Ring

(Redirected from Ring with identity)
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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]
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.

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

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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

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 $\forall a,b\in R[a+b=b+a]$ 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 [ilmath]a+b+c[/ilmath] 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 [ilmath]e[/ilmath] forall [ilmath]x\in R[/ilmath]" is important, there exists a single [ilmath]e[/ilmath] 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 [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 $\forall a,b,c\in R[(ab)c=a(bc)]$
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 [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 $\forall x,y\in R[xy=yx]$ 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 "[ilmath]1[/ilmath]"

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

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:

$0+0'=0$ by the property of [ilmath]0[/ilmath]
$0+0'=0'+0$ by the commutivity of addition
$0'+0=0'$ by the property of [ilmath]0'[/ilmath]
Thus $0=0'$
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 $c+a=c+b\implies a=b$ too)

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

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 [/itex]b+c=c+b[/itex] 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: $\forall x\in R[0x=x0=0]$ - 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 $\forall x\in R[0x=0\wedge x0=0]$ or simply $\forall x\in R[0x=x0=0]$

This completes the proof.