avewright/proofwiki-math huggingface dataset:
Theorem: Inverse for Complex Multiplication

Each element $z = x + i y$ of the set of non-zero complex numbers $\C_{\ne 0}$ has an inverse element $z^{-1}$ under the operation of complex multiplication:
:$\forall z \in \C_{\ne 0}: \exists z^{-1} \in \C_{\ne 0}: z \times z^{-1} = 1 + 0 i = z^{-1} \times z$
This inverse can be expressed as:
:$\dfrac 1 z := \dfrac {x - i y} {x^2 + y^2} = \dfrac {\overline z} {z \overline z}$
where $\overline z$ is the complex conjugate of $z$.

Proof:
{{begin-eqn}}
{{eqn | l = \paren {x + i y} \frac {x - i y} {x^2 + y^2}
      | r = \frac {\paren {x \cdot x - y \cdot \paren {-y} } + i \paren {x \cdot \paren {-y} + x \cdot y} } {x^2 + y^2}
      | c = 
}}
{{eqn | r = \frac {\paren {x^2 + y^2} + 0 i} {x^2 + y^2}
      | c = 
}}
{{eqn | r = 1 + 0 i
      | c = 
}}
{{end-eqn}}
Similarly for $\dfrac {x - i y} {x^2 + y^2} \paren {x + i y}$.
So the inverse of $x + i y \in \struct {\C_{\ne 0}, \times}$ is $\dfrac {x - i y} {x^2 + y^2}$.
As $x^2 + y^2 > 0 \iff x, y \ne 0$ the inverse is defined for all $z \in \C: z \ne 0 + 0 i$.
{{qed|lemma}}
From the definition, the complex conjugate $\overline z$ of $z = x + i y$ is $x - i y$.
From the definition of the modulus of a complex number, we have:
:$\cmod z = \sqrt {a^2 + b^2}$
From Modulus in Terms of Conjugate, we have that:
:$\cmod z^2 = z \overline z$
Hence the result:
:$\dfrac 1 z = \dfrac {\overline z} {z \overline z}$
{{qed}}
{{expand|Add the polar and exponential forms of this}}
Theorem: Additive Group of Integers is Normal Subgroup of Complex

Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\C, +}$ be the additive group of complex numbers.
Then $\struct {\Z, +}$ is a normal subgroup of $\struct {\C, +}$.

Proof:
From Additive Group of Integers is Normal Subgroup of Reals, $\struct {\Z, +} \lhd \struct {\R, +}$.
From Additive Group of Reals is Subgroup of Complex, $\struct {\R, +} \lhd \struct {\C, +}$.
Thus $\struct {\Z, +} \le \struct {\C, +}$.
From Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is abelian.
From Subgroup of Abelian Group is Normal, it follows that $\struct {\Z, +} \lhd \struct {\C, +}$.
{{qed}}
Category:Additive Group of Integers
Category:Additive Group of Complex Numbers
Category:Examples of Normal Subgroups
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Theorem: Additive Group of Rationals is Normal Subgroup of Complex

Let $\struct {\Q, +}$ be the additive group of rational numbers.
Let $\struct {\C, +}$ be the additive group of complex numbers.
Then $\struct {\Q, +}$ is a normal subgroup of $\struct {\C, +}$.

Proof:
From Additive Group of Rationals is Normal Subgroup of Reals, $\struct {\Q, +} \lhd \struct {\R, +}$.
From Additive Group of Reals is Normal Subgroup of Complex, $\struct {\R, +} \lhd \struct {\C, +}$.
Thus $\struct {\Q, +} \le \struct {\C, +}$.
From Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is abelian.
From Subgroup of Abelian Group is Normal, it follows that $\struct {\Q, +} \lhd \struct {\C, +}$.
{{qed}}
Category:Additive Group of Rational Numbers
Category:Additive Group of Complex Numbers
Category:Examples of Normal Subgroups
lx3nwn86n5l4jlf10twpq4quv94r0ag
Theorem: Multiplicative Group of Rationals is Normal Subgroup of Complex

Let $\struct {\Q, \times}$ be the multiplicative group of rational numbers.
Let $\struct {\C, \times}$ be the multiplicative group of complex numbers.
Then $\struct {\Q, \times}$ is a normal subgroup of $\struct {\C, \times}$.

Proof:
From Multiplicative Group of Rationals is Normal Subgroup of Reals, $\struct {\Q, \times} \lhd \struct {\R, \times}$.
From Multiplicative Group of Reals is Normal Subgroup of Complex, $\struct {\R, \times} \lhd \struct {\C, \times}$.
Thus $\struct {\Q, \times} \le \struct {\C, \times}$.
From Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group, $\struct {\C, \times}$ is abelian.
From Subgroup of Abelian Group is Normal it follows that $\struct {\Q, \times} \lhd \struct {\C, \times}$.
{{qed}}
Category:Multiplicative Group of Rational Numbers
Category:Multiplicative Group of Complex Numbers
Category:Examples of Normal Subgroups
nrq6pniyfe5acbc8h5pkecacxmlshy4
Theorem: Rational Numbers form Subfield of Complex Numbers

Let $\struct {\Q, +, \times}$ denote the field of rational numbers.
Let $\struct {\C, +, \times}$ denote the field of complex numbers.
$\struct {\Q, +, \times}$ is a subfield of $\struct {\C, +, \times}$.

Proof:
From Rational Numbers form Subfield of Real Numbers, $\struct {\Q, +, \times}$ is a subfield of $\struct {\R, + \times}$.
From Real Numbers form Subfield of Complex Numbers, $\struct {\R, +, \times}$ is a subfield of $\struct {\C, + \times}$.
Thus from Subfield of Subfield is Subfield $\struct {\Q, +, \times}$ is a subfield of $\struct {\C, + \times}$.
{{qed}}
Theorem: Subfield of Subfield is Subfield

Let $R$ be a ring with unity.
Let $K_1, K_2$ be fields, such that:
: $K_1$ is a subfield of $R$
: $K_2$ is a subfield of $K_1$
Then $K_2$ is a subfield of $R$.

Proof:
Let $K_1$ be a subfield of $R$ and $K_2$ be a subfield of $K_1$.
Then by definition:
:$K_1 \subseteq R$
:$K_2 \subseteq K_1$
From Subset Relation is Transitive it follows that $K_2 \subseteq R$
So by definition $K_2$ is a subfield of $R$.
{{qed}}
Category:Subfields
g8qj1gxkqsgvlury7jpgaahfpiooiz9
Theorem: Nakayama's Lemma

Let $A$ be a commutative ring with unity.
Let $M$ be a finitely generated $A$-module. 
Let $\map {\operatorname{Jac} } A$ be the Jacobson radical of $A$.
Let $\mathfrak a \subseteq \map {\operatorname{Jac} } A$ be an ideal of $A$.
Suppose $\mathfrak a M = M$.
Then:
:$M = 0$

Proof:
We induct on the number of generators of $M$. 
;Basis for the Induction
Let $M$ have a single generator $m_1 \in M$.
Then $\mathfrak a m_1 = M$.
So:
:$m_1 \in \mathfrak a m_1$
That is:
:$m_1 = a m_1$
for some $a \in \mathfrak a$.
By Characterisation of Jacobson Radical, $1 - a$ is a unit in $A$.
So:
:$\paren {1 - a}^{-1} \paren {1 - a} m = 0$
thus $m = 0$.
;Inductive Step
Suppose that $M$ is generated by $n$ elements:
:$M = A m_1 + \dotsb + A m_n$
for some $m_1, \dotsc, m_n \in M$.
Then we have:
:$M = \mathfrak a M = \mathfrak a m_1 + \dotsb + \mathfrak a m_n $
Thus for some $a_1, \dotsc, a_n \in \mathfrak a$:
:$ m_1 = a_1 m_1 + \dotsb + a_n m_n $
Then:
:$\paren {1 - a_1} m_1 = a_2 m_2 + \dotsb + a_n m_n$
By Characterisation of Jacobson Radical, $1 - a$ is a unit in $A$.
Multiplying both sides by $\paren {1 - a_1}^{-1}$ gives:
:$m_1 = \paren {1 - a_1}^{-1} a_2 m_2 + \dotsb + \paren {1 - a_1}^{-1} a_n m_n$
so we have:
:$m_1 \in A m_2 + \cdots + A m_n$
Therefore $M$ has $n - 1$ generators $m_2, \dotsc, m_n$.
By the induction hypothesis:
:$\mathfrak a M = M \implies M = 0$
{{qed}}
Theorem: Nakayama's Lemma

Let $A$ be a commutative ring with unity.
Let $M$ be a finitely generated $A$-module. 
Let $\map {\operatorname{Jac} } A$ be the Jacobson radical of $A$.
Let $\mathfrak a \subseteq \map {\operatorname{Jac} } A$ be an ideal of $A$.
Suppose $\mathfrak a M = M$.
Then:
:$M = 0$

Proof:
Let $\phi : M \to M$ be the identity mapping on $M$, i.e.:
:$\forall x \in M : \map \phi x = x$
Since $\mathfrak a M = M$ {{hypothesis}}, $\phi$ is an endomorphism of $M$ such that:
:$\map \phi M \subseteq a M$
By Cayley-Hamilton Theorem, there exist $a_0, \ldots , a_{n-1} \in \mathfrak a$ such that:
:$(1):\quad \phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$
Evaluating $(1)$ at $x=1$, especially, we have:
:$(2):\quad 1 + a= 0$
where:
:$a := a_{n - 1} + \cdots + a_1 + a_0$
Since:
:$a \in \mathfrak a \subseteq \map {\operatorname {Jac} } A$
by Characterisation of Jacobson Radical, $1 + a$ is a unit in $A$.
Thus for each $x \in M$, we have:
{{begin-eqn}}
{{eqn | l = x
      | r = \paren {1 + a}^{-1} \paren {1 + a} x
}}
{{eqn | r = \paren {1 + a}^{-1} 0
      | c =  by $(2)$
}}
{{eqn | r = 0
}}
{{end-eqn}}
That is:
:$M = 0$
{{qed}}
Theorem: Characterisation of Jacobson Radical

Let $A$ be a commutative ring with unity.
Let $A^\times$ be the group of units of $A$.
Let $\map {\operatorname {Jac} } A$ be the Jacobson radical of $A$.
Then:
:$\map {\operatorname {Jac} } A = \set {a \in A: 1_A - a x \in A^\times \text{ for all } x \in A}$
where $1_A$ is the unity of $A$.

Proof:
Recall the definition of Jacobson radical of $A$:
:$\map {\operatorname {Jac} } A$ is the intersection of all maximal ideals of $R$.
Theorem: Prime Ideal iff Quotient Ring is Integral Domain

Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $J$ be an ideal of $R$.
Then $J$ is a prime ideal of $R$ {{iff}} the quotient ring $R / J$ is an integral domain.

Proof:
Since $J \subset R$, it follows from:
:Quotient Ring of Commutative Ring is Commutative
and:
:Quotient Ring of Ring with Unity is Ring with Unity
that $R / J$ is a commutative ring with unity. 
Let $0_{R / J}$ be the zero of $R / J$.
Theorem: Maximal Ideal of Commutative and Unitary Ring is Prime Ideal

Let $R$ be a commutative ring with unity.
Let $M$ be a maximal ideal of $R$.
Then $M$ is a prime ideal of $R$.

Proof:
From Maximal Ideal iff Quotient Ring is Field:
:the quotient ring $R / M$ is a field.
It follows from Field is Integral Domain that $R / M$ is an integral domain.
Finally it follows from Prime Ideal iff Quotient Ring is Integral Domain that $M$ is a prime ideal.
{{qed}}
Theorem: Radical Ideal iff Quotient Ring is Reduced

Let $\left({R, +, \circ}\right)$ be a commutative ring with unity.
Let $J$ be an ideal of $R$.
Then $J$ is a radical ideal {{iff}} the quotient ring $R / J$ is a reduced ring.

Proof:
Since $J \subset R$, it follows from:
:Quotient Ring of Commutative Ring is Commutative
and:
: Quotient Ring of Ring with Unity is Ring with Unity
that $R / J$ is a commutative ring with unity.
Let $0_{R / J}$ be the zero of $R / J$.
Theorem: Set of Monomials is Closed Under Multiplication

Let $M$ be the set of all monomials on the set $\set {X_j: j \in J}$, with multiplication $\circ$ defined by:
:$\ds \paren {\prod_{j \mathop \in J} X_j^{k_j} } \circ \paren {\prod_{j \mathop \in J} X_j^{k_j'} } = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$
Then $M$ is closed under $\circ$.

Proof:
Let $\ds m_1 = \prod_{j \mathop \in J} X_j^{k_j}, m_2 = \prod_{j \mathop \in J} X_j^{k_j'}$ be two monomials.
Their product is:
:$\ds m_1 \circ m_2 = \paren {\prod_{j \mathop \in J} X_j^{k_j + k_j'} }$
If $k_j + k_j' \ne 0$ then either $k_j \ne 0$ or $k_j' \ne 0$ (or both are nonzero).
Therefore if $k_j + k_j' \ne 0$ for infinitely many $j$, then either $m_1$ or $m_2$ is not a monomial.
{{qed}}
Category:Monomials
h2lb3nxhe53agesh51ic1k8xmovh20p
Theorem: Free Commutative Monoid is Commutative Monoid

The free commutative monoid on a set $\family {X_j: j \in J}$ is a commutative monoid.

Proof:
Let $M$ be the set of all monomials on the indexed set $\family {X_j: j \in J}$.
We are required to show that the following properties hold:
{{begin-axiom}}
{{axiom | q = \forall m_1, m_2 \in M
        | m = m_1 \circ m_2 \in M
        | t = Closure
}}
{{axiom | q = \forall m_1, m_2, m)3 \in M
        | m = \paren {m_1 \circ m_2} \circ m_3 = m_1 \circ \paren {m_2 \circ m_3}
        | t = Associativity
}}
{{axiom | q = \forall m_1, m_2 \in M
        | m = m_1 \circ m_2 = m_2 \circ m_1
        | t = Commutativity
}}
{{axiom | q = \exists e_m \in M: \forall m_1 \in M
        | m = m_1 \circ e_M = m_1
        | t = Identity
}}
{{end-axiom}}
First note that using the multiindex notation described in the definition of monomials, for $r \in \N$, $m_i = \mathbf X^{k^i} \in M$, $i = 1, \ldots, r$, the product of the $m_i$ is given by:
:$m_1 \circ \cdots \circ m_r = \mathbf X^{k^1 + \cdots + k^r}$
Here the superscripts enumerate the multiindices, and do not indicate raising to a power.
So to show the closure, associativity and commutativity of monomials under $\circ$, it is sufficient to show the corresponding properties for multiindices under addition defined by:
:$\family {k^1 + k^2}_j := k^1_j + k^2_j$
In the following $k^1, k^2, k^3$ are multiindices, that is, families of non-negative integers indexed by $J$ such that only finitely many entries are non-zero.
Theorem: Polynomial Addition is Associative

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:
{{begin-eqn}}
{{eqn | l = \paren {f + g} + h
      | r = \sum_{k \mathop \in Z} \paren {\paren {a_k + b_k} + c_k} \mathbf X^k
      | c = {{Defof|Addition of Polynomial Forms}} twice
}}
{{eqn | r = \sum_{k \mathop \in Z} \paren {a_k + \paren {b_k + c_k} } \mathbf X^k
      | c = because $+$ in $R$ is associative
}} 
{{eqn | r = f + \paren {g + h}
      | c = {{Defof|Addition of Polynomial Forms}} twice
}} 
{{end-eqn}}
{{qed}}
Category:Polynomial Theory
s5vmb3sb1rzk72i24nlxl74im8g0yj5
Theorem: Null Polynomial is Additive Identity

The set of polynomial forms has an additive identity.
{{explain|Context}}

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:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
be an arbitrary polynomial form in the indeterminates $\set {X_j: j \in J}$ over $R$.
Let:
:$\ds N = \sum_{k \mathop \in Z} 0_R \mathbf X^k$ 
be the null polynomial.
Then:
{{begin-eqn}}
{{eqn | l = f + N
      | r = \sum_{k \mathop \in Z} \paren {a_k + 0_R} \mathbf X^k
      | c = {{Defof|Polynomial Addition}}
}}
{{eqn | r = \sum_{k \mathop \in Z} a_k \mathbf X^k
      | c = 
}} 
{{eqn | r = f
      | c = {{Defof|Polynomial Addition}}
}} 
{{end-eqn}}
Therefore, $N + f = f$ for all polynomial forms $f$.
Therefore, $N$ is an additive identity for the set of polynomial forms.
{{qed}}
Category:Polynomial Theory
26haui82r2ie3ormeskfn4xjtd9un5b
Theorem: Polynomial Addition is Commutative

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:
{{begin-eqn}}
{{eqn | l = f + g
      | r = \sum_{k \mathop \in Z} \paren {a_k + b_k} \mathbf X^k
      | c = {{Defof|Addition of Polynomial Forms}}
}}
{{eqn | r = \sum_{k \mathop \in Z} \paren {b_k + a_k} \mathbf X^k
      | c = because addition in $R$ is commutative
}} 
{{eqn | r = g + f
      | c = {{Defof|Addition of Polynomial Forms}}
}} 
{{end-eqn}}
So $f + g = g + f$ for all polynomials $f$ and $g$.
Therefore polynomial addition is commutative.
{{qed}}
Category:Polynomial Theory
ewos7jvvxd8s5rrhmh2878fvghxjafc
Theorem: Multiplication of Polynomials is Associative

Multiplication of polynomials is associative.

Proof:
Let $\struct {R, +, \circ}$ be a commutative ring with unity with zero $0_R$.
To improve readability of the expressions used, we will write the ring product $\circ$ in multiplicative notation.
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 it follows from Polynomials Closed under Ring Product that:
:$\ds f \circ \paren {g \circ h} = \sum_{k \mathop \in Z} m_k \mathbf X^k$
for some $m_k \in R$, and:
:$\ds \paren {f \circ g} \circ h = \sum_{k \mathop \in Z} n_k \mathbf X^k$
for some $n_k \in R$.
To establish associativity of $\circ$ we compute $m_k$ and $n_k$, and check that they are equal. 
We have:
{{begin-eqn}}
{{eqn | l = m_k
      | r = \sum_{r + s \mathop = k} a_r \sum_{p + q \mathop = s} b_p c_q
      | c = {{Defof|Multiplication of Polynomial Forms}}
}}
{{eqn | r = \sum_{r + s \mathop = k} \, \sum_{p + q \mathop = s} a_r b_p c_q
      | c = Distributive property of finite sums
}} 
{{eqn | r = \sum_{p + q + r \mathop = k} a_r b_p c_q
      | c = Properties of finite sums
}} 
{{end-eqn}}
{{explain|Sorry, but what I meant by "what properties?" was: Links are needed to the appropriate pages, in accordance with the philosophy of {{ProofWiki}}.}}
Similarly we compute:
{{begin-eqn}}
{{eqn | l = n_k
      | r = \sum_{r + s \mathop = k} \left({ \sum_{p + q \mathop = r} a_p b_q }\right) c_s
      | c = {{Defof|Multiplication of Polynomial Forms}}
}}
{{eqn | r = \sum_{r + s \mathop = k} \, \sum_{p + q \mathop = r} a_p b_q c_s
      | c = Distributive property of finite sums
}} 
{{eqn | r = \sum_{p + q + s \mathop = k} a_p b_q c_s
      | c = Properties of finite sums
}} 
{{end-eqn}}
{{explain|again, links are needed to the appropriate pages}}
Since $p$, $q$, $r$ and $s$ are all dummy variables, it follows that $m_k = n_k$ for all $k \in Z$.
Therefore, $f \circ \paren {g \circ h} = \paren {f \circ g} \circ h$ for all polynomials $f$, $g$ and $h$.
Hence multiplication of polynomials is associative.
{{qed}}
Category:Polynomial Theory
6zi6uvkk8ski5ks39fradwncsink88h
Theorem: Polynomials Contain Multiplicative Identity

The set of polynomials has a multiplicative identity.

Proof:
Let $\struct {R, +, \circ}$ be a commutative ring with unity with multiplicative identity $1_R$ and additive identity $0_R$.
Let $\set {X_j: j \in J}$ be a set of indeterminates.
Let $Z$ be the set of all multiindices indexed by $\left\{{X_j: j \in J}\right\}$.
Let:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$ 
be an arbitrary polynomial in the indeterminates $\set {X_j: j \in J}$ over $R$.
Let:
:$\ds N = 1_R \mathbf X^0 = \sum_{k \mathop \in Z} b_k \mathbf X^k$ 
where $b_k = 0_R$ if $k \ne 0$ and $b_0 = 1_R$.
Then:
{{begin-eqn}}
{{eqn | l = f \circ N
      | r = \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_p b_q} \mathbf X^k
      | c = {{Defof|Polynomial Multiplication}}
}}
{{eqn | r = \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + 0 \mathop = k} a_p 1_R} \mathbf X^k
      | c = by the definition of $b_k$
}} 
{{eqn | r = \sum_{k \mathop \in Z} a_p \mathbf X^k
      | c = because $1_R$ is a multiplicative identity
}} 
{{eqn | r = f
      | c = 
}} 
{{end-eqn}}
Therefore:
:$f \circ N = f$
for all polynomials $f$.
Therefore $N$ is a multiplicative identity for the set of polynomials.
{{qed}}
Category:Polynomial Theory
8py1n1dhghz1sjuntk63qp4ybg07rmb
Theorem: Multiplication of Polynomials is Commutative

Multiplication of polynomials is commutative.

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:
:$\ds f = \sum_{k \mathop \in Z} a_k \mathbf X^k$
:$\ds g = \sum_{k \mathop \in Z} b_k \mathbf X^k$
be arbitrary polynomials in the indeterminates $\set {X_j: j \in J}$ over $R$.
Then:
{{begin-eqn}}
{{eqn | l = f \circ g
      | r = \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_p b_q} \mathbf X^k
      | c = {{Defof|Polynomial Multiplication}}
}}
{{eqn | r = \sum_{k \mathop \in Z} \paren {\sum_{p \mathop + q \mathop = k} a_q b_p} \mathbf X^k
      | c = Addition of multiindices is commutative, and $p$ and $q$ are bound variables
}} 
{{eqn | r = g \circ f
      | c = {{Defof|Polynomial Multiplication}}
}} 
{{end-eqn}}
Therefore, $f \circ g = g \circ f$ for all polynomials $f$ and $g$.
Therefore, polynomial multiplication is commutative.
{{qed}}
Category:Polynomial Theory
pp12fxdbboearvwy3e6lgc0i4jt6jkl
Theorem: Multiplication of Polynomials Distributes over Addition

Multiplication of polynomials is left- and right- distributive over addition.

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 $R$.
By Multiplication of Polynomials is Commutative, it is sufficient to prove that $\circ$ is left distributive over addition only.
Then
{{begin-eqn}}
{{eqn | l = f \circ \paren {g + h}
      | r = \sum_{k \mathop \in Z} \sum_{p \mathop + q \mathop = k} a_p \paren {b_q + c_q} \mathbf X^k
      | c = by the definitions of polynomial multiplication and addition
}}
{{eqn | r = \sum_{k \mathop \in Z} \sum_{p \mathop + q \mathop = k} \paren {a_p b_q + a_p c_q} \mathbf X^k
      | c = by the properties of finite sums
}} 
{{eqn | r = \sum_{k \mathop \in Z} \sum_{p \mathop + q \mathop = k} a_p b_q \mathbf  X^k + \sum_{k \mathop \in Z} \sum_{p + q = k} a_p c_q \mathbf X^k
      | c = by the properties of finite sums
}} 
{{eqn | r = f \circ g + f \circ h
      | c = by the definitions of polynomial multiplication and addition
}} 
{{end-eqn}}
Therefore, $f \circ \paren {g + h} = f \circ g + f \circ h$ for all polynomials $f, g, h$.
Therefore, polynomial multiplication is distributive over addition.
{{qed}}
Category:Polynomial Theory
Category:Examples of Distributive Operations
6bfrmhnd2zcsvefaz8tnmulsno267jk
Theorem: Distributivity is Preserved in Induced Structure

Let $\struct {T, \oplus, \otimes}$ be an algebraic structure, and let $S$ be a set.
Let $T^S$ denote the set of all mappings from $S$ to $T$.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\oplus$.
Let $\struct {T^S, \otimes}$ be the structure on $T^S$ induced by $\otimes$.
If $\otimes$ is distributive over $\oplus$, then the pointwise operation induced on $T^S$ by $\otimes$ is distributive over the operation induced by $\oplus$.

Proof:
Let $\struct {T, \oplus, \otimes}$ be an algebraic structure in which $\otimes$ distributes over $\oplus$.
Let $f, g, h \in T^S$. 
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \otimes \paren {g \oplus h} } } x
      | r = \map f x \otimes \paren {\map g x \oplus \map h x}
      | c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \paren {\map f x \otimes \map g x} \oplus \paren {\map f x \otimes \map h x}
      | c = because $\otimes$ distributes over $\oplus$ in $T$
}}
{{eqn | r = \map {\paren {\paren {f \otimes g} \oplus \paren {f \otimes h} } } x
      | c = {{Defof|Pointwise Operation}}
}}
{{end-eqn}}
This shows left distributivity.
The proof for right distributivity is identical.
{{qed}}
Category:Pointwise Operations
Category:Distributive Operations
c2unl9bb8a9qwficzc2xs3madjanz72
Theorem: Dirichlet Series Absolute Convergence Lemma

Let $\ds \map f s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series.
Suppose that $f$ converges absolutely at $s_0 = \sigma_0 + i t_0 \in \C$.
Then $f$ converges absolutely at all points $s = \sigma + i t \in \C$ with $\sigma \ge \sigma_0$.

Proof:
Suppose that $f$ converges absolutely at $\sigma_0 + i t_0$.
If $\sigma \ge \sigma_0$, then:
{{begin-eqn}}
{{eqn | l = \size {\frac {a_n} {n^s} }
      | r = \frac {\size {a_n} } {n^\sigma}
}}
{{eqn | o = \le
      | r = \frac {\size {a_n} } {n^{\sigma_0} }
}}
{{eqn | r = \size {\frac {a_n} { n^{s_0} } }
}}
{{end-eqn}}
Therefore absolute convergence of $\map f {s_0}$ directly implies absolute convergence of $\map f s$ for all $s = \sigma + i t$ with $\sigma > \sigma_0$.
{{qed}}