Computation of extension groups in the category of pairs $(M,f)$

Question

Let $A$ be a unitary commutative ring, and let $B$ be an $A$-algebra. We consider the category whose objects are pairs $\textbf{M}=(M,f)$ where $M$ is an $A$-module and where $f$ is a $B$-linear endomorphism of $M\otimes_A B$, and whose morphisms are $A$-linear maps whose scalar extension to $B$ make the obvious diagram commute, i.e. a morphism between $\textbf{M}$ and $(N,g)$ is a $A$-linear morphism $h:M\to N$ such that $g\circ (h\otimes_A 1)=(h\otimes_A 1)\circ f$. Let $\textbf{1}$ be the object $(A,\operatorname{id}_B)$. I want to compute the extension group of $\textbf{1}$ by $(M,f)$ in the latter category. Hereafter, I sketched my computation:

The object $\textbf{P}=(A[X],X)$ is a projective object in this category, hence $0\to \textbf{P}\to \textbf{P}\to \textbf{1}\to 0$ where the first (nontrivial) arrow is multiplication by $1-X$ and the second one is the evaluation at $1$ is a projective resolution of $\textbf{1}$.

Let us compute $\operatorname{Hom}(\textbf{P},\textbf{M})$. We consider the $A$-linear morphism $\operatorname{Hom}(\textbf{P},\textbf{M})\to M$ which maps a morphism $h:A[X]\to M$ to its value $m=h(1)$. By commutativity of the "obvious" diagram $h(X^n)=f^n(m)$ which indicates that $f^n(m)$ belongs to $M$ for all positive $n$. Thanks to this remark, we easily obtain: $$\operatorname{Hom}(\textbf{P},\textbf{M})=\{m\in M~|~\forall n\geq 0:~f^n(m\otimes_A 1)\in M\otimes_A A\}=:f^{\bullet}(M).$$ Hence, $\operatorname{RHom}(\textbf{1},\textbf{M})$ is given by the complex $[f^{\bullet}(M)\to f^{\bullet}(M)]$ in degree $0$ and $1$ where the arrow is $1-f$.

We conclude that $\operatorname{Ext^1(\textbf{1},\textbf{M})}$, which is given by the first cohomology group of this complex, is isomorphic to $\operatorname{coker}(1-f|f^{\bullet}(M))$.

However, let $m\in \operatorname{Hom}_B(B,M\otimes_A B)=M\otimes_A B$. The object $(M\oplus A, \begin{pmatrix} f & m \\ 0 & 1 \end{pmatrix})$ is an extension of $\textbf{M}$ by $\textbf{1}$. Besides, it is equivalent to $(M\oplus A, \begin{pmatrix} f & m' \\ 0 & 1 \end{pmatrix})$ whenever $$\begin{pmatrix} f & m' \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & u \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & u \\ 0 & 1 \end{pmatrix}\begin{pmatrix} f & m \\ 0 & 1 \end{pmatrix} $$ for some $u\in \operatorname{Hom}_A(A,M)=M$, which boils down to $m'=m+u-f(u)$. Thanks to this argument, it seems that the extension group is rather given by $$\operatorname{coker}(M\stackrel{1-f}{\longrightarrow} M\otimes_A B). $$ Can anyone point me out my mistake? Many thanks!

Answer 1

I have finally found an answer to my problem: in fact, the object $\textbf{P}$ is not a projective object in my category and hence, there is no clear projective resolution of $\textbf{1}$. The complex $\operatorname{RHom}(\textbf{1},\textbf{M})$, however, can still be computed using "rigid glued categories" as presented in Proposition 4.2.3 of
A. Huber, Mixed Motives and Their Realization in Derived Categories, Lecture Notes in Mathematics 1604. One easily find out that $\operatorname{RHom}(\textbf{1},\textbf{M})$ is naturally (in $\textbf{M}$) quasi-isomorphic to the complex of $A$-modules $[M\stackrel{1-f}{\longrightarrow} M\otimes_A B]$ placed in degree $0$ and $1$, which explains my second computation.