Mathematics / Mathematik / Matemática

Mathematics / Mathematik / Matemática

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  • Winfried Weber
    Winfried Weber    Group moderator
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    construction of "new" fields
    In this forum I would like to start a discussion about the following construction of “new” fields:

    I is an index set with indices i. (It should I be an infinite set, see example 1)
    Let (R,m,k) a local rings with maximal ideal m, k:= R/m its residue field.
    Define P := Π R = R x R x ... the cartesian product of R with
    addition: (a_i) + (b_i) = (a_i + b_i) { ( i in I) }
    multiplication: (a_i) * (b_i) = (a_i * b_i) {„komponentenweise“}

    Hint: If I is a infinite countable index set, then P is the “formal” ring of sequences with values in R (see example 3).

    Objective: How do the residue fields P/M for the maximal ideals M of P look like?

    Remark:
    Define M_i := Π A_n with A_n=R for n ≠ i , A_i = m
    Then M_i €Max P and P/M_i ≈ R/m = k, so the M_i are not relevant to investigate.

    Define T :=Max P\{M_i}

    Questions:
    Q1: Under which conditions for the (R,m,k) and the index set I we have T ≠ Ø?

    1. Example:
    If I is finite with n elements, then P=R^n=R x .. x R has exakt the n different maximal ideals M_i = R x ... x R x m x R x ... R , and P/M_i ≈ R/m = k
    So it is only interesting, if the index set I is infinite.

    Conjecture:
    T ≠ Ø <=> the Index set I and the ring R have infinite elements.
    Proof with the Lemma of Zorn. (?!)

    Let now T ≠ Ø (and so I infinite)
    Take a maximal ideal M € T and regard K:=P/M.

    Q2: Do “new” fields occure? What are the properties of K, if the k is well-known?

    Q3: In which cases are K ≈ k (≈ means “isomorphic”) . How is this depending on I and M?

    Q4: How many different (that means not-isomorphic) fields K can be generated?

    2. Example and motivation of the discussion
    Let Q the field of rational numbers {a/b, with a,b integers}, with the “ordinary” topology.
    Let C the space of Cauchy-sequences in Q, subring of the sequences ring P=Q^N
    Let M the subspace of the 0-convergent Cauchy-sequences (to is shown: it is an ideal and it is maximal)
    Then K=C/M is isomorphic to the field of real numbers.

    Q5: Are there other “interesting” maximal ideals M° in C, so C/M° generates a “new”field, not isomorphic to the real numbers?


    Q6: Are there other possibilities to define a subring B of P, with a “new” field B/M’ (M’ € Max B)?

    3.Example to Q6 + idea
    Let B the subring of bounded sequences in Q
    Let N the subspace of the 0-convergent sequences (to show: it is an ideal, but it is not maximal)
    Take a maximal ideal M’, containing N and regard K:=B/M’

    Q7: Is K isomorphic to Q? Or to the real number field (should not!).


    Best regards and good discussion,
    Winfried
    This post was modified on 02 Jan 2007 at 09:25 am.
    • 30 Dec 2006, 3:12 pm
  • Dr. Andreas Ecker
    Dr. Andreas Ecker
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    Re: construction of "new" fields
    Hi Winfried,

    just some thougts ...

    example 2 may be changed slightly with the use of ultra-filters:
    Let B be the space of all bounded sequences in Q (as a subring of Q^N) and F be a free ultra-filter on N.
    Let B_0 be the subring of all sequences in B that converge to 0 respectively to F
    (that means for (b_n) in B_0 and all r>0 the set {n; |b_n| < r} is in F).
    Then B/B_0 is isomorphic to the real numbers.

    If R is finite and P=R^N, then one could define similarly
    R_0 = {(a_n) in P; {n; a_n in m} in F}
    and R/R_0 isomorphic to k.

    The existence of free ultra-filters is proven with the Lemma of Zorn.

    I don't know, if that helps, but I am courious to hear more in this discussion.

    regards,
    Andreas
    This post was modified on 03 Jan 2007 at 09:56 pm.
    • 02 Jan 2007, 3:53 pm
  • Dr. Andreas Ecker
    Dr. Andreas Ecker
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    Re^2: construction of "new" fields
    If R is finite and P=R^N, then one could define similarly
    R_0 = {(a_n) in P; {n; a_n in m} in F}
    and R/R_0 isomorphic to k.
    This should also work for infinite R, but R/R_0 may not be isomorphic to k.
    If R = Q and m = {0}, then the factor should be some kind of nonstandard reals.
    In fact, nonstandard analysis works basically this way.

    If (m_n) is a sequence of maximal ideals, one could define
    R_0 = {(a_n) in P; {n; a_n in m_n} in F}
    similarly.

    The method with ultra-products and ultra-filters has been used in model theory to create saturated (rich) models. See for example Philipp Rothmaler, Einführung in die Modelltheorie.

    best regards,
    Andreas
    This post was modified on 03 Jan 2007 at 10:08 pm.
    • 03 Jan 2007, 10:07 pm
  • Winfried Weber
    Winfried Weber    Group moderator
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    Re^2: construction of "new" fields
    Dr. Andreas Ecker schrieb:
    Hi Winfried,
     
    just some thougts ...
     
    example 2 may be changed slightly with the use of ultra-filters:     Hello Andreas,
    you are right: using ultra-filters + methods of non-standard analysis is sureley the right methode.

    Let B be the space of all bounded sequences in Q (as a subring of Q^N) and F be a free ultra-filter on N.
    Let B_0 be the subring of all sequences in B that converge to 0 respectively to F     B_0 is an ideal, but not a subring, since 1 is not in B_0.
    You should take a amixaimal Ideal containig B_0 (and a OR b).

    (that means for (b_n) in B_0 and all r>0 the set {n; |b_n| < r} is in F).
    Then B/B_0 is isomorphic to the real numbers.

    I think B_0 is not a maximal ideal, since with

    a : = (1,0,1,0,1,0,1,0,1,...) in B
    b:= (0,1,0,1,0,1,0,1,...) in B

    then in B:
    a+b = 1, a*b = 0

    but a and b are not in B_0
    so B/B_0 is not a field (it remains a*b = 0)


    If R is finite and P=R^N, then one could define similarly
    R_0 = {(a_n) in P; {n; a_n in m} in F}
    and R/R_0 isomorphic to k.     yes, this should be right

    The existence of free ultra-filters is proven with the Lemma of Zorn. ok

    I don't know, if that helps, but I am courious to hear more in this discussion. intersting theme , but too far away my horizon ...

    regards,
    Andreas
    • 04 Jan 2007, 1:55 pm
  • Dr. Andreas Ecker
    Dr. Andreas Ecker
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    Re^3: construction of "new" fields
    Let B be the space of all bounded sequences in Q (as a subring of Q^N) and F be a free ultra-filter on N.
    Let B_0 be the subring of all sequences in B that converge to 0 respectively to F
    B_0 is an ideal, but not a subring, since 1 is not in B_0.
    You should take a amixaimal Ideal containig B_0 (and a OR b).
     
    (that means for (b_n) in B_0 and all r>0 the set {n; |b_n| < r} is in F).
    Then B/B_0 is isomorphic to the real numbers.
     
     
    I think B_0 is not a maximal ideal, since with
     
    a : = (1,0,1,0,1,0,1,0,1,...) in B
    b:= (0,1,0,1,0,1,0,1,...) in B
     
    then in B:
    a+b = 1, a*b = 0
     
    but a and b are not in B_0
    No. As F is an ultra-filter, either the set of all even numbers or the set of all od numbers is in F. That means, either a is in B_0 or b is in B_0.
    • 04 Jan 2007, 6:03 pm
  • Winfried Weber
    Winfried Weber    Group moderator
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    Re^2: construction of "new" fields
    Dr. Andreas Ecker schrieb:
    Hi Winfried,
     
    just some thougts ...     also I have only a question to understand the situation:

     
    example 2 may be changed slightly with the use of ultra-filters:
    Let B be the space of all bounded sequences in Q (as a subring of Q^N) and F be a free ultra-filter on N.     could you give a description of this ultra filter F and perhaps an ordinary filter which is dominated by F?
    And why "N"? Do you mean I=N (the index set) or N as a subset of K=Q?

    Let B_0 be the subring of all sequences in B that converge to 0 respectively to F Does an u l t r a -filter with this property really exist?
    (that means for (b_n) in B_0 and all r>0 the set {n; |b_n| < r} is in F).
    Then B/B_0 is isomorphic to the real numbers.     But is has to be shown!
    why not B/B_0 = Q ???

    best regards,
    Winfried

    EDIT: Typo
    This post was modified on 06 Jan 2007 at 11:06 pm.
    • 06 Jan 2007, 11:04 pm
  • Dr. Andreas Ecker
    Dr. Andreas Ecker
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    Re^3: construction of "new" fields
    Ok, I think I should clarify some things:

    A filter F on a set M is a subset of the power set of M (the set of all subsets of M) that fulfills the following conditions (I hope I get it right):
    - the empty set is not in F
    - if A and B in F, then also the intersection of A and B
    - If A is in F and A is a subset of B, then B is also in F

    An ultra-filter is a maximal filter.

    I always used (ultra-)filters on the index set.
    If a is in M, then {A; A is subset of M and a is in A} is an ultra-filter.
    More interesting are free ultra-filters. These are not generated by a single element, but contain the Frechet Filter.
    The Frechet-Filter on M is:
    {A; A is a subset of M and M-A (the complement of A in M) is finite} .

    Winfried Weber schrieb:
    Dr. Andreas Ecker schrieb:
    Hi Winfried,
     
    just some thougts ...
    also I have only a question to understand the situation:
     
     
    example 2 may be changed slightly with the use of ultra-filters:
    Let B be the space of all bounded sequences in Q (as a subring of Q^N) and F be a free ultra-filter on N.
    could you give a description of this ultra filter F and perhaps an ordinary filter which is dominated by F?     The existence of free ultra-filters is proven with the axiom or choice (or Zorn's Lemma). (In fact the existence of free ultra-filters is weaker than the axiom of choice.) That means the free ultra-filters are not constructible. For the most use cases it suffices to know that the Frechet Filter is a subfilter.

    And why "N"? Do you mean I=N (the index set) or N as a subset of K=Q? Yes, I meant N as index set, for example the natural numbers.

    Let B_0 be the subring of all sequences in B that converge to 0 respectively to F
    Does an u l t r a -filter with this property really exist?     For each ultra-filter F and each bounded sequence there is a limit to which the sequence converges. The proof is quite similar to the proof that each bounded sequence has a convergent subsequence.

    (that means for (b_n) in B_0 and all r>0 the set {n; |b_n| < r} is in F).
    Then B/B_0 is isomorphic to the real numbers.
    But is has to be shown!
    why not B/B_0 = Q ???     For a non-free ultra-filter it would be Q.
    But a sequence, that converges to pi (in the usual notion) converges to pi in respect to every free ultra-filter.

     
    best regards,
    Winfried
     
    Another access:
    I start with your definition:
    Define P := Π R = R x R x ... the cartesian product of R with
    addition: (a_i) + (b_i) = (a_i + b_i) { ( i in I) }
    multiplication: (a_i) * (b_i) = (a_i * b_i) {„komponentenweise“}

    Suppose R is a field, and J an ideal in P, then define
    F := {A; A is subset of I and there is (a_i) in J so that A = {i in I; a_i = 0}}
    Claim: F is a filter, and if J is a maximal ideal, then F is an ultra-filter.

    regards,
    Andreas
    • 07 Jan 2007, 9:06 pm
  • Winfried Weber
    Winfried Weber    Group moderator
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    Re^4: construction of "new" fields
    Hello,

    EDIT. Resume of the theory:

    Let I an (infinite) index set , R_i fields
    Define P := Π R_i the cartesian product of R_i with
    addition: (a_i) + (b_i) = (a_i + b_i)
    multiplication: (a_i) * (b_i) = (a_i * b_i)

    For a proper ideal J in P, we define
    F := {A; A is subset of I and there is (a_i) in J so that A = {i in I; a_i = 0}}

    Andreas has shown:
    F is a filter, and if J is a maximal ideal, then F is an ultra-filter.

    On the other way: For a filter F of I, we define
    J:={(a_i)_i in P with property : { {i€ I : a_i =0} € F}
    J is an ideal of P, and if F is an ultra-filter then J is a maximal ideal.

    So there is a bijection between
    the filters of the index set I <-> the proper ideals of the Ring P ,
    the ultrafilters <-> the maximal ideals of P.

    Definition:
    An ultrafilter dominating the Frechet-filter is called a free ultrafilter.
    For a free ultrafilter, we call the corresponding ideal a "free maximal ideal of P".

    Now it is necessary to show that the field P/M does not dependent on the choice of the free maximal ideal M of P OR to give the conditions that

    (*) for two free maximal ideals L and M of P we have P/L isomorphic to P/M

    LG,
    Winfried
    This post was modified on 15 Feb 2007 at 09:20 am.
    • 13 Feb 2007, 1:12 pm
  • Winfried Weber
    Winfried Weber    Group moderator
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    Re^5: construction of "new" fields
    Hello,

    I would like to discuss some examples.

    In this article we always write:
    The product ring is (f.e.) R
    M is a free maximal ideal of R and we assume
    (*) the field *R=R/M does not dependent on the choice of the free maximal ideal M of R
    or if we choose different free maximal ideals M, then we add an extra-index, f.e. *R_M

    We want to analyse the properties of the residue field *R (resp. or *R_m)

    Let F_(p^n) the field with p^n elements with discrete metric/topology [that means d(a,b) =0 <=> a=b; d(a,b)=1 <=> a unequal b]

    remember: F_(p^n) = Z/(p)[X] / (f), {with p prime number, f irreducible polynomial of Z/(p)[X], deg f = n }
    The set { F_(p^n) } is exactly the set of all finite fields.
    note: [F_(p^n) : F_p] = n

    I.) Example
    Let n and p be fixed and define for an infinite set I
    P_(p^n) := Π_i F_(p^n) , *F_(p^n) := P_(p^n) /M

    Proposal (Andreas): *F_(p^n) = F_(p^n) (!)

    II.) Example
    Let n be fixed and define for I:={p, p prime number}
    S_n := Π_p F_(p^n) , *S_n := S_n/M

    especially very interesting:
    *S:= *S_1 =(F_2 x F_3 x F_5 x...) / M

    III.) Example
    Let p be fixed and define for I=N
    T_p := Π_n F_(p^n) , *T_p := T_p/M

    IV.) "Giant" Examples
    Variation of n and p

    i) I=N ; U := Π_n *S_n ; *U := U/M

    ii) I:={p prime number} ; V := Π_p *T_p ; *V := V/M

    iii) let now the index set I = {p prime number} x N
    W := Π_(p,n) F_(p^n) , *W := W/M

    - Relationship between *U , *V and *W ? Perhaps *U= *V = *W ??? (All this fields have an image of all finite fields!)


    Regard *R (resp. *R_M) one of the fields *F_(p^n) , *S_n , *T_p , *U , *V , *W (resp. added with index M) and let k its prime field.

    Questions: Find the properties of *R (resp. *R_M) !!!
    - char *R = ?
    - *R has uncountable elements? Set-theoretic bijection to the field of real numbers possible ?
    - Trdeg [*R:k] is finite or unfinite ?
    - Do elements x in *R and irreducible polynomials f in k [X] , deg f > 1 , exist with f(x) = 0 ?
    - How does the induced product/residue topology of *R look like ?
    - Metric space ?
    - Has *R an ordering (like Q )?
    - Compare : [F_(p^n) : F_p] = n => [*S_n : *S] = n (?)

    Good investigations ;-)

    Winfried
    This post was modified on 16 Feb 2007 at 11:49 am.
    • 14 Feb 2007, 2:20 pm
  • Dr. Andreas Ecker
    Dr. Andreas Ecker
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    Re^6: construction of "new" fields
    Hi,

    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
    or (in German)
    http://de.wikipedia.org/wiki/Ultraprodukt

    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.

    Regards,
    Andreas
    • 21 Feb 2007, 9:17 pm