Let $A,B$ be modules. An extension of $A$ by $B$ is a short exact sequence $$B\to E\to A$$. Let $Ext(A,B)$ be a set of equivalence classes of such modules $E$. The book said that this admit an abelian group structure. I would like to confirm is it true that the zero object is the split short exact sequence $B\to A\oplus B\to A$? I wanna prove this, but I am stuck at making an isomorphism between $E$ and $E\oplus A\oplus B$. And I have no clue about the inverse element of $B\to E\to A$: that is $B\to X\to A$ such that $E\oplus X$ is isomorphic to $A\oplus B$. Any help is appreciated.

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    $\begingroup$ You can't get an isomorphism between $E$ and $E\oplus A\oplus B$ since these will rarely be isomorphic. You need to look at what the actual sum is on the extensions (it is not just direct sum). $\endgroup$ – Tobias Kildetoft Oct 3 '16 at 9:58
  • $\begingroup$ My thesis contains just this and all the proofs :-) $\endgroup$ – Zelos Malum Oct 3 '16 at 10:13
  • $\begingroup$ @TobiasKildetoft, yes. I think that does not work $\endgroup$ – Chen M Ling Oct 4 '16 at 5:12

For two extensions $$A\to E \to B$$ and $$A\to F \to B$$ we define the sum of them to be be the quotient of the pullback $X$ on the morphisms $$\pi_e: E\to B$$ $$\pi_f: F\to B$$ with the module $$N=\{\imath_e(a)\oplus-\imath_f(a):a\in A\}$$

Then the baer sum is $X/N$. Let $A\oplus B$ be given and $E$ an extension, then we have. $$X=\{(a\oplus b)\oplus e:\pi(a\oplus b)=\pi_e(e)\}$$ $$=\{a\oplus b\oplus e:b=\pi_e(e)\}$$ $$=\{(a\oplus b)\oplus e:b=\pi_e(e)\}$$ $$\cong\{a \oplus e\}$$ $$=A\oplus E$$ and for the submodule we have through a similar reasoning that $N\cong A\cong A\oplus 0$ and taking their quotient you get the rest. I recommend you look into MacLanes book on it.

  • $\begingroup$ I do not get what is wrong with my definition and why it does not work. I think something should have been missing. But the way you define the zero element works. Thanks. $\endgroup$ – Chen M Ling Oct 4 '16 at 5:11
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    $\begingroup$ The inverse is just the same chain but the negative epimorphism. What is wrong is that you cannot establish them being equivalent through commutative diagrams, like this we can. $\endgroup$ – Zelos Malum Oct 4 '16 at 5:32
  • $\begingroup$ I see. Thanks for your explanation :) $\endgroup$ – Chen M Ling Oct 5 '16 at 8:23
  • $\begingroup$ @ChenMLing welcome, it all really boils down to generalizations and commutative diagrams, as i mentioned i did an entire thesis on this :-) $\endgroup$ – Zelos Malum Oct 5 '16 at 8:35

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