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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 **Z**_{6} 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**

- The ring
**Z**is an integral domain. (This explains the name.) - The polynomial rings
**Z**[*x*] and**R**[*x*] are integral domains.

(Look at the degree of a polynomial to see how to prove this.) - The ring {
*a*+*b*√2 |*a*,*b*∈**Z**} is an integral domain.

(Proof?) - 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**

- 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:

- (
*F*, +) is an abelian group

- (
*F*- {0}, . ) is an abelian group

- The distributive law.

The example

**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*, *a*^{2}, *a*^{3}, ... Since there are only finitely many elements we must have *a*^{m} = *a*^{n} for some *m* < *n*(say).

Then 0 = *a*^{m} - *a*^{n} = *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**

- If
*p*is prime**Z**_{p}is a field. It has*p*elements. - Consider the set of things of the form {
*a*+*bx*|*a*,*b*∈**Z**_{2}} with*x*an "indeterminate".

Use arithmetic modulo 2 and multiply using the "rule"*x*^{2}=*x*+ 1.

Then we get a field with 4 elements: {0, 1,*x*, 1 +*x*}.

For example:*x*(1 +*x*) =*x*+*x*^{2}=*x*+ (1 +*x*) = 1 (since we work modulo 2). Thus every non-zero element has a multiplicative inverse. - Consider the set of things of the form {
*a*+*bx*+*cx*^{2}|*a*,*b*,*c*∈**Z**_{2}} where we now use the rule*x*^{3}= 1 +*x*.

This gives a field with 8 elements: {0, 1,*x*, 1 +*x*,*x*^{2}, 1 +*x*^{2},*x*+*x*^{2}, 1 +*x*+*x*^{2}}.

For example, (1 +*x*^{2})(*x*+*x*^{2}) =*x*+*x*^{2}+*x*^{3}+*x*^{4}=*x*+*x*^{2}+ (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*x*^{3}+*x*= 1 we have*x*(*x*^{2}+ 1) = 1 and*x*^{-1}= 1 +*x*^{2}. - Consider the set of things of the form {
*a*+*bx*|*a*,*b*∈**Z**_{3}} with arithmetic modulo 3 and the "rule"*x*^{2}= -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*, 2*x*, 1 + 2*x*, 2 + 2*x*}.

**Exercise:**Find mutiplicative inverses.

More generally, using "tricks" like the above one can construct a finite field with

**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 **Z**_{2} × **Z**_{2}).

The multiplicative group {1, *x*, 1 + *x*} is a cyclic group of order 3 (generated by *x* since *x*^{2} = 1 + *x* and *x*^{3} = *x*(1 + *x*) = *x* + *x*^{2} = *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.

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