Ring

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Not to be confused with a ring of sets

Definition

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

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

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

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

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

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

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

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Further properties of elementary rings

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

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

Giving us the following 4 types of elementary rings:

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

• Types of ring
• Ring morphism
  • Ring homomorphism
    • Kernel of a ring homomorphism - see also: kernel
    • Image of a ring homomorphism - see also: image
  • Ring isomorphism
• Unit of a ring
• Division ring
• Ring ideal
• Quotient ring
• Fundamental ring homomorphism theorem
• Ring isomorphism theorems
• Module

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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:

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:

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

It is that simple.

Immediate theorems

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

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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 "$0$"

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(Cancellation laws) Theorem: if $a+c=b+c$ then $a=b$ (and due to commutivity of addition

$$c+a=c+b\implies a=b$$ too)

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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.

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

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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"

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Theorem:

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

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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
• Direct sum of rings
• Integral domain

See also

• Group
• Category for various types of rings