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  • $\begingroup$ 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.) $\endgroup$ Commented Jul 28, 2011 at 19:00
  • $\begingroup$ 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. $\endgroup$ Commented Jul 29, 2011 at 21:11
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    $\begingroup$ 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! $\endgroup$ Commented Jul 29, 2011 at 21:14
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    $\begingroup$ 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?" $\endgroup$ Commented Jul 24, 2024 at 0:56