# Smallness condition for augmented algebras

MathOverflow Asked by TTip on November 28, 2020

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 $$Aotimes_{k}A^{op}$$. It is naturally augmented $$k$$-algebra.

assumptions

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