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Alternate definition of field?

  1. Jun 20, 2008 #1
    Is there an easy way to define a field by only using one operation?

    It's easy to see that a cyclic abelian group of prime order uniquely defines a field if you consider multiplication as just repeated addition. Ie, consider such a group [tex]G[/tex] with generator [tex] g [/tex], identity [tex] e [/tex] with order [tex] p [/tex]. Then, we can say [tex]G = \{g,2g,3g,\ldots,(p-1)g,pg = e\}[/tex].

    Just define multiplication as [tex]x\cdot y = \sum_{i=1}^jx[/tex] where [tex] j [/tex] is the order of y with respect to the generator [tex] _g [/tex]. This is clearly commutative, and inverses exist because [tex] p [/tex] is prime.

    Is there a similarly easy way to extend this to polynomial fields as well?

    Also, why are my singleton letters so high up?
     
    Last edited: Jun 20, 2008
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  3. Jun 20, 2008 #2

    matt grime

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    No. Is the too short answer to be accepted by the forum.

    Your definition (of x.y) doesn't make sense, either: you've just written that x.y is jx for all x,y and j=ord(y), though I have no idea what 'the order of y with respect to g' means, admittedly. At the very least it would seem to vary as I changed g! That can't be right.

    To be more specific, your attempt to define x.y depends only on y (and g) so can't possibly be commutative as you claim.
     
    Last edited: Jun 20, 2008
  4. Jun 20, 2008 #3
    Sorry maybe I wasn't clear. I meant to begin by "fixing" a particular generator g. We then can call this [itex]g[/itex] by the name "1". We'll call [itex](g+g) = 2g[/itex] by the name "2", [itex](g+g+g) = 3g[/itex] by the name "3", and so on, with [itex]pg = 0[/itex].

    To see that it's commutative, say we want to multiply 2*3 in the group of order 5.

    Then, as I've defined it, we have [itex](g+g) + (g+g) + (g+g) = 6g = g = [/itex]"1".
    Alternatively, we have defined 3*2 by [itex](g+g+g) + (g+g+g) = 6g = g = [/itex]"1".

    Note that we can pick g to be *any* generator of the group, and all the resulting fields will be isomorphic.
     
    Last edited: Jun 20, 2008
  5. Jun 20, 2008 #4

    gel

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    So a finite cyclic group defines a field up to isomorphism, and will uniquely define it once you've picked the multiplicative identity.

    You can also say that any finite field is uniquely defined up to isomorphism by its additive structure (because all finite fields of the same size are isomorphic). However, choosing the multiplicative identity isn't enough to define the multiplicative stucture if the size isn't prime.

    You can't say the same about polynomial fields. Q(X) and Q(X,Y) will both be vector spaces over Q with a countably infinite basis, so are isomorphic as additive groups, but not as fields.

    However, you can say the following.

    The field Q is uniquely defined up to isomorphism by addition, and its multiplicative structure is fixed once you've chosen the multiplicative identity.

    The ordered field R is uniquely defined up to isomorphism by addition and its order, and uniquely defined once you've picked the multiplicative identity.

    I think the last example is quite interesting. R is often defined as the unique complete ordered field, defined up to (unique) isomorphism.
    You could equally define it as the unique complete ordered abelian group with a distinguished positive element (i.e., 1).
     
    Last edited: Jun 20, 2008
  6. Jun 21, 2008 #5

    matt grime

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    Your last example causes me concern: Q is an ordered field, and nothing you said about R doesn't apply to Q as well. Unless you're using 'order' to mean 'cardinality' as well as '<'.

    The problem with the idea of just using (integer) multiplies of the identity is that you can never generate more than an isomorphic the integers inside a field of characteristic zero just using addition.

    Say you wanted to define R like this: how do you assert that something like pi exists if all you've allowed is 0,1, and how to add these together and multiply the results? And how are you going to evaluate pi^2?
     
  7. Jun 21, 2008 #6

    matt grime

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    When you say 'order with respect to generator', you're better off saying 'discrete log with respect to g'.

    What is the log of pi with respect to 1 (considering R as an additive group)?
     
  8. Jun 21, 2008 #7

    gel

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    I should have said, the unique complete ordered non-discrete abelian group.

    The only (Dedekind) complete ordered abelian groups are the trivial group {0}, the integers Z and the reals R.
     
  9. Jun 21, 2008 #8

    gel

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    How do you define pi with any other definition of the reals? It's no different with my definition, except you will probably need to show that multiplication can be well defined first.
     
  10. Jun 21, 2008 #9

    matt grime

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    This isn't about your definition, gel; it's about the OP's definition - that one doesn't need multiplication only addition. In it, he wishes to add up pi, pi times, essentially.
     
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