[FOM] Algebraic closure of Q/difficulties

[Harvey Friedman]

On 5/14/06 3:17 AM, "Robert M. Solovay" <solovay at Math.Berkeley.EDU> wrote:

> Harvey,
> 
> Of course, none of this shows that the result you are trying to
> prove isn't true. Indeed, perhaps there is a correct proof along the rough
> lines of your current attempt.
> 

Of course, as usual, you are quite right. There are several holes in my
"argument". 

I look forward to somebody filling them in.

To reacap Chow's question:

PROBLEM: Is the uniqueness of algebraic closures of Q equivalent to some
simple form of the axiom of choice, over ZF?

CANDIDATE: A countable union of finite sets is countable.

This easily implies uniqueness of algebraic closures of Q, since every
algebraic closure of Q is a countable union of finite sets.

Harvey Friedman