Difference between revisions of "Integral domain"

From Maths
Jump to: navigation, search
m
m (No idea what todo note was about)
Line 32: Line 32:
  
 
{{Definition|Abstract Algebra}}
 
{{Definition|Abstract Algebra}}
{{Todo|Cancellation laws of multiplication in a ring, page 54 Neal H McCoy - Fundamentals of Abstract Algebra is a good place to start}}
 

Revision as of 18:51, 28 August 2015

Definition

Given a ring [ilmath](D,+,\times)[/ilmath], it is called an integral domain[1] if it is:

  • A commutative ring, that is: [math]\forall x,y\in D[xy=yx][/math]
  • Contains no non-zero divisors of zero
    • An element [ilmath]a[/ilmath] of a ring [ilmath]R[/ilmath] is said to be a divisor of zero in [ilmath]R[/ilmath] if:
      • [math]\exists c\in R[c\ne e_+\wedge ac=e_+][/math] or if (by writing [ilmath]e_+[/ilmath] as [ilmath]0[/ilmath] we can say: [math]\exists c\in R[c\ne 0\wedge ac=0][/math])
      • [math]\exists d\in R[d\ne e_+\wedge da=e_+][/math] (by writing [ilmath]e_+[/ilmath] as [ilmath]0[/ilmath] we can say: [math]\exists d\in R[d\ne 0\wedge da=0][/math])
      • We can write this as: [math]\exists c\in R[c\ne 0\wedge(ac=0\vee ca=0)][/math]

As the integral domain is commutative we don't need both [ilmath]ac[/ilmath] and [ilmath]ca[/ilmath].

Shorter definition

We can restate this as[2] a ring [ilmath]D[/ilmath] is an integral domain if:

  • [math]\forall x,y\in D[xy=yx][/math]
  • [math]\forall a,b\in D[(a\ne 0,b\ne 0)\implies(ab\ne 0)][/math]

Example of a ring that isn't an integral domain

Consider the ring [ilmath]\mathbb{Z}/6\mathbb{Z} [/ilmath], the ring of integers modulo 6, notice that [ilmath][2][3]=[6]=[0]=e_+[/ilmath].

This means both [ilmath][2][/ilmath] and [ilmath][3][/ilmath] are non-zero divisors of zero.

Examples of rings that are integral domains

  • The integers
  • [ilmath]\mathbb{Z}/p\mathbb{Z} [/ilmath] where [ilmath]p[/ilmath] is prime

See next

See also


References

  1. Fundamentals of Abstract Algebra - An Expanded Version - Neal H. McCoy
  2. My (Alec's) own work