Difference between revisions of "Ring"
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{{Definition|Abstract Algebra}} | {{Definition|Abstract Algebra}} | ||
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Revision as of 13:08, 19 February 2016
Not to be confused with rings of sets which are a topic of algebras of sets and thus σ-Algebras and σ-rings
Contents
Definition
A set R and two binary operations + and × such that the following hold[1]:
Rule | Formal | Explanation |
---|---|---|
Addition is commutative | ∀a,b∈R[a+b=b+a] | It doesn't matter what order we add |
Addition is associative | ∀a,b,c∈R[(a+b)+c=a+(b+c)] | Now writing a+b+c isn't ambiguous |
Additive identity | ∃e∈R∀x∈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∈R" is important, there exists a single e that always works |
Additive inverse | ∀x∈R∃y∈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∈R there exists" states that for a given x∈R a y exists. Not a y exists for all x |
Multiplication is associative | ∀a,b,c∈R[(ab)c=a(bc)] | |
Multiplication is distributive | ∀a,b,c∈R[a(b+c)=ab+ac] ∀a,b,c∈R[(a+b)c=ac+bc] |
Is a ring, which we write: (R,+:R×R→R,×:R×R→R) but because Mathematicians are lazy we write simply:
- (R,+,×)
Subring
If (S,+,×) is a ring, and every element of S is also in R (for another ring (R,+,×)) 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 ×
Properties
Name | Statement | Explanation |
---|---|---|
Commutative Ring | ∀x,y∈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 | ∃e×∈R∀x∈R[xe×=e×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:
- ∀x,y∈R[xy=yx]
- ∃e×∈R∀x∈R[xe×=e×x=x]
It is that simple.
Immediate theorems
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∈R is such that ∀x∈R[0+x=x+0=x]
- Suppose that 0′∈R with 0′≠0 and also such that: ∀x∈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≠0′ so the claim they are distinct cannot be, we have only one "zero element", which herein we shall denote as "0"
(Cancellation laws) Theorem: if a+c=b+c then a=b (and due to commutivity of addition c+a=c+b⟹a=b too)
Suppose that a+c=b+c
- By the additive inverse property, ∃x∈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
- Let us take (a+c)+x
- We see that a=a+c+x=b+c+x=b
- First notice that (a+c)+x=(b+c)+x (using a+c=b+c)
- Which is indeed just a=b
As claimed.
Note:
- Note that c+a=b+c⟹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.
Theorem: The additive inverse of an element is unique (and herein, for a given x∈R shall be denoted −x)
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: ∀x∈R[0x=x0=0] - an interesting result, in line with what we expect from our number system
Let x∈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
- xx=x(x+0)=xx+x0 by distributivity
- By the cancellation laws: ⟹0=x0
- So we have shown ∀x∈R[x0=0]
- Note that x=x+0 then
- 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
- xx=(x+0)x=xx+0x by distributivity
- By the cancellation laws: ⟹0=0x
- So we have shown ∀x∈R[0x=0]
- Note that x=x+0 then
- So ∀x∈R[0x=0∧x0=0] or simply ∀x∈R[0x=x0=0]
This completes the proof.
See next
See also
References
- ↑ Fundamentals of abstract algebra - an expanded version - Neal H. McCoy