I've been working through Auslander, Reiten, and Smalø's Representation Theory of Artin Algebras. In reading through the basic setup relevant to the transpose operation, I've come across two statements that are described as "clear" and "not hard to see", but which I don't find to be either, which suggests I'm missing something significant.
For a finitely generated module $C$ over an Artin algebra $\Lambda$, we let
$$
P_1\to P_0\to C\to 0
$$
be a minimal projective presentation, and then define the transpose $\operatorname{Tr}(C)$ to make this sequence exact:
$$
P_0^*\to P_1^*\to \operatorname{Tr}(C)\to 0
$$
where $P_i^* := \operatorname{Hom}_\Lambda(P_i, \Lambda)$ is the duality on projective modules.
The book first says that
Note that if $C$ is indecomposable nonprojective, then it is clear that $f:P_1\to P_0$ is also an indecomposable map which is not an isomorphism. Hence $f^*:P_0^*\to P_1^*$ is also an indecomposable map which is not an isomorphism, so that $\operatorname{coker}(f^*) = \operatorname{Tr}(C)$ is clearly indecomposable.
First, I don't think it's explicitly stated what an "indecomposable map" is in this context, but I'm assuming it's an indecomposable object in the category of morphisms between projectives, i.e., there's no simultaneous decomposition of the domain and codomain which is respected by the map. If that's the case, I understand that $f^*$ will also be indecomposable, since if it decomposed we could just apply the duality again and contradict the indecomposability of the original map. However, I don't understand why the cokernel of an indecomposable non-isomorphism will be indecomposable.
The book continues
It is also not hard to see that $P_0^*\to P_1^*\to \operatorname{Tr}(C)\to 0$ is a minimal projective presentation of $\operatorname{Tr}(C)$.
If I knew why this were true, I would also be satisfied with the previous statement: if $\operatorname{Tr}(C)$ were decomposable, then its minimal projective presentation $P_0^*\to P_1^*\to \operatorname{Tr}(C)$ would also be decomposable, which we know won't happen. However, I can't see why this is true either.
This latter statement has already been asked about on this site: Question about minimal projective presentations of a module. The idea behind the approach in Assem, Simson, and Skowronski's book mentioned in that question makes sense to me, but I get hung up on the same point that the question asks about, and I don't understand the answer it received or the ensuing discussion.
My understanding of the argument is as follows. Suppose $P_0^*\to P_1^*\to \operatorname{Tr}(M)\to 0$ is not a minimal presentation. Suppose that $E_0'\to E_1'\to \operatorname{Tr}(M)$ is a minimal presentation. Then since $E_1'\to \operatorname{Tr}(M)$ is right minimal, we can obtain a decomposition $P_1^*\cong E_1'\oplus E_1''$ such that the cover $\pi: P_1^*\to \operatorname{Tr}(M)$ corresponds to the cover $\pi': E_1'\to \operatorname{Tr}(M)$ on $E_1'$ and is 0 on $E_1''$. For the same reason, we can write $P_0^*\cong E_0'\oplus E_0''$ such that the composition $P_0^*\to \ker(\pi)\cong \ker(\pi')\oplus E_1''\to \ker(\pi')$ corresponds to the cover $\rho:E_0'\to \ker(\pi')$ on $E_0'$ and is 0 on $E_0''$.
Rewriting $P_0^*$ and $P_1^*$ using these decompositions, our original presentation becomes
$$
E_0'\oplus E_0''\xrightarrow{\left(\begin{smallmatrix} u & 0 \\ f & v \end{smallmatrix}\right)} E_1'\oplus E_1''\to \operatorname{Tr}(M)\to 0
$$
Assem-Simson-Skowronski's argument appears to suggest that either $f$ is 0, or we can in some other way set up our decompositions to make $f$ 0. I don't understand why this must be true. If it is, then it implies $P_0^*\to P_1^*$ is decomposable, a contradiction.
Could someone provide clarification on the argument given in the answer to the linked question, or suggest a simpler way of understanding why the quoted statements are true? I feel like I'm missing something very basic here.
Best Answer
In your argument at the end of the post, instead of taking a projective cover $E'_0\to\ker(\pi')$ of $\ker(\pi')$, take a projective cover $F'_0\to\ker(\pi)$ of $\ker(\pi)=\ker(\pi')\oplus E''_1$, so $F'_0=E'_0\oplus E''_1$.
Then the same argument you used before shows that $P^\ast_0=F'_0\oplus F''_0$ for some $F''_0\leq\ker(\pi)$ and the original presentation becomes $$E'_0\oplus E''_1\oplus F''_0\xrightarrow{\left(\begin{smallmatrix}u&0&0\\0&1&0 \end{smallmatrix}\right)} E'_1\oplus E''_1\to\operatorname{Tr}(M).$$