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Say that $A$ is a $K$-algebra and that $M$ is an $(A-A)$-bimodule. Let $\Lambda = A \ltimes M$ be a trivial extension of $A$, that is, the algebra with subjacent module $A \oplus M$, with multiplication defined as $$(a,x)(a',x') = (aa', ax' + xa')$$ in which $a,a' \in A$ and $x,x' \in M$. I am trying to solve the following problem (question 23 of the fourth chapter of Ibrahim Assem's "Algèbres et Modules: cours et exercices"):

Show that $\mathscr{M}(\Lambda)$ is equivalent to $\text{Mod}(\Lambda)$.

Here, $\mathscr{M}(\Lambda)$ is the category defined as follows: The objects are pairs $(X,\varphi)$ such that $X$ is an $A$-module and $\varphi: X \otimes_A M \rightarrow X$ satisfies $\varphi(\varphi \otimes 1_M) = 0$, and a morphism $(X, \varphi) \mapsto (X', \varphi')$ is an $A$-linear aplication $f:X \rightarrow X'$ such that $\varphi'(f \otimes 1_M) = f \varphi$.

I have tried to define functors between $\mathscr{M}(\Lambda)$ and $\text{Mod}(\Lambda)$ whose compositions are isomorphic to identities, but it leads me nowhere. Could anyone give me some pointers on how to proceed?

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I think the idea is the same as the identification of $A[x]$-modules with pairs $(X,\varphi)$ of a module $X$ and an $A$-linear map $\varphi:X\rightarrow X$.

Let $F:Mod(\Lambda) \rightarrow Mod(A)$ be the forgetful functor. Define $G: Mod(\Lambda) \rightarrow \mathcal{M}(\Lambda)$ by $G(X)=(F(X),\varphi_X)$, for a right $\Lambda$-module $X$, where $\varphi_X :X\otimes_A M \rightarrow X$ is defined by $\varphi_X (x\otimes m) := x\cdot(0,m)$, for all $x\in X$ and $m\in M$. If $f:X\rightarrow Y$ is a homomorphism of right $\Lambda$-modules, then $$f\varphi_X(x\otimes m) = f(x\cdot(0,m)) = f(x)\cdot(0,m) = \varphi_Y(f(x)\otimes m), \forall x\in X, m\in M.$$ Thus, $f$ is also a morphism between $(X,\varphi_X)$ and $(Y,\varphi_Y)$ in $\mathcal{M}(\Lambda)$.

Define $H:\mathcal{M}(\Lambda) \rightarrow Mod(\Lambda)$, by $H((X,\varphi))= X_\Lambda$, where $X_\Lambda = X$ is a right $\Lambda$-module by the action: $$ x\cdot(a,m) := \varphi(xa \otimes m), \qquad \forall x\in X, (a,m)\in \Lambda.$$ In particular, if $f:(X,\varphi)\rightarrow (Y,\varphi')$ is a morphism in $\mathcal{M}(\Lambda)$, then $f:X_\Lambda \rightarrow Y_\Lambda$ is also right $\Lambda$-linear, since $$f(x\cdot(a,m)) = f(\varphi(xa\otimes m)) = \varphi'(f\otimes id_M)(xa\otimes m)=\varphi'(f(x)a\otimes m) = f(x)\cdot(a,m)$$ Then one should be able of checking that $G$ and $H$ are functors and mutual inverses.

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  • $\begingroup$ Thank you. It is not clear to me why $G(X)$ is an element of $\mathcal{M}(\Lambda)$, specificaly, why $\varphi_X(\varphi_X, Id_M) = 0$. Could you please clarify? $\endgroup$
    – Brass One
    May 27, 2021 at 14:24
  • $\begingroup$ If $X\in Mod(\Lambda)$ and $\varphi_X(x\otimes m):=x\cdot (0,m)$, then by the associativity condition of the right $\Lambda$-module structure on $X$ we get $$\varphi_X(\varphi_X\otimes Id_M)(x\otimes m \otimes m') = \varphi_X(\varphi_X(x\otimes m)\otimes m') = \left(x\cdot(0,m)\right)\cdot(0,m') = x\cdot\left(0,m)(0,m')\right) = 0.$$ $\endgroup$ May 28, 2021 at 23:48

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