[FOM] Choice axioms and degree axioms
joeshipman@aol.com
joeshipman at aol.com
Sun Jan 1 16:00:12 EST 2006
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
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