Timeline for Unnecessary uses of the axiom of choice
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 18, 2022 at 21:53 | comment | added | LSpice | @tj_, if we are allowed to work with a counting of our countable set (that doesn't use choice, right?), then I think the greedy algorithm (toss new elements into a proposed transcendence basis if they are transcendental over the extension so far constructed) will produce a transcendence basis. | |
| Feb 18, 2022 at 21:37 | comment | added | tj_ | @LSpice: One can compose the field as a transcendental extension of Q followed by an algebraic extension. The trans. ext. can be embedded into C and since C is alg. closed, the whole field can be embedded into C. But probably this approach needs countable choice for obtaining a transcendence base of the countable field. | |
| Feb 18, 2022 at 21:17 | comment | added | LSpice | @user21820, of course $\mathbb Q(t)$ doesn't really give the full power of this result, since it embeds in $\mathbb C$. (Is there a characteristic-$0$, countable field that doesn't embed in $\mathbb C$?) | |
| Feb 18, 2022 at 20:53 | history | edited | Pace Nielsen | CC BY-SA 4.0 |
added 144 characters in body
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| Feb 18, 2022 at 15:39 | comment | added | user21820 | @TomLeinster: You're not missing anything at all. $ℝ$ and $ℂ$ are very special; you don't need AC to prove that $ℂ$ is an algebraic closure of $ℝ$ either! What is more interesting is that any arbitrary countable field (e.g. $ℚ(t)$ where $t$ is an indeterminate) has an algebraic closure (i.e. an algebraic extension that is closed under algebraic extension), without relying on choice. | |
| Feb 18, 2022 at 12:17 | comment | added | Tom Leinster | Maybe I'm overlooking something obvious, but to construct an algebraic closure of $\mathbb{Q}$ without using either transfinite induction or choice, can't I just take the set of elements of $\mathbb{C}$ algebraic over $\mathbb{Q}$? | |
| Feb 18, 2022 at 6:29 | comment | added | David Roberts♦ | Chasing references back, I get to this posting of Stephen Simpson cs.nyu.edu/pipermail/fom/2006-May/010538.html claiming the result, where he refers back to his RM book. | |
| Feb 18, 2022 at 5:13 | comment | added | Robert Furber | @DavidRoberts perhaps you are recalling this answer? math.stackexchange.com/a/115060 | |
| Feb 18, 2022 at 4:52 | comment | added | David Roberts♦ | What about the algebraic closure of a countable field? I have a vague memory that this is another case that doesn't need as much AC as generally advertised. | |
| S Feb 18, 2022 at 2:07 | history | answered | Pace Nielsen | CC BY-SA 4.0 | |
| S Feb 18, 2022 at 2:07 | history | made wiki | Post Made Community Wiki by Pace Nielsen |