# Smallness condition for augmented algebras

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 $$\langle A\rangle$$ ?

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 $$t\geq0$$.