Difference between revisions of "Ring"

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{{Definition|Abstract Algebra}}
 
{{Definition|Abstract Algebra}}
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[[Category:First-year friendly]]

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


Definition

A set R and two binary operations + and × such that the following hold[1]:

Rule Formal Explanation
Addition is commutative a,bR[a+b=b+a] It doesn't matter what order we add
Addition is associative a,b,cR[(a+b)+c=a+(b+c)] Now writing a+b+c isn't ambiguous
Additive identity eRxR[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 xR" is important, there exists a single e that always works

Additive inverse xRyR[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 xR there exists" states that for a given xR a y exists. Not a y exists for all x

Multiplication is associative a,b,cR[(ab)c=a(bc)]
Multiplication is distributive a,b,cR[a(b+c)=ab+ac]

a,b,cR[(a+b)c=ac+bc]

Is a ring, which we write: (R,+:R×RR,×:R×RR) 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,yR[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×RxR[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:

  1. x,yR[xy=yx]
  2. e×RxR[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 0R is such that xR[0+x=x+0=x]

Suppose that 0R with 00 and also such that: xR[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 00 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+ba=b too)


Suppose that a+c=b+c

By the additive inverse property, xR: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+ca=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 xR 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: xR[0x=x0=0] - an interesting result, in line with what we expect from our number system


Let xR 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: 0=x0
So we have shown xR[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: 0=0x
So we have shown xR[0x=0]
So xR[0x=0x0=0] or simply xR[0x=x0=0]

This completes the proof.


See next

See also

References

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