paper-machine comments on The Pascal's Wager Fallacy Fallacy - Less Wrong

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Comment author: Sniffnoy 28 April 2011 09:50:43PM 1 point [-]

Gal(C/R) = Z/2Z

I'm confused; this is true for any real closed field. What are you getting at with this?

Comment author: [deleted] 28 April 2011 10:05:25PM *  0 points [-]

A mistake. I was thinking of C as the so-called "generic complex numbers." You're right that if you replace C with the algebraic closure of whatever countable model's been dreamed up, then C = R[i] and that's it.

Admittedly I'm only conjecturing that Gal(C/K) will be different for some K countable, but I think there's good evidence in favor of it. After all, if K is the algebraic closure of Q, then Gal(C/K) is gigantic. It doesn't seem likely that one could "fix" the other "degrees of freedom" with only countably many irrationals.