Let $A$ be an algebra over a field k. $D$ is the standard duality functor. A module $_AM$ is called a generator if $add(A) \subseteq add(M)$, a cogenerator if $add(D(A)) \subseteq add(M)$. $M$ is n-rigid if $Ext_A^i(M,M)=0$ for $1 \leq i \leq n$. Now suppose $_AM$ is a generator-cogenerator which is n-rigid and neither projective nor injective. $\Lambda := End_A(M)$.

Suppose there is an long exact sequence of $\Lambda^{op}$-modules $$0 \rightarrow Hom_A(M^-,M) \rightarrow D(E_{n+1}) \rightarrow \cdots \rightarrow D(E_1) \rightarrow D(E_0) \rightarrow D(_{\Lambda} \Lambda) \rightarrow 0$$ such that $D(E_i)$ are projective-injective $\Lambda$-modules for $0 \leq i \leq n+1$. Let $\mathfrak{C}(Y)$ be the smallest triangulated subcategory of $\mathfrak{D}^b(\Lambda^{op})$ which is closed under direct summands and contains $Y$ and all projective-injective modules (here $\mathfrak{D}^b(\Lambda^{op})$ is the bounded derived category of complexes over $\Lambda ^{op}$-modules). Then $\mathfrak{C}(D(\Lambda))= \mathfrak{C}(Hom_A(M^-,M))$. Up to multiplicity and isomorphism, we know all projective-injective modules occur as direct summands of all tilting modules. Then how to get that $D(\Lambda)$ is a tilting module if and only if so is $Hom_A(M^-,M)$?

  • $\begingroup$ @Matthew Pressland Sorry, I thought $M^-$ is useless. $\tau ^{-}=TrD$ is the Auslander-Reiten translation,$\Omega^{-1}$ is the cosyzygy functor. $\tau^- _{n+1} :=\tau^- \Omega^{-n}_A$, then $M^- :=A \oplus \tau^- _{n+1}(M)$. $\endgroup$ – Xiaosong Peng Mar 18 '17 at 0:49
  • $\begingroup$ @Matthew Pressland The precise definition and question can be seen at this paper: arxiv.org/abs/1506.03337, on page 12 and lemma 3.13. $\endgroup$ – Xiaosong Peng Mar 18 '17 at 2:10
  • $\begingroup$ @Matthew Pressland By the long exact sequence, I have that the projective dimension $proj.dim(Hom_A(M^-,M)) < \infty$ iff $proj.dim(D(\Lambda)) < \infty$. Also I have that $Ext^i(DHom_A(M^-,M),DHom_A(M^-,M))=0$ for all$i >0$, so $Ext^i(Hom_A(M^-,M),Hom_A(M^-,M))=0$. By the injectivity of $D(\Lambda)$, we also have $Ext^i(D(\Lambda),D(\Lambda))=0$ for all $i >0$. So the last question is how to get the third condition of tilting modules is held by $Hom_A(M^-,M)$ iff it is held by $D(\Lambda)$. Thank you. $\endgroup$ – Xiaosong Peng Mar 18 '17 at 2:10
  • $\begingroup$ Thanks—my comment was slightly wrong, but I hope the answer is helpful. It does seem to be important what $M^-$ is (or at least that it has $A$ as a summand) in order to know that $\operatorname{Hom}(M^-,M)$ contains all indecomposable projective-injectives as summands. $\endgroup$ – Matthew Pressland Mar 18 '17 at 12:13

From the comments, it seems that you know how to prove equivalence for the first two conditions for being a tilting module.

For the last condition you can use $\mathfrak{C}(D(\Lambda))= \mathfrak{C}(\operatorname{Hom}_A(M^-,M))$, since the final condition for a $\Lambda^{\mathrm{op}}$-module $T$ to be tilting is that $\Lambda^{\mathrm{op}}\in\mathfrak{C}(T)$ and $\operatorname{add}{T}$ contains all projective-injective modules. (This second part is true automatically for both $D(\Lambda)$ and $\operatorname{Hom}_A(M^-,M)$.)

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