Timeline for Is there a particular field that cannot be proven to have an algebraic closure in ZF?
Current License: CC BY-SA 3.0
19 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Mar 21, 2016 at 0:05 | comment | added | Goldstern | (A generalisation of @JulianRosen's comment:) For any ordered (i.e., linearly ordered) field $K$, the ordered finite extensions form a directed system. (The embeddings between them are canonical, because of the order.) Their naturally defined limit (without AC, I think) is a real closed field. Adjoin an imaginary unit $i$ to get an algebraically closed field. | |
| Mar 18, 2016 at 14:11 | comment | added | Julian Rosen | *slaps forehead* For some reason I was thinking of $\mathcal{P}(\mathbb{R})$ as finite subsets of $\mathbb{R}$, which can be totally ordered. | |
| Mar 18, 2016 at 12:33 | comment | added | Emil Jeřábek | Yes, indeed, that was the reason. $\mathcal P(\mathbb R)$ looks like one of the simplest sets that can’t be proved totally ordered in ZF. | |
| Mar 18, 2016 at 10:11 | comment | added | Jeremy Rickard | @JulianRosen It's consistent with ZF that $\mathcal{P}(\mathbb{R})$ can't be totally ordered. I suspect that's why Emil chose that example. | |
| Mar 18, 2016 at 0:26 | comment | added | Julian Rosen | @EmilJeřábek For any finite ordered set of variables, we can build an algebraically closed field of iterated Puiseax series in those variables. Enlarging the set of variables gives an extension of Puiseax series fields, so if $S$ is a totally ordered set, we can embed $\mathbb{Q}(\{x_s:s\in S\})$ into the directed union of finite iterated Puiseax series fields, which is algebraically closed. Maybe we could choose a set on which a total order cannot be constructed in ZF | |
| Mar 17, 2016 at 18:59 | comment | added | Julian Rosen | I edited the title to reflect the question in the body. | |
| Mar 17, 2016 at 18:58 | history | edited | Julian Rosen | CC BY-SA 3.0 |
Edited title to match question in body
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| Mar 17, 2016 at 18:00 | comment | added | Todd Trimble | Ah, so it's open whether existence of an algebraic closure implies BPIT, according to Caicedo's answer. Thanks, @BenjaminSteinberg. | |
| Mar 17, 2016 at 17:15 | comment | added | Benjamin Steinberg | Seems like mathoverflow.net/questions/46566/… is closelY related this question. | |
| Mar 17, 2016 at 17:04 | comment | added | Todd Trimble | Well, to be clear, the word "require" needs some disambiguation: my point is that AC is not a necessary consequence of existence of an algebraic closure. (I would guess that BPIT is necessary as well as sufficient, but our resident expert Asaf could probably say for sure.) However, it is a side comment in the sense that the OP was very clear that his question is about working in ZF, and the comment doesn't answer his question. | |
| Mar 17, 2016 at 17:03 | comment | added | Asaf Karagila♦ | @Emil: That's a good idea, actually, since it is consistent that there is an amorphous set which is a subset of $\mathcal P(\Bbb R)$. It might actually work. | |
| Mar 17, 2016 at 17:00 | comment | added | Asaf Karagila♦ | @katz: Because it's a side comment. It doesn't answer the question if there is an explicit field whose algebraic closure requires the axiom of choice to exist. | |
| Mar 17, 2016 at 16:59 | comment | added | Mikhail Katz | @ToddTrimble, I am not sure why this is a "side comment". It seems like an answer to me :-) | |
| Mar 17, 2016 at 16:58 | comment | added | Todd Trimble | A side comment -- it's fairly well known, but still -- is that the ultrafilter principle or BPIT, which is strictly weaker than AC, suffices to prove existence (and uniqueness up to isomorphism) of algebraic closures. | |
| Mar 17, 2016 at 16:46 | comment | added | Emil Jeřábek | A rational function field over an unsightly set of variables, such as $\mathbb Q(\{x_a:a\in\mathcal P(\mathbb R)\})$, might do the trick. This is pretty explicit, IMHO. | |
| Mar 17, 2016 at 16:45 | comment | added | Wojowu | @AsafKaragila I suspect the OP expects an answer analoguous to $\Bbb R$ over $\Bbb Q$ or $\Bbb F_2^\infty$ over $\Bbb F_2$ as examples of explicit vector spaces without bases constructible (in the sense of proving existence) in ZF. | |
| Mar 17, 2016 at 16:16 | comment | added | Asaf Karagila♦ | The main problem with "explicit examples" is that most of the examples of non-AC tend to be non-explicit. | |
| Mar 17, 2016 at 16:13 | history | asked | Julian Rosen | CC BY-SA 3.0 |