A Couple of Interesting Problems

1) Linear Disjointness

Consider the following tower of fields:

    L
    | algebraic,
    | of degree n
    K
    | transcendental
    |
    F

Suppose b_1, b_2, ..., b_n forms a basis for L/K such that each
b_i is algebraic over F.

Let c_1, c_2, ..., c_n be elements of K. Is it true that whenever
c_1 b_1 + c_2 b_2 + ... + c_n b_n is algebraic over F,
the c_i themselves are algebraic over F?


2) Symmetric Bases

Given a field K, let x_1, x_2, ..., x_n denote indeterminates
and s_1, s_2, ..., s_n the induced elementary symmetric polynomials 
in K[x_1, x_2, ..., x_n].

We then have the following diagram: (alg denotes algebraic closure)

       alg(K[x_1, x_2, ..., x_n]) = alg(K[s_1, s_2, ..., x_n])
                                  |
                                  |  oo
                                  |
                       K(x_1, x_2, ..., x_n)
                               /   \
                              /     \  n!
                         oo  /       \
                            /     K(s_1, s_2, ..., s_n)
           K[x_1, x_2, ..., x_n]       | 
                             \         |  oo
                              \        |
                            n  \  K[s_1, s_2, ..., s_n]
                                \ /
                                 K  n

Let F be a field in the lattice above. Is it possible to find
a basis for F/K that is set-wise invariant under permutation of the
indeterminates x_1, x_2, ..., x_n?

Notes:

  - For the case where F is the uppermost field, this requires
    a rigorous definition of permutation of indeterminates.

    Assuming such a rigorous definition were to exist, it is
    unlikely that a desired basis would exist.
    Consider the extension:

         F = alg(K(x_1, x_2, ..., x_n)) = alg(K(s_1, s_2, ..., s_n))
                               |
                               |  oo
                               |
                     L = K(s_1, s_2, ..., s_n) .

    Suppose there exists a set-wise symmetric basis of F/L.
    If u is such a basis element and f_u is a minimal polynomial
    for u over L, then one would expect (assuming a rigorous
    definition) that all roots of f_u be in the basis as well,
    due to symmetry. However, the sum of these roots is the
    second coefficient of f_u, which is an element of L.

    Since f_u has finitely many roots, this would mean the assumed
    basis is L-linearly dependent, a contradiction. Note further
    that this does not preclude a basis for F over K, but the
    existence of such seems highly unlikely.

  - For the case where F = K(x_1, x_2, ..., x_n), this reduces
    to finding a set-symmetric basis for F over
    K(s_1, s_2, ... s_n), and then multiplying this basis by
    a basis of K(s_1, s_2, ..., s_n), whose elements are clearly
    all invariant under permutation.

    By Galois theory, we have that the index

         [ K(x_1, x_2, ..., x_n) : K(s_1, s_2, ..., s_n) ] = n!

    so it suffices to find a linearly independent set of
    size n! that is set-wise invariant under permutation.

    If we let the symbol "o" range over the elements of the
    symmetric group S_n (acting on {1,2,...,n}), then
    elements of one such set would take the form:

         (x_1)^(o(1)) * (x_2)^(o(2)) * ... * (x_n)^(o(n)) .

  - The remaining cases are trivial.

Jonathan Lee