Difference between revisions of "Ring"

From Maths
Jump to: navigation, search
(Created page with "Not to be confused with rings of sets which are a topic of algebras of sets and thus Algebras}} and Sigma-ring...")
 
(Starting refactoring (on a subpage Ring/New page))
 
(11 intermediate revisions by the same user not shown)
Line 1: Line 1:
 +
{{Refactor notice|grade=A|msg=Page was ancient, mostly written in May 2015}}
 +
__TOC__
 +
{{/New page}}
 +
<hr/><br/><hr/><br/><hr/>
 +
=OLD PAGE=
 +
{{Requires references|The content here has 1 reference and is accurate, but I should be able to (easily) find many more, it also requires updating, as most of it is over 9 months old, before the new-style indexing and formatting of the project}}
 +
 
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}}]]
 
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==
 
==Definition==
−
A set {{M|R}} and two [[Binary operation|binary operations]] {{M|+}} and {{M|\times}} such that the following hold:
+
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"
 
{| class="wikitable" border="1"
 
|-
 
|-
Line 37: Line 44:
 
|}
 
|}
  
 +
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}}
 
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}}
+
{{Definition|Abstract Algebra|Ring Theory}}
 +
{{Theorem Of|Abstract Algebra|Ring Theory}}
 +
[[Category:First-year friendly]]

Latest revision as of 05:02, 16 October 2016

Grade: A
This page is currently being refactored (along with many others)
Please note that this does not mean the content is unreliable. It just means the page doesn't conform to the style of the site (usually due to age) or a better way of presenting the information has been discovered.
The message provided is:
Page was ancient, mostly written in May 2015

Not to be confused with a ring of sets

Definition

Let [ilmath]R[/ilmath] be a non-empty set, let there be two binary operations (a kind of map where rather than [ilmath]f(a,b)[/ilmath] we write [ilmath]afb[/ilmath]):

  1. [ilmath]\oplus:R\times R\rightarrow R[/ilmath] - called "addition", [ilmath]\oplus:(a,b)\mapsto a\oplus b[/ilmath]
  2. [ilmath]\odot:R\times R\rightarrow R[/ilmath] - called "multiplication", [ilmath]\odot:(a,b)\mapsto a\odot b[/ilmath]

and let there be elements [ilmath]0_R\in R[/ilmath] and [ilmath]1_R\in R[/ilmath] (not necessarily distinct)[Note 1] such that we have the following 7 properties[1]:

TODO: This would be much nicer as a table....

  • [ilmath](R,\oplus,0_R)[/ilmath] is an abelian group
    • Group definition:
      1. [ilmath]\forall a,b,c\in R[(a\oplus b)\oplus c=a\oplus(b\oplus c)][/ilmath] - associativity
      2. [ilmath]\exists e\in R\ \forall a\in R[e\oplus a=a\oplus e=a][/ilmath] - existence of identity, on the group page we show it is unique[Note 2], we denote it by [ilmath]0_R[/ilmath], so: [ilmath]\forall a\in R[a\oplus 0_R=0_R\oplus a=a][/ilmath]
      3. [ilmath]\forall a\in R\ \exists b\in R[a\oplus b=b\oplus a=0_R][/ilmath] - existence of inverse, on the group page we show it is unique[Note 3]. Denoted by [ilmath]-a[/ilmath] as we're using additive notation[Note 4]
    • Being an Abelian group adds an additional property:
      1. [ilmath]\forall a,b\in R[a\oplus b=b\oplus a][/ilmath] - commutivity
  • [ilmath](R,\odot)[/ilmath] is a semigroup
    • Semigroup definition:
      1. [ilmath]\forall a,b,c\in R[(a\odot b)\odot c=a\odot(b\odot c)][/ilmath]
  • There is distributivity in play in.
    • [ilmath]\odot[/ilmath] distributes across [ilmath]\oplus[/ilmath] Caution:I think... it might be the other way around... the following 2 rules are certainly correct however:
      1. [ilmath]\forall a,b,c\in R[a\odot(b\oplus c)=(a\odot b)\oplus(a\odot c)][/ilmath] and
      2. [ilmath]\forall a,b,c\in R[(a+b)c=ac+bc][/ilmath]

Then [ilmath](R,\oplus:R\times R\rightarrow R,\odot:R\times R\rightarrow R,0_R)[/ilmath] is a ring, but as mathematicians are lazy we just write [ilmath](R,\oplus,\odot,0_R)[/ilmath], [ilmath](R,\oplus,\odot)[/ilmath] or even just "Let [ilmath]R[/ilmath] be a ring".

TODO: Be more formal about distributivity, I've checked my books, no one specified, they just say "it is distributive: "

Further properties of elementary rings

There are 2 more additional properties we can apply to define rings:

  1. [ilmath]\exists e_\odot\ \forall a\in R[a\odot e_\odot=e_\odot\odot a=a][/ilmath] - a multiplicative identity, this element if it exists is unique and denoted [ilmath]1_R[/ilmath] or just [ilmath]1[/ilmath]
  2. [ilmath]\forall a,b\in R[a\odot b=b\odot a][/ilmath] - commutative with respect to [ilmath]\odot[/ilmath]

Giving us the following 4 types of elementary rings[Note 5]:

  1. Ring - properties 1-7
  2. Ring with unity (AKA: u-ring, ring with identity) - properties 1-8
  3. Commutative ring (AKA: c-ring) - properties 1-7 and 9
  4. Commutative ring with unity (AKA: cu-ring or q-ring - properties 1-9

Caveats

Some authors define a ring to be what we would call a ring with unity (which we shall call a u-ring throughout the site). Especially if the book covers the topics of rings and modules. We defined "commutative ring" and "ring with unity" above.

See next

Notes

  1. ↑ So we could have [ilmath]0_R=1_R[/ilmath] or we could have [ilmath]0_R\ne 1_R[/ilmath]
  2. ↑ there is only one inverse
  3. ↑ there is only one inverse for an element
  4. ↑ For multiplicative notation we'd use [ilmath]a^{-1} [/ilmath]
  5. ↑ field, integral domain are also all rings, there's like 6 kinds. We call "Elementary ring" just the ones listed

References

  1. ↑ Fundamentals of Abstract Algebra - Neal H. McCoy


OLD PAGE

(Unknown grade)
This page requires references, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:
The content here has 1 reference and is accurate, but I should be able to (easily) find many more, it also requires updating, as most of it is over 9 months old, before the new-style indexing and formatting of the project

Not to be confused with rings of sets which are a topic of algebras of sets and thus [ilmath]\sigma[/ilmath]-Algebras and [ilmath]\sigma[/ilmath]-rings


Definition

A set [ilmath]R[/ilmath] and two binary operations [ilmath]+[/ilmath] and [ilmath]\times[/ilmath] such that the following hold[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 [ilmath]a+b+c[/ilmath] 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"

The "exists [ilmath]e[/ilmath] forall [ilmath]x\in R[/ilmath]" is important, there exists a single [ilmath]e[/ilmath] 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 [ilmath]-x[/ilmath], just that it exists

The "forall [ilmath]x\in R[/ilmath] there exists" states that for a given [ilmath]x\in R[/ilmath] a y exists. Not a y exists for all [ilmath]x[/ilmath]

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]

[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 [ilmath](S,+,\times)[/ilmath] is a ring, and every element of [ilmath]S[/ilmath] is also in [ilmath]R[/ilmath] (for another ring [ilmath](R,+,\times)[/ilmath]) and the operations of addition and multiplication on [ilmath]S[/ilmath] are the same as those on [ilmath]R[/ilmath] (when restricted to [ilmath]S[/ilmath] of course) then we say "[ilmath]S[/ilmath] is a subring of [ilmath]R[/ilmath]"


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

Properties

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 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 "[ilmath]1[/ilmath]"

Using properties

A commutative ring with unity is a ring with the additional properties of:

  1. [math]\forall x,y\in R[xy=yx][/math]
  2. [math]\exists e_\times\in R\forall x\in R[xe_\times=e_\times x=x][/math]

It is that simple.

Immediate theorems

Theorem: The additive identity of a ring [ilmath]R[/ilmath] is unique (and as such can be denoted [ilmath]0[/ilmath] unambiguously)

This is a classic "suppose there are two" proof, and we will do the same.

Suppose that [ilmath]0\in R[/ilmath] is such that [ilmath]\forall x\in R[0+x=x+0=x][/ilmath]

Suppose that [ilmath]0'\in R[/ilmath] with [ilmath]0'\ne 0[/ilmath] and also such that: [ilmath]\forall x\in R[0'+x=x+0'=x][/ilmath]

We will show that [ilmath]0=0'[/ilmath], contradicting them being different! Thus showing there is no other "zero"

Proof:

[math]0+0'=0[/math] by the property of [ilmath]0[/ilmath]
[math]0+0'=0'+0[/math] by the commutivity of addition
[math]0'+0=0'[/math] by the property of [ilmath]0'[/ilmath]
Thus [math]0=0'[/math]
This contradicts that [ilmath]0\ne 0'[/ilmath] so the claim they are distinct cannot be, we have only one "zero element", which herein we shall denote as "[ilmath]0[/ilmath]"

(Cancellation laws) Theorem: if [ilmath]a+c=b+c[/ilmath] then [ilmath]a=b[/ilmath] (and due to commutivity of addition [math]c+a=c+b\implies a=b[/math] too)

Suppose that [ilmath]a+c=b+c[/ilmath]

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.

Theorem: The additive inverse of an element is unique (and herein, for a given [ilmath]x\in R[/ilmath] shall be denoted [ilmath]-x[/ilmath])


TODO:


Important theorems

These theorems are "two steps away" from the definitions if you will, they are not immediate things like "the identity is unique"

Theorem: [math]\forall x\in R[0x=x0=0][/math] - an interesting result, in line with what we expect from our number system

Let [ilmath]x\in R[/ilmath] be given.

Proof of: [ilmath]x0=0[/ilmath]
Note that [ilmath]x=x+0[/ilmath] then
[ilmath]xx=x(x+0)=xx+x0[/ilmath] by distributivity
Note that [ilmath]xx=xx+0[/ilmath] then
[ilmath]xx+0=xx+x0[/ilmath]
By the cancellation laws: [ilmath]\implies 0=x0[/ilmath]
So we have shown [ilmath]\forall x\in R[x0=0][/ilmath]
Proof of: [ilmath]0x=0[/ilmath]
Note that [ilmath]x=x+0[/ilmath] then
[ilmath]xx=(x+0)x=xx+0x[/ilmath] by distributivity
Note that [ilmath]xx=xx+0[/ilmath] then
[ilmath]xx+0=xx+0x[/ilmath]
By the cancellation laws: [ilmath]\implies 0=0x[/ilmath]
So we have shown [ilmath]\forall x\in R[0x=0][/ilmath]
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.


See next

See also

References

  1. ↑ Fundamentals of abstract algebra - an expanded version - Neal H. McCoy
Retrieved from "https://wiki.unifiedmathematics.com/index.php?title=Ring&oldid=3341"