Let $A$ be an associative algebra over field $k$, and $V_0$ is a finite dimensional $A$-module. Set $D=k[\epsilon]/(\epsilon^2)$. A first order deformation of $V_0$ is a module $V$ over $A \otimes_k D$ flat as $D$-module and such that $V \otimes_D k \cong V_0$.
Suppose $V_0$ is a cyclic module, in other words it is generated by one element $v_0 \in V_0$: $V_0=Av_0$. Is it true that any first order deformation of a cyclic module is cyclic?
To me, it looks like $V$ is cyclic and for any $v_1 \in V_0$ element $v=v_0+\epsilon v_1 \in V$ is a generator of $V$.
Let $p:A \to \operatorname{End}_D(P)$ be the deformed action map. Set $W=p(A)v \subset V$. Our goal is to show that $W = V$. We start with short exact sequence $$ 0 \to W \to V \to V/W \to 0, $$ after taking tensor product with $k$ we get $$ 0 \to \operatorname{Tor}_1^D(V/W,k) \to W_0 \to V_0 \to (V/W) \otimes_D k \to 0, $$ where $W_0 = W \otimes_D k$. Key observation is $W_0 = p_0(A)v_0=V_0$, hence $W_0 = V_0$ and $\operatorname{Tor}_1^D(V/W,k) \cong 0 \cong (V/W) \otimes_D k$. Condition $\operatorname{Tor}_1^D(V/W,k) \cong 0$ imply that $V/W$ is a free module over $D$, then by $(V/W) \otimes_D k \cong 0$ we conclude that $V/W \cong 0$.