[FOM] A question on fields

[Joe Shipman]

This is easy, pick a prime p greater than d and adjoin algebraic numbers to the field starting with Q and using an enumeration of all algebraic numbers whose degree over Q is not a multiple of p. At each stage you have a finite extension whose degree over q is not a multiple of p, and whose Galois group over Q contains no elements of order p, so none of the algebraic numbers whose degree is a multiple of p will ever get in.

I already published a nonredundant axiomatization for algebraically closed fields in my 2007 Math. Intelligencer  paper "Improving the Fundamental Theorem of Algebra"--take the conjunction of the axioms for ordered fields, and an axiom for each prime p saying that polynomials of degree p have roots. Adding any axiom settling the characteristic of the field gives the complete theory for algebraically closed fields of that characteristic; adding an axiom for each prime p that p is not the characteristic gives the complete theory for characteristic 0.

-- JS

Sent from my iPhone

On Aug 4, 2013, at 12:11 PM, SHASHI SRIVASTAVA <smohan53 at gmail.com> wrote:

 How does one prove the following?
 
For every $d \geq 2$, there is a field $K$ such that every polynomial in $K[X]$ of degree $\leq d$ has a root in $K$ but $K$ is not algebraically closed.
 
This will imply that the theory of algebraically closed fields is not finitely axiomatizable, which is my main interest.
 
Shashi M. Srivastava
Indian Statistical Institute
Kolkata     
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