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Mar 7, 2018 at 20:18 vote accept CommunityBot
Mar 7, 2018 at 20:18 vote accept CommunityBot
Mar 7, 2018 at 20:18
Mar 6, 2018 at 19:49 answer added Will Sawin timeline score: 11
Mar 1, 2018 at 15:30 comment added Emil Jeřábek There is also vast literature on quantifier elimination in henselian valued fields (or valuation rings, for that matter), which I am mostly unfamiliar with, but very likely some of it can be applied here.
Mar 1, 2018 at 15:26 comment added Emil Jeřábek Hmm. So, that means the question is cross-posted from math.stackexchange.com/questions/2667763/… .
Mar 1, 2018 at 15:10 comment added user111524 @EmilJeřábek: But even that might not be so easy ... see math.stackexchange.com/questions/2669359/… and math.stackexchange.com/questions/2669549/…
Mar 1, 2018 at 15:06 comment added Emil Jeřábek Nevertheless, I think that some kind of automorphism argument might work. In particular, since $\mathbb C$ has infinite transcendence degree, I would expect that even if you fix finitely many parameters, there should be an automorphism of $\mathbb C[[X]]$ that moves some element of $\mathbb C$ outside $\mathbb C[X]$.
Mar 1, 2018 at 15:02 comment added user111524 @EmilJeřábek : Exactly ... thank you for getting my point ... I also explained this in my comment above
Mar 1, 2018 at 15:00 comment added Emil Jeřábek @PiotrAchinger Note that the OP is interested in definability with parameters. Your argument is blocked if you take e.g. $X$ as a parameter.
Mar 1, 2018 at 14:09 comment added Piotr Achinger The answer to the second question is no: pick any power series $Y = X + \ldots$, then there is an automorphism of $\mathbb{C}[[X]]$ sending $X$ to $Y$, and hence $\mathbb{C}[X]$ to $\mathbb{C}[Y]$.
Mar 1, 2018 at 13:43 history asked user111524 CC BY-SA 3.0