Timeline for Are $\mathbb C$ , $\mathbb C[X]$ definable in $\mathbb C[[X]]$?
Current License: CC BY-SA 3.0
13 events
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| Mar 9, 2018 at 15:30 | comment | added | Alex Kruckman | 1. Ah, boundary points makes a lot more sense, thanks. 2. You're right - in the Cluckers paper he explicitly allows parameters in the quantifier-free formulas. I'm surprised that this is necessary. In this case, I would prefer to say "QE in the Macintyre language together with constants naming every element of the field". | |
| Mar 9, 2018 at 6:17 | comment | added | Will Sawin | @AlexKruckman Sorry, I meant finitely many boundary points. The boundary of a logical combination of sets is contained in the union of the boundary of the sets, and each formula defines a set with finitely many boundary points, but all of $\mathbb C$ and $\mathbb C[X]$ are boundary. | |
| Mar 9, 2018 at 6:10 | comment | added | Will Sawin | @AlexKruckman How do you define elements with the first $n$ terms zero without parameters or quantifiers? They are clearly definable with quantifiers, as the products of $n$ non-units. | |
| Mar 9, 2018 at 5:40 | comment | added | Alex Kruckman | Also "all elements of the field are in the language" - this is not what's meant by "quantifier elimination in the MacIntyre language". The MacIntyre language only has constant symbols for $0$ and $1$, and every formula without parameters should have an equivalent quantifier-free formula without parameters. But still it's easy to define the elements with zero constant term - they're exactly the non-units in $\mathbb{C}[[X]]$! | |
| Mar 9, 2018 at 5:36 | comment | added | Alex Kruckman | Can you explain a bit more the assertion about finitely many isolated points? Do you mean in the topology induced by the valuation? Doesn't $\mathbb{C}[X]$ have no isolated points in this topology? | |
| 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 7, 2018 at 18:17 | comment | added | Will Sawin | @MattF. All elements of the field are in the language. | |
| Mar 7, 2018 at 18:14 | comment | added | user44143 | So then X (or t in your answer) is in the language too? | |
| Mar 7, 2018 at 18:13 | comment | added | Will Sawin | @Matt F. If you want to define the elements with zero constant term, you just need to state that $X^{-1} x$ is an element of $C[[x]]$. I think this is in fact equivalent to your statement. | |
| Mar 7, 2018 at 18:06 | comment | added | user44143 | What do you get from eliminating quantifiers in "x is the difference of two non-squares in C[[x]]"? I.e. $\exists y \in C[[x]] \exists z \in C[[x]] \forall w \in C[[x] \ x = y - z\ \&\ y \neq w^2\ \&\ z \neq w^2$. That should, at least roughly, define the elements with 0 constant term. | |
| Mar 6, 2018 at 19:57 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 12 characters in body
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| Mar 6, 2018 at 19:49 | history | answered | Will Sawin | CC BY-SA 3.0 |