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 $\langle A\rangle$ ?



Smallness condition for augmented algebras
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 $t\geq0$.


