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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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Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Julian Kuelshammer Nov 25 '12 at 23:15
Sorry about that. I initially tried to give better title, but could not make it short and precise. – Pooya Nov 26 '12 at 4:05

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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Thank you for the kind answer. I don't think our exact sequence splits in general and so am still in trouble understanding the action of $\mathrm{Aut}(B)$ in the context. Anyways, thank you for your assistance. I will think about this more. – Pooya Nov 26 '12 at 4:09
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. – Berci Nov 26 '12 at 11:43

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