Suppose $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is a short exact sequence of $R$-modules, where $R$ is a ring with unity.

If $A$ and $B$ are known, then $C$ is unique up to isomorphism, because by the definition of short exact sequences, $C \cong B/A,$ where $A$ is considered to be a submodule of $B$.

But what about the other cases?

Suppose $A$ and $C$ are known. What choices do we have for the module $B$? I know that $B$ is the direct sum of $A$ and $C$ if and only if the sequence at the top is a split exact sequence. But otherwise, is there anything we can say about the structure of $B$?

Or if $B$ and $C$ are known, what can we derive about the structure of $A$?

(I feel like there was a specific name for this type of problem which I once saw on Wikipedia, but have forgotten it, so it would be great if somebody could remind me of it!)

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    $\begingroup$ The case that $A$ and $C$ are known is referred to as an extension problem. $\endgroup$
    – Dan Rust
    Sep 26, 2017 at 0:31
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    $\begingroup$ What do you mean by ring? If a "ring" must include 1, then $A$ cannot be a ring (unless $C = 0$). In other words, the category of commutative rings with 1 is not an abelian category. (Kernels don't exist in general, since they usually do not have 1) $\endgroup$
    – oxeimon
    Sep 26, 2017 at 0:32
  • $\begingroup$ @DanRust right, yes, thank you very much! $\endgroup$ Sep 26, 2017 at 1:37
  • $\begingroup$ @oxeimon oops, I meant to say modules, not rings. I will edit the question accordingly. $\endgroup$ Sep 26, 2017 at 1:37

1 Answer 1


Certainly if $B$ and $C$ are known, then $A$ is just the kernel of $B\rightarrow C$.

If only $A$ and $C$ are known, then $B$ is an extension of $C$ by $A$, and cannot in general be recovered from $A$ and $C$ alone. For example, one can choose $A = C = \mathbb{Z}/2$, then $B$ could either be $\mathbb{Z}/2\times\mathbb{Z}/2$ or $\mathbb{Z}/4$. In this example you can take $R = \mathbb{Z}$.

The set of all extensions of $C$ by $A$ (up to isomorphism, see this wikipedia article on the Ext functor) forms an abelian group $Ext^1(A,C)$.

If we're working instead in the category of groups, and $A$ is abelian, then the set of possible choices of $B$ form the group $H^2(C,A)$, where $H^2$ denotes the second group cohomology. However, note that in $H^2(C,A)$, one must specify an action of $C$ on $A$. Given your exact sequence, this action is given by conjugating elements of $A$ by lifts of elements of $C$ to $B$. Since $A$ is abelian, you can easily prove that this action is well-defined (does not depend on the choice of lift).

In general, describing the groups $Ext^1(A,C)$ or $H^2(C,A)$ are difficult problems, though there are many cases where they are understood.

Note that you may have heard of (c)homology from a topological setting, and group cohomology, despite the (initially) strange definition, is very much related to cohomology in topology.

In any case, from what I can tell of your background, cohomology is probably going to seem a bit arcane to you (it certainly did to me). However, if you keep doing math, especially at the graduate level and beyond, it is going to be pretty difficult to avoid, so the earlier you start reading about it, the more familiar it will be when you finally need to learn it!

  • $\begingroup$ I did think they would be difficult problems in general, but it helps to know in which area of mathematics they belong to; thank you! You mentioned cohomology is related to homology, but is it usually the case that the former is studied after the latter, or does it not matter much? $\endgroup$ Sep 26, 2017 at 8:19
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    $\begingroup$ I suppose in topology, homology might seem more natural at first compared to cohomology, and since you can define cohomology as the "homology of the dual complex", it would make sense to study homology first. On the other hand, from what I can see, cohomology seems to be far more pervasive (e.g. in algebraic geometry). I wonder if this is because algebraic geometry places more emphasis on functions as opposed to points. $\endgroup$
    – oxeimon
    Sep 26, 2017 at 16:28

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