Timeline for Does k(X) have a k-basis for every set X, without AC?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 21, 2015 at 16:10 | vote | accept | Jeremy Rickard | ||
| Jan 21, 2015 at 9:15 | history | edited | Jeremy Rickard | CC BY-SA 3.0 |
Corrected $k^\omega$ to $k^{\oplus \omega}$
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| Jan 21, 2015 at 1:26 | comment | added | Eric Wofsey | This argument also works for any well-orderable $k$; you just have to find $|k|$ copies of the regular representation in each $r(I)$. In particular, assuming choice, this shows $k(X)$ has a symmetric basis for any $k$ of characteristic zero. | |
| Jan 20, 2015 at 21:49 | comment | added | Emil Jeřábek | The argument I gave in essence exploits the field structure of $W=k(x_1,\dots,x_n)$ to provide an explicit isomorphism $W\simeq W_\lambda\otimes kS_n$, where $\lambda=(n)$ corresponds to the trivial representation. I don’t know whether such a thing stands a chance of working for $r(I)$, but if so, it would considerably alleviate the need for $k$ to be countable. | |
| Jan 20, 2015 at 17:54 | comment | added | Emil Jeřábek | Very nice! ---- | |
| Jan 20, 2015 at 17:49 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
added 49 characters in body
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| Jan 20, 2015 at 16:01 | history | edited | François G. Dorais | CC BY-SA 3.0 |
fixed small typos
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| Jan 20, 2015 at 13:47 | comment | added | David E Speyer | Okay, I made the easy switch and imposed $k$ counteable. Now to think about the better switch, where we deal with any characteristic zero field. | |
| Jan 20, 2015 at 13:46 | history | edited | David E Speyer | CC BY-SA 3.0 |
added 15 characters in body
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| Jan 20, 2015 at 13:45 | comment | added | David E Speyer | Oh, uggh, you're right. I thought through a lot of this with $\mathbb{Q}$ and then switched at the last moment. Editing now... | |
| Jan 20, 2015 at 13:43 | comment | added | Emil Jeřábek | $k(x)$ has an independent subset $\{(x-a)^{-1}:a\in k\}$ of size $|k|$, hence $k(x_1,\dots,x_n)$ only has a countable basis if $k$ itself is countable. Or am I missing something? | |
| Jan 20, 2015 at 13:37 | history | answered | David E Speyer | CC BY-SA 3.0 |