Timeline for Does k(X) have a k-basis for every set X, without AC?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 21, 2015 at 2:28 | comment | added | David E Speyer | @MattF. It's not. We choose one $f$, once and for all, and then define $c(x,y)$ using it. | |
| Jan 21, 2015 at 1:49 | comment | added | user44143 | Why is $c(x,y)$ independent of $f$? Suppose the basis includes 1, $x$, $x/(x+y)$, $xy/(x+y)$. Then $c(x,y\,|\,f=1) = 0$ but $c(x,y\,|\,f=x) = 1$. | |
| Jan 20, 2015 at 11:55 | comment | added | Eric Wofsey | Essentially the same argument also gives a direct proof that if $k$ is a field of positive characteristic, then $k(x_1,x_2,\dots)$ does not have an almost symmetric basis. If $f$ is any basis element of such an almost symmetric basis, pick $p$ variables not occuring in $f$ (or the finite set your permutations must fix) and apply this argument to get a contradiction. | |
| Jan 20, 2015 at 10:02 | comment | added | Jeremy Rickard | Very nice! More generally this shows (if we stick to characteristic 2) that if there's no choice function for pairs of elements of $X$, but there is for pairs of elements of $k$, then $k(X)$ can't have a basis. I wonder what happens if there's no choice function for pairs of elements of $k$? It would seem a bit weird if $k$ having bad choice properties made it easier to find a basis, although I suppose that for particular (related) $k$ and $X$, there might be a construction of a basis that avoids choice. | |
| Jan 20, 2015 at 2:46 | comment | added | Eric Wofsey | A similar argument shows that for $k=\mathbb{F}_3$, you get choice for 3-element sets (if three elements of $\mathbb{F}_3$ add to 1, two must be equal and the other one is a canonical choice). More generally, for $k=\mathbb{F}_p$ you get a choice of a nontrivial partition of every $p$-element set, which surely requires choice. | |
| Jan 20, 2015 at 1:30 | history | answered | David E Speyer | CC BY-SA 3.0 |