avewright/proofwiki-math huggingface dataset:
Theorem: Commutative Semigroup is Entropic Structure

A commutative semigroup is an entropic structure.

Proof:
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $a, b, c, d \in S$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ \paren {c \circ d}
      | r = a \circ \paren {b \circ \paren {c \circ d} }
      | c = {{Semigroup-axiom|1}}
}}
{{eqn | r = a \circ \paren {\paren {b \circ c} \circ d}
      | c = {{Semigroup-axiom|1}}
}}
{{eqn | r = a \circ \paren {\paren {c \circ b} \circ d}
      | c = Commutativity of $\circ$
}}
{{eqn | r = a \circ \paren {c \circ \paren {b \circ d} }
      | c = {{Semigroup-axiom|1}}
}}
{{eqn | r = \paren {a \circ c} \circ \paren {b \circ d}
      | c = {{Semigroup-axiom|1}}
}}
{{end-eqn}}
{{qed}}
Theorem: Abelian Group Induces Entropic Structure

Let $\struct {G, \circ}$ be an abelian group.
Let the operation $*$ be defined on $G$ such that:
:$\forall x, y \in G: x * y = x \circ y^{-1}$
Then $\struct {G, *}$ is an entropic structure.

Proof:
We need to prove that $\forall a, b, c, d \in G: \paren {a * b} * \paren {c * d} = \paren {a * c} * \paren {b * d}$.
So:
{{begin-eqn}}
{{eqn | l = \paren {a * b} * \paren {c * d}
      | r = \paren {a \circ b^{-1} } \circ \paren {c \circ d^{-1} }^{-1}
      | c = Definition of $*$
}}
{{eqn | r = \paren {a \circ b^{-1} } \circ \paren {\paren {d^{-1} }^{-1} \circ c^{-1} }
      | c = Inverse of Group Product
}}
{{eqn | r = \paren {a \circ b^{-1} } \circ \paren {d \circ c^{-1} }
      | c = Inverse of Group Inverse
}}
{{eqn | r = \paren {a \circ c^{-1} } \circ \paren {d \circ b^{-1} }
      | c = {{Defof|Abelian Group}}
}}
{{eqn | r = \paren {a \circ c^{-1} } \circ \paren {\paren {d^{-1} }^{-1} \circ b^{-1} }
      | c = Inverse of Group Inverse
}}
{{eqn | r = \paren {a \circ c^{-1} } \circ \paren {b \circ d^{-1} }^{-1}
      | c = Inverse of Group Product
}}
{{eqn | r = \paren {a * c} * \paren {b * d}
      | c = Definition of $*$
}}
{{end-eqn}}
{{qed}}
Theorem: Cancellable Semiring with Unity is Additive Semiring

Let $\struct {S, *, \circ}$ be a cancellable semiring with unity $1_S$.
Then the distributand $*$ is commutative.
That is to say, $\struct {S, *, \circ}$ is also an additive semiring.

Proof:
Let $\struct {S, *, \circ}$ be a semiring, all of whose elements of $S$ are cancellable for $*$.
We expand the expression $\paren {a * b} \circ \paren {c * d}$ using the distributive law in two ways:
{{begin-eqn}}
{{eqn | l = \paren {a * b} \circ \paren {c * d}
      | r = \paren {\paren {a * b} \circ c} * \paren {\paren {a * b} \circ d}
      | c = 
}}
{{eqn | r = \paren {a \circ c} * \paren {b \circ c} * \paren {a \circ d} * \paren {b \circ d}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {a * b} \circ \paren {c * d}
      | r = \paren {a \circ \paren {c * d} } * \paren {b \circ \paren {c * d} }
      | c = 
}}
{{eqn | r = \paren {a \circ c} * \paren {a \circ d} * \paren {b \circ c} * \paren {b \circ d}
      | c = 
}}
{{end-eqn}}
So, by the fact that all elements of $\struct {S, *}$ are cancellable (and thus, a fortiori, are $a \circ c$ and $b \circ d$), we have:
{{begin-eqn}}
{{eqn | l = \paren {a \circ c} * \paren {b \circ c} * \paren {a \circ d} * \paren {b \circ d}
      | r = \paren {a \circ c} * \paren {a \circ d} * \paren {b \circ c} * \paren {b \circ d}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {b \circ c} * \paren {a \circ d}
      | r = \paren {a \circ d} * \paren {b \circ c}
      | c = 
}}
{{end-eqn}}
As this is true for all $a, b, c, d \in S$, it is true in particular if $c = d = 1_S$.
Thus it is clear that $b * a = a * b$, which is exactly to say that $*$ is commutative.
Hence the result, by definition of additive semiring.
{{Qed}}
Category:Cancellable Semirings
Category:Additive Semirings
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Theorem: Ring is not Empty

A ring cannot be empty.

Proof:
In a ring $\struct {R, +, \circ}$, $\struct {R, +}$ forms a group.
From Group is not Empty, the group $\struct {R, +}$ contains at least the identity, so cannot be empty.
So every ring $\struct {R, +, \circ}$ contains at least the identity for ring addition.
{{qed}}
Theorem: Ring Product with Zero

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Then:
:$\forall x \in R: 0_R \circ x = 0_R = x \circ 0_R$
That is, the zero is a zero element for the ring product, thereby justifying its name.

Proof:
Because $\struct {R, +, \circ}$ is a ring, $\struct {R, +}$ is a group.
Since $0_R$ is the identity in $\struct {R, +}$, we have $0_R + 0_R = 0_R$.
From the Cancellation Laws, all group elements are cancellable, so every element of $\struct {R, +}$ is cancellable for $+$.
Thus:
{{begin-eqn}}
{{eqn | l = x \circ \paren {0_R + 0_R}
      | r = x \circ 0_R
      | c = {{Defof|Ring Zero}}
}}
{{eqn | ll= \leadsto
      | l = \paren {x \circ 0_R} + \paren {x \circ 0_R}
      | r = x \circ 0_R
      | c = {{Ring-axiom|D}}
}}
{{eqn | ll= \leadsto
      | l = \paren {x \circ 0_R} + \paren {x \circ 0_R}
      | r = \paren {x \circ 0_R} + 0_R
      | c = {{Defof|Ring Zero}}
}}
{{eqn | ll= \leadsto
      | l = x \circ 0_R
      | r = 0_R
      | c = Cancellation Laws
}}
{{end-eqn}}
Next:
{{begin-eqn}}
{{eqn | l = \paren {0_R + 0_R} \circ x
      | r = 0_R \circ x
      | c = {{Defof|Ring Zero}}
}}
{{eqn | ll= \leadsto
      | l = \paren {0_R \circ x} + \paren {0_R \circ x}
      | r = 0_R \circ x
      | c = {{Ring-axiom|D}}
}}
{{eqn | ll= \leadsto
      | l = \paren {0_R \circ x} + \paren {0_R \circ x}
      | r = 0_R + \paren {0_R \circ x}
      | c = {{Defof|Ring Zero}}
}}
{{eqn | ll= \leadsto
      | l = 0_R \circ x
      | r = 0_R
      | c = Cancellation Laws
}}
{{end-eqn}}
{{Qed}}
Theorem: Product with Ring Negative

Let $\struct {R, +, \circ}$ be a ring.
Then:
:$\forall x, y \in \struct  {R, +, \circ}: \paren {-x} \circ y = -\paren {x \circ y} = x \circ \paren {-y}$
where $\paren {-x}$ denotes the negative of $x$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \paren {x + \paren {-x} } \circ y
      | r = 0_R \circ y
      | c = {{Defof|Ring Zero}}
}}
{{eqn | r = 0_R
      | c = Ring Product with Zero
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x \circ y} + \paren {\paren {-x} \circ y}
      | r = 0_R
      | c = {{Ring-axiom|D}}
}}
{{end-eqn}}
So from {{Group-axiom|3}} as applied to $\struct {R, +}$:
:$\paren {-x} \circ y = -\paren {x \circ y}$
The proof that $x \circ \paren {-y} = -\paren {x \circ y}$ follows identical lines.
{{qed}}
Theorem: Unity of Ring is Unique

A ring can have no more than one unity.

Proof:
Let $\struct {R, +, \circ}$ be a ring.
If $\struct {R, \circ}$ has an identity, then it is a monoid.
From Identity of Monoid is Unique, it follows that such an identity is unique.
{{qed}}
Theorem: Trivial Ring from Abelian Group

An abelian group $\struct {G, +}$ may be turned into a trivial ring by defining the ring product to be:
:$\forall x, y \in G: x \circ y = e_G$

Proof:
Follows directly from the definition of a trivial ring.
{{qed}}
Theorem: Null Ring iff Zero and Unity Coincide

The null ring is the only ring in which the unity and zero coincide.

Proof:
The single element of the null ring serves as an identity for both of the operations.
So, in this particular ring, the unity and the zero are the same element.
A non-null ring contains some non-zero element $a$.
Since $1_R \circ a = a \ne 0_R = 0_R \circ a$, then $1_R \ne 0_R$.
So if a ring is non-null, its unity cannot be zero.
So a ring is null {{iff}} its unity is also its zero.
{{qed}}
Theorem: Unity is Unit

The unity in a ring is a unit.

Proof:
This is a special case of Direct Product of Unitary Modules is Unitary Module.
{{qed}}
Theorem: Unity is Unit

The unity in a ring is a unit.

Proof:
This is a special case of the Unitary Module of All Mappings is Unitary Module, where $S$ is the set $\closedint 1 n \subset \N_{>0}$.
{{qed}}
Theorem: Unity is Unit

The unity in a ring is a unit.

Proof:
This is a special case of a Finite Direct Product of Unitary Modules is Unitary Module where each of the $G_k$ is the $R$-module $R$.
{{qed}}
Theorem: Unity is Unit

The unity in a ring is a unit.

Proof:
Let $\left({R, +, \circ}\right)$ be a ring with unity $1_R$.
From Identity is Self-Inverse:
:$1_R^{-1} = 1_R \in R \implies 1_R \in U_R$
{{qed}}
Category:Rings with Unity
1kgq0vcsz8xqaswu08wx5qsyj7zb6a3
Theorem: Unity and Negative form Subgroup of Units

Let $\struct {R, +, \circ}$ be a ring with unity.
Then:
:$\struct {\set {1_R, -1_R}, \circ} \le U_R$
That is, the set consisting of the unity and its negative forms a subgroup of the group of units.

Proof:
From Unity is Unit:
:$1_R \in U_R$
It remains to be shown that $-1_R \in U_R$.
From {{Ring-axiom|A}}, $\struct {R, +}$ is an abelian group.
Therefore:
:$1_R \in R \implies -1_R \in R$.
From Product of Ring Negatives:
:$-1_R \circ -1_R = 1_R \circ 1_R = 1_R$
Thus $-1_R$ has a ring product inverse (itself) and therefore $-1_R \in U_R$.
We have:
{{begin-eqn}}
{{eqn | l = 1_R \circ 1_R
      | r = 1_R
      | c = 
}}
{{eqn | l = 1_R \circ -1_R
      | r = -1_R
      | c = 
}}
{{eqn | l = -1_R \circ 1_R
      | r = -1_R
      | c = 
}}
{{eqn | l = -1_R \circ -1_R
      | r = 1_R
      | c = 
}}
{{end-eqn}}
This exhausts all the ways we can form a ring product between $1_R$ and $-1_R$.
That is:
:$\forall x, y \in \set {1_R, -1_R}: x \circ y \in \set {1_R, -1_R}$
Hence the result from the Two-Step Subgroup Test.
{{qed}}
Category:Rings with Unity
Category:Subgroups
4xfqsv3mwxeuhx44xgisxjrxwdhk8k6
Theorem: Negative of Product Inverse

Let $\struct {R, +, \circ}$ be a ring with unity.
Let $z \in U_R$, where $U_R$ is the set of units.
Then:
:$\paren {-z}^{-1} = -\paren {z^{-1} }$
where $z^{-1}$ is the ring product inverse of $z$.

Proof:
Let the unity of $\struct {R, +, \circ}$ be $1_R$.
{{begin-eqn}}
{{eqn | l = \paren {-\paren {z^{-1} } } \circ \paren {-z}
      | r = z^{-1} \circ z
      | c = Product of Ring Negatives
}}
{{eqn | r = 1_R
      | c = Inverse under $\circ$
}}
{{eqn | r = z \circ z^{-1}
      | c = Inverse under $\circ$
}}
{{eqn | r = \paren {-z} \circ \paren {-\paren {z^{-1} } }
      | c = Product of Ring Negatives
}}
{{end-eqn}}
Thus:
:$\paren {-z}^{-1} = -\paren {z^{-1} }$
{{qed}}
Theorem: Product of Negative with Product Inverse

Let $\struct {R, +, \circ}$ be a ring with unity.
Let $z \in U_R$, where $U_R$ is the set of units.
Then:
:$(1): \quad \forall x \in R: -\paren {x \circ z^{-1} } = \paren {-x} \circ z^{-1} = x \circ \paren {\paren {-z}^{-1} }$
:$(2): \quad \forall x \in R: -\paren {z^{-1} \circ x} = z^{-1} \circ \paren {-x} = \paren {\paren {-z}^{-1} } \circ x$

Proof:
{{begin-eqn}}
{{eqn | n = 1
      | l = -\paren {x \circ z^{-1} }
      | r = \paren {-x} \circ z^{-1}
      | c = Product with Ring Negative
}}
{{eqn | r = x \circ \paren {-\paren {z^{-1} } }
      | c = Product with Ring Negative
}}
{{eqn | r = x \circ \paren {\paren {-z}^{-1} }
      | c = Negative of Product Inverse
}}
{{end-eqn}}
{{qed|lemma}}
{{begin-eqn}}
{{eqn | n = 2
      | l = -\paren {z^{-1} \circ x}
      | r = z^{-1} \circ \paren {-x}
      | c = Product with Ring Negative
}}
{{eqn | r = \paren {-\paren {z^{-1} } } \circ x
      | c = Product with Ring Negative
}}
{{eqn | r = \paren {\paren {-z}^{-1} } \circ x
      | c = Negative of Product Inverse
}}
{{end-eqn}}
{{qed}}
Theorem: Negative of Division Product

Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Then:
:$\forall x \in R: -\dfrac x z = \dfrac {-x} z = \dfrac x {-z}$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is the division product of $x$ by $z$.

Proof:
Follows directly from Product of Negative with Product Inverse and the definition of division product.
{{qed}}
Theorem: Addition of Division Products

Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, c \in R, b, d \in U_R$.
Then:
:$\dfrac a b + \dfrac c d = \dfrac {a \circ d + b \circ c} {b \circ d}$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.

Proof:
First we demonstrate the operation has the specified property:
{{begin-eqn}}
{{eqn | l = \frac a b + \frac c d
      | r = a \circ b^{-1} + c \circ d^{-1}
      | c = {{Defof|Division Product}}
}}
{{eqn | r = a \circ b^{-1} \circ d \circ d^{-1} + c \circ d^{-1} \circ b \circ b^{-1}
      | c = {{Defof|Inverse Element}} and {{Defof|Identity Element}} under $\circ$
}}
{{eqn | r = \paren {a \circ d} \circ \paren {d^{-1} \circ b^{-1} } + \paren {b \circ c} \circ \paren {d^{-1} \circ b^{-1} }
      | c = {{Defof|Commutative Operation}}
}}
{{eqn | r = \paren {a \circ d + b \circ c} \circ \paren {b \circ d}^{-1}
      | c = {{Defof|Distributive Operation}} $\circ$ over $+$
}}
{{eqn | r = \frac {a \circ d + b \circ c} {b \circ d}
      | c = {{Defof|Division Product}}
}}
{{end-eqn}}
Notice that this works only if $\struct {R, +, \circ}$ is commutative.
{{Qed}}
Theorem: Equality of Division Products

Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, b \in R, c, d \in U_R$.
Then:
:$\dfrac a c = \dfrac b d \iff a \circ d = b \circ c$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.

Proof:
{{begin-eqn}}
{{eqn | l = \frac a c
      | r = \frac b d
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = a \circ c^{-1}
      | r = b \circ d^{-1}
      | c = {{Defof|Division Product}}
}}
{{eqn | ll= \leadstoandfrom
      | l = a \circ c^{-1} \circ c \circ d
      | r = b \circ d^{-1} \circ c \circ d
      | c = {{Defof|Cancellable Element}} of $U_R$
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {a \circ d} \circ \paren {c^{-1} \circ c}
      | r = \paren {b \circ c} \circ \paren {d^{-1} \circ d}
      | c = {{Defof|Commutative Operation}} and {{Defof|Associative Operation}}
}}
{{eqn | ll= \leadstoandfrom
      | l = a \circ d
      | r = b \circ c
      | c = {{Defof|Identity Element}} and {{Defof|Inverse Element}}
}}
{{end-eqn}}
{{qed}}
Theorem: Product of Division Products

Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, b \in R, c, d \in U_R$.
Then:
:$\dfrac a c \circ \dfrac b d = \dfrac {a \circ b} {c \circ d}$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.

Proof:
{{begin-eqn}}
{{eqn | l = \frac a c \circ \frac b d
      | r = a \circ c^{-1} \circ b \circ d^{-1}
      | c = {{Defof|Division Product}}
}}
{{eqn | r = \paren {a \circ b} \circ \paren {d^{-1} \circ c^{-1} }
      | c = {{Defof|Commutative Operation}}
}}
{{eqn | r = \paren {a \circ b} \circ \paren {c \circ d}^{-1}
      | c = Inverse of Product
}}
{{eqn | r = \frac {a \circ b} {c \circ d}
      | c = {{Defof|Division Product}}
}}
{{end-eqn}}
{{qed}}
Theorem: Inverse of Division Product

Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $\struct {U_R, \circ}$ be the group of units of $\struct {R, +, \circ}$.
Let $a, b \in U_R$.
Then:
:$\paren {\dfrac a b}^{-1} = \dfrac {1_R} {\paren {a / b}} = \dfrac b a$
where $\dfrac x z$ is defined as $x \circ \paren {z^{-1} }$, that is, $x$ divided by $z$.

Proof:
{{begin-eqn}}
{{eqn | l = \frac {1_R} {\paren {a / b} }
      | r = 1_R / \paren {a \circ b^{-1} }
      | c = {{Defof|Division Product}}
}}
{{eqn | r = 1_R \circ \paren {a \circ b^{-1} }^{-1}
      | c = {{Defof|Division Product}}
}}
{{eqn | r = \paren {a \circ b^{-1} }^{-1}
      | c = {{Defof|Identity Element}} of $\circ$
}}
{{eqn | r = \paren {\frac a b}^{-1}
      | c = {{Defof|Division Product}}
}}
{{eqn | r = \paren {a \circ b^{-1} }^{-1}
      | c = {{Defof|Division Product}}
}}
{{eqn | r = \paren {b^{-1} }^{-1} \circ a^{-1}
      | c = Inverse of Group Product
}}
{{eqn | r = b \circ a^{-1}
      | c = Inverse of Group Inverse
}}
{{eqn | r = \frac b a
      | c = {{Defof|Division Product}}
}}
{{end-eqn}}
{{qed}}
Theorem: Zero Product with Proper Zero Divisor is with Zero Divisor

Let $\struct {R, +, \circ}$ be a ring.
Let $x \in R$ be a proper zero divisor of $R$.
Then:
:$\paren {x \divides 0_R} \land \paren {x \circ y = 0_R} \land \paren {y \ne 0_R} \implies y \divides 0_R$
That is, if $x$ is a proper zero divisor, then whatever non-zero element you form the product with it by to get zero must itself be a zero divisor.

Proof:
Follows directly from the definition of proper zero divisor.
If $y \ne 0_R$ and $x \circ y = 0_R$ and $x \in R^*$ (which is has to be if it's a proper zero divisor), then all the criteria of being a zero divisor are fulfilled by $y$.
{{qed}}
Category:Zero Divisors
ggtpz8ldh9x4o8hpodh433socagnn6f
Theorem: Unit of Ring is not Zero Divisor

Let $\struct {R, +, \circ}$ be a non-null ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $x$ be a unit of $\struct {R, +, \circ}$.
Then $x$ is neither a left zero divisor nor a right zero divisor of $\struct {R, +, \circ}$.

Proof:
{{AimForCont}} $x$ is either a left zero divisor or a right zero divisor of $\struct {R, +, \circ}$.
{{WLOG}}, suppose $x$ is a left zero divisor of $\struct {R, +, \circ}$.
That is:
:$x \circ y = 0_R$
for some $y \in R \setminus \set {0_R}$.
Then:
{{begin-eqn}}
{{eqn | l = y
      | r = 1_R \circ y
      | c = {{Defof|Unity of Ring}}
}}
{{eqn | r = x^{-1} \circ x \circ y
      | c = {{Defof|Unit of Ring}}
}}
{{eqn | r = x^{-1} \circ 0_R
      | c = $x \circ y = 0_R$
}}
{{eqn | r = 0_R
      | c = Ring Product with Zero
}}
{{end-eqn}}
This contradicts the deduction that $y \ne 0_R$.
Thus by Proof by Contradiction $x$ is neither a left zero divisor nor a right zero divisor of $\struct {R, +, \circ}$.
{{qed}}
Category:Rings with Unity
Category:Units of Rings
Category:Zero Divisors
q8zaysenljyelt0dzz17brllw3bdj9z
Theorem: Zero Divisor Product is Zero Divisor

The ring product of a zero divisor with any ring element is a zero divisor.

Proof:
Let $\struct {R, +, \circ}$ be a ring.
Let $x \divides 0_R$ in $R$.
Then:
{{begin-eqn}}
{{eqn | q = \exists y \in R, y \ne 0_R
      | l = x \circ y
      | r = 0_R
      | c = {{Defof|Zero Divisor of Ring}}
}}
{{eqn | ll= \leadsto
      | q = \forall z \in R
      | l = z \circ \paren {x \circ y}
      | r = z \circ 0_R
}}
{{eqn | r = 0_R
      | c = Ring Product with Zero
}}
{{eqn | ll= \leadsto
      | q = \forall z \in R
      | l = \paren {z \circ x} \circ y
      | r = 0_R
      | c = Associativity of $\circ$
}}
{{end-eqn}}
So $z \circ x \divides 0_R$ in $R$.
The same thing happens if we form the product $\paren {x \circ y} \circ z$.
{{Qed}}
Category:Zero Divisors
06evfrnyvh4kiah8gc0ovf54xf1rixb
Theorem: Product is Zero Divisor means Zero Divisor

If the ring product of two elements of a ring is a zero divisor, then one of the two elements must be a zero divisor.

Proof:
{{begin-eqn}}
{{eqn | l = \paren {x \circ y}
      | o = \divides
      | r = 0_R
}}
{{eqn | ll= \leadsto
      | q = \exists z \divides 0_R \in R
      | l = \paren {x \circ y} \circ z
      | r = 0_R, x \ne 0_R, y \ne 0_R
      | c = {{Defof|Zero Divisor of Ring}}
}}
{{eqn | ll= \leadsto
      | l = x \circ \paren {y \circ z}
      | r = 0_R
      | c = {{Ring-axiom|M1}}
}}
{{eqn | ll= \leadsto
      | l = x \divides 0_R
      | o = \lor
      | r = \paren {y \circ z} \divides 0_R
      | c = {{Defof|Zero Divisor of Ring}}
}}
{{eqn | ll= \leadsto
      | l = x \divides 0_R
      | o = \lor
      | r = y \divides 0_R
      | c = Zero Product with Proper Zero Divisor is with Zero Divisor applies, as $z \divides 0_R$
}}
{{end-eqn}}
{{qed}}
Category:Ring Theory
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Theorem: Ring Element is Zero Divisor iff not Cancellable

Let $\struct {R, +, \circ}$ be a ring which is not null.
Let $z \in R^*$.
Then $z$ is a zero divisor {{iff}} $z$ is not cancellable for $\circ$.

Proof:
=== Sufficient Condition ===
Let $z$ be a zero divisor.
Then either $z \circ x = 0_R$ or $x \circ z = 0_R$ for some $x \in R^*$.
{{Recall|Cancellable Element}}
{{:Definition:Cancellable Element}}
{{WLOG}}, suppose $z \circ x = 0_R$ for some $x \in R^*$.
Then:
{{begin-eqn}}
{{eqn | l = z \circ x
      | r = 0_R
      | c = 
}}
{{eqn | r = 0_R \circ x
      | c = {{Defof|Zero Element}}
}}
{{end-eqn}}
That is, we have:
:$z \circ x = 0_R \circ x$
but $z \ne 0_R$.
Hence, by definition, $z$ is not cancellable.
Similarly, {{mutatis}}, if $x \circ z = 0_R$ for some $x \in R^*$.
{{qed|lemma}}
Theorem: Ring Less Zero is Semigroup for Product iff No Proper Zero Divisors

Let $\struct {R, +, \circ}$ be a non-null ring.
Then $R$ has no zero divisors {{iff}} $\struct {R^*, \circ}$ is a semigroup.

Proof:
=== Necessary Condition ===
Let $\struct {R, +, \circ}$ be a non-null ring with no zero divisors.
The set $R^* = R \setminus \set {0_R} \ne \O$ as $\struct {R, +, \circ}$ is non-null.
All elements of $R$ are not zero divisors, and therefore are cancellable.
$\struct {R, +, \circ}$ is closed under $\circ$, from the fact that there are no zero divisors, and also that $\struct {R, \circ}$ is also closed.
From Restriction of Associative Operation is Associative, ring product is associative on $\struct {R^*, +, \circ}$, as it is associative on $\struct {R, +, \circ}$.
Thus $\struct {R^*, \circ}$ is a semigroup.
{{qed|lemma}}
Theorem: Idempotent Elements of Ring with No Proper Zero Divisors

Let $\struct {R, +, \circ}$ be a non-null ring with no (proper) zero divisors.
Let $x \in R$.
Then:
:$x \circ x = x \iff x \in \set {0_R, 1_R}$
That is, the only elements of $\struct {R, \circ}$ that are idempotent are zero and unity.

Proof:
We have $0_R \circ 0_R = 0_R$, so that sorts out one element.
Let $R^*$ be the ring $R$ without the zero: $R^* = R \setminus \set {0_R}$.
By Ring Element is Zero Divisor iff not Cancellable, all elements of $R^*$ that are not zero divisors are cancellable.
Therefore all elements of $R^*$ are cancellable.
Suppose $x \circ x = x$ where $x \ne 0_R$.
Then $x \circ x = x = x \circ 1_R$.
As $x$ is cancellable, the result follows.
{{Qed}}
Theorem: Non-Zero Elements of Division Ring form Group

Let $\struct {R, +, \circ}$ be a division ring.
Then $\struct {R^*, \circ}$ is a group.

Proof:
A division ring by definition is a ring with unity, and therefore not null.
A division ring by definition has no zero divisors, so $\struct {R^*, \circ}$  is a semigroup.
$1_R \in \struct {R^*, \circ}$ and so the identity of $\circ$ is in $\struct {R^*, \circ}$.
By the definition of a division ring, each element of $\struct {R^*, \circ}$ is a unit, and therefore has a unique inverse in $\struct {R^*, \circ}$.
Thus $\struct {R^*, \circ}$ is a semigroup with an identity and inverses and so is a group.
{{qed}}
Category:Division Rings
Category:Group Theory
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Theorem: Subring Test

Let $S$ be a subset of a ring $\struct {R, +, \circ}$.
Then $\struct {S, +, \circ}$ is a subring of $\struct {R, +, \circ}$ {{iff}} these all hold:
:$(1): \quad S \ne \O$
:$(2): \quad \forall x, y \in S: x + \paren {-y} \in S$
:$(3): \quad \forall x, y \in S: x \circ y \in S$

Proof:
=== Necessary Condition ===
If $S$ is a subring of $\struct {R, +, \circ}$, the conditions hold by virtue of the ring axioms as applied to $S$.
Theorem: Subdomain Test

Let $S$ be a subset of an integral domain $\struct {R, +, \circ}$.
Then $\struct {S, + {\restriction_S}, \circ {\restriction_S} }$ is a subdomain of $\struct {R, +, \circ}$ {{iff}} these conditions hold:
:$(1): \quad \struct {S, + {\restriction_S}, \circ {\restriction_S} }$ is a subring of $\struct {R, +, \circ}$
:$(2): \quad$ The unity of $R$ is also in $S$, that is $1_R = 1_S$.

Proof:
By Idempotent Elements of Ring with No Proper Zero Divisors, it follows that the unity of a subdomain is the unity of the integral domain it's a subdomain of.
{{qed}}