Mathematics / Mathematik / Matemática

Mathematics / Mathematik / Matemática

Posts 11-12 of 12
  • Winfried Weber
    Winfried Weber    Group moderator
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    Re^7: construction of "new" fields
    Dr. Andreas Ecker schrieb:
    the question posed by Winfried reminded me to some model theoretic approaches, especially to ultraproducts.
    And Wikipedia has some information:
    http://en.wikipedia.org/wiki/Ultraproduct
    Hi Adreas,
    you are right: "my new construction" is simple exact the same what is called "ultraproduct", which was introduced by the American mathematician H. Jerome Keisler in 1961 (46 years ago!). So not a completely new theme, but nevertheless interesting.


    If the R_i are all fields, then P is an ultraproduct of the R_i. yes, but the discussion is now starting: what other properties do the ultraproduct have?

    In the theory it is seems to be very important to understand the Łoś' theorem (due to Jerzy Łoś)
    Łoś' theorem states that any “first-order formula” is true in the ultraproduct if and only if the set of indices i such that the formula is true in M_i is a member of U. More precisely: see http://en.wikipedia.org/wiki/Los%27_theorem

    Questions to the consequences of the Łoś' theorem :
    Which "well-known" algebraic or analytic sets can be formulated in a “first order model”?
    Which "well-known" properties can be formulated in a “first order logic”?

    References: http://en.wikipedia.org/wiki/First-order_predicate_calculus

    Sorry, but a don't understand this! Not yet, hopefully!
    Ah, I hear, you are saying: "This is quite easy"? OK, then you should follow my ideas:

    I.) Let C a set of well-known classified objects (f.e. C is a category), then we build the ultraproduct with all objects of C. Do we remain in the same category?

    examples:
    - set all abelian groups A_i, then П_i A_i/M is also abelian? (I think so.)
    - set all cyclic groups Z_i, then П_i Z_i/M is a cyclic group? (mmmhh, no?!)

    - set all fields A_i, then П_i A_i/M is also a field? Yes, it is!
    - “your” field *S = П_p F_p/M , the ultraproduct of all prime fields, is also a prime field?
    - “my” field *W = П_(pxn) F_(p^n)/M , the ultraproduct of all finite fields, is also a finite field?

    II.) Let A a set with property P, let A_I the subsets of A with the same property P.
    Question: Is the ultraproduct П_i A_i/M a subset of A? With the same property P?

    Examples:
    - let V a vector spaces and U_i the sub-vectorspaces of V.
    Is П_i U_i/M a vector spaces? (yes!), specially a sub-vectorspace of V (?!)

    - let L/K a infinite field extension, let A_i the intermediate fields of L/K.
    Is П_i A_i/M a field? (yes!), specially a intermediate field of L/K (?!)

    - let L the algebraic closure of K, [L:K] is infinite, let A_i the intermediate fields of L/K. Is П_i A_i/M a algebraic field extension of K (and so equal to one of the A_i)?
    Concrete example to this is (p a fixed prime number) “my” field *T_p= П_n F_(p^n)/M , because the F_(p^n) are exact the intermediate fields between F_p and its agebraic closure.


    If the R_i are all fields, then P is an ultraproduct of the R_i.
    But I have to think over the general case, when the R_i are rings.
    If the R_i are not fields, then there is not an easy bijection between the (ultra)filters of I and the (maximal)ideals of the product ring П_i R_i . But not easy means not impossible!

    III.) We choose two (or even more!) different families of objects to the same index set I, and we should have morphism between the objects of the families. Now let the ultrafilter M fixed and we craete the two ultraproducts . This should induce a morphism with the ultraproducts?

    example:
    F_(p^n) is a finite field extension of F_p
    so *S_n is a finite field extendion of *S?

    etc.

    Best Regards,
    Winfried
    • 28 Feb 2007, 1:34 pm
  • Dr. Andreas Ecker
    Dr. Andreas Ecker
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    Re^8: construction of "new" fields
    Hi Winfried,

     
    Questions to the consequences of the Łoś' theorem : Which "well-known" algebraic or analytic sets can be formulated in a “first order model”?
    Which "well-known" properties can be formulated in a “first order logic”?     Not an easy question ...
    This also dependent on the language you use, that means, what functions, relations and constants are in the language.

    I.) Let C a set of well-known classified objects (f.e. C is a category), then we build the ultraproduct with all objects of C. Do we remain in the same category?
     
    examples: - set all abelian groups A_i, then П_i A_i/M is also abelian? (I think so.)     Yes. The group axioms can be defined in first order:
    (i) for all a,b,c: (a * b) * c = a * (b * c)
    (ii) for all a: a * 1 = 1 * a = a
    (iii) for all a: a^(-1) * a = a * a^(-1) = 1
    (You could also define the group axioms in a language without the constant 1 and the function ^(-1).)
    Abelian means:
    (iv) for all a,b: a * b = b * a

    - set all cyclic groups Z_i, then П_i Z_i/M is a cyclic group? (mmmhh, no?!) No, "cyclic" is not a definable (in first order logic) property.
    The property "has at least n elements" is definable, which means that П_i Z_i/M is infinite. Other properties will depend heavily on M.

    - set all fields A_i, then П_i A_i/M is also a field? Yes, it is!
    - “your” field *S = П_p F_p/M , the ultraproduct of all prime fields, is also a prime field?     The property "for all i<k: i * 1 != 0" is first order definable (but not like written in the quotes!).
    That means *S has characteristic 0.

    - “my” field *W = П_(pxn) F_(p^n)/M , the ultraproduct of all finite fields, is also a finite field? Certainly not finite, but could have any characteristic, depending on M.

    II.) Let A a set with property P, let A_I the subsets of A with the same property P.
    Question: Is the ultraproduct П_i A_i/M a subset of A? With the same property P?     Depends, if the property is definable.

    Examples:
    - let V a vector spaces and U_i the sub-vectorspaces of V. Is П_i U_i/M a vector spaces? (yes!), specially a sub-vectorspace of V (?!)     Well, that depends how you define your language. Vector spaces are not so easy to handle with first order logic.
    As far as I know there are two approaches:
    Define the base set of the strucure as the union of F (the field) and V (the vector space) and add a relation saying x is an element of the field. That does not give you much control of the field.
    Or define the base set as V and add a function to the language for each element of F as scalar multiplication.

    - let L/K a infinite field extension, let A_i the intermediate fields of L/K.
    Is П_i A_i/M a field? (yes!), specially a intermediate field of L/K (?!)     Not likely. I would guess that (if L is countable) the cardinality of an ultra product is in most cases higher than L.

    - let L the algebraic closure of K, [L:K] is infinite, let A_i the intermediate fields of L/K. Is П_i A_i/M a algebraic field extension of K (and so equal to one of the A_i)? I would guess, you find elements that are transcendent over K.

    Concrete example to this is (p a fixed prime number) “my” field *T_p= П_n F_(p^n)/M , because the F_(p^n) are exact the intermediate fields between F_p and its agebraic closure. *T_p should be an algebraic closed field with characteristic p, but there should also be transcendent elements.

    If the R_i are not fields, then there is not an easy bijection between the (ultra)filters of I and the (maximal)ideals of the product ring П_i R_i . But not easy means not impossible! I am trying to prove an relation:
    In a former post you used something like
    J_F:={(a_i)_i in П_i R_i with property : { {i€ I : a_i =0} € F}
    If M is a maximal ideal, there should be a J_F a subset of M and F an ultrafilter.
    (just an idea)

    Best regards, Andreas
    • 28 Feb 2007, 8:42 pm