How can you compute a set of extensions up to isomorphism from Ext?

Question

Given $A$ and $B$ in an abelian $\mathbf{k}$-linear category, let $\mathrm{E_{A,B}}$ denote the set of all isomorphism classes of objects that fit in the middle of a short exact sequence $B \hookrightarrow \_\_ \twoheadrightarrow A$. Can we explicitly describe the set $\mathrm{E_{A,B}}$ if we know $\mathrm{Ext}^1_\mathbf{k}(A,B)$? I'm hoping that if $\{X_1, \dotsc, X_n\}$ is a basis for $\mathrm{Ext}^1_\mathbf{k}(A,B)$ as a $\mathbf{k}$-vector space, then we can say $$ \mathrm{E_{A,B}} \;\;=\;\; \big\{[X_1], \dotsc, [X_n], [A\oplus B]\big\}\,, $$ but this seems too hopeful.

Answer 1

No. As Roland commented, there is a simple counterexample in the category of chain complexes of $k$-vector spaces: let $B$ be $k^n$ concentrated in degree $1$ and $A$ be $k^m$ concentrated in degree $0$. An extension of $A$ by $B$ is then just a chain complex of the form $0\to k^n\to k^m\to 0$ and so $\operatorname{Ext}^1(A,B)\cong \operatorname{Hom}(k^n,k^m)\cong k^{mn}$. Up to isomorphism, though, such a chain complex is determined by the rank of the map $k^n\to k^m$, and so there are $\min(m,n)+1$ isomorphism classes. Since $\min(m,n)+1$ is not determined by the product $mn$, the cardinality of $\mathrm{E_{A,B}}$ is not determined by $\operatorname{Ext}^1(A,B)$ up to isomorphism.

Here are some things you can say. The automorphism groups of $A$ and $B$ each act on $\operatorname{Ext}^1(A,B)$ as isomorphisms of the middle objects, so $\mathrm{E_{A,B}}$ is no larger than the quotient of $\operatorname{Ext}^1(A,B)$ by these actions. However, it may be even smaller, since there can be extensions whose middle objects are isomorphic but such an isomorphism cannot preserve the subobject $B$. For instance, in the category of $k[x]$-modules, consider $A=B=C=(k[x]/(x))^{\oplus \mathbb{N}}\oplus (k[x]/(x^2))^{\oplus \mathbb{N}}$. Then there are lots of short exact sequences $0\to B\to C\to A\to 0$ which have different images of $B\to C$ even up to automorphisms of $C$, since you can have different numbers of $k[x]/(x)$ summands that map into a $k[x]/(x^2)$ summand to form a nontrivial extension.

Note moreover that $\mathrm{E_{A,B}}$ can also be larger than your guess $\{[X_1], \dotsc, [X_n], [A\oplus B]\}$. For instance, in the category of $k[x,y]$-modules, let $A=B=k[x,y]/(x,y)$. Then $\operatorname{Ext}^1(A,B)\cong k^2$ is finite-dimensional, but if $k$ is infinite, then $\mathrm{E_{A,B}}$ is infinite. Indeed, an element $(a,b)\in k^2$ corresponds to the extension with $k$-basis $\{e_1,e_2\}$ in which $xe_1=ae_2$ and $ye_1=be_2$ (and $x$ and $y$ both annihilate $e_2$). For $(a,b)\neq (0,0)$, the annihilator of this module is the ideal generated by $bx-ay$. In particular, such modules can only be isomorphic when their $(a,b)$'s are scalar multiples of each other. So in this case, $\mathrm{E_{A,B}}$ is actually the projectivization of $\operatorname{Ext}^1(A,B)$ together with the trivial extension, which is larger than just a basis together with the trivial extension.

(Note that in general, $k^\times$ acts by automorphisms of $A$ and $B$ and this induces the scalar action of $k^\times$ on $\operatorname{Ext}^1(A,B)$, so $\mathrm{E_{A,B}}$ will always be no larger than the projectivization together with the trivial extension.)