the same rank become isomorphic after tensoring with $R({\mathscr{S}})$. However, it is intuitively clear that the situation for modules with connections is quite different, which is indeed confirmed by:

\[th:rational-section\] Let ${\mathscr{F}}$ be a finitely generated ${\mathscr{R}}$-module with a connection. Then, if $f \in {\mathscr{F}}\otimes_{\mathscr{R}}R({\mathscr{S}})$ is a rational solution of $\nabla_{\mathscr{F}}(f) = 0$, it automatically lies in ${\mathscr{F}}$ itself.

Suppose that $f \neq 0$, and choose a point on our curve. Take a function $r \in {\mathscr{R}}$ which vanishes (to order $1$) at this point. There is a unique $m \in {\mathbb{Z}}$ such that $r^m f$ takes a nonzero finite value at our point. By assumption $$\nabla_{\mathscr{F}}( r^m f) = m (r^{-1} dr) r^m f,$$ but while the left hand side is regular, the right hand one has a pole unless $m = 0$.

One can extend the notion of connection to the derived category (see for instance [@markarian09; @huybrechts-thomas10]), as follows. Any complex of modules ${\mathscr{F}}$ sits in a short exact sequence generalizing . Define a [*homotopy connection*]{} to be a splitting ${\mathscr{F}}\rightarrow J^1({\mathscr{F}})$ in the derived category. If this exists, it induces connections on all the cohomology modules ${\mathscr{H}}^i({\mathscr{F}})$. The obstruction to the existence of a homotopy connection 