Difference between revisions of "Ring"

Revision as of 22:21, 24 May 2015

Not to be confused with rings of sets which are a topic of algebras of sets and thus $\sigma$ -Algebras and $\sigma$ -rings

Definition

A set $R$ and two binary operations $+$ and $\times$ such that the following hold^[1]:

Rule	Formal	Explanation
Addition is commutative	$\forall a,b\in R[a+b=b+a]$	It doesn't matter what order we add
Addition is associative	$\forall a,b,c\in R[(a+b)+c=a+(b+c)]$	Now writing $a+b+c$ isn't ambiguous
Additive identity	$\exists e\in R\forall x\in R[e+x=x+e=x]$	We do not prove it is unique (after which it is usually denoted 0), just "it exists" The "exists $e$ forall $x\in R$ " is important, there exists a single $e$ that always works
Additive inverse	$\forall x\in R\exists y\in R[x+y=y+x=e]$	We do not prove it is unique (after we do it is usually denoted $-x$ , just that it exists The "forall $x\in R$ there exists" states that for a given $x\in R$ a y exists. Not a y exists for all $x$
Multiplication is associative	$\forall a,b,c\in R[(ab)c=a(bc)]$
Multiplication is distributive	$\forall a,b,c\in R[a(b+c)=ab+ac]$ $\forall a,b,c\in R[(a+b)c = ac+bc]$

Is a ring, which we write: $(R,+:R\times R\rightarrow R,\times:R\times R\rightarrow R)$ but because Mathematicians are lazy we write simply:

$(R,+,\times)$

Subring

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

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 $(G,+)$ with another operation on $G$ called $\times$

Properties

Name	Statement	Explanation
Commutative Ring	$\forall x,y\in R[xy=yx]$	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	$\exists e_\times\in R\forall x\in R[xe_\times=e_\times x=x]$	The existence of a multiplicative identity, once we have proved it is unique we often denote this " $1$ "

Using properties

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

$\forall x,y\in R[xy=yx]$
$\exists e_\times\in R\forall x\in R[xe_\times=e_\times x=x]$

It is that simple.

Immediate theorems

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

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

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

Suppose that

$0'\in R$ with

$0'\ne 0$ and also such that:

$\forall x\in R[0'+x=x+0'=x]$

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

Proof:

$0+0'=0$ by the property of

$0$

$0+0'=0'+0$ by the commutivity of addition

$0'+0=0'$ by the property of

$0'$

Thus

$0=0'$

This contradicts that

$0\ne 0'$ so the claim they are distinct cannot be, we have only one "zero element", which herein we shall denote as "

$0$ "

[Expand]
Theorem: if $a+c=b+c$ then $a=b$ (and due to commutivity of addition $c+a=c+b\implies a=b$ too)

Suppose that $a+c=b+c$

By the additive inverse property,

$\exists x\in R:c+x=0$

First notice that

$(a+c)+x=(b+c)+x$ (using

$a+c=b+c$ )

Let us take $(a+c)+x$
By associativity of addition, $(a+c)+x=a+(c+x)=a+0=a$
Let us take $(b+c)+x$
By associativity of addition, $(b+c)+x=b+(c+x)=b+0=b$

We see that

$a=a+c+x=b+c+x=b$

Which is indeed just

$a=b$

As claimed.

Note:

Note that

$c+a=b+c\implies a=b$ , this can be proved identically to the above (but adding x to the left) or by:

$c+a=a+c$ and </math>b+c=c+b</math> and then apply the above.

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

Important theorems

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

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

Let $x\in R$ be given.

Proof of: $x0=0$

Note that

$x=x+0$ then

$xx=x(x+0)=xx+x0$ by distributivity

Note that

$xx=xx+0$ then

$xx+0=xx+x0$

By the cancellation laws:

$\implies 0=x0$

So we have shown

$\forall x\in R[x0=0]$

Proof of: $0x=0$

Note that

$x=x+0$ then

$xx=(x+0)x=xx+0x$ by distributivity

Note that

$xx=xx+0$ then

$xx+0=xx+0x$

By the cancellation laws:

$\implies 0=0x$

So we have shown

$\forall x\in R[0x=0]$

So

$\forall x\in R[0x=0\wedge x0=0]$ or simply

$\forall x\in R[0x=x0=0]$

This completes the proof.

See next

Examples of rings

References

Jump up ↑ Fundamentals of abstract algebra - an expanded version - Neal H. McCoy

[1] Jump up ↑ Fundamentals of abstract algebra - an expanded version - Neal H. McCoy

[1]

@@ Line 67: / Line 67: @@
 # <math>\exists e_\times\in R\forall x\in R[xe_\times=e_\times x=x]</math>
 It is that simple.
-==Important theorem==
-a0=0a=0
-use a(a+0)=aa and go from there.
+==Immediate theorems==
-==Important theorems==
 {{Begin Theorem}}
 Theorem: The additive identity of a ring {{M|R}} is unique (and as such can be denoted {{M|0}} unambiguously)
@@ Line 115: / Line 111: @@
 {{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}}

Difference between revisions of "Ring"

Revision as of 22:21, 24 May 2015

Contents

Definition

Subring

Properties

Using properties

Immediate theorems

Important theorems

See next

See also

References

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