# How to show that ${}_{B}T_{A} \otimes DM \in \operatorname{Gen}({}_{B} T)$?

I am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1: Techniques of Representation Theory by Ibrahim Assem, Daniel Simson, Andrzej Skowronski.

Let $A$ be a $K$-algebra and $T_A$ be a tilting module. Let $B=\operatorname{End}_A(T_A)$. Let $D$ be the standard dual. That is $D(M)=\operatorname{Hom}_{K}(M, K)$.

I have a question on the proof of Theorem 3.8. On page 207, line 3 of the proof of Theorem 3.8, it is said that ${}_{B}T_{A} \otimes DM \in \operatorname{Gen}({}_{B} T)$. How to show that ${}_{B}T_{A} \otimes DM \in \operatorname{Gen}({}_{B} T)$?

Since $$\mathcal{F}({}_{B}T) = \operatorname{Cogen} \tau ({}_{B} T) = \{{}_{B}V \mid \operatorname{Hom}_{B}(T, V) = 0\},$$

$$\mathcal{T}({}_{B}T) = \operatorname{Gen} ({}_{B} T) = \{{}_{B}V \mid \operatorname{Ext}_{B}^{1}(T, U) = 0\},$$ we have to show that $\operatorname{Ext}_{B}^{1}(T, {}_{B}T_{A} \otimes DM ) =0$.

Do we have $\operatorname{Ext}_{B}^{1}(T, {}_{B}T_{A} \otimes DM ) = \operatorname{Ext}_{B}^{1}(T, {}_{B}T_{A}) \otimes \operatorname{Ext}_{B}^{1}(T, DM) =0$? $\operatorname{Ext}_{B}^{1}(T, {}_{B}T_{A})$ is zero since $T$ is a tilting module.

How to show that $\operatorname{Ext}_{B}^{1}(T, {}_{B}T_{A} \otimes DM ) =0$ and $\operatorname{Ext}_{B}^{1}(T, DM) =0$? Thank you very much. • One way to show something is in $Gen(T)$, is to give a surjective ($B$-module) homomorphism from some direct copies of $T$ to it. So, in this situation, first you need to understand how $B$ acts on $T\otimes DM$; second, figure out how this tell you the number of direct copies of $T$ to gives such a surjection, as well as writing down this surjection. – Aaron Jul 8 '13 at 22:28

The (left) module $DM$ is finitely generated, hence there is a left module epimorphism $A^n\twoheadrightarrow DM$. Now you know that $_BT_A\otimes_A -$ is right exact, hence we have a left $B$-epimorphism $T\otimes_A A^n\twoheadrightarrow T\otimes DM$. Now use that $T\otimes_A A^n\cong T^n$, hence there is a left $B$-epimorphism $T^n\twoheadrightarrow T\otimes DM$, so $T\otimes DM\in \operatorname{Gen}(_BT)$ by definition.