Timeline for Can a module be an extension in two really different ways?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 9, 2022 at 5:41 | comment | added | Mariano Suárez-Álvarez | The paper by Striuli is at doi.org/10.1016/j.jalgebra.2004.11.002 | |
| Oct 30, 2022 at 18:00 | history | edited | Glorfindel | CC BY-SA 4.0 |
broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594
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| Oct 18, 2010 at 1:38 | vote | accept | Anton Geraschenko | ||
| Oct 18, 2010 at 1:28 | comment | added | Anton Geraschenko | Ah, I think I see the problem. You have answered my original question (I think my first comment really does work). Greg has answered my "alternative formulation". The problem is that the two formulations are not equivalent after all. I overlooked the fact that it's possible to have an automorphism of $E$ as an extension ... an automorphism which induces the identity maps on both $M$ and $N$. So you (and Miyata) are correct: Greg's extension does split. Greg is also correct: the isomorphism with the direct sum is not induced by isomorphisms of $M$ and $N$. | |
| Oct 18, 2010 at 0:30 | comment | added | Hailong Dao | Hmm, that does seem a little strange. Let me think... | |
| Oct 17, 2010 at 14:50 | comment | added | Anton Geraschenko | When he takes $C=[0\ 0]$, it's exactly giving $M\oplus N$, right? On the other hand, $C=[0\ 1]$ gives a non-split extension which is abstractly isomorphic. | |
| Oct 16, 2010 at 7:22 | comment | added | Hailong Dao | @Anton: to be precise, I don't think your first comment work. The fact that the middle modules of $\alpha, \beta$ are isomorphic does not imply that $\alpha -\beta$ is isomorphic to the trivial extension. If that is the case we can concludes that $\alpha = \beta$ already! | |
| Oct 16, 2010 at 7:06 | comment | added | Hailong Dao | @Anton: Miyata result only applies for $E=M\oplus N$. It does not prove it for all $E$ over Noetherian commutative ring. | |
| Oct 16, 2010 at 6:51 | comment | added | Anton Geraschenko | Wait a second. Greg Muller's example $R=\mathbb{C}[x]$ is commutative and noetherian, and the modules are finitely generated. Am I misunderstanding Miyata's result, or am I misunderstanding Greg's example? | |
| Oct 15, 2010 at 20:59 | history | edited | Hailong Dao | CC BY-SA 2.5 |
added 330 characters in body
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| Oct 15, 2010 at 18:03 | comment | added | Anton Geraschenko | I see ... my $E$ represents two different elements $E_1$ and $E_2$ of $Ext^1(M,N)$ which happen to be abstractly isomorphic. The abstract isomorphism induces an isomorphism between the trivial extension and the "Yoneda quotient" (difference in $Ext^1$) of $E_1$ and $E_2$. Applying Miyata's result, you get that the "Yoneda quotient" of $E_1$ and $E_2$ splits, and this splitting induces the isomorphism of exact sequences I was looking for. This is great! Thank you for this answer! | |
| Oct 15, 2010 at 17:09 | history | answered | Hailong Dao | CC BY-SA 2.5 |