I am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1 , and I have a question on its proof of main properties of transpose $Tr$, appearing in Auslander-Reiten theory.
Here we assume $A$ is a finite dimensional algebra over an algebraically closed field $K$, and $mod A$ is a category of finitely generated modules of $A$. Also, consider a functor $(−)^t = Hom_A(−,A) : mod A → mod A^{op}$.
For an object $M$ in $mod A$, we define $Tr M$ as follows:

If $P_1→P_0→M→0$ is a minimal projective presentation, then $Tr M=Coker(P_0^t→P_1^t)$. In other words, $P_0^t→P_1^t→TrM→0$ is exact.

Proposition 2.1.(b) in the textbook p.107 is about showing that

If M is not projective, then the exact sequence $P_0^t→P_1^t→TrM→0$ is a minimal projective presentation of an $A^{op}$ module $Tr M$.

In the proof, author assumes $P_0^t→P_1^t→TrM→0$ is not a minimal projective presentation. Then there exist non-trivial decompositions (by projective modules) $P_0^t=E_0'⊕E_0''$, $P_1^t=E_1'⊕E_1''$ and an isomorphism $v:E_0''→E_1''$ and a homomorphism $u:E'_0→E'_1$ such that the previous exact sequence is expressed as $E_0'⊕E_0''\overset{\begin{pmatrix} u & 0 \\ 0 & v \end{pmatrix}}{\rightarrow}E_1'⊕E_1''→Tr M→0$.
(I think the author assumed $E'_0\overset{u}{\rightarrow}E'_1→Tr M→0$ is a minimal projective presentation.)

But I couldn't quite understand the next step. It is written as,

But then applying $(−)^t$ yields a projective presentation of M of the form ${E'_1}^t\overset{u^t}{\rightarrow}{E'_0}^t→M→0$

But I couldn't figure out why we get $M$ as a cokernel of $u^t$. Any helps will be appreciated. Thank you for reading.


For finitely generated projective modules $X$, the evaluation map $X\to X^{tt}$ (defined by $x\mapsto[\varphi\mapsto\varphi(m)]$) is a natural isomorphism.

So $P_1\to P_0$ is isomorphic to $P_1^{tt}\to P_0^{tt}$, which in turn is isomorphic to the direct sum of $E_1'^t\xrightarrow{u^t}E_0'^t$ and $E_1''^t\xrightarrow{v^t}E_0''^t$.

But $v$ is an isomorphism, so $v^t$ is an isomorphism and so its cokernel is zero. So $M$, which is the cokernel of $P_1\to P_0$, is also the cokernel of the first summand $E_1'^t\xrightarrow{u^t}E_0'^t$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.