Timeline for Uncountable counterexamples in algebra
Current License: CC BY-SA 4.0
30 events
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| May 10, 2021 at 12:36 | comment | added | KConrad | @David Handelman all that you wrote is true in characteristic $p$ too: for each prime $p$, uncountable algebraically closed fields of characteristic $p$ are (by Zorn’s Lemma) determined up to field isomorphism by their cardinality and this is false for countable algebraically closed fields of characteristic $p$. In general, an algebraically closed field is determined up to field isomorphism by its characteristic and its transcendence degree over its prime subfield ($\mathbf Q$ or $\mathbf F_p$). Its transcendence degree equals its cardinality for uncountable algebraically closed fields. | |
| May 10, 2021 at 5:04 | comment | added | YCor | @bof thanks, indeed, I wasn't aware of the nuance. So I mean atomless, i.e. without atoms. An atom in a Boolean algebra is a nonzero element $x$ such that for no nonzero $y,z$ we have $yz=0$ and $y+z=x$. | |
| May 10, 2021 at 4:19 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added the (counterexamples) tag
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| May 10, 2021 at 4:08 | answer | added | Alessandro Codenotti | timeline score: 9 | |
| May 10, 2021 at 3:59 | history | became hot network question | |||
| May 10, 2021 at 2:15 | answer | added | Pace Nielsen | timeline score: 13 | |
| May 10, 2021 at 1:53 | comment | added | bof | @YCor I believe you're using "non-atomic Boolean algebra" to mean what I'd call an atomless Boolean algebra, but linguistically it should mean "a Boolean algebra which is not an atomic Boolean algebra", which could have some atoms. | |
| May 9, 2021 at 22:16 | answer | added | Jeremy Rickard | timeline score: 14 | |
| May 9, 2021 at 22:04 | comment | added | David Handelman | On the other hand, things can be nicer in the uncountable case: if $F$ is a characteristic zero field of cardinality equal to that of the reals, and $F$ is algebraically closed, then $F$ is isomorphic to the complexes. On the other hand, there are countably infinitely many isomorphism classes of countable characteristic zero algebraically closed fields (isomorphism depends only on transcendence degree). More generally, in characteristic zero, for any uncountable cardinal, there exists a unique algebraically closed field of that cardinality, up to isomorphism. | |
| May 9, 2021 at 21:52 | comment | added | Nik Weaver | @YemonChoi I don't know if they count as algebra but I'd be interested in that kind of example too. | |
| May 9, 2021 at 21:13 | comment | added | Christian Remling | @NikWeaver: You do have an ordered spectral representation when $H$ is separable, but that's probably just rephrasing your original comment. | |
| May 9, 2021 at 20:58 | answer | added | Benjamin Steinberg | timeline score: 4 | |
| May 9, 2021 at 20:56 | history | edited | Stefan Kohl♦ |
edited tags
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| May 9, 2021 at 20:56 | history | made wiki | Post Made Community Wiki by Stefan Kohl♦ | ||
| May 9, 2021 at 20:46 | comment | added | Arno Fehm | Speaking of $\omega$-categoricity, of course in algebra there is also the opposite phenomenon - that some things are nice in the uncountable case but go wrong in the countable case: vector spaces of the same cardinality over the same (countable) field being isomorphic, algebraically closed fields of the same cardinality and characteristic being isomorphic, etc. | |
| May 9, 2021 at 20:43 | comment | added | Yemon Choi | Would posets count as algebra, or is that getting more towards "abstract set theory"? | |
| May 9, 2021 at 20:35 | comment | added | YCor | A keyword for many examples might be "$\omega$-categorical". For instance, the property "any non-atomic Boolean algebra is free" is true for countable Boolean algebras but not in general (the countable case is a restatement [thru Stone duality] of the fact that any nonempty metrizable totally disconnected compact space is homeomorphic to the Cantor set). | |
| May 9, 2021 at 20:34 | answer | added | Benjamin Steinberg | timeline score: 12 | |
| May 9, 2021 at 20:27 | answer | added | Arno Fehm | timeline score: 8 | |
| May 9, 2021 at 20:23 | comment | added | Nik Weaver | @ChristianRemling sure, but I like the bundle version better because it's canonical --- there are lots of ways to split $H$ up as $\bigoplus L^2(\sigma(A), \rho_\alpha)$. | |
| May 9, 2021 at 20:20 | comment | added | Christian Remling | @NikWeaver: But that just seems a (perhaps unnecessarily) fancy version of the theorem on spectral representations ($A$ is unitarily equivalent to multiplication by the variable in $\bigoplus L^2(\sigma(A),\rho_{\alpha})$), which works on any Hilbert space. | |
| May 9, 2021 at 20:18 | comment | added | Nik Weaver | It fails in the nonseparable case because one doesn't have a good notion of "measurable Hilbert bundle". E.g. take $A$ to be multiplication by $x$ on $L^2[0,1]$ direct sum multiplication by $x$ on the nonseparable space $l^2[0,1]$. | |
| May 9, 2021 at 20:17 | comment | added | Nik Weaver | @AndréHenriques: Let $A$ be a bounded self-adjoint operator on a separable Hilbert space. Then there is a measurable bundle $X$ of Hilbert spaces over ${\rm sp}(A)$ and an isomorphism $H\cong L^2({\rm sp}(A), X)$ which takes $A$ to multiplication by $x$. | |
| May 9, 2021 at 20:14 | answer | added | André Henriques | timeline score: 5 | |
| May 9, 2021 at 20:09 | answer | added | YCor | timeline score: 24 | |
| May 9, 2021 at 20:08 | comment | added | André Henriques | @NikWeaver What's your "favorite version of the spectral theorem"? | |
| May 9, 2021 at 20:06 | comment | added | Benjamin Steinberg | That said my feeling is that in algebra I think finitely generated versus non finitely generated is really the bigger dividing line. | |
| May 9, 2021 at 20:04 | comment | added | Benjamin Steinberg | It's also true that countable dimensional Leavitt path algebras and more generally Hausdorff etale groupoid algebras over fields are prime iff primitive but this is false in the uncountable case. | |
| May 9, 2021 at 20:01 | comment | added | Benjamin Steinberg | If you a countable dimensional $\mathbb C$-algebra then the endomorphism algebra of a simple module is $\mathbb C$ but this is not true for uncountable dimensional ones | |
| May 9, 2021 at 19:47 | history | asked | Nik Weaver | CC BY-SA 4.0 |