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Not to be confused with a ring of sets


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


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


  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


  1. Fundamentals of Abstract Algebra - Neal H. McCoy


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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


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]


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]"

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]


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"


[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 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])


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
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
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


  1. Fundamentals of abstract algebra - an expanded version - Neal H. McCoy