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Math Help - Dense rings.

  1. #1
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    Dense rings.

    Hi:

    First, I must make some definitions, following Neal H.
    McCoy, The Theory of Rings.

    Def.1: A vector space V over a division ring D is a
    D-module with the additional requirement that if 1
    is the unity of D and x \in V, then x1 = x.

    Def.2: Let k be a positive integer. Then the ring R of
    linear transformations is said to be k-fold
    transitive if given any ordered set x_1, x_2,...,x_k
    of k linearly independent vectors and any ordered
    set y_1, y_2,...,y_k of k arbitrary vectors, there
    exists a \in R such that x_i a = y_i (i=1,2,...,k).
    If R is k-fold transitive for every positive
    interger k, R is said to be a dense ring of linear
    transformations.

    Now, I have this theorem:
    Theor.1: Let T be the ring of all linear
    transformations of a vector space V with a denumerable
    basis. Then...

    At a certain point in the proof, he says: Since Ve is
    now assumed to have infinite dimension, it has a
    denumerable basis X = {x_1, x_2,... } [because V has a
    denumerable basis; e \in T; that is, Ve is a subspace
    of V].

    [T being the ring of ALL linear transformations of V, T
    is dense.]

    Further on he says: Let Z = {z_1, z_2, ...} be a basis
    of V. Since there exists an element of T which maps the
    basis elements z_i (i = 1, 2,...) onto any specified
    elements of V, there exists an elemnt d of T such that
    Code:
           z_i d = x_i     (i = 1, 2, ...)
    .

    Here is what I do not understand. Def.2 says that
    given any ordered set x_1, x_2,..., x_k of k lin. ind.
    vectors in V and a set y_1, y_2,..., y_k of vectors in
    V, there exists a\in T such that x_i a = y_i for i = 1,
    2, ..., k, and that this is true for any positive k.
    An entirely different thing is to say that given a set
    x_1, x_2,... of lin. ind. vectors and a set y_1,
    y_2,... there is a\in T such that x_i a = y_i for i =1,
    2,... That is to say, now he is saying T is w-fold
    transitive, where w is omega, the set of natural
    numbers. Any suggestion will be welcome.
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  2. #2
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    Hi:

    First, I must make some definitions, following Neal H.
    McCoy, The Theory of Rings.

    Def.1: A vector space V over a division ring D is a
    D-module with the additional requirement that if 1
    is the unity of D and x \in V, then x1 = x.

    Def.2: Let k be a positive integer. Then the ring R of
    linear transformations is said to be k-fold
    transitive if given any ordered set x_1, x_2,...,x_k
    of k linearly independent vectors and any ordered
    set y_1, y_2,...,y_k of k arbitrary vectors, there
    exists a \in R such that x_i a = y_i (i=1,2,...,k).
    If R is k-fold transitive for every positive
    interger k, R is said to be a dense ring of linear
    transformations.

    Now, I have this theorem:
    Theor.1: Let T be the ring of all linear
    transformations of a vector space V with a denumerable
    basis. Then...

    At a certain point in the proof, he says: Since Ve is
    now assumed to have infinite dimension, it has a
    denumerable basis X = {x_1, x_2,... } [because V has a
    denumerable basis; e \in T; that is, Ve is a subspace
    of V].

    [T being the ring of ALL linear transformations of V, T
    is dense.]

    Further on he says: Let Z = {z_1, z_2, ...} be a basis
    of V. Since there exists an element of T which maps the
    basis elements z_i (i = 1, 2,...) onto any specified
    elements of V, there exists an elemnt d of T such that
    Code:
           z_i d = x_i     (i = 1, 2, ...)
    .

    Here is what I do not understand. Def.2 says that
    given any ordered set x_1, x_2,..., x_k of k lin. ind.
    vectors in V and a set y_1, y_2,..., y_k of vectors in
    V, there exists a\in T such that x_i a = y_i for i = 1,
    2, ..., k, and that this is true for any positive k.
    An entirely different thing is to say that given a set
    x_1, x_2,... of lin. ind. vectors and a set y_1,
    y_2,... there is a\in T such that x_i a = y_i for i =1,
    2,... That is to say, now he is saying T is w-fold
    transitive, where w is omega, the set of natural
    numbers. Any suggestion will be welcome.



    I think you're right: the leap from "for any k" to "infinite many", whether denumerable or not, seems too high to give it just like this and, perhaps, it doesn't work according to the given definition.

    Nevertheless the general claim is true: there's ALWAYS a UNIQUE linear transformation (operator, in fact) that carries ANY basis to ANY set with the same cardinal as a basisin your case, simple define v_id:=x_i and extend this definition by linearity.

    Tonio
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  3. #3
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    Thank you very much, Tonio.
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  4. #4
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    the important point you're missing is that we're dealing with T, the ring of all linear transformations of V, here. T is k-fold transitive for any k, finite or infinite. this is easy to see:
    for any set I, any linearly independent vectors z_i, \ i \in I, and any vectors x_i, \ i \in I, you can easily define \left(\sum_{i \in I} \alpha_i z_i \right)d=\sum_{i \in I} \alpha_i x_i, where all but finitely many of the scalars \alpha_i are zero. the map d is well-defined because z_i are linearly independent and it is obviously linear. thus d \in T. but if R is some subring of T, then d might not be in R.
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  5. #5
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    Yes, in particular, if I = the natural numbers, then I have the result I was looking for. And "where all but finitely many of the scalars are zero", because any element of V is a lin. comb. of a finite subset of the basis {z_1, z_2, ...}.

    This is a generalization of Example 1 on page 96 (section 27) of the book. Or better: there is a condition that is stronger than density although, in the case of T, both are equivalent.
    Thank you so much.
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