Timeline for Is it possible to construct algebraic numbers from $\mathbb{Q}$ without using polynomials? [closed]
Current License: CC BY-SA 4.0
15 events
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| Sep 27 at 6:43 | comment | added | Emil Jeřábek | As a group, $\mathbb A$ is the same as $\mathbb Q^{(\omega)}$ (the direct sum of countably many copies of $\mathbb Q$). Thus, polynomials are irrelevant anyway. | |
| Sep 26 at 18:43 | history | closed |
Yemon Choi Todd Trimble Steven Landsburg Will Sawin Sam Hopkins |
Needs details or clarity | |
| Sep 26 at 16:43 | comment | added | Will Sawin | Of course the answer to your final question is positive in any case: Fix a bijection $f\colon \mathbb Q \to \mathbb A$ and consider the binary operation $(x,y) \mapsto f(x)$. | |
| Sep 26 at 16:43 | comment | added | Will Sawin | Limits are not a binary operation, they are a partial function on infinitely many variables. So why not allow a multivalued function on finitely many variables, i.e. extracting roots of a polynomial from its coefficients, which $\mathbb A$ is the smallest subfield of $\mathbb R$ containing $\mathbb Q$ and closed under? | |
| Sep 26 at 16:16 | comment | added | takeyoi | @Wojowu I’ve tried to update the question based on your feedback. Please let me know if the clarification makes things clearer. | |
| Sep 26 at 16:16 | comment | added | takeyoi | @AlexKruckman I’ve tried to update the question based on your feedback. Please let me know if the clarification makes things clearer. | |
| Sep 26 at 16:12 | history | edited | takeyoi | CC BY-SA 4.0 |
added 380 characters in body
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| Sep 26 at 16:10 | review | Close votes | |||
| S Sep 26 at 18:47 | |||||
| Sep 26 at 15:39 | comment | added | Alex Kruckman | One way to make the question non-trivial is to ask whether the ring $\mathbb{A}$ is interpretable in the ring $\mathbb{Q}$ in the sense of first-order logic. But this might not be what you really intended to ask, since the construction of $\mathbb{R}$ from $\mathbb{Q}$ by Cauchy sequences is not a first-order interpretation. | |
| Sep 26 at 15:37 | comment | added | Alex Kruckman | But now to make the question non-trivial, you need to put some restriction on how you want the operations to be defined. After all, any countable set $X$ can be put in bijection with $\mathbb{A}$ (since $\mathbb{A}$ is countable), and this bijection can be turned into an isomorphism of rings by transferring the ring operations from $\mathbb{A}$ to $X$ along the isomorphism. | |
| Sep 26 at 15:37 | comment | added | Wojowu | "isomorphic" in what sense? As sets, you can even take $n=1$ and $\sim$ to be trivial; this is indeed just a statement of countability. If you mean as a group, then this is not possible, since $\mathbb A$ is infinite-dimensional over $\mathbb Q$. | |
| Sep 26 at 15:36 | comment | added | Alex Kruckman | There's an issue here about what you mean by "isomorphic to". In both of your examples, the quotient sets are isomorphic to the structures after defining the operations of $+$ and $\times$ in some way. e.g. there is a natural bijection $\mathbb{N}^2/{\sim}\to \mathbb{Z}$, and there is a natural way to define operations on $\mathbb{N}^2/{\sim}$ in such a way that this bijection turns into an isomorphism of rings. So I think what you really want to ask is whether we can define operations on $\mathbb{Q}^n/{\sim}$ in such a way that it is isomorphic to $\mathbb{A}$. | |
| Sep 26 at 15:21 | history | edited | takeyoi | CC BY-SA 4.0 |
edited title
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| S Sep 26 at 15:19 | review | First questions | |||
| S Sep 26 at 18:47 | |||||
| S Sep 26 at 15:19 | history | asked | takeyoi | CC BY-SA 4.0 |