# Formal deformations of modules

This question is based on an exercise in Etingof's Introduction to Representation Theory. Let $A$ be an associative algebra (say over a field $k$) and $M$ an $A$-module. Write $\rho : A \to \text{End}_k(M)$ for the action map. By a formal deformation of $M$ I mean a formal power series $\widetilde{\rho} = \rho + \rho_1t + \rho_2t^2 + \cdots$, where each $\rho_i : A \to \text{End}_k(M)$ is a $k$-linear map, satisfying $\widetilde{\rho}(ab) = \widetilde{\rho}(a) \widetilde{\rho}(b)$ for all $a,b \in A$. Two formal deformations are equivalent if they are conjugate by a power series $b = 1 + b_1t + b_2t^2 + \cdots$ where $b_i \in \text{End}_k(M)$.

My question: why does $\text{Ext}^1(M,M) = 0$ imply that any formal deformation $\widetilde{\rho}$ is equivalent to $\rho$? I can see that $\rho_1$ is a $1$-cocycle, so if we are trying to achieve $b\widetilde{\rho} = \rho b$ we should choose $b_1 \in \text{End}_k(M)$ so that $db_1 = \rho_1$. But what happens at the higher levels?

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Just to clarify, do you intend $b$ to be a formal power series $1+b_1t+b_2t^2+\cdots$, or just a quadratic expression? – Alex Becker Jul 5 '12 at 17:07
Thanks Alex, I have fixed the typo. – Justin Campbell Jul 5 '12 at 17:37

Here is a hint. Let $X$ be a module with underlying vector space $M\oplus M$ where the second summand is a submodule isomorphic to $M$, and the action on the first is $a(m,0)=(am, \rho_1(a)(m))$. If this really is a module structure $\rho_1$ must be a cocycle and the fact that $X$ splits is equivalent to your "there exists $b_1$ such that $db_1=\rho_1$". To work with the higher $b_i$s, you can look at modules with underlying vector space $M\oplus M \oplus \cdots$.
e.g. $M\oplus M\oplus M$ with the last summand having the standard $M$-action, $a(m,0,0)=(am, \rho_1(a)(m), \rho_2(a)(m))$ and $a(0,m,0)=(0,am, \rho_1(a)(m))$. The condition for this to be a module is exactly that coming from $\tilde{\rho}(ab)=\tilde{\rho}(a)\tilde{\rho}(b)$. Now, $(m,n,l)\mapsto m$ must split via $m \mapsto (m, \phi(m), \psi(m))$ for some $\phi, \psi$...
Excellent! I figured it out. This is much better than the horrible inductive argument which initially occurred to me. I noticed there is an easy induction with long exact sequences involved here, namely to show that $\text{Ext}^1$ of $M$ and the $n^{\text{th}}$ module you construct vanishes. – Justin Campbell Jul 5 '12 at 23:14