The simple module over dual numbers is not perfect

For $a+b\epsilon \in A$ one has $\epsilon (a+b\epsilon )=a\epsilon$, so multiplication by $\epsilon$ has $\ker (\epsilon \cdot )=(\epsilon )=\mathrm{im}(\epsilon \cdot )$: the kernel consists exactly of the multiples of $\epsilon$, and it equals the image. The quotient augmentation $\pi :A\to S$ is surjective with kernel $(\epsilon )$, equal to the image of the differential into the degree-zero copy of $A$. Thus the sequence is exact at every copy of $A$ and at $S$, and is a free resolution $Q_{\bullet }\to S$ with every term $A$ finitely generated free.

Since $A$ is commutative, the resolution of step 1.1 supplies both a left and a right projective resolution of $S$. Applying $S{\otimes }_A(-)$ to its unaugmented complex $Q_{\bullet }$ gives a complex with $S{\otimes }_{A}A\cong S$ in every nonnegative degree and induced differentials equal to multiplication by $\epsilon$ on $S$, which is zero because $\epsilon S=0$; hence its homology is $S\cong k$ in every degree $i\ge 0$. By [F4] the specified-resolution Tor is ${\mathrm{Tor}}_i^{A,Q}(S,S)\cong k$ for every $i\ge 0$; under the DC supplied by AC [F3], the balanced bifunctor [F5] identifies this with ${\mathrm{Tor}}_i^A(S,S)\cong k$, and [F7] then gives $H^{-n}(S[0]{\otimes }_A^{\mathbf{L}}S[0])\cong {\mathrm{Tor}}_n^A(S,S)\cong k$ for every $n\ge 0$, in particular $H^{-n}\mathrm{\ne }0$ for all $n\ge 0$.

Suppose $S[0]$ were perfect; then [F1] supplies a bounded cochain complex $P$ of finitely generated projective left $A$-modules together with an isomorphism $P\cong S[0]$ in $D(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$. Both $P$ and $S[0]$ are bounded above, so [F9] lifts this isomorphism to $D^{-}(A\text{-}\mathrm{M}\mathrm{o}\mathrm{d})$. Since the bounded-above derived tensor is a bifunctor in its second variable [F6], the lifted isomorphism gives $S[0]{\otimes }_A^{\mathbf{L}}S[0]\cong S[0]{\otimes }_A^{\mathbf{L}}P$. The identity $P\stackrel{\mathrm{i}\mathrm{d}}{\to }P$ is a quasi-isomorphism from a bounded-above complex of projective modules, so it is a supplied projective replacement as required by [F6]. Thus $S[0]{\otimes }_A^{\mathbf{L}}P$ is represented by $\mathrm{Tot}(S[0]{\otimes }_{A}P)$, which by [F8] is the bounded complex $S{\otimes }_{A}P$: its differential is $1\otimes d_P$ since the first factor is in degree zero, and it vanishes outside the finite support of $P$. Therefore $H^{-n}(S[0]{\otimes }_A^{\mathbf{L}}S[0])\cong H^{-n}(S{\otimes }_{A}P)=0$ for all sufficiently large $n$, contradicting step 2.1, which gives the nonzero $k$ in every degree $n\ge 0$. Hence $S[0]$ is not perfect, and since it is a complex concentrated in degree $0$ with $H^0(S[0])=S\cong k$ finite dimensional over $k$ and all other cohomology zero, this failure of perfectness is not detected by boundedness or by finite-dimensional cohomology.