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

Hypothesis

  1. $k$ is a commutative ring.
  2. $A$ is an augmented $k$-algebra.
  3. $A^e$ is defined as the $k$-algebra $A\otimes_{k}A^{op}$. It is naturally augmented $k$-algebra.

assumptions

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

Question Let $\langle A\rangle$ 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 $A^e\rightarrow k$ ) is an object of $\langle A\rangle$ ?

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1 Answer 1

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No.

Let $k$ be a field, and let $A$ be the algebra of upper triangular $2\times 2$ 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 $$k^\perp=\{X\in\mathcal{D}\mid \operatorname{Hom}_{\mathcal{D}}(k,X[t])=0\text{ for all $t\in\mathbb{Z}$}\}$$ is a thick subcategory of $\mathcal{D}$. I claim (proof below) that $A\in k^{\perp}$. So $\langle A\rangle\subseteq k^{\perp}$ for all $t\in\mathbb{Z}$. But clearly $k\not\in k^{\perp}$.

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 $$0\to e_{22}A\to e_{11}A\to k\to0,$$ where the first nonzero map is $\pmatrix{0&0\\0&c}\mapsto\pmatrix{0&c\\0&0}$. So as an $A^e$-module, $k$ has a projective resolution $$0\to Ae_{11}\otimes_k e_{22}A\to Ae_{11}\otimes_k e_{11}A\to k\to0.$$ 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 $\pmatrix{a&0\\0&0}\mapsto\pmatrix{0&a\\0&0}$. The kernel and cokernel of this map are both zero, so $\operatorname{Ext}^t(k,A)=0$ for all $t\geq0$.

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