Reference request - existence of formal solutions for integrable connections Let $K$ be a field of characteristic $0$, let $A = K[[t_1, \ldots, t_n]]$ be a power series ring over $K$, and let $V$ be a free $A$-module. Let $\nabla \colon V \rightarrow V \otimes_A \Omega^1_{A/K} = \bigoplus_{k=1}^n V \ dt_k$ be an integrable connection, so $\nabla^2 = 0$. I want to show that there always exist solutions $s \in V$ to the equation $\nabla(s) = 0$ subject to any initial condition $s(0) = \alpha$ for $\alpha \in V(0) = V \otimes_A A/\mathfrak{m}_A$. 
More concretely, letting $e_1, \ldots, e_r$ be a basis for $V$, I have a system of $n \times r$ linear first-order homogeneous partial differential equations
$$
\frac{\partial}{\partial t_k} s_i = \sum_j M^j_{ik} s_j
$$
For some $M_{ik}^j \in A$. Or, writing $M_k$ for the matrix $(M_{i,k}^j)_{i,j}$, 
$$
\frac{\partial}{\partial t_k} s = M_k s
$$
How can I show that integrability implies there are always solutions? Concretely, integrability is exactly the statement that for all $i,j,k, \ell$, we have
$$
\frac{\partial}{\partial t_\ell} M_{k} - \frac{\partial}{\partial t_k} M_{\ell} = M_k M_\ell - M_\ell M_k
$$ 
 A: I hope you'll allow one or two slight adjustments to your question:


*

*I think you are interested in $(t_1, \dots, t_n)$-adically continuous connections rather than arbitrary connections (for which the desired solution principle would not generally hold).

*Judging from how you use "$r$," I think you intend $V$ to be a finite-rank free $A$-module (rather than an arbitrary free $A$-module).
You will find a very nice treatment of your question thus modified in Proposition 8.9 of Nicholas Katz's paper "Nilpotent connections and the monodromy theorem: Applications of a result of Turrittin."
Here is the statement of that proposition:

Let $K$ be a field of characteristic zero, $K[[t_1, \dots, t_n]]$ the
  ring of formal power series over $K$ in $n$ variables. Let $M$ be a
  finitely generated module over $K[[t_1, \dots, t_n]]$, given with an
  integrable connection $\nabla$ (for the continuous $K$-derivations of
  $K[[t_1, \dots, t_n]]$ to itself). Then $M^\nabla$, the $K$-space of
  horizontal elements of $M$, is finite-dimensional over $K$, and the
  pair $(M, \nabla)$ is isomorphic to the pair
  $(M^\nabla\otimes_K K[[t_1, \dots, t_n]], 1\otimes d)$
  where $d$ denotes the ``identical'' connection on $K[[t_1, \dots, t_n]]$.

Note that in this proposition, one can even assume only that the module $M$ with integrable connection is finitely generated (rather than free of finite rank). As a consequence of fact that $M^\nabla \otimes_K K[[t_1, \dots, t_n]] \to M$ is an isomorphism, one can then conclude that $M$ is necessarily free.
