Timeline for Does k(X) have a k-basis for every set X, without AC?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 21, 2015 at 16:10 | vote | accept | Jeremy Rickard | ||
| Jan 20, 2015 at 13:37 | answer | added | David E Speyer | timeline score: 6 | |
| Jan 20, 2015 at 3:40 | answer | added | Eric Wofsey | timeline score: 8 | |
| Jan 20, 2015 at 1:30 | answer | added | David E Speyer | timeline score: 8 | |
| Jan 19, 2015 at 13:06 | answer | added | Emil Jeřábek | timeline score: 6 | |
| Jan 17, 2015 at 20:11 | comment | added | Emil Jeřábek | @PaceNielsen: Define the monoid $M_X$ of monomials to consist of functions $X\to\mathbb N$ that are $0$ for all but finitely many $x\in X$, with pointwise addition, and then the ring $k[X]$ as the set of functions $M\to k$ that are $0$ for all but finitely many arguments, with pointwise addition, and convolution as multiplication. No choice is involved. | |
| Jan 17, 2015 at 19:22 | comment | added | Pace Nielsen | Do we even know that $k(X)$ exists, without the axiom of choice? Don't we need some sort of choice axiom to define arbitrary polynomial rings (and hence their quotient field), or am I just misremembering? | |
| Jan 13, 2015 at 15:12 | comment | added | Emil Jeřábek | I’m not an expert on these matters, however “every set has a total order” must be strictly stronger, as it in fact implies the axiom of choice for (unbounded) finite sets. The axiom of choice for pairs doesn’t even imply the axiom of choice for triples, see e.g. rjlipton.wordpress.com/2010/04/12/… . | |
| Jan 13, 2015 at 15:02 | comment | added | Jeremy Rickard | @EmilJeřábek Do you know how that compares with "every set has a total order", which clearly implies it? | |
| Jan 13, 2015 at 14:50 | comment | added | Emil Jeřábek | @YCor: A shorter way of saying that is that every set carries a tournament. I believe it’s equivalent to the axiom of choice for 2-element sets: on the one hand, it allows you to select one element from each unordered pair from a given set (e.g., the first in the ordered pair given by your section); in the opposite direction, you can take a selector on $\{\{(x,y),(y,x)\}:x\ne y\in X\}$. | |
| Jan 13, 2015 at 14:45 | comment | added | Jeremy Rickard | @YCor I was just wondering exactly the same thing! | |
| Jan 13, 2015 at 14:26 | comment | added | YCor | Is it known how the axiom "every set $X$ satisfies $Or(X)$: there is a section of the projection $(x,x')\mapsto \{x,x'\}$ from the set of pairs of $X$ to the set of 2-elements subsets" is related to AC? of course $Or(X)$ is strictly weaker than the existence of a well-ordering on $X$. | |
| Jan 13, 2015 at 13:30 | answer | added | user44143 | timeline score: 1 | |
| Jan 9, 2015 at 16:39 | comment | added | user44143 | So we need to choose canonically whether $1/(x-y)$ or $1/(y-x)$ goes in the basis. That may take the axiom of choice after all. | |
| Jan 8, 2015 at 12:38 | history | asked | Jeremy Rickard | CC BY-SA 3.0 |