Ring

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:

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)

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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)

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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])

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TODO:

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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

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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

• Examples of rings

See also

• Group