Graded Bimodules and Tensor Functors — Examples

In $A\{2\}$ the homogeneous piece of degree $d$ is $A\{2{\}}_d=A_{d-2}$, so the monomial $x^j\in A_j$ is a homogeneous element of degree $j+2$ in $A\{2\}$, and the element $1\in A_0$ has degree $2$.

In the published twist, $A(2{)}_d=A_{d+2}$, so $x^j\in A_j$ has degree $j-2$ in $A(2)$, while $A(-2{)}_d=A_{d-2}$ gives $x^j$ degree $j+2$ in $A(-2)$; hence $A\{2{\}}_d=A_{d-2}=A(-2{)}_d$ for every $d$.

The equality of piecewise $k$-modules in step 1.2 holds for every degree, and both modules carry the same scalar action inherited from $A$; hence $A\{2\}=A(-2)$ as graded $A$-modules. The twist $A(2)$ is a genuinely different grading, not the same one written with the opposite sign: $A(2{)}_0=A_2=k{x}^2\mathrm{\ne }0$, while $A(-2{)}_0=A_{-2}=0$, so $A(2)$ and $A(-2)$ have different degree-zero pieces.

Shifting changes no underlying map. The multiplication of $A$, the action of $A$ on these graded modules, and every degree-zero $A$-linear map are the same functions before and after the shift; only the degree attached to a homogeneous element changes, by the fixed amount $r$. In particular the shift of a complex would move each homogeneous piece to the $r$-fold shifted degree without introducing a sign in any differential. Since $A$ carries no differential, no sign is introduced here at all.

The computation exhibits both claims: $A\{2\}=A(-2)$ as graded modules, so the internal shift reverses the published twist parameter, and the shift does not alter multiplication or differential signs. ∎

The two actions on $M$ commute: $(b\cdot m)\cdot \lambda =\pi (b)m\lambda =b\cdot (m\lambda )$, and each is additive and unital, so $M$ is a $(B,A)$-bimodule; both actions preserve the degree-$0$ part because $\pi$ and the scalar action do, so $M$ is a graded bimodule.

$M=k$ is a free right $k$-module of rank one, hence flat, so $M{\otimes }_k-$ is exact; equivalently the functor is $k{\otimes }_k-$, which is naturally the identity on $k$-vector spaces.

The unit isomorphism $M{\otimes }_{A}A\to M$, $m\otimes \lambda \mapsto m\lambda$, identifies $M{\otimes }_{k}k$ with $k$, and under this identification the left $B$-action is $b\cdot (m\otimes \lambda )=bm\otimes \lambda$, i.e. the action of $B$ on $k$ through $\pi$; so $M{\otimes }_{k}k\cong k$ as graded left $B$-modules.

The module $k=B\mathrm{/}(\epsilon )$ is not projective as a left $B$-module. The quotient map $\pi :B\twoheadrightarrow k=B\mathrm{/}(\epsilon )$ is $B$-linear and degree-zero; if $k$ were projective, its lifting property against $\pi$ and the identity of $k$ would produce a $B$-linear section $s:k\to B$ with $\pi s=1_k$. Writing $u:=s(1)$ one has $\pi (u)=1$, so $u=1+c\epsilon$ for some $c\in k$, and $B$-linearity gives $\epsilon u=s(\epsilon \cdot 1)=s(0)=0$, whereas $\epsilon u=\epsilon +c{\epsilon }^2=\epsilon \mathrm{\ne }0$. This contradiction shows that no such section exists, so $k$ is not projective over $B$.

Steps 1.2 and 1.3 give a right-flat bimodule $M$ whose tensor functor is exact and whose value on the finite graded projective left $A$-module $A=k$ is $M{\otimes }_{k}k\cong k$; by step 1.4 that output is not projective as a left $B$-module, and it is not a finite graded projective module either. Hence the exactness hypothesis of [L3] does not deliver its projectivity conclusion, which is why that conclusion carries the separate hypothesis that $M$ be finite graded projective over $B$ — a hypothesis $M$ fails by step 1.4.

The example therefore exhibits a right-flat tensor bimodule whose tensor functor is exact but which produces a nonprojective, non-finite-projective output from a finite graded projective input. ∎

The two actions on $M$ commute, since $(b\cdot m)\cdot a=\pi (a)bm=b\cdot (m\cdot a)$, and are additive and unital, so $M$ is a $(B,A)$-bimodule; both preserve degree $0$, so $M$ is graded.

$M=k=B$ is the free left $B$-module of rank one, hence a finite direct sum of shifts $B\{0\}$ and therefore a finite graded projective left $B$-module.

In $A=k[\epsilon ]\mathrm{/}({\epsilon }^2)$ the ideal $(\epsilon )=k\epsilon$ has ${\epsilon }^2=0$, so $a\epsilon =\pi (a)\epsilon$ for every $a\in A$; hence $(\epsilon )$ is isomorphic to $k=A\mathrm{/}(\epsilon )$ as a left $A$-module by $1\mapsto \epsilon$, and the sequence $0\to (\epsilon )\to A\stackrel{\pi }{\to }k\to 0$ is exact with all maps degree-zero and $A$-linear.

Tensoring the sequence of step 1.3 with $M$: the unit isomorphisms identify $M{\otimes }_{A}A\cong M=k$ and $M{\otimes }_{A}k\cong k$, and by step 1.3 also $M{\otimes }_A(\epsilon )\cong M{\otimes }_{A}k\cong k$. The induced map $M{\otimes }_A(\epsilon )\to M{\otimes }_{A}A$ sends $1\otimes \epsilon$ to $1\otimes \epsilon =(1\cdot \epsilon )\otimes 1=0$, by the balancing relation and the right action $1\cdot \epsilon =\pi (\epsilon )=0$ on $M$. So the induced map is zero while its source is $k\mathrm{\ne }0$, and it is not injective.

By step 2.1 the functor $M{\otimes }_A-$ fails to preserve the monomorphism $(\epsilon )\to A$, so it is not exact and $M$ is not flat as a right $A$-module; by step 1.2 $M$ is nevertheless finite graded projective over $B$. Hence finite left $B$-projectivity of a bimodule does not imply right $A$-flatness, the hypothesis that the exactness clause of [L4] requires.

The example therefore exhibits a bimodule that is finite projective on the tensoring-out side but whose tensor functor is not exact. ∎