Rings and Fields

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## 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 Zn 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, bZ} is an integral domain.
(Proof?)

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

Definition

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.

2. 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.

The example Z shows that some integral domains are not fields.

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 am = an for some m < n(say).
Then 0 = am - an = am(1 - an-m). Since there are no zero-divisors we must have am ≠ 0 and hence 1 - an-m = 0 and so 1 = a(an-m-1) and we have found a multiplicative inverse for a. More examples

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

2. Consider the set of things of the form {a + bx | a, bZ2} 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, cZ2} 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, bZ3} 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.

More generally, using "tricks" like the above one can construct a finite field with pk elements for any prime p and positive integer k. This is called GF(pk) which stands for Galois Field named after the French mathematician Évariste Galois (1811 - 1832).

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 pk is a direct sum of k copies of Zp , while the multiplicative group F - {0} is a cyclic group of order pk - 1.

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JOC/EFR 2004