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The question is natural. Let's just work in the category of modules over a ring. Pick three modules $M_1, M_2, M_3$. Consider consecutive extensions of these modules, i.e., consider M, such that we have short exact sequences $0 \to M_1 \to M \to M' \to 0$, and $0 \to M_2 \to M' \to M_3 \to 0$. This is a naturally defined functor right? And do all these extensions form an abelian group? (as for the extension of two modules). And what are other properties for these extensions?

I suppose this should be some standard materials in homological algebra, but I could not find any references.

Thanks!

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    $\begingroup$ What exactly are you claiming is functorial in what? $\endgroup$ Dec 28, 2014 at 16:25
  • $\begingroup$ I was guessing, functorial with respect to each M_i?? $\endgroup$
    – tqvb
    Dec 28, 2014 at 16:34
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    $\begingroup$ Just ask yourself whether you want it to be covariantly or contravariantly functorial in the middle term $M_2$. $\endgroup$ Dec 28, 2014 at 17:00
  • $\begingroup$ OK...I guess my main concern is that if the space of all extensions is an abelian group.. $\endgroup$
    – tqvb
    Dec 28, 2014 at 17:01

1 Answer 1

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If $M$ is a consecutive extension then $M/M_1$ ia an extension of $M_3$ by $M_2$ and $Ker(M \to M_3)$ is an extension of $M_2$ by $M_1$. This defines a map from the space of consecutive extensions into $Ext^1(M_3,M_2) \times Ext^1(M_2,M_1)$. In fact this is an embedding and the image is the subset of pairs $(e_1,e_2)$ which compose to zero in $Ext^2(M_3,M_1)$. I am not sure this has a natural structure of an abelian group.

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  • $\begingroup$ Thanks for the first hint. Do you have reference for your result (embedding, and the image)? $\endgroup$
    – tqvb
    Dec 28, 2014 at 16:35
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    $\begingroup$ This is a simple exercise. $\endgroup$
    – Sasha
    Dec 28, 2014 at 16:56
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    $\begingroup$ I upvoted this answer, repeating the mistake contained in it; now my vote is "locked", as the time has passed. If one understands "consecutive extensions" as three-term filtrations with the successive quotients $M_i$, then, in fact, the map from the space of consecutive extensions into $Ext^1(M_3,M_2)\times Ext^1(M_2,M_1)$ is not an embedding. E.g., when $M_2=0$, the space of consecutive extensions is identified with $Ext^1(M_3,M_1)$. $\endgroup$ Jan 2, 2015 at 14:09

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