TITLE: Recovering an abelian category from the Ext of its simple objects
QUESTION [14 upvotes]: Let $C$ be an abelian category, assume for simplicity that $C$ is enriched over $Vect_k$ (vector spaces over $k$) for some fixed field $k$.
Suppose also that $C$ is both Artinian and Noetherian, so that for any object $X$ there is a sequence of objects $0=X_0 \hookrightarrow \ldots \hookrightarrow X_n = X$ with $X_i/X_{i-1}$ simple. Finally suppose that $C$ has enough injective/projective objects so that $\operatorname{Ext}_C$ can be defined.
Given $C$, we build a new category $S$, enriched over graded $k$-vector spaces, in the following way:

The objects of $S$ are the simple objects of $C$
If $X,Y\in Ob(S)$ then $\operatorname {Hom}_S(X,Y) = \bigoplus_{n\geq0}\operatorname{Ext}^n_C(X,Y)$
Compositions of morphisms are defined using the natural maps $\operatorname{Ext}^n_C(X,Y)\otimes \operatorname{Ext}^m_C(Y,Z)\to\operatorname{Ext}^{n+m}_C(X,Z)$

My question is: Can we recover $C$ from $S$ (say up to equivalence)?
Assuming the answer is "yes", I guess that there is an analogue for when $C$ is only enriched over $Ab$, maybe if we redefine $S$ so that $\operatorname{Hom}_S(X,Y)=\operatorname{Hom}_{D(C)}(X,Y)$ or something

REPLY [18 votes]: Here's a counterexample that appears in nature.
Fix a prime $p$ and a field $k$ of characteristic $p$, and let
$G=C_{p^{n}}$ be a cyclic group of order $p^{n}$ (where $n\geq1$ if
$p$ is odd, and $n\geq2$ if $p=2$).
Then the category $\operatorname{mod}kG$ of finitely generated
$kG$-modules has only one simple module: the trivial module $k$.
As graded $k$-algebras
$$\operatorname{Ext}^{\ast}_{kG}(k,k)=H^{\ast}(G,k)\cong k[s,t]/(s^{2}),$$
with $s$ and $t$ in degree $1$ and $2$ respectively, independent of
$n$.
But $\operatorname{mod}kG$ determines $n$, as $p^n$ is the length of an
indecomposable projective module.
So, for example, $\operatorname{mod}\mathbb{F}_{2}C_{4}$ and
$\operatorname{mod}\mathbb{F}_{2}C_{8}$ are not equivalent, but have
the same $\operatorname{Ext}$-algebra.
Alternatively, this generalizes to looking at
$\operatorname{mod}k[x]/(x^{m})$ for any field $k$ and varying values
of $m\geq3$.
So a slightly smaller counterexample is
$\operatorname{mod}k[x]/(x^{3})$ and $\operatorname{mod}k[x]/(x^{4})$.
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