Timeline for Non-degenerate alternating bilinear form on a finite abelian group
Current License: CC BY-SA 2.5
9 events
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| Jul 24 at 0:56 | comment | added | David Roberts♦ | The continuation of user16796's comment (converted to a comment from an answer a long time ago, and slightly messed up in the process) is this: $${}$$ "...Davydov states $$ A \cong \langle a \rangle^{\perp}\big/\langle a \rangle \oplus \langle a \rangle \oplus \widehat{\langle a \rangle}. $$ Could anyone help me by understanding Davydovs proof or do you have an alternative source (proof) for me?" | |
| Jul 29, 2011 at 21:14 | comment | added | user16796 | Thank you so much! You saved my day! That was the last gap in my diploma thesis :) If I have an acknowlegdement in my thesis, you will be mentionned! | |
| Jul 29, 2011 at 21:11 | comment | added | Francesco Polizzi | No. This means that both $0→⟨a⟩→A→A/⟨a⟩→0$ and $0→⟨a⟩^{\perp}→A→\widehat{⟨a⟩}$ split. Now it is easy to see (for instance by using the Snake Lemma) that $A/⟨a⟩=⟨a⟩^{\perp}/⟨a⟩⊕ \widehat{⟨a⟩}$, so the claim follows. | |
| Jul 28, 2011 at 19:00 | comment | added | user16796 | I'm very glad about this topic, because I have a similar problem and Lemma 5.2 by Davydov would be the solution. But I have a problem with the proof. Davydov says that the inclusion $\langle a \rangle \to A$ and the surjection $A \to \widehat{\langle a \rangle}$ split. Does that mean that the short exact sequence $$ 0 \to \langle a \rangle \to A \to \widehat{\langle a \rangle} \to 0 $$ splits? But then the Splitting Lemma provides an isomorphism $A \cong \langle a \rangle \oplus \widehat{\langle a \rangle}$, but Davydov states that (cont.) | |
| Mar 21, 2011 at 10:40 | vote | accept | Giuseppe | ||
| Mar 18, 2011 at 11:58 | history | edited | Francesco Polizzi |
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| Mar 18, 2011 at 11:47 | answer | added | Francesco Polizzi | timeline score: 27 | |
| Mar 18, 2011 at 11:26 | history | edited | Qfwfq | CC BY-SA 2.5 |
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| Mar 18, 2011 at 11:10 | history | asked | Giuseppe | CC BY-SA 2.5 |