MathOverflow Asked by TTip on November 28, 2020

I’m not sure this question is research level question. Sorry in advance.

**Hypothesis**

- $k$ is a commutative ring.
- $A$ is an augmented $k$-algebra.
- $A^e$ is defined as the $k$-algebra $Aotimes_{k}A^{op}$. It is naturally augmented $k$-algebra.

**assumptions**

- $k$ (as left $A$-module) is quasi-isomorphic to a perfect complex. $kin mathbf{Perf}(A)$.
- $k$ (as left $A^e$-module) is quasi-isomorphic to a perfect complex. $kin mathbf{Perf}(A^e)$.
- $A$ (viewed as left $A^e$-module in a standard way) is quasi-isomorphic to a perfect complex. $Ain mathbf{Perf}(A^e)$.

**Question**

Let $langle Arangle$ be the thick subcategory of the category of perfect complexes $mathbf{Perf}(A^e)$ generated by the left $A^e$-module $A$ (where $A$ is viewed as $A^e$-module in standard way). Is it clear that $k$ (viewed as $A^e$-module via the augmentation $A^erightarrow k$ ) is an object of $langle Arangle$ ?

No.

Let $k$ be a field, and let $A$ be the algebra of upper triangular $2times 2$ matrices over $k$, with augmentation map $pmatrix{a&b\0&c}mapsto a$.

$A$ and $A^e$ have finite global dimension, so all modules have finite projective dimension, and are therefore quasi-isomorphic to perfect complexes.

Let $mathcal{D}=mathcal{D}(A^e)$ be the derived category of $A^e$-modules. The category $$k^perp={Xinmathcal{D}mid operatorname{Hom}_{mathcal{D}}(k,X[t])=0text{ for all $tinmathbb{Z}$}}$$ is a thick subcategory of $mathcal{D}$. I claim (proof below) that $Ain k^{perp}$. So $langle Aranglesubseteq k^{perp}$ for all $tinmathbb{Z}$. But clearly $knotin k^{perp}$.

*Proof of claim:* Let $e_{11}=pmatrix{1&0\0&0}$ and $e_{22}=pmatrix{0&0\0&1}$, idempotent elements of $A$. Then $k=Ae_{11}$ is projective as a left $A$-module, and as a right $A$-module $k$ has a projective resolution
$$0to e_{22}Ato e_{11}Ato kto0,$$
where the first nonzero map is $pmatrix{0&0\0&c}mapstopmatrix{0&c\0&0}$.
So as an $A^e$-module, $k$ has a projective resolution
$$0to Ae_{11}otimes_k e_{22}Ato Ae_{11}otimes_k e_{11}Ato kto0.$$
Applying the functor $operatorname{Hom}_{A^e}(-,A)$ to the projective terms of this resolution gives the map
$e_{11}Ae_{11}to e_{11}Ae_{22}$ where $pmatrix{a&0\0&0}mapstopmatrix{0&a\0&0}$. The kernel and cokernel of this map are both zero, so $operatorname{Ext}^t(k,A)=0$ for all $tgeq0$.

Correct answer by Jeremy Rickard on November 28, 2020

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