[FOM] A question on fields
Joe Shipman
JoeShipman at aol.com
Mon Aug 5 16:59:09 EDT 2013
Correction – I shouldn't have said the axioms for ordered fields, just the axioms for fields are enough. If you take away the axiom that polynomials of degree 2 have roots, leaving in the axioms for odd primes, then you need to add the order relation and its axioms so that you can get a complete axiomatization of real closed fields by adding the axiom that positive elements have square roots.
-- JS
Sent from my iPhone
On Aug 5, 2013, at 1:57 AM, Joe Shipman <JoeShipman at aol.com> wrote:
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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