The diagonal bimodule $k[x]$ is projective on both sides but not over its enveloping algebra

The regular left $A$-module is $A\{0\}$, generated by its degree-zero unit; [L10] makes it finite graded projective. Thus the sole nonzero term of $F$ meets the left projectivity hypothesis.

By [L2, L6] with one and two indeterminates, $A=k[x]$ and $R:=k[x,y]$ are domains, hence commutative rings by [L3]. Define the enveloping action by $(a\otimes b^{\mathrm{o}\mathrm{p}})\cdot m=amb$. The map $R\to A{\otimes }_k{A}^{\mathrm{o}\mathrm{p}}$ sending $x^i{y}^j$ to $x^i\otimes (x^j{)}^{\mathrm{o}\mathrm{p}}$ is multiplicative because $A$ is commutative; its inverse sends $f(x)\otimes g(x{)}^{\mathrm{o}\mathrm{p}}$ to $f(x)g(y)$. This inverse is balanced over $k$, and the maps are inverse on the monomials and elementary tensors that span their modules by [L4, L5, L7, L8]. Thus $A^e$ is isomorphic to the commutative ring $R$.

The underlying right regular module is projective: given a surjection $q:E\twoheadrightarrow M$ of right $A$-modules and a map $g:A\to M$, choose $e\in E$ with $q(e)=g(1)$ and define $\tilde{g}(a)=ea$; then $q\tilde{g}=g$. Since the complex $F$ is concentrated in degree zero, $A{\otimes }_{A}X\to X$, $a\otimes z\mapsto az$, is a natural chain isomorphism for every bounded left $A$-complex $X$, with inverse $z\mapsto 1\otimes z$. Under the standing size convention in [L11], its derived tensor functor is represented by this ordinary tensor operation as well.

In these coordinates the action on $A$ is induced by the surjective ring map $\mu :R\to A$, $f(x,y)\mapsto f(x,x)$; it is surjective since every $g(x)\in A$ is $\mu (g(x))$. For a monomial $x^i{y}^j$ with $j\ge 1$, $x^i{y}^j-x^{i+j}=x^i(y-x){\sum }_{r=0}^{j-1}{y}^{j-1-r}{x}^r$, while the difference is zero for $j=0$; by finite support [L4], $f-f(x,x)\in (x-y)$ for every $f$. Hence $\ker \mu =(x-y)$. This ideal is nonzero because the distinct monomials $x$ and $y$ have nonzero coefficients, and proper because $\mu (1)=1\mathrm{\ne }0$.

Suppose for contradiction that $A$ is projective as a left $R$-module. Since $\mu$ is a surjection, [L9] lifts ${\mathrm{id}}_A$ to an $R$-linear map $s:A\to R$ with $\mu s={\mathrm{id}}_A$. Then $p={\mathrm{id}}_R-s\mu$ is an $R$-linear projection onto $I:=\ker \mu$: $p(R)\subseteq I$ and $p(i)=i$ for $i\in I$. Every $R$-linear map $p:R\to R$ is multiplication by $e:=p(1)$, so $I=Re$; also $p^2=p$ gives $e^2=e$.

By [L2, L3, and L6], $R$ is a commutative domain. Thus $e^2=e$ implies $e(e-1)=0$, so $e=0$ or $e=1$. Then $I=Re$ is either zero or all of $R$, contradicting Step 2.2, where $I$ was shown nonzero and proper. Therefore $A$ is not projective over $A^e$.

The zero polynomial has empty support and lies in $\ker \mu$; the kernel calculation includes empty finite sums. By [L2], $1\mathrm{\ne }0$ in $k$, so the zero algebra is excluded. In characteristic two, $x-y=x+y$ still has two distinct monomials, so $I$ remains nonzero and proper. The complex $F$ has exactly one nonzero cochain term in degree zero, with zero differential, so both endpoints are covered by the unit calculation. The only lift used is the single lift of ${\mathrm{id}}_A$ supplied by projectivity; no family of choices or AC is used. This example proves no iff statement. [L2, L4, L9, L10, step 1.1, step 2.1, step 2.2, step 4.1, given, algebra] $\mathrm{\square }$