# A category equivalence on the modules over a trivial extension

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?

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.
• 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? May 27, 2021 at 14:24
• 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 $$\varphi_X(\varphi_X\otimes Id_M)(x\otimes m \otimes m') = \varphi_X(\varphi_X(x\otimes m)\otimes m') = \left(x\cdot(0,m)\right)\cdot(0,m') = x\cdot\left(0,m)(0,m')\right) = 0.$$ May 28, 2021 at 23:48