Questions about $T^{\perp}$

Question

Let $\Lambda$ be an algebra. For an $\Lambda$-module $M$, denote by $M^{\perp}$ the subcategory consisting of modules $N$ such that $\mathrm{Ext}_{\Lambda}^i(M,N)=0$ for $i \geq 1$. Similarly, we can define $\mathcal{C}^{\perp}$ for a subcategory of $\mathrm{mod} \Lambda$. Dually, we have $^{\perp}M$ and $^{\perp} \mathcal{C}$.

Now let $\mathcal{C}$ be a resolving subcategory of $\mathrm{mod} \Lambda$ such that each object of $\mathcal{C}$ has finite projective dimension. Let $T \in \mathcal{C}$ be a tilting module. Then obviously, we have $$\mathcal{C}^{\perp} \subseteq T^{\perp}$$.

1. How to get $^{\perp}(T^{\perp}) \subseteq \mathcal{C}$?
2. Is there any references proving that $T^{\perp}$ is an exact category with enough projectives $\mathrm{add}T$?

Answer 1

Here is a more general answer that will incorporate your question.

Let $R$ be any ring, $\mathcal{C}$ a class of modules and $\kappa$ a cardinal. Let $\mathcal{C}^{<\kappa}$ denote the subclass of $\mathcal{C}$ containing all modules with a projective resolution of $(<\kappa)$-generated projective modules. Also for $n<\omega$ let $\mathcal{P}_{n}$ be the class of modules of projective dimension at most $n$.

So, for example, the class $\mathcal{P}_{0}^{<\omega}$ is the class of projective $R$-modules with projective resolutions consisting of finitely generated projective modules, which is just the finitely generated projective modules if $R$ is coherent.

Recall that a class $\mathcal{C}$ is a resolving subcategory of $\text{mod}\,R$ if

• $\mathcal{P}_{0}^{<\omega}\subseteq\mathcal{C}\subseteq\text{mod}\,R$
• $\mathcal{C}$ is closed under extensions, direct summands and kernels of epimorphisms

Now we can get to tilting.

Theorem [13.49] Let $R$ be a ring and $n<\omega$. Then there is a bijection between $n$-tilting classes $\mathcal{T}$ and resolving subcategories $\mathcal{C}$ of $\text{mod}\,R$ such that $\mathcal{S}\subseteq \mathcal{P}_{n}^{<\omega}$. This is given by mutually inverse bijections $$\mathcal{C}\mapsto \mathcal{C}^{\perp}$$ and $$\mathcal{T}\mapsto (^{\perp}\mathcal{T})^{<\omega}$$

This answers the first part of your question: if $\mathcal{C}$ is a resolving subcategory such that every element has finite projective dimension, we can use the theorem so $\mathcal{C}^{\perp}$ is a tilting class. Therefore there is a tilting module $T$ such that $\mathcal{C}^{\perp}=T^{\perp}$. Consequently $^{\perp}(T^{\perp})^{<\omega}=\,^{\perp}(\mathcal{C}^{\perp})=\mathcal{C}$ by the above bijection.

That $T^{\perp}$ is an exact category is trivial from its definition. That it has enough projectives follows from the following fact.

Proposition [13.14(b)] Let $T$ be a tilting module and $\mathcal{T}$ the corresponding $n$-tilting class. Then $T$ consists of all $R$-modules $M$ such that there is an exact sequence $$T^{(\lambda_{n})}\rightarrow \cdots \rightarrow T^{(\lambda_{1})}\rightarrow M\rightarrow 0,$$ with each $\lambda_{j}$ a cardinal.

In particular, every element of $T^{\perp}$ is an epimorphic image of a module in $\text{Add}(T)$, and every element of the form $T^{(\kappa)}$ is projective in $\text{Add}(T)$ since

Lemma [13.10(c)] $\text{Add}(T)$ is the kernel of the cotorsion pair $(\,^{\perp}(T^{\perp}),T^{\perp})$.

All the references in the above are to Approximations and Endomorphism Algebras of Modules by Göbel and Trlifaj.

If you are only interested in finitely generated tilting modules (which rules out many interesting cases such as commutative noetherian rings), the corresponding results are almost certainly in Tilting theory and homologically finite subcategories with applications to quasihereditary algebras by I. Reiten.