Timeline for Homomorphisms from powers of Z to Z
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| S Apr 2, 2019 at 17:31 | history | suggested | user26857 | CC BY-SA 4.0 |
impoved format
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| Apr 2, 2019 at 17:08 | review | Suggested edits | |||
| S Apr 2, 2019 at 17:31 | |||||
| May 29, 2013 at 0:38 | history | edited | Ralph | CC BY-SA 3.0 |
Incorporated Andreas Blass' comments
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| May 28, 2013 at 21:11 | comment | added | Todd Trimble | @Ralph: Right, this is what I thought was going on (I say this even though I'm not sure what is meant by the "free complete product"). So this means a clarification is in order for the other question; I'll refer to your answer here. What is somewhat annoying is that the notation used in the Shelah-Strüngmann paper, referring to free complete products, looks identical to cartesian product notation, and this looks like a likely source of confusion. Thanks for your attention to this matter! | |
| May 28, 2013 at 21:03 | comment | added | Ralph | ... direct product. This may also be the reason why the title of their paper includes the words "non-commutative". In particular, they refer to the Eklof-Mekler book for generalizations of the Specker phenomenon on uncountable direct products (p. 420, before Def. 2.1). To summarize: 1. The isomorphism in my 1st comment is correct (reference: Eklof-Mekler, Cor. III.3.7). 2. $\phi$ is an isomorphism if $I$ is not $\omega$-measurable (references in my answer). | |
| May 28, 2013 at 20:34 | comment | added | Ralph | Todd, if it seemed that I didn't take your comment/question seriously, I apologize. I just had a look into the Shelah-Strüngmann paper (it's freely available on degruyter.com/view/j/jgt.2013.16.issue-3/issue-files/…). Unfortunately Mariano didn't give a precise reference within the paper where it is shown that $\phi$ (the map from my answer) fails to be an isomorphism for uncountable cardinals. But it seems to me that the point is that Shelah-Strüngmann consider homomorphisms from the free complete product of groups into the integers while I take the ... | |
| May 28, 2013 at 19:39 | vote | accept | Michael Barr | ||
| May 28, 2013 at 19:35 | comment | added | Todd Trimble | @Ralph: perhaps I'm being thick, but I'm having a hard time seeing how you're addressing my query. Martin's question (the page I linked to) asked about general uncountable $I$. For the moment, let's say that $I$ is less than the first measurable cardinal (where Andreas's remark would not apply). Then you seem be be asserting that the map $\phi$ is an isomorphism. Whereas Mariano gave the opposite answer. Am I missing something? | |
| May 28, 2013 at 19:09 | comment | added | Ralph | @Todd: As explained by Andreas Blass, not all homs factor when I isn't measurable (I'll correct my remark above later). I guess the Shelah-Strüngmann paper in the question you linked, is based on a non-measurable index set. However, it should be pointed out that in the measurable as well as in the non-measurable case $Hom(\prod_I \mathbb{Z},\mathbb{Z})\cong \bigoplus_D Hom(\mathbb{Z},\mathbb{Z})$ is a free abelian group with basis a set $D$ of ultrafilters on I. | |
| May 28, 2013 at 15:51 | comment | added | Todd Trimble | I'm having some trouble reconciling this answer with the answer given here: mathoverflow.net/questions/12586/dual-of-zi-for-uncountable-i/… (This is not to say that I disbelieve the present answer. I'm guessing that something might have gotten lost in the translation with the other answer.) | |
| May 28, 2013 at 13:06 | comment | added | Andreas Blass | The Eklof-Mekler book is also the first reference that I'd suggest. I believe, though, that the theorem the OP asked about is entirely due to Łoś. Eda's contribution concerned what happens when $I$ is greater than or equal to the first measurable cardinal. For such $I$ it's clear that non-principal countably complete ultrafilters on $I$ give rise to non-trivial (i.e., not factoring through projections to finite products) homomorphisms, but it takes some work to show that these together with the trivial homomorphisms actually generate all the homomorphisms. | |
| May 28, 2013 at 11:38 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
fix name
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| May 28, 2013 at 7:23 | history | edited | Ralph | CC BY-SA 3.0 |
Added the non-measurable case.
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| May 28, 2013 at 7:09 | history | answered | Ralph | CC BY-SA 3.0 |