Timeline for answer to Can one show that the real field is not interpretable in the complex field without the axiom of choice? by Erik Walsberg
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| Jul 19, 2021 at 4:29 | comment | added | Erik Walsberg | thanks Alex, as usual I see you were less lazy than me. | |
| Jul 18, 2021 at 19:04 | history | edited | Erik Walsberg | CC BY-SA 4.0 |
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| Jul 18, 2021 at 14:28 | comment | added | Alex Kruckman | @tomasz I gave a detailed argument directly from QE that $\mathbb{C}$ does not have the strict order property in this Math Stackexchange answer (points 5 and 6). I had totally forgotten about it until Asaf linked to it in the comments above. | |
| Jul 15, 2021 at 19:44 | comment | added | Erik Walsberg | That was actually my first approach. I think it's a little easier to get type bounds for $\mathbb{C}$ than it is to get failure of the order property. But yeah, this should work as well. | |
| Jul 15, 2021 at 19:12 | comment | added | tomasz | Correct me if I'm wrong, but isn't it far easier to do this via (maybe even strict, why not) order property instead of explicit stability? You obviously have the order property in the reals, its lack is obviously preserved by interpretations, so the main difficulty lies in showing it fails in the complex field, which seems to me like the sort of statement that would be easily shown to be absolute (unfortunately, I have little experience with these sorts of arguments, so I'm not quite sure about this), and probably easy via QE. Strong minimality seems OK for this kind of argument, too, no? | |
| Jul 15, 2021 at 17:12 | history | edited | Erik Walsberg | CC BY-SA 4.0 |
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| Jul 15, 2021 at 10:34 | comment | added | Emil Jeřábek | Concerning the last edit, without AC it's not clear to me how witnesses to non-$\omega$-stability lift under interpretations: if the interpretation has non-absolute equality, this requires choosing a selector in $\omega$ many equivalence classes. I think it's easier to formulate the argument for preservation of the property that there are countably many types over finite sets of parameters (i.e., smallness). This still works as a real-closed field has uncountably many parameter-free types (rational constants are definable). | |
| Jul 15, 2021 at 0:17 | history | edited | Erik Walsberg | CC BY-SA 4.0 |
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| Jul 13, 2021 at 22:12 | comment | added | Erik Walsberg | Now, one could also apply the fact that a theory is stable if every formula $\delta(x,y)$ is stable where $y$ is a single variable, this reduces to one-variable equations, which are easy. But this would involve using more stability. | |
| Jul 13, 2021 at 22:10 | comment | added | Erik Walsberg | it should follows from Noetherianness of the Zariski topology that there is $n$ such that any finite conjunction of instances of $\delta$ is equivalent to a conjunction of $n$ instances. It should be easy to see that this contracts the order property. Now that I've written this down it seems a lot more involved then the argument on your blog! | |
| Jul 13, 2021 at 22:04 | comment | added | Erik Walsberg | It should be possible to do this using QE for algebraically closed fields + some algebraic geometry. At this point it would help to know that a boolean combination of stable formulas is stable. I forgot the proof of this but it should just be finite combinatorics. Now we just need to show that $f(x,y) = 0$ does not have the order property, where $x = (x_1,\ldots,x_n)$, $y = (y_1,\ldots,y_m)$, and $f \in \mathbb{C}[x,y]$. Let $\delta(x,y)$ be this formula. So each instance $\delta(a,x)$ of $\delta$ defines a Zariski closed set. | |
| Jul 13, 2021 at 22:01 | comment | added | Erik Walsberg | I see. I think one can probably get the stability argument down to something pretty basic that doesn't actually require the stability-theoretic machinery. Let's use the order property definition of stability - that's the most elementary. It's obvious that $\mathbb{R}$ has the order property, and the absence of the order property is easily seen to be preserved under interpretations - if $M$ interprets $N$ then a witness of the order property in $N$ lifts to one in $M$. So it should just come down to showing that $\mathbb{C}$ does not have the order property. | |
| Jul 13, 2021 at 21:21 | comment | added | Joel David Hamkins | +1 Thank you for your answer. Stability theory is great, amazing. Nevertheless, people outside model theory often find it mysterious, if not impenetrable. This was why I had sought an elementary argument not requiring any stability theory. But I'm glad to know now that this part of model theory does not make use of AC. | |
| Jul 13, 2021 at 21:13 | history | edited | Erik Walsberg | CC BY-SA 4.0 |
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| Jul 13, 2021 at 21:02 | history | edited | Erik Walsberg | CC BY-SA 4.0 |
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| Jul 13, 2021 at 20:16 | history | answered | Erik Walsberg | CC BY-SA 4.0 |