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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 [itex]0\neq1[/itex] or [itex]|G|\ge2[/itex]) even though fields are not varieties? Is there any value to the structure defined above (the smallest variety containing all rings)?
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 [itex]0\neq1[/itex] or [itex]|G|\ge2[/itex]) even though fields are not varieties? Is there any value to the structure defined above (the smallest variety containing all rings)?