# Varieties, rings, fields

1. Jun 23, 2008

### CRGreathouse

A commutative ring is a variety, because its definition consists only of universally quantified identities:
g+(h+k) = (g+h)+k
g+0=g
(-g) + g = 0
g + h = h + g
g(hk) = (gh)k
g(h+k) = gh+gk
1g = g
gh = hg
where (-g) denotes the additive inverse of g.

Adding a new predicate symbol "(1/...)" defined by
(1/g)g = 1

and adding this identity to the list, a new variety is created, whose members are fields along with the trivial ring.

Why is it so important to exclude the trivial ring in the definition of a field (by the requirement $0\neq1$ or $|G|\ge2$) even though fields are not varieties? Is there any value to the structure defined above (the smallest variety containing all rings)?

2. Jun 23, 2008

### matt grime

Um, isn't a variety an algebraic subset of P^n or A^n?

3. Jun 23, 2008

### CRGreathouse

4. Jun 23, 2008

### Hurkyl

Staff Emeritus
Not true. Adding this gives you the degenerate variety: your axioms allow you to prove the identity x=0 as follows:

0 = (x (1/0))0 = x ((1/0) 0) = x 1 = 1 x = x.

Fields are not a universal algebra. This can be seen, for example, by noting that the Cartesian product of two fields is not a field.

5. Jun 23, 2008

### CRGreathouse

Ah, right. That's what I was looking for, thanks!