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I am reading an article on elementary homological algebra and have a trouble understanding one statement. Let $R$ be a ring and $A,B,C$ modules over $R$. Let $S$ be a set of exact sequences of the form $ 0\rightarrow A\rightarrow B\rightarrow C \rightarrow 0 $ The article says

$\operatorname{Aut}(B)$ acts on $S$ with stabilizer $1+\alpha \operatorname{Hom}(C,A)\beta$ where $\alpha,\beta$ are the maps fitting in the short exact sequence of the trivial extension $ 0 \rightarrow A\stackrel{\alpha}{\rightarrow} A\oplus C \stackrel{\beta}{\rightarrow} C \rightarrow 0 $

Firstly I don't quite understand what $1+\alpha \operatorname{Hom}(C,A)\beta$ means (what is $1+\dots$?) and secondly don't see why the stabilizer of $\operatorname{Aut}(B)$ is identified with the above set.

Could anyone kindly explain what is going on? Thank you very much.

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    Sorry about that. I initially tried to give better title, but could not make it short and precise.2012-11-26

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Well, if $\gamma\in\hom(C,A)$, then $\alpha\gamma\beta:A\oplus C\to A\oplus C$: $(a,c)\mapsto (\gamma(c),0)$.

I think, in the highlighted part $B=A\oplus C$, then it makes sense, and $1$ means its identity, so that all $1+\alpha\gamma\beta$ will be an automorphism that fixes the exact sequence $(\alpha,\beta)$.

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    It seems like a simple example with the explicit role that $B=A\oplus C$. It also well might be a typo only. For me, this is the only context that makes it meaningful.2012-11-26