Timeline for Are the natural powers of two conservatively embedded in $\mathbb{C}$?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| 21 mins ago | comment | added | Noah Schweber | In particular, while $\mathcal{X}$'s $+$-part is trivial, $\mathbb{C}$'s isn't and could conceivably add some information (although I don't see how). | |
| 23 mins ago | comment | added | Noah Schweber | @tomasz That's incorrect. As I said in the italicized part of my question, I'm looking at $\mathbb{C}$ with the graphs of $+$ and $\times$, which are (ternary) relations rather than functions so every (nonempty) subset is the underlying set of a substructure. See also cody's comment above. | |
| 45 mins ago | comment | added | tomasz | (cont.) Then stable embeddedness should be resolved by q.e. in abelian groups, I think. But I guess you don't mean to only look at the multiplication monoid, but rather, a multiplication-substructure as compared to the induced structure from the field. | |
| 45 mins ago | comment | added | tomasz | $\newcommand{\bZ}{\mathbf Z}$I'm a bit confused: the natural powers of $2$ are not a substructure of the field of complex numbers (they are not closed under addition). So I guess you mean that you consider them as a substructure of the multiplication monoid? When we just forget about addition, then $(\mathbf C,\cdot)$ is (I think) isomorphic to ${\mathbf Q}^{\oplus \mathfrak c}\oplus(\mathbf Q/\bZ)\cup \{0\}$, with $X$ being isomorphic to $\mathbf N$ embedded in one of the $\mathbf Q$-s. | |
| 2 days ago | comment | added | James E Hanson | @NoahSchweber Oh, you're right. I was forgetting that automorphisms of $\mathbb{C}$ are going to fix the set of powers of $2$. | |
| 2 days ago | comment | added | Noah Schweber | @pastebee The quantifier elimination that occurs is in an expanded language, where the additional definitions do use quantifiers. So I don't think it quite gets what we want here. | |
| 2 days ago | comment | added | Noah Schweber | @JamesEHanson Why do you say it's probably bi-interpretable with it? $\mathbb{R}$ has an ordering after all ... | |
| 2 days ago | comment | added | paste bee | The paper shows more than just "it's decidable" - it uses quantifier elimination, and I think a corollary of the QE result is that the definable subsets of $\mathcal{X}$ are exactly the subsets definable in $\mathcal{X}$. If I'm right, that gives an answer to this question as well, because anything definable in the structure in the question would also be definable in this structure that interprets it. | |
| 2 days ago | comment | added | Noah Schweber | @JamesEHanson Ooh, I didn't know that - that's neat (although I don't think it immediately helps here)! | |
| 2 days ago | comment | added | James E Hanson | Minor note which you may already know, van den Dries showed in this paper that the theory of $(\mathbb{R},+,\cdot,\{2^n : n \in \mathbb{N}\})$ is decidable. This interprets the structure you're talking about (and is probably biinterpretable with it), so $\mathbb{Z}$ is not definable in the structure in this question. | |
| 2 days ago | comment | added | Noah Schweber | @cody Yes, that's correct. | |
| 2 days ago | comment | added | cody | I admit I'm a little confused about what signature we're talking about: $\mathbb{C}$ with only the relations of $+$ and $\times$ and equality? | |
| 2 days ago | history | edited | Alec Rhea | CC BY-SA 4.0 |
added 10 characters in body
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| 2 days ago | history | asked | Noah Schweber | CC BY-SA 4.0 |