Comment Re:This doesn't seem likely (Score 1) 482
Is AC really needed? Put any total ordering on the complex plane, for example by ordering everything first by magnitude and then by going around circles counterclockwise starting from the positive real axis. This isn't a well-ordering on C of course but it induces one on every finite subset of C. So for some p(x) over C we can just take its smallest root with respect to this ordering. I don't believe we need AC to guarantee a choice function in this circumstance.
Compare the usual proof that well-ordering implies AC: let X_alpha be a collection of sets indexed by I and let X be the union over I of the X_alpha. Put a well-ordering on X and then define a function f : I --> X by taking f(alpha) to be the minimal element of X_alpha as a subset of X. This constructs a choice function. But we don't actually use exactly that X is well-ordered, just that there is an ordering under which every subset we're interested in has a minimal element.
Compare the usual proof that well-ordering implies AC: let X_alpha be a collection of sets indexed by I and let X be the union over I of the X_alpha. Put a well-ordering on X and then define a function f : I --> X by taking f(alpha) to be the minimal element of X_alpha as a subset of X. This constructs a choice function. But we don't actually use exactly that X is well-ordered, just that there is an ordering under which every subset we're interested in has a minimal element.
