[FOM] Algebraic closure of Q

[Timothy Y. Chow]

Andreas Blass wrote:
>As far as I can see, avoiding this problem requires replacing the  
>sequence of polynomials z^{p_i} - 2 with some sequence such that the  
>Galois group of each one is the full symmetric group of permutations  
>of all its roots; furthermore, the Galois group must be this big not  
>only over Q but over the extension obtained by adjoining roots of  
>previous polynomials in our list.

There's a general method for constructing extensions whose Galois group is 
the symmetry group, described for example in Lang's Algebra, 3rd ed., VII, 
section 2.  It might work here, though some details need to be worked out.

The basic idea is that the "general" polynomial has the full Galois group, 
and by the Hilbert irreducibility theorem you can specialize to get a 
"specific" polynomial with the same group.  In more detail, let k be the 
base field and let

  f(X) = X^n + w_{n-1} X^(n-1) + ... + w_0

where the w_i are algebraically independent over k.  Letting

  K = k(w_0, ..., w_{n-1})

and letting L be the splitting field, we know that Gal(L/K) is the 
symmetric group.  Now let alpha generate L over K (we might need 
separability here, I'm not sure), where alpha is integral over 
k[w_0,...,w_{n-1}].  The irreducible polynomial g(X) of alpha over K has 
coefficients that are polynomials in the w_i.  Then it's a theorem that if 
we substitute elements a_i in k for the w_i and the polynomial g(X) is 
still irreducible (over k, now), then the Galois group is still the 
symmetric group.  Finding appropriate a_i should be guaranteed by some 
kind of Hilbert irreducibility theorem.

Tim