avewright/proofwiki-math huggingface dataset:
Theorem: Idempotent Elements form Submonoid of Commutative Monoid

Let $\struct {S, \circ}$ be a commutative monoid.
Let $e \in S$ be the identity element of $\struct {S, \circ}$.
Let $I$ be the set of all elements of $S$ that are idempotent under $\circ$.
That is:
:$I = \set {x \in S: x \circ x = x}$
Then $\struct {I, \circ}$ is a submonoid of $\struct {S, \circ}$ with identity $e$.

Proof:
By Idempotent Elements form Subsemigroup of Commutative Semigroup, $\struct {I, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
By Identity Element is Idempotent:
:$e \in I$
By Identity of Algebraic Structure is Preserved in Substructure, $e$ is an identity of $\struct {I, \circ}$.
Since $\struct {T, \circ}$ is a semigroup and has an identity, $\struct {T, \circ}$ is a monoid.
Since $T \subseteq S$, $\struct {T, \circ}$ is a submonoid of $\struct {S, \circ}$ with identity $e$.
{{qed}}
Category:Commutative Monoids
nmay5kem8mwbx18o79suxw80kywkmx4
Theorem: Inverse Image under Order Embedding of Strict Upper Closure of Image of Point

Let $\struct {S, \preceq}$ and $\struct {T, \preceq'}$ be ordered sets.
Let $\phi: S \to T$ be an order embedding of $\struct {S, \preceq}$ into $\struct {T, \preceq'}$
Let $p \in S$.
Then:
:$\map {\phi^{-1} } {\map \phi p^{\succ'} } = p^\succ$
where $\cdot^\succ$ and $\cdot^{\succ'}$ represent strict upper closure with respect to $\preceq$ and $\preceq'$, respectively.

Proof:
Let $x \in \map {\phi^{-1} } {\map \phi p^{\succ'} }$.
By the definition of inverse image:
:$\map \phi x \in \map \phi p^{\succ'}$
By the definition of strict upper closure:
:$\map \phi p \prec' \map \phi x$
Since $\phi$ is an order embedding:
:$p \prec x$
Thus by the definition of strict upper closure:
:$x \in p^\succ$
and so:
:$\map {\phi^{-1} } {\map \phi p^{\succ'} } \subseteq p^\succ$
Let $x \in p^\succ$.
By the definition of strict upper closure:
:$p \prec x$
Since $\phi$ is an order embedding:
:$\map \phi p \prec' \map \phi x$
Thus by the definition of strict upper closure:
:$\map \phi x \in \map \phi p^{\succ'}$
Thus by the definition of inverse image:
:$x \in \map {\phi^{-1} } {\map \phi p^{\succ'} }$
and so:
:$p^\succ \subseteq \map {\phi^{-1} } {\map \phi p^{\succ'} }$
Thus by definition of set equality:
:$\map {\phi^{-1} } {\map \phi p^{\succ'} } = p^\succ$
{{qed}}
Category:Order Embeddings
Category:Upper Closures
io0oltfff5r2qj4nelhv9cohyuvf1sj
Theorem: Equivalence of Definitions of Order Embedding/Definition 1 implies Definition 3

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: S \to T$ be a mapping.
Let $\phi: S \to T$ be an order embedding by Definition 1:
{{:Definition:Order Embedding/Definition 1}}
Then $\phi: S \to T$ is an order embedding by Definition 3:
{{:Definition:Order Embedding/Definition 3}}

Proof:
Let $\phi$ be an order embedding by definition 1.
Then by definition:
:$\forall x, y \in S: x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
$\phi$ is injective by Order Embedding is Injection.
It remains to be shown that:
: $x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$
Suppose first that $x \prec_1 y$.
Then $x \preceq_1 y$ and $x \ne y$.
Thus by the premise:
:$\map \phi x \preceq_2 \map \phi y$
Since $\phi$ is injective:
:$\map \phi x \ne \map \phi y$
Therefore:
:$\map \phi x \prec_2 \map \phi y$
Suppose instead that $\map \phi x \prec_2 \map \phi y$
Then:
:$\map \phi x \preceq_2 \map \phi y$
and:
:$\map \phi x \ne \map \phi y$
By the premise:
:$x \preceq_1 y$
By the substitutive property of equality:
:$x \ne y$
Thus:
:$x \prec_1 y$
Thus $\phi$ is an order embedding by definition 3.
Theorem: Equivalence of Definitions of Order Embedding/Definition 3 implies Definition 1

Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: S \to T$ be a mapping.
Let $\phi: S \to T$ be an order embedding by Definition 3:
{{:Definition:Order Embedding/Definition 3}}
Then $\phi: S \to T$ is an order embedding by Definition 1:
{{:Definition:Order Embedding/Definition 1}}

Proof:
Let $\phi$ be an order embedding by definition 3.
Then by definition:
:$(1): \quad \phi$ is injective
:$(2): \quad \forall x, y, \in S: x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$
Let $x \preceq_1 y$.
Then $x \prec_1 y$ or $x = y$.
If $x \prec_1 y$, then by hypothesis:
:$\map \phi x \prec_2 \map \phi y$
Thus:
:$\map \phi x \preceq_2 \map \phi y$
If $x = y$, then:
:$\map \phi x = \map \phi y$
Thus:
:$\map \phi x \preceq_2 \map \phi y$
Thus it has been shown that:
:$x \preceq_1 y \implies \map \phi x \preceq_2 \map \phi y$
{{qed|lemma}}
Let $\map \phi x \preceq_2 \map \phi y$.
Then:
:$\map \phi x \prec_2 \map \phi y$
or:
:$\map \phi x = \map \phi y$
Suppose $\map \phi x \prec_2 \map \phi y$.
Then by hypothesis:
:$x \prec_1 y$
and so:
:$x \preceq_1 y$
Suppose $\map \phi x = \map \phi y$.
Then since $\phi$ is injective:
:$x = y$
and so:
:$x \preceq_1 y$
Thus in both cases:
:$x \preceq_1 y$
and so:
:$\map \phi x \preceq_2 \map \phi y \implies x \preceq_1 y$
{{qed|lemma}}
Hence the result:
:$x \preceq_1 y \iff \map \phi x \preceq_2 \map \phi y$
and so $\phi$ is an order embedding by definition 1.
Theorem: Inverse Image under Embedding of Image under Relation of Image of Point

Let $S$ and $T$ be sets.
Let $\RR_S$ and $\RR_t$ be relations on $S$ and $T$, respectively.
Let $\phi: S \to T$ be a mapping with the property that:
:$\forall p, q \in S: \paren {p \mathrel {\RR_S} q \iff \map \phi p \mathrel {\RR_T} \map \phi q}$
Then for each $p \in S$:
:$\map {\RR_S} p = \phi^{-1} \sqbrk {\map {\RR_T} {\map \phi p} }$

Proof:
Let $p \in S$.
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \map {\RR_S} p
}}
{{eqn | ll= \leadstoandfrom
      | l = p
      | o = \mathrel {\RR_S}
      | r = x
      | c = {{Defof|Image of Element under Relation|Image}} of $p$ under $\RR_S$
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \phi p
      | o = \mathrel {\RR_T}
      | r = \map \phi x
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \phi x
      | o = \in
      | r = \map {\RR_T} {\map \phi p}
      | c = Definition of the image of $\map \phi x$ under $\RR_T$
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | o = \in
      | r = \phi^{-1} \sqbrk {\map {\RR_T} {\map \phi p} }
      | c = {{Defof|Inverse Image|inverse image}}
}}
{{end-eqn}}
Thus by definition of set equality:
:$\map {\RR_S} p = \phi^{-1} \sqbrk {\map {\RR_T} {\map \phi p} }$
{{qed}}
Category:Relation Theory
3tcgjbam7j35iw7y9qwcearneit0lhs
Theorem: Path as Parameterization of Contour

Let $\closedint a b$ be a closed real interval.
Let $\gamma: \closedint a b \to \C$ be a path.
Let there exist $n \in \N$ and a subdivision $\set {a_0, a_1, \ldots, a_n}$ of $\closedint a b$ such that:
:$\gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } }$ is a smooth path for all $k \in \set {1, \ldots, n}$
where $\gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } }$ denotes the restriction of $\gamma$ to $\closedint {a_{k - 1} } {a_k}$.
Then there exists a contour $C$ with parameterization $\gamma$.

Proof:
Put $\gamma_k = \gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } } : \closedint {a_{k - 1} } {a_k} \to \C$.
By definition, it follows that there exists a directed smooth curve $C_k$ with parameterization $\gamma_k$.
For all $k \in \set {1, \ldots, n - 1}$, we have:
:$\map {\gamma_k} {a_k} = \map {\gamma_{k + 1} } {a_k}$
Define the contour $C$ as the concatenation $C_1 \cup C_2 \cup \ldots \cup C_n$.
Then, it follows by definition that $\gamma$ is a possible parameterization of $C$.
{{qed}}
Category:Path as Parameterization of Contour
Category:Contour Integrals
Category:Complex Contour Integrals
4v08nd00aemoffzypror98nsso7svah
Theorem: Kernel of Induced Homomorphism of Polynomial Forms

Let $R$ and $S$ be commutative rings with unity.
Let $\phi: R \to S$ be a ring homomorphism.
Let $K = \ker \phi$.
Let $R \sqbrk X$ and $S \sqbrk X$ be the rings of polynomial forms over $R$ and $S$ respectively in the indeterminate $X$.
Let $\bar \phi: R \sqbrk X \to S \sqbrk X$ be the induced morphism of polynomial rings.
Then the kernel of $\bar \phi$ is:
:$\ker \bar \phi = \set {a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X: \map \phi {a_i} = 0 \text{ for } i = 0, \ldots, n}$
Or, more concisely:
:$\ker \bar \phi = \paren {\ker \phi} \sqbrk X$

Proof:
Let $\map P X = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$.
Suppose first that $\map \phi {a_i} = 0$ for $i = 0, \ldots, n$.
We have by definition that:
:$\map {\bar \phi} {a_0 + a_1 X + \cdots + a_n X^n} = \map \phi {a_0} + \map \phi {a_1} X + \cdots + \map \phi {a_n} X^n = 0$
That is to say, $\map P X \in \ker \bar \phi$.
Conversely, suppose that $\map P X \in \ker \bar \phi$.
That is, $\map {\bar \phi} {\map P X} = \map \phi {a_0} + \map \phi {a_1} X + \cdots + \map \phi {a_n} X^n$ is the null polynomial.
This by definition means that for $i = 0, \ldots, n$ we have $\map \phi {a_i} = 0$.
Hence, $\map P X \in \paren {\ker \phi} \sqbrk X$.
This concludes the proof.
{{qed}}
Category:Polynomial Theory
0r61yqcjnkjefm8swmyae2e1p4a8rr9
Theorem: Boundary of Polygon as Contour

Let $P$ be a polygon embedded in the complex plane $\C$.
Denote the boundary of $P$ as $\partial P$.
Then there exists a simple closed contour $C$ such that:
:$\Img C = \partial P$
where $\Img C$ denotes the image of $C$.

Proof:
Let $n \in \N$ be the number of sides of $P$.
Denote the vertices of $P$ as $A_1, \ldots, A_n$.
From Complex Plane is Metric Space, it follows that $\C$ is homeomorphic to $\R^2$.
Then, we can consider $\partial P$ as a subset of $\R^2$.
From Boundary of Polygon is Jordan Curve, it follows that there exists a Jordan curve $\gamma: \closedint 0 1 \to \R^2$ such that $\Img \gamma = \partial P$.
The same theorem shows that $\gamma$ is a concatenation of $n$ line segments, parameterized as:
:$\map {\gamma_k} t = \paren {1 - t} A_k + t A_{k + 1}$
where $k \in \set {1, 2, \ldots, n}$.
We identify $A_{n + 1}$ with $A_1$.
Then, $\gamma_k: \closedint {a_{k - 1} } {a_k} \to \R^2$, where $a_{k - 1}, a_k \in \closedint 0 1$.
As $\gamma$ is a concatenation of $\gamma_1, \ldots, \gamma_n$, it follows that $\set {a_0, a_1, \ldots, a_n}$ is a subdivision of $\closedint 0 1$.
We have:
:$\dfrac \d {\d t} \map {\gamma_k} t = A_{k + 1} - A_k \ne \mathbf 0$
as $A_k$ and $A_{k + 1}$ are two different vertices.
As $\C$ is homeomorphic to $\R^2$, we can consider $\gamma$ as a continuous complex function $\gamma: \closedint 0 1 \to \C$.
Then $\gamma_k$ is complex-differentiable for all values of $t \in \openint {a_{k - 1} } {a_k}$ with its derivative $\gamma_k'$ defined by:
:$\map {\gamma_k'} t = \map x {\dfrac \d {\d t} \map {\gamma_k} t} + i \map y {\dfrac \d {\d t} \map {\gamma_k} t}$
where:
:$\map x {\dfrac \d {\d t} \map {\gamma_k} t}$ is the real part of $\dfrac \d {\d t} \map {\gamma_k} t$
:$\map y {\dfrac \d {\d t} \map {\gamma_k} t}$ is the imaginary part of $\dfrac \d {\d t} \map {\gamma_k} t$
As shown above:
:$\map {\gamma_k'} t \ne 0$
By definition, it follows that $\gamma_k = \gamma {\restriction_{\closedint {a_{k - 1} } {a_k} } }$ is a smooth path.
From Path as Parameterization of Contour, it follows that there exists a simple closed contour $C$ with $\Img C = \partial P$.
{{qed}}
Category:Complex Contour Integrals
17ugvfeau38frbjx317wb8djy17f7hu
Theorem: Zero Simple Staircase Integral Condition for Primitive

Let $f: D \to \C$ be a continuous complex function, where $D$ is a connected domain.
Let $\ds \oint_C \map f z \rd z = 0$ for all simple closed staircase contours $C$ in $D$.
Then $f$ has a primitive $F: D \to \C$.

Proof:
Let $C$ be a closed staircase contour in $D$, not necessarily simple.
If we show that $\ds \oint_C \map f z \rd z = 0$, then the result follows from Zero Staircase Integral Condition for Primitive.
The staircase contour $C$ is a concatenation of $C_1, \ldots, C_n$, where the image of each $C_k$ is a line segment parallel with either the real axis or the imaginary axis.
Denote the parameterization of $C$ as $\gamma: \closedint a b \to \C$, where $\closedint a b$ is a closed real interval.
Denote the parameterization of $C_k$ as $\gamma_k: \closedint {a_k} {b_k} \to \C$.
Theorem: Preordering induces Equivalence Relation

Let $\struct {S, \RR}$ be a relational structure such that $\RR$ is a preordering.
Let $\sim$ be the relation induced by $\RR$:
:$x \sim y$ {{iff}} $x \precsim y$ and $y \precsim x$.
Then $\sim$ is an equivalence relation.

Proof:
To show that $\sim$ is an equivalence relation, we must show that it is reflexive, transitive, and symmetric.
By the definition of preordering, $\precsim$ is transitive and reflexive.
Theorem: Content of Rational Polynomial is Multiplicative

Let $h \in \Q \sqbrk X$ be a polynomial with rational coefficients.
Let $\cont h$ denote the content of $h$.
Then for any polynomials $f, g \in \Q \sqbrk X$ with rational coefficients:
:$\cont {f g} = \cont f \cont g$

Proof:
From Rational Polynomial is Content Times Primitive Polynomial, let $\map f X$ and $\map g X$ be expressed as:
:$\map f X = \cont f \cdot \map {f^*} X$
:$\map g X = \cont g \cdot \map {g^*} X$
where:
:$\cont f, \cont g$ are the content of $f$ and $g$ respectively
:$f^*, g^*$ are primitive.
We have, by applications of Rational Polynomial is Content Times Primitive Polynomial:
:$\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \, \map {g^*} X$
By Gauss's Lemma on Primitive Rational Polynomials we have that $\map {f^*} X \, \map {g^*} X$ is primitive.
As $\cont f > 0$ and $\cont g > 0$, then so is $\cont f \cont g > 0$.
By the uniqueness clause in Rational Polynomial is Content Times Primitive Polynomial, the result follows.
{{qed}}
Theorem: Gauss's Lemma on Irreducible Polynomials

Let $\Z$ be the ring of integers.
Let $\Z \sqbrk X$ be the ring of polynomials over $\Z$.
Let $h \in \Z \sqbrk X$ be a polynomial.
{{TFAE}}
:$(1): \quad h$ is irreducible in $\Q \sqbrk X$ and primitive
:$(2): \quad h$ is irreducible in $\Z \sqbrk X$.

Proof:
{{explain|Needs to be made explicit as to exactly what is being assumed and what follows as a consequence. As it stands, the consequence of 1 implies 2, for example, still needs to be completed so as to explain exactly how that show $h$ is irreducible -- it relies upon the implicit understanding of what irreducible means.}}
Theorem: Antisymmetric Quotient of Preordered Set is Ordered Set

Let $\struct {S, \precsim}$ be a preordered set.
Let $\sim$ be the equivalence relation $S$ induced by $\precsim$.
Let $\struct {S / {\sim}, \preceq}$ be the antisymmetric quotient of $\struct {S, \precsim}$.
Then:
:$\struct {S / {\sim}, \preceq}$ is an ordered set.
:$\forall P, Q \in S / {\sim}: \paren {P \preceq Q} \land \paren {p \in P} \land \paren {q \in Q} \implies p \precsim q$
This second statement means that we could just as well have defined $\preceq$ by letting $P \preceq Q$ {{iff}}:
:$\forall p \in P: \forall q \in Q: p \precsim q$

Proof:
By the definition of equivalence relation, $\sim$ is transitive, reflexive, and symmetric.
By the definition of preordering, $\precsim$ is transitive and reflexive.
From Preordering induces Ordering, $\preceq$ is an ordering.
Let $P, Q \in S / {\sim}$ with $P \preceq Q$.
Then by the definition of $\preceq$, there are $p \in P$ and $q \in Q$ such that $p \precsim q$.
Let $p' \in P$ and $q' \in Q$.
By the definition of quotient set:
:$p' \sim p$
:$q \sim q'$
Thus by the definition of $\sim$:
:$p' \precsim p$
:$q \precsim q'$
Since $p \precsim q$ and $\precsim$ is transitive:
:$p' \precsim q'$
We have shown that:
:$\forall P, Q \in S / {\sim}: \paren {P \preceq Q} \land \paren {p \in P} \land \paren {q \in Q} \implies p \precsim q$
{{qed}}
Category:Order Theory
Category:Preorder Theory
Category:Quotient Sets
21so65b0axq4rohinhidjkotgq6grbs
Theorem: Ordering on Partition Determines Preordering

Let $S$ be a set.
Let $\PP$ be a partition of $S$.
Let $\phi: S \to \PP$ be the quotient mapping.
Let $\preceq$ be a ordering of $\PP$.
Define a relation $\precsim$ on $S$ by letting $p \precsim q$ {{iff}}:
:$\map \phi p \preceq \map \phi q$
Then:
:$\precsim$ is a preordering on $S$.
:$\precsim$ is the only preordering on $S$ that induces the $\preceq$ ordering on $\PP$.

Proof:
To show that $\precsim$ is a preordering we must show that it is reflexive and transitive.
Theorem: Units of Ring of Polynomial Forms over Commutative Ring

Let $\struct {R, +, \circ}$ be a non-null commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $R \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $R$.
Let $\map P X = a_0 + a_1 X + \cdots + a_n X^n \in R \sqbrk X$.
Then:
:$\map P X$ is a unit of $R \sqbrk X$
{{iff}}:
:$a_0$ is a unit of $R$
Also, for $i = 1, \ldots, n$, $a_i$ is nilpotent in $R$.

Proof:
=== Necessary condition ===
Let $a_0$ be a unit of $R$.
For $i = 1, \ldots, n$, let $a_i$ be nilpotent in $R$.
Because the nilradical is an ideal of $R$, it follows that:
:$Q = -a_1 X + \dotsb + -a_n X^n$
is nilpotent.
Moreover, multiplying through by $a_0^{-1}$ we may as well assume that $a_0 = 1_R$.
Then:
:$P = 1_R - Q$
and from Unity plus Negative of Nilpotent Ring Element is Unit $P$ is a unit of $R \sqbrk X$.
{{qed|lemma}}
Theorem: Polynomials Closed under Addition/Polynomials over Integral Domain

Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is closed under the operation $+$.

Proof:
Let $p, q$ be polynomials in $x$ over $D$.
We can express them as:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k$
:$\ds q = \sum_{k \mathop = 0}^n b_k \circ x^k$
where:
:$(1): \quad a_k, b_k \in D$ for all $k$
:$(2): \quad m, n \in \Z_{\ge 0}$, that is, are non-negative integers.
Suppose $m = n$.
Then:
:$\ds p + q = \sum_{k \mathop = 0}^n a_k \circ x^k + \sum_{k \mathop = 0}^n b_k \circ x^k$
Because $\struct {R, +, \circ}$ is a commutative ring, it follows that:
:$\ds p + q = \sum_{k \mathop = 0}^n \paren {a_k + b_k} \circ x^k$
which is also a polynomials in $x$ over $D$.
{{explain|Establish whether $R$ actually needs to be commutative for the above to hold.}}
{{WLOG}}, suppose $m > n$.
Then we can express $q$ as:
:$\ds \sum_{k \mathop = 0}^n b_k \circ x^k + \sum_{k \mathop = n \mathop + 1}^m 0_D \circ x^k$
Thus:
:$\ds p + q = \sum_{k \mathop = 0}^n \paren {a_k + b_k} \circ x^k + \sum_{k \mathop = n \mathop + 1}^m a_k \circ x^k$
which is also a polynomials in $x$ over $D$.
Thus the sum of two polynomials in $x$ over $D$ is another polynomial in $x$ over $D$.
Hence the result.
{{qed}}
Theorem: Polynomials Closed under Addition/Polynomials over Integral Domain

Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.
Then $\forall x \in R$, the set $D \sqbrk x$ of polynomials in $x$ over $D$ is closed under the operation $+$.

Proof:
A commutative ring with unity is a ring.
An integral subdomain of a commutative ring with unity $R$ is a subring of $R$.
The result then follows as a special case of Polynomials Closed under Addition: Polynomials over Ring
{{qed}}
Theorem: Polynomials Closed under Addition/Polynomials over Ring

Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +, \circ}$ be a subring of $R$.
Then $\forall x \in R$, the set $S \sqbrk x$ of polynomials in $x$ over $S$ is closed under the operation $+$.

Proof:
Let $p, q$ be polynomials in $x$ over $S$.
We can express them as:
:$\ds p = \sum_{k \mathop = 0}^m a_k \circ x^k$
:$\ds q = \sum_{k \mathop = 0}^n b_k \circ x^k$
where:
:$(1): \quad a_k, b_k \in S$ for all $k$
:$(2): \quad m, n \in \Z_{\ge 0}$, that is, are non-negative integers.
Suppose $m = n$.
Then:
:$\ds p + q = \sum_{k \mathop = 0}^n a_k \circ x^k + \sum_{k \mathop = 0}^n b_k \circ x^k$
Because $\struct {R, +, \circ}$ is a ring, it follows that:
:$\ds p + q = \sum_{k \mathop = 0}^n \paren {a_k + b_k} \circ x^k$
which is also a polynomial in $x$ over $S$.
{{handwaving|What needs to be done here is to establish that each of the terms is an element of $R$ and that each of the coefficients is an element of $S$. Trivial, but important.}}
{{WLOG}}, suppose $m > n$.
Then we can express $q$ as:
:$\ds \sum_{k \mathop = 0}^n b_k \circ x^k + \sum_{k \mathop = n \mathop + 1}^m 0_D \circ x^k$
Thus:
:$\ds p + q = \sum_{k \mathop = 0}^n \paren {a_k + b_k} \circ x^k + \sum_{k \mathop = n \mathop + 1}^m a_k \circ x^k$
which is also a polynomial in $x$ over $S$.
Thus the sum of two polynomials in $x$ over $S$ is another polynomial in $x$ over $S$.
Hence the result.
{{qed}}
Category:Polynomial Theory
e58nmyk1e81j1j39ol5dkek421b2qqb
Theorem: Polynomials Closed under Addition/Polynomial Forms

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 polynomials in the indeterminates $\set {X_j: j \in J}$ over the ring $R$.
Then the operation of polynomial addition on $f$ and $g$:
Define the sum:
:$\ds f \oplus g = \sum_{k \mathop \in Z} \paren {a_k + b_k} \mathbf X^k$
Then $f \oplus g$ is a polynomial.
That is, the operation of polynomial addition is closed on the set of all polynomials on a given set of indeterminates $\set {X_j: j \in J}$.

Proof:
It  is immediate that $f \oplus g$ is a map from the free commutative monoid to $R$, so we need only prove  that $f \oplus g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$.
Suppose that for some $k \in Z$, $a_k + b_k \ne 0$
This forces at least one of $a_k$ and $b_k$ to be non-zero.
This can only be true for a finite number of terms because $f$ and $g$ are polynomials.
The result follows.
{{qed}}
Category:Polynomial Theory
{{proofread}}
sc92g1wkqvfxictf67zgoeew0su5i9k
Theorem: Fuzzy Intersection is Commutative

Fuzzy intersection is commutative.

Proof:
Let $\textbf A = \struct{A, \mu_A}$ and $\textbf B = \struct{B, \mu_B}$ be fuzzy sets.
Theorem: Nilpotent Element is Zero Divisor

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Suppose further that $R$ is not the null ring.
Let $x \in R$ be a nilpotent element of $R$.
Then $x$ is a zero divisor in $R$.

Proof:
First note that when $R$ is the null ring the result is false.
This is because although $0_R$ is nilpotent element in the null ring, it is not actually a zero divisor.
Hence in this case $0_R$ is both nilpotent and a zero divisor.
So, let $R$ be a non-null ring.
By hypothesis, there exists $n \in \Z_{>0}$ such that $x^n = 0_R$.
If $n = 1$, then $x = 0_R$.
By hypothesis, $R$ is not the null ring, so we may choose $y \in R \setminus \set 0$.
By Ring Product with Zero:
:$y \circ x = y \circ 0_R = 0_R$
Therefore $x$ is a zero divisor in $R$.
If $n \ge 2$, define $y = x^{n - 1}$.
Then:
:$y \circ x = x^{n - 1} \circ x = x^n = 0_R$
so $x$ is a zero divisor in $R$.
{{Qed}}
Category:Nilpotent Ring Elements
Category:Zero Divisors
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Theorem: Integral Domain is Reduced Ring

Let $\left({D, +, \circ}\right)$ be an integral domain.
Then $D$ is reduced.

Proof:
Let $x \in D$ be a nilpotent element.
Then by Nilpotent Element is Zero Divisor, $x$ is a zero divisor in $D$.
By the definition of an integral domain, this means that $x = 0$.
Therefore the only nilpotent element of $D$ is $0$.
That is, $D$ is reduced.
{{Qed}}
Category:Ring Theory
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Theorem: Units of Ring of Polynomial Forms over Integral Domain

Let $\struct {D, +, \circ}$ be an integral domain.
Let $D \sqbrk X$ be the ring of polynomial forms in an indeterminate $X$ over $D$.
Then the group of units of $D \sqbrk X$ is precisely the group of elements of $D \sqbrk X$ of degree zero that are units of $D$.

Proof:
It is immediate that a unit of $D$ is also a unit of $D \sqbrk X$.
Let $P$ be a unit of $D \sqbrk X$.
Then there exists $Q \in D \sqbrk X$ such that $P Q = 1$.
By Corollary 2 to Degree of Product of Polynomials over Ring we have:
:$0 = \map \deg 1 = \map \deg P + \map \deg Q$
Therefore:
:$\map \deg P = \map \deg Q = 0$
That is, $P \in R$ and $Q \in R$.
Moreover $P Q = 1$ in $R$, so it follows that $P$ is a unit of $R$.
{{Qed}}
Category:Polynomial Theory
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