an integral domain has no proper zero divisors:
:$a_m \circ b_n \ne 0_F$.
So, by definition, $f \otimes g$ has a leading coefficient which is not $0_F$.
That is, $f \otimes g$ is not the null polynomial
The result follows by definition of integral domain.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 25$. Polynomials
Category:Polynomial Forms over Field form Integral Domain
Polynomial Forms over Field form Principal Ideal Domain

== Theorem ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Then $F \sqbrk X$ is a principal ideal domain.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
=== Converse ===
A converse to this result is given by Polynomial Forms is PID Implies Coefficient Ring is Field:
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
Category:Field Theory
Category:Polynomial Theory
Category:Principal Ideal Domains
Category:Polynomial Forms over Field form Principal Ideal Domain
Polynomial Forms over Field form Principal Ideal Domain/Corollary 1

== Corollary to Polynomial Forms over Field form Principal Ideal Domain ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Let $f$ be an irreducible element of $F \sqbrk X$.
Then $F \sqbrk X / \ideal f$ is a field, where $\ideal f$ denotes the ideal generated by $f$.
== Proof ==
It follows from Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal that $\ideal f$ is maximal for irreducible $f$.
Therefore by Maximal Ideal iff Quotient Ring is Field, $F \sqbrk X / \ideal f$ is a field.
== Sources ==
* : $\S 65.3$ Some properties of $F \sqbrk X$, where $F$ is a field
Category:Polynomial Forms over Field form Principal Ideal Domain
Polynomial Forms over Field form Principal Ideal Domain/Corollary 2

== Corollary to Polynomial Forms over Field form Principal Ideal Domain ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Let $f \in F \sqbrk X$.
Then $\ideal f$ is a maximal ideal of $F \sqbrk X$ $f$ is irreducible.
== Proof ==
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Theorem $57$
Category:Polynomial Forms over Field form Principal Ideal Domain
Polynomial Forms over Field form Principal Ideal Domain/Corollary 3

== Corollary to Polynomial Forms over Field form Principal Ideal Domain ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental over $F$.
Let $F \sqbrk X$ be the ring of polynomials in $X$ over $F$.
Then $F \sqbrk X$ is a unique factorization domain.
== Proof ==
We have the result Principal Ideal Domain is Unique Factorization Domain.
The result then follows from Polynomial Forms over Field form Principal Ideal Domain.
== Sources ==
* : $\S 65.2$ Some properties of $F \sqbrk X$, where $F$ is a field
Category:Polynomial Forms over Field form Principal Ideal Domain
Category:Unique Factorization Domains
Polynomial Forms over Field form Principal Ideal Domain/Proof 1

== Theorem ==
== Proof ==
For any $d \in F \sqbrk X$, let $\ideal d$ denote the principal ideal of $F \sqbrk X$ generated by $d$.
Let $J$ be any ideal of $F \sqbrk X$. What we need to prove is that $J$ is a principal ideal.
Let us first distinguish the following two cases for $J$:
:If $J = \set {0_F}$, then by Zero Element Generates Null Ideal $J = \ideal {0_F}$, and hence is a principal ideal.
:If $J = F \sqbrk X$, then by Ideal of Unit is Whole Ring: Corollary $J = \ideal {1_F}$, and hence is a principal ideal. 
Now suppose $J \ne \set {0_F}$ and $J \ne F \sqbrk X$.
Then $J$ necessarily contains a non-zero element.
By the Well-Ordering Principle, we can introduce the lowest degree of a non-zero element of $J$.
Denote this degree by $n$.
If $n = 0$, then $J$ contains a polynomial of degree $0$.
This is a non-zero element of $F$.
As $F$ is a field, this is therefore a unit of $F$, and thus by Ideal of Unit is Whole Ring, $J = F \sqbrk X$.
Because the degree of a non-zero element is a natural number, we conclude that $n \ge 1$.
Now let $d$ be a polynomial of degree $n$ in $J$, and let $f \in J$.
By Division Theorem for Polynomial Forms over Field, $f = q \circ d + r$ for some $q, r \in F \sqbrk X$ where either:
:$r = 0_F$
or:
:$r$ is a polynomial of degree smaller than $n$.
Because $J$ is an ideal and $d \in J$, it follows that:
:$q \circ d \in J$
Since $f \in J$, we also conclude:
:$r = f - q \circ d \in J$
From the construction of $d$, it follows that we must have $r = 0_F$.
Therefore:
:$f = q \circ d$
and thus:
:$f \in \ideal d$.
This reasoning shows that:
:$J \subseteq \ideal d$
From property $(3)$ of the principal ideal $\ideal d$, we conclude that:
:$\ideal d \subseteq J$
as $d \in J$.
Hence $J = \ideal d$.
These $2$ distinguished cases cover all of the possible ideals of $F \sqbrk X$. 
Hence $F \sqbrk X$ is a principal ideal domain.
== Sources ==
* : $\S 65.2$ Some properties of $F \sqbrk X$, where $F$ is a field
Category:Polynomial Forms over Field form Principal Ideal Domain
Polynomial Forms over Field is Euclidean Domain

== Theorem ==
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $X$ be transcendental in $F$.
Let $F \sqbrk X$ be the ring of polynomial forms in $X$ over $F$.
Then $F \sqbrk X$ is a Euclidean domain.
== Proof ==
From Degree of Product of Polynomials over Integral Domain not Less than Degree of Factors we have that:
:$\forall a, b \in F \sqbrk X, a \ne 0_F, b \ne 0_F: \map \deg {a b} \ge \map \deg a$
where $\map \deg a$ denotes the degree of $a$.
From Division Theorem for Polynomial Forms over Field:
:$\forall a, b \in F \sqbrk X, b \ne 0_F: \exists q, r \in F \sqbrk X: a = q b + r$
where $\map \deg r < \map \deg b$ (or $r = 0$), and $q$ and $r$ are unique.
So $\deg$ is a Euclidean valuation on $F \sqbrk X$.
Hence the result.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 27$. Euclidean Rings: Example $53$
* 
Category:Field Theory
Category:Examples of Euclidean Domains
Category:Polynomial Theory
Polynomial Functions form Submodule of All Functions

== Theorem ==
Let $K$ be a commutative ring with unity.
Let $K^K$ be the $K$-module mappings $f: K \to K$.
Let $\map P K \subseteq K^K$ be the set of all polynomial functions on $K$.
Then $\map P K$ is a $K$-submodule of $K^K$.
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.4$
Category:Module Theory
Category:Polynomial Theory
Polynomial Long Division

== Technique ==
Let $\map {P_n} x$ be a polynomial in $x$ of degree $n$.
Let $\map {Q_m} x$ be a polynomial in $x$ of degree $m$ where $m \le n$.
Then $\map {P_n} x$ can be expressed in the form:
:$\map {P_n} x \equiv \map {Q_m} x \map {D_{n - m} } x + \map {R_k} x$
where:
:$\map {D_{n - m} } x$ is a polynomial in $x$ of degree $n - m$
:$\map {R_k} x$ is a polynomial in $x$ of degree $k$, where $k < m$, or may be null.
Hence we can define $\dfrac {\map {P_n} x} {\map {Q_m} x}$:
:$\dfrac {\map {P_n} x} {\map {Q_m} x} = \map {D_{n - m} } x + \dfrac {\map {R_k} x} {\map {Q_m} x}$
The polynomial $\map {R_k} x$ is called the remainder.
The procedure for working out what $\map {D_{n - m} } x$ and $\map {R_k} x$ are is called (polynomial) long division.
== Proof ==
Let $\ds \map {P_n} x = \sum_{j \mathop = 0}^n p_j x^j$.
Let $\ds \map {Q_m} x = \sum_{j \mathop = 0}^m q_j x^j$.
First calculate $\map {Q'_m} x = \map {Q_m} x \times \dfrac {p_n} {q_m} x^{n - m}$.
This gives:
Then evaluate:
:$\map {P'_{n - 1} } x = \map {P_n} x - \map {Q'_m} x$
which (after some algebra) works out as:
:$\ds \map {P_n} x - \map {Q'_m} x = \sum_{j \mathop = n - m}^{n - 1} \frac {p_n q_{j - n + m} } {q_m} x^j + \sum_{j \mathop = 0}^{n - m - 1} p_j x^j$
So we see that $\map {P_n} x - \map {Q'_m} x$ is a polynomial in $x$ of degree $n - 1$.
Let $\dfrac {p_n} {q_m} = d_{n - m}$.
Hence we have:
:$\map {P_n} x = d_{n - m} x^{n - m} \map {Q_m} x + \map {P'_{n - 1} } x$
We can express $\map {P'_{n - 1} } x$ as:
:$\ds \map {P'_{n - 1} } x = \sum_{j \mathop = 0}^{n - 1} p'_j x^j$
Repeat the above by subtracting $\ds \frac {p'_{n - 1} } {q_m} x^{n - m - 1} \map {Q_m} x$ from $\map {P'_{n - 1} } x$, and letting $\dfrac {p'_{n - 1} } {q_m} = d_{n - m - 1}$.
Hence:
:$\map {P'_{n - 1} } x = d_{n - m - 1} x^{n - m - 1} \map {Q_m} x + \map {P_{n - 2} } x$
The process can be repeated $n - m$ times.
It can be seen that after the last stage, we have:
:$\map {P_n} x = \map {D_{n - m} } x \map {Q_m} x + \map {R_k} x$
where:
:$\ds \map {D_{n - m} } x = \sum_{j \mathop = 0}^{n - m} d_j x^j$
:$\map {R_k} x$ is a polynomial of degree at most $m - 1$.
Category:Proof Techniques
Polynomial Ring is Generated by Indeterminate over Ground Ring

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $R \sqbrk X$ be a polynomial ring over $R$.
Let $\iota: R \to R \sqbrk X$ be the embedding.
Then $R \sqbrk X$ is generated by $X$ over $R$.
== Proof ==
Category:Polynomial Theory
Polynomial Ring of Sequences is Ring

== Theorem ==
Let $R$ be a ring.
Let $P \sqbrk R$ be the polynomial ring over sequences in $R$.
Then $P \sqbrk R$ is itself a ring.
== Proof ==
We have by definition of polynomial ring over sequences in $R$ that:
:$P \sqbrk R = \set {\sequence {r_0, r_1, r_2, \ldots} }$
where each $r_i \in R$, and all but a finite number of terms is zero.
=== Proof that Operations are Closed ===
We need to ensure that the operations as defined are closed.
Let $r = \sequence {r_0, r_1, r_2, \ldots}, s = \sequence {s_0, s_1, s_2, \ldots} \in P \sqbrk R$.
As all but a finite number of terms of $r$ and $s$ are zero, there exist $m, n \ge 0$ such that:
:$\forall i > m: r_i = 0$
:$\forall j > n: s_j = 0$
Let $l = \max \set {m, n}$.
We can express the operations on $P \sqbrk R$ as:
We have that:
:$\forall i > l: r_i + s_i = 0$
and so 
:$r + s \in P \sqbrk R$
Equally clearly:
:$-r \in P \sqbrk R$
Now consider:
:$\ds \paren {r s}_i = \sum_{j \mathop + k \mathop = i}r_j s_k$
Let $i > m + n$.
Then in any $r_j s_k$ such that $j + k = i$, either $j > m$ or k > n.
In the first case $r_j = 0$ and in the second $s_j = 0$.
In either case $r_j s_k = 0$.
So:
:$\ds \forall i > m + n: \paren {r s}_i = \sum_{j \mathop + k \mathop = i} r_j s_k = 0$
So:
:$r s \in P \sqbrk R$
=== Proof of Additive Group ===
The addition operation $r + s$ is clearly commutative and associative, and:
:$\sequence {0, 0, \ldots} = 0_{P \sqbrk R}$
Equally clearly:
:$\forall r \in P \sqbrk R: r + \paren {-r} = \sequence {r_i + \paren {-r_i} } = \sequence {0, 0, \ldots} = 0_{P \sqbrk R}$
and so $-r$ is the inverse of $r$ for addition.
So $\struct {P \sqbrk R, +}$ is an abelian group, as it needs to be for $P \sqbrk R$ to be a ring.
=== Proof of Ring Product ===
We need to establish that the ring product is associative.
Let $r = \sequence {r_0, r_1, \ldots}, s = \sequence {s_0, s_1, \ldots}, t = \sequence {t_0, t_1, \ldots} \in P \sqbrk R$.
Then:
Similarly for $\paren {r \paren {s t} }_n$.
So the ring product is associative, and so forms a semigroup.
=== Proof of Distributivity ===
Finally we need to show that the ring product is distributive over ring addition.
Similarly for $\paren {t \paren {r + s} }_n$.
Hence the result.
== Sources ==
* : $\S 3.2$: Polynomial rings: Theorem $3.5$
Category:Polynomial Theory
Category:Ring Theory
Polynomial Time/Examples/Squaring an Integer

== Example of Polynomial Time ==
Consider the process of squaring an $n$-digit integer using the basic computational steps:
:addition of $2$ digits
:multiplication of $2$ digits.
This squaring process runs in polynomial time.
== Proof ==
By the classical multiplication algorithm it is seen that the squaring process takes fewer than $4 n^2$ steps.
Hence the result by definition of polynomial time: $A = 4$ and $k = 2$.
== Sources ==
* 
* 
Category:Examples of Polynomial Time