Lemma 109.68.1. There exists a differential graded algebra $(A, \text{d})$ and a compact object $E$ of $D(A, \text{d})$ such that $E$ cannot be represented by a finite and graded projective differential graded $A$-module.

## 109.68 An interesting compact object

Let $R$ be a ring. Let $(A, \text{d})$ be a differential graded $R$-algebra. If $A = R$, then we know that every compact object of $D(A, \text{d}) = D(R)$ is represented by a finite complex of finite projective modules. In other words, compact objects are perfect, see More on Algebra, Proposition 15.78.3. The analogue in the language of differential graded modules would be the question: “Is every compact object of $D(A, \text{d})$ represented by a differential graded $A$-module $P$ which is finite and graded projective?”

For general differential graded algebras, this is not true. Namely, let $k$ be a field of characteristic $2$ (so we don't have to worry about signs). Let $A = k[x, y]/(y^2)$ with

$x$ of degree $0$

$y$ of degree $-1$,

$\text{d}(x) = 0$, and

$\text{d}(y) = x^2 + x$.

Then $x : A \to A$ is a projector in $K(A, \text{d})$. Hence we see that

in $K(A, \text{d})$, see Differential Graded Algebra, Lemma 22.5.4 and Derived Categories, Lemma 13.4.14. It is clear that $A$ is a compact object of $D(A, \text{d})$. Then $\mathop{\mathrm{Ker}}(x)$ is a compact object of $D(A, \text{d})$ as follows from Derived Categories, Lemma 13.37.2.

Next, suppose that $M$ is a differential graded (right) $A$-module representing $\mathop{\mathrm{Ker}}(x)$ and suppose that $M$ is finite and projective as a graded $A$-module. Because every finite graded projective module over $k[x, y]/(y^2)$ is graded free, we see that $M$ is finite free as a graded $k[x, y]/(y^2)$-module (i.e., when we forget the differential). We set $N = M/M(x^2 + x)$. Consider the exact sequence

Since $x^2 + x$ is of degree $0$, in the center of $A$, and $\text{d}(x^2 + x) = 0$ we see that this is a short exact sequence of differential graded $A$-modules. Moreover, as $\text{d}(y) = x^2 + x$ we see that the differential on $N$ is linear. The maps

are isomorphisms as $H^*(M) = H^0(M) = k$ since $M \cong \mathop{\mathrm{Ker}}(x)$ in $D(A, \text{d})$. A computation of the boundary map shows that $H^*(N) = k[x, y]/(x, y^2)$ as a graded module; we omit the details. Since $N$ is a free $k[x, y]/(y^2, x^2 + x)$-module we have a resolution

compatible with differentials. Since $N$ is bounded and since $H^0(N) = k[x,y]/(x, y^2)$ it follows from Homology, Lemma 12.25.3 that $H^0(N/Ny) = k[x]/(x)$. But as $N/Ny$ is a finite complex of free $k[x]/(x^2 + x) = k \times k$-modules, we see that its cohomology has to have even dimension, a contradiction.

**Proof.**
See discussion above.
$\square$

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