Integral domains and Fields

These are two special kinds of ring

Definition

If a, b are two ring elements with a, b ≠ 0 but ab = 0 then a and b are called zero-divisors.

Example

In the ring Z₆ we have 2.3 = 0 and so 2 and 3 are zero-divisors.
More generally, if n is not prime then Z_n contains zero-divisors.

Definition

An integral domain is a commutative ring with an identity (1 ≠ 0) with no zero-divisors.
That is ab = 0 ⇒ a = 0 or b = 0.

Examples

1. The ring Z is an integral domain. (This explains the name.)

2. The polynomial rings Z[x] and R[x] are integral domains.
   (Look at the degree of a polynomial to see how to prove this.)

3. The ring {a + b√2 | a, b ∈ Z} is an integral domain.
   (Proof?)

4. If p is prime, the ring Z_p is an integral domain.
   (Proof?)
   

A field is a commutative ring with identity (1 ≠ 0) in which every non-zero element has a multiplicative inverse.

Examples

The rings Q, R, C are fields.

Remarks

1. If a, b are elements of a field with ab = 0 then if a ≠ 0 it has an inverse a^-1 and so multiplying both sides by this gives b = 0. Hence there are no zero-divisors and we have:
   
   Every field is an integral domain.
   
   
   

   The axioms of a field F can be summarised as:
   

   1. (F, +) is an abelian group
      
   2. (F - {0}, . ) is an abelian group
      
   3. The distributive law.
      

Theorem

Every finite integral domain is a field.

Proof

The only thing we need to show is that a typical element a ≠ 0 has a multiplicative inverse.
Consider a, a², a³, ... Since there are only finitely many elements we must have a^m = aⁿ for some m < n(say).
Then 0 = a^m - aⁿ = a^m(1 - a^n-m). Since there are no zero-divisors we must have a^m ≠ 0 and hence 1 - a^n-m = 0 and so 1 = a(a^n-m-1) and we have found a multiplicative inverse for a.

More examples

1. If p is prime Z_p is a field. It has p elements.

2. Consider the set of things of the form {a + bx | a, b ∈ Z₂} with x an "indeterminate".
   Use arithmetic modulo 2 and multiply using the "rule" x² = x + 1.
   Then we get a field with 4 elements: {0, 1, x, 1 + x}.
   For example: x(1 + x) = x + x² = x + (1 + x) = 1 (since we work modulo 2). Thus every non-zero element has a multiplicative inverse.

3. Consider the set of things of the form {a + bx + cx² | a, b, c ∈ Z₂} where we now use the rule x³ = 1 + x.
   This gives a field with 8 elements: {0, 1, x, 1 + x, x², 1 + x², x + x², 1 + x + x²}.
   For example, (1 + x²)(x + x²) = x + x² + x³ + x⁴ = x + x² + (1 + x) + x(1 + x) = 1 + x since we work modulo 2.

   Exercise: Experiment by multiplying together elements to find multiplicative inverses.
   (e.g. Since x³ + x = 1 we have x(x² + 1) = 1 and x^-1 = 1 + x².

4. Consider the set of things of the form {a + bx | a, b ∈ Z₃} with arithmetic modulo 3 and the "rule" x² = -1 (so its a bit like multiplying in C !).
   Then we get a field with 9 elements: {0, 1, 2, x, 1 + x, 2 + x, 2x, 1 + 2x, 2 + 2x}.
   Exercise: Find mutiplicative inverses.
   

Remark

If F is a field then both (F, +) and (F - {0}, . ) are abelian groups.
For the field of order 4 {0, 1, x, 1 + x} above, under addition each element has order 2 and so the additive group is the Klein 4-group (isomorphic to Z₂ × Z₂).
The multiplicative group {1, x, 1 + x} is a cyclic group of order 3 (generated by x since x² = 1 + x and x³ = x(1 + x) = x + x² = x + 1 + x = 1)
In general the additive group of a finite field F of order p^k is a direct sum of k copies of Z_p , while the multiplicative group F - {0} is a cyclic group of order p^k - 1.