Timeline for A conceptual proof that bounded index subgroups of a bounded torsion abelian group contain bounded index complemented subgroups
Current License: CC BY-SA 4.0
10 events
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| Dec 6, 2023 at 1:23 | comment | added | Denis T | @TerryTao Obviously, the fact that every irreducible module is cyclic (aka classification of finite abelian groups) is NECESSARY to prove the bound you asked for; I've provided a counterexample that explains necessity of this property. I guess, you can try to hide using this fact in a proof, but it will still be there. // Essentiall hull is uniquely defined (as a submodule) in uniserial case; in general it's not even unique up to abstract isomorphism. I'm not claiming that my post contains all details — usually it takes ~40 pages in a book to state and prove properties of Nakayama algebras. | |
| Dec 3, 2023 at 22:17 | history | edited | Denis T | CC BY-SA 4.0 |
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| Dec 3, 2023 at 21:27 | comment | added | Terry Tao | From what I can read up on about injective hulls, it seems that in order to embed the essential submodule of $N$ into $M$, it is necessary that $M$ is itself an injective $R$-module? This is basically the point in my proof where the classification of finite abelian groups came in. | |
| Dec 3, 2023 at 21:09 | comment | added | Terry Tao | I'm sorry, I cannot follow the construction of the essential submodule and will probably need an explicit example. Let's take the one from my previous comment: $R = {\bf Z}/p^2$, $M = ({\bf Z}/p^2)^2$, and $N = \langle (p,0) \rangle$ is the size one submodule of $M$ generated by $(p,0)$. What is $E_M(N)$, viewed as a submodule of $M$? Is it only unique up to isomorphism? As I said my previous remark, both $\langle (1,0)\rangle$ and $\langle (1,p)\rangle$ seem to be candidates for this envelope. | |
| Dec 3, 2023 at 19:50 | comment | added | Denis T | @TerryTao I hope that after complete rewriting my answer makes more sense. | |
| Dec 3, 2023 at 19:49 | history | edited | Denis T | CC BY-SA 4.0 |
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| Dec 3, 2023 at 19:27 | history | edited | Denis T | CC BY-SA 4.0 |
Finally a readable full answer.
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| Dec 3, 2023 at 18:56 | comment | added | Terry Tao | Can you explain more about how every group gets a unique radical? The non-uniqueness of roots was the biggest difficulty I had when trying to resolve this problem. For instance, in $({\bf Z}/p^2{\bf Z})^2$, it seems to me that the order $p$ subgroup $\langle (p,0) \rangle$ has multiple candidates for such a "radical"; there is $\langle (1,0) \rangle$, of course, but also $\langle (1,p) \rangle$ for instance. Also, do you mean $p^{dk}$ instead of $dp^k$? | |
| Dec 3, 2023 at 17:25 | history | edited | Denis T | CC BY-SA 4.0 |
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| Dec 3, 2023 at 17:17 | history | answered | Denis T | CC BY-SA 4.0 |