[FOM] Choice axioms and degree axioms

[joeshipman@aol.com]

In 1970, Conway identified the necessary and sufficient condition for 
effective implications between "finite choice axioms".  
  
Let AC(n) be the statement "every collection of n-element sets has a 
choice function."  
  
Then the conjunction AC(i1)&...&AC(i_m) effectively implies AC(n) if 
and only if,  
  
for every subgroup G of Sn that acts on {1,2,...,n} without fixed 
points, there exists a subgroup H of G and subgroups H1,...,H_k of H, 
with indices [H:H1]=j1,,...,[H:H_k]=j_k, such that one of the i's can 
be expressed as a sum of some of the j's.  
  
("Effective implication" here means the usual logical implication, but 
it is to be distinguished from an additional notion of "ineffective 
implication", where a conjunction of choice axioms implies a weaker 
version of AC(n), restricted to well-ordered collections of n-element 
sets. We have no need of this weaker notion since we will actually be 
strengthening the necessary and sufficient condition).  
  
Now, let D(n) be the statement "every polynomial of degree n has a 
root", which is expressible in the first-order language of field 
theory.  
  
Then the conjunction D(i1)&...&D(i_m) implies D(n) for all fields K if 
and only if,  
  
for every subgroup H of Sn that acts on {1,2,...,n} without fixed 
points, there exist subgroups H1,...,H_k of H, with indices 
[H:H1]=j1,,...,[H:H_k]=j_k, such that one of the i's can be expressed 
as a sum of some of the j's.  
  
In other words, the condition for "degree axioms" simply strengthens 
the condition for "choice axioms" by requiring that H=G.  
  
All the previous results I have announced follow straightforwardly from 
this necessary and sufficient condition, plus a little Galois theory 
(the Sylow theorems are necessary, but the combinatorics of my earlier 
proof underlie a standard proof of the Sylow theorems so it is "really" 
the same proof, reduced to its algebraic essence).  
  
-- JS