# [FOM] some references?

joeshipman@aol.com joeshipman at aol.com
Tue Feb 14 22:17:27 EST 2006

```Salehi asks for my proof.  There is no room here for the full proof,
but I will send anyone who requests it the manuscript I submitted to
the Bulletin of the AMS.

If we restrict to characteristic 0 and show just the "if" direction,
and omit all the applications, a proof of this weaker theorem can be
presented very concisely:

Theorem. Let [n] denote the statement "All polynomials of degree n have
a root". The implication ([i1]&[i2]&...&[i_m])-->[n] is true in all
fields of characteristic 0 if the following condition holds:

For any subgroup G of Sn that acts without fixed points on {1,2,...,n},
the additive semigroup generated by the indexes in G of its proper
subgroups contains one of the i's.

Proof: Fix a char 0 field K. Assume that the condition is true, and
that [n] is false; we need to falsify one of the degree axioms on the
left hand side. If [n] is false there is a polynomial of f(x) degree n
with no roots, with splitting field L. Its Galois group G acts on the
set {r1,...,r_n} of roots of f without fixed points (multiple roots of
f appearing the appropriate number of times with different labels). For
each proper subgroup H of G if index h, there is a corresponding
intermediate field between K and L of degree h, and since we are in
characteristic 0 this intermediate field has a primitive element so
there is an irreducible polynomial of degree h. By multiplying such
irreducible polynomials together, we can generate a rootless polynomial
of any degree in the semigroup; but the semigroup contains one if the
i's by assumption so the corresponding degree axiom on the
left-hand-side must fail.

-- Joe Shipman
```

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