Timeline for For a ring $k$ and a set $X$, what are the $k$-algebra homomorphisms $k^X \to k$?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 6, 2015 at 23:40 | vote | accept | Tom Leinster | ||
| Jun 4, 2015 at 19:43 | comment | added | Avshalom | The JSL 2014 paper by George Bergman might be helpful: arxiv.org/abs/1301.6383 | |
| Jun 4, 2015 at 18:49 | answer | added | Eric Wofsey | timeline score: 4 | |
| Jun 4, 2015 at 18:32 | history | edited | Arturo Magidin | CC BY-SA 3.0 |
latex title
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| Jun 4, 2015 at 17:56 | history | edited | Tom Leinster | CC BY-SA 3.0 |
improved wording
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| Jun 4, 2015 at 15:18 | comment | added | Tom Leinster | @TomGoodwillie: Thanks! Your argument implies that if $|X| \leq |k|$ then there are no nontrivial homomorphisms $k^X \to k$, doesn't it? | |
| Jun 4, 2015 at 13:45 | comment | added | YCor | @TomLeinster sure, but in the usual sense of "large cardinal", something as $2^{2^{2^{\aleph_0}}}$ is minute! The point is that if it exists (it's consistent with ZFC that not), the smallest large cardinal $\kappa$ is inaccessible (that is, $2^\alpha<\kappa$ for every $\alpha<\kappa$ and every $\sup$ of $<\kappa$ cardinals all $<\kappa$ is $<\kappa$); moreover $\kappa$ is the $\kappa$-th accessible cardinal... and on the other hand $\kappa$ admits a nonprincipal ultrafilter stable by all $<\kappa$-intersections. So it possibly makes the question more complicated. | |
| Jun 4, 2015 at 13:39 | comment | added | Tom Goodwillie | A given function $\phi:X\to k$ will represent an element outside $k$ in the residue field associated with the ultrafilter $U$ if and only if no level set $\phi^{-1}(c)$ belongs to $U$. So for example if both $X$ and $k$ are countably infinite then there are no more examples of such algebra homomorphisms. | |
| Jun 4, 2015 at 13:38 | comment | added | Tom Leinster | @YCor: I don't have a particular interest in large cardinals, thanks, but neither do I particularly want to rule out sets of large cardinality. | |
| Jun 4, 2015 at 13:33 | comment | added | YCor | If $X$ is huge enough to have nonprincipal ultrafilters stable under infinite countable intersections, then assuming that $k$ is not to big (say $k$ infinite countable, to begin with), you can compute the limit wrt this ultrafilter. Maybe it's reasonable to assume that $X$ has no such ultrafilter unless you have a specific interest in large cardinals. | |
| Jun 4, 2015 at 13:20 | comment | added | Tom Goodwillie | If $k$ is a field then the ideals in $k^X$ are in bijection with the filters, with the maximal ideals corresponding to the ultrafilters. So the question comes down to: for which ultrafilters is the residue field a trivial extension of $k$? | |
| Jun 4, 2015 at 11:51 | history | asked | Tom Leinster | CC BY-SA 3.0 |