Timeline for Direct product decomposition for infinite abelian groups with constrained torsion
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jun 29, 2013 at 17:53 | vote | accept | Pete L. Clark | ||
| Jun 29, 2013 at 15:30 | answer | added | YCor | timeline score: 3 | |
| Jun 29, 2013 at 5:34 | comment | added | Kevin Ventullo | Ah, I see my mistake. I think I should have said "of bounded $\ell$-exponent for any $\ell$". | |
| Jun 29, 2013 at 4:53 | history | edited | Pete L. Clark | CC BY-SA 3.0 |
added 258 characters in body
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| Jun 29, 2013 at 4:29 | comment | added | Pete L. Clark | @Kevin: Here, by "Mordell-Weil group" I mean any group of the form $A(K)$ for an abelian variety over any field $K$. (And it is not enough to assume that the torsion subgroup is finite. In fact that case is significantly easier than the general case, because the torsion subgroup is necessarily a direct factor.) | |
| Jun 29, 2013 at 4:27 | comment | added | Pete L. Clark | @Qiaochu: indeed the torsion subgroup need not be a direct summand. I asked an MO question about this a while back. The easiest example is probably $\prod_{\ell} \mathbb{Z}/\ell \mathbb{Z}$. The precise result is a theorem of Baer: for a torsion commutative group $T$, every (commutative!) extension of a torsionfree group by $T$ is split iff $T$ is the direct sum of a divisible $T_1$ and a group $T_2$ of bounded exponent (i.e., $T_2 = T_2[n]$ for some $n$). | |
| Jun 29, 2013 at 4:03 | comment | added | Kevin Ventullo | I'm a bit confused about the motivation; in the Mordell-Weil situation the group is finitely generated, right? Or is the idea that you want to show any such group can arise as the $K$-points of an ab. var. for some possibly infinite extension $K$? In that case, you may assume $G[tors]$ is finite, basically since $\mathbb{Q}_p/\mathbb{Z}_p$ is divisible. | |
| Jun 29, 2013 at 4:02 | comment | added | Kevin Ventullo | @QiaochuYuan That's impossible. $(\mathbb{Q}/\mathbb{Z})^3$ is divisible and hence injective as $\mathbb{Z}$-module. | |
| Jun 29, 2013 at 4:01 | comment | added | Qiaochu Yuan | I've been told the torsion subgroup of an abelian group need not be a direct summand, but I don't think I actually know any examples. It seems like a reasonable strategy here to construct a counterexample would be to find a group with torsion subgroup, say, $(\mathbb{Q}/\mathbb{Z})^3$ which isn't a direct summand. | |
| Jun 29, 2013 at 3:43 | history | edited | Pete L. Clark | CC BY-SA 3.0 |
edited title
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| Jun 28, 2013 at 22:25 | history | asked | Pete L. Clark | CC BY-SA 3.0 |