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Varieties, rings, fields

  1. Jun 23, 2008 #1

    CRGreathouse

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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)?
     
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  3. Jun 23, 2008 #2

    matt grime

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    Um, isn't a variety an algebraic subset of P^n or A^n?
     
  4. Jun 23, 2008 #3

    CRGreathouse

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  5. Jun 23, 2008 #4

    Hurkyl

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    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.
     
  6. Jun 23, 2008 #5

    CRGreathouse

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