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Not to be confused with [[Ring of sets|rings of sets]] which are a topic of [[Algebra of sets|algebras of sets]] and thus [[Sigma-algebra|{{Sigma|Algebras}}]] and [[Sigma-ring|{{Sigma|rings}}]]


==Definition==
A set {{M|R}} and two [[Binary operation|binary operations]] {{M|+}} and {{M|\times}} such that the following hold<ref>Fundamentals of abstract algebra - an expanded version - Neal H. McCoy</ref>:
{| class="wikitable" border="1"
|-
! Rule
! Formal
! Explanation
|-
| Addition is commutative
| <math>\forall a,b\in R[a+b=b+a]</math>
| It doesn't matter what order we add
|-
| Addition is associative
| <math>\forall a,b,c\in R[(a+b)+c=a+(b+c)]</math>
| Now writing {{M|a+b+c}} isn't ambiguous
|-
| Additive identity
| <math>\exists e\in R\forall x\in R[e+x=x+e=x]</math>
| We do not prove it is unique (after which it is usually denoted 0), just "it exists"<br/>
The "exists {{M|e}} forall {{M|x\in R}}" is important, there exists a single {{M|e}} that always works
|-
| Additive inverse
| <math>\forall x\in R\exists y\in R[x+y=y+x=e]</math>
| We do not prove it is unique (after we do it is usually denoted {{M|-x}}, just that it exists<br/>
The "forall {{M|x\in R}} there exists" states that for a given {{M|x\in R}} a y exists. Not a y exists for all {{M|x}}
|-
| Multiplication is associative
| <math>\forall a,b,c\in R[(ab)c=a(bc)]</math>
|
|-
| Multiplication is distributive
| <math>\forall a,b,c\in R[a(b+c)=ab+ac]</math><br/>
<math>\forall a,b,c\in R[(a+b)c = ac+bc]</math>
|}

Is a ring, which we write: <math>(R,+:R\times R\rightarrow R,\times:R\times R\rightarrow R)</math> but because [[Mathematicians are lazy]] we write simply:
* <math>(R,+,\times)</math>

===Subring===
If {{M|(S,+,\times)}} is a ring, and every element of {{M|S}} is also in {{M|R}} (for another ring {{M|(R,+,\times)}}) and the operations of addition and multiplication on {{M|S}} are the same as those on {{M|R}} (when restricted to {{M|S}} of course) then we say ''"{{M|S}} is a subring of {{M|R}}"''


'''Note:'''<br/>
Some books introduce rings first, I do not know why. A ring is an additive [[Group|group]] (it is commutative making it an Abelian one at that), that is a ring is just a group {{M|(G,+)}} with another operation on {{M|G}} called {{M|\times}}

==Properties==
{| class="wikitable" border="1"
|-
! Name
! Statement
! Explanation
|-
! Commutative Ring
| <math>\forall x,y\in R[xy=yx]</math>
| The order we multiply by does not matter. Calling a ring commutative isn't ambiguous because by definition addition in a ring is [[Commutative|commutative]] so when we call a ring commutative we must mean "it is a ring, and also multiplication is commutative". 
|-
! Ring with Unity
| <math>\exists e_\times\in R\forall x\in R[xe_\times=e_\times x=x]</math>
| The existence of a multiplicative identity, once we have proved it is unique we often denote this "{{M|1}}"
|}
===Using properties===
A ''commutative ring with unity'' is a ring with the additional properties of:
# <math>\forall x,y\in R[xy=yx]</math>
# <math>\exists e_\times\in R\forall x\in R[xe_\times=e_\times x=x]</math>
It is that simple.

==Immediate theorems==
{{Begin Theorem}}
Theorem: The additive identity of a ring {{M|R}} is unique (and as such can be denoted {{M|0}} unambiguously)
{{Begin Proof}}
This is a classic "suppose there are two" proof, and we will do the same.

Suppose that {{M|0\in R}} is such that {{M|1=\forall x\in R[0+x=x+0=x]}} 
: Suppose that {{M|0'\in R}} with {{M|1=0'\ne 0}} and also such that: {{M|1=\forall x\in R[0'+x=x+0'=x]}}

We will show that {{M|1=0=0'}}, contradicting them being different! Thus showing there is no other "zero"

'''Proof:'''
: <math>0+0'=0</math> by the property of {{M|0}}
: <math>0+0'=0'+0</math> by the commutivity of addition
:: <math>0'+0=0'</math> by the property of {{M|0'}}
: Thus <math>0=0'</math>
:: This contradicts that {{M|0\ne 0'}} so the claim they are distinct cannot be, we have only one "zero element", which herein we shall denote as "{{M|0}}"
{{End Proof}}{{End Theorem}}
{{Begin Theorem}}
''(Cancellation laws)'' Theorem: if {{M|1=a+c=b+c}} then {{M|1=a=b}} (and due to commutivity of addition <math>c+a=c+b\implies a=b</math> too)
{{Begin Proof}}
Suppose that {{M|1=a+c=b+c}}
: By the ''additive inverse'' property, <math>\exists x\in R:c+x=0</math>
:: First notice that <math>(a+c)+x=(b+c)+x</math> (using <math>a+c=b+c</math>)
::* Let us take <math>(a+c)+x</math>
::*: By associativity of addition, <math>(a+c)+x=a+(c+x)=a+0=a</math>
::* Let us take <math>(b+c)+x</math>
::*: By associativity of addition, <math>(b+c)+x=b+(c+x)=b+0=b</math>
:: We see that <math>a=a+c+x=b+c+x=b</math>
: Which is indeed just <math>a=b</math>

As claimed.



'''Note:'''
: Note that <math>c+a=b+c\implies a=b</math>, this can be proved identically to the above (but adding x to the left) or by:
:: <math>c+a=a+c</math> and </math>b+c=c+b</math> and then apply the above.
{{End Proof}}{{End Theorem}}
{{Begin Theorem}}
Theorem: The additive inverse of an element is unique (and herein, for a given {{M|x\in R}} shall be denoted {{M|-x}})
{{Begin Proof}}
{{Todo}}
{{End Proof}}{{End Theorem}}

==Important theorems==
These theorems are "two steps away" from the definitions if you will, they are not immediate things like "the identity is unique"
{{Begin Theorem}}
Theorem: <math>\forall x\in R[0x=x0=0]</math> - an interesting result, in line with what we expect from our number system
{{Begin Proof}}
Let {{M|x\in R}} be given.
: '''Proof of: {{M|1=x0=0}}'''
:: Note that {{M|1=x=x+0}} then
::: {{M|1=xx=x(x+0)=xx+x0}} by ''distributivity''
:::: Note that {{M|1=xx=xx+0}} then
::: {{M|1=xx+0=xx+x0}}
:: By the cancellation laws: {{M|1=\implies 0=x0}}
:: So we have shown {{M|1=\forall x\in R[x0=0]}}
: '''Proof of: {{M|1=0x=0}}'''
:: Note that {{M|1=x=x+0}} then
::: {{M|1=xx=(x+0)x=xx+0x}} by ''distributivity''
:::: Note that {{M|1=xx=xx+0}} then
::: {{M|1=xx+0=xx+0x}}
:: By the cancellation laws: {{M|1=\implies 0=0x}}
:: So we have shown {{M|1=\forall x\in R[0x=0]}}
: So <math>\forall x\in R[0x=0\wedge x0=0]</math> or simply <math>\forall x\in R[0x=x0=0]</math>
This completes the proof.
{{End Proof}}{{End Theorem}}

==See next==
* [[Examples of rings]]
* [[Direct sum (ring)|Direct sum of rings]]
* [[Integral domain]]

==See also==
* [[Group]]
* [[:Category:Types of rings|Category for various types of rings]]

==References==
<references/>

{{Definition|Abstract Algebra|Ring Theory}}
{{Theorem Of|Abstract Algebra|Ring Theory}}
[[Category:First-year friendly]]