Lemma 57.5.1. With $k$, $n$, and $R$ as above, for an object $K$ of $D(R)$ the following are equivalent

1. $\sum _{i \in \mathbf{Z}} \dim _ k H^ i(K) < \infty$, and

2. $K$ is a compact object.

Proof. If $K$ is a compact object, then $K$ can be represented by a complex $M^\bullet$ which is finite projective as a graded $R$-module, see Differential Graded Algebra, Lemma 22.36.6. Since $\dim _ k R < \infty$ we conclude $\sum \dim _ k M^ i < \infty$ and a fortiori $\sum \dim _ k H^ i(M^\bullet ) < \infty$. (One can also easily deduce this implication from the easier Differential Graded Algebra, Proposition 22.36.4.)

Assume $K$ satisfies (1). Consider the distinguished triangle of trunctions $\tau _{\leq m}K \to K \to \tau _{\geq m + 1}K$, see Derived Categories, Remark 13.12.4. It is clear that both $\tau _{\leq m}K$ and $\tau _{\geq m + 1} K$ satisfy (1). If we can show both are compact, then so is $K$, see Derived Categories, Lemma 13.37.2. Hence, arguing on the number of nonzero cohomology modules of $K$ we may assume $H^ i(K)$ is nonzero only for one $i$. Shifting, we may assume $K$ is given by the complex consisting of a single finite dimensional $R$-module $M$ sitting in degree $0$.

Since $\dim _ k(M) < \infty$ we see that $M$ is Artinian as an $R$-module. Thus it suffices to show that every simple $R$-module represents a compact object of $D(R)$. Observe that

$I = \left( \begin{matrix} 0 & S_1 & S_2 & \ldots & \ldots \\ 0 & 0 & S_1 & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & \ldots & \ldots & \ldots & 0 \end{matrix} \right)$

is a nilpotent two sided ideal of $R$ and that $R/I$ is a commutative $k$-algebra isomorphic to a product of $n + 1$ copies of $k$ (placed along the diagonal in the matrix, i.e., $R/I$ can be lifted to a $k$-subalgebra of $R$). It follows that $R$ has exactly $n + 1$ isomorphism classes of simple modules $M_0, \ldots , M_ n$ (sitting along the diagonal). Consider the right $R$-module $P_ i$ of row vectors

$P_ i = \left( \begin{matrix} 0 & \ldots & 0 & S_0 & \ldots & S_{i - 1} & S_ i \end{matrix} \right)$

with obvious multiplication $P_ i \times R \to P_ i$. Then we see that $R \cong P_0 \oplus \ldots \oplus P_ n$ as a right $R$-module. Since clearly $R$ is a compact object of $D(R)$, we conclude each $P_ i$ is a compact object of $D(R)$. (We of course also conclude each $P_ i$ is projective as an $R$-module, but this isn't what we have to show in this proof.) Clearly, $P_0 = M_0$ is the first of our simple $R$-modules. For $P_1$ we have a short exact sequence

$0 \to P_0^{\oplus n + 1} \to P_1 \to M_1 \to 0$

which proves that $M_1$ fits into a distinguished triangle whose other members are compact objects and hence $M_1$ is a compact object of $D(R)$. More generally, there exists a short exact sequence

$0 \to C_ i \to P_ i \to M_ i \to 0$

where $C_ i$ is a finite dimensional $R$-module whose simple constituents are isomorphic to $M_ j$ for $j < i$. By induction, we first conclude that $C_ i$ determines a compact object of $D(R)$ whereupon we conclude that $M_ i$ does too as desired. $\square$

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