Reference: p-adic differential equations by Kiran S. Kedlaya
We briefly cover the most basic concepts of differential algebra. For this part we are not relating to \(D\)-modules, but that would certainly be beneficial to think at at some moment.
A differential ring is a commutative ring \(R\) equipped with a given derivation \(\partial: R \to R\). Also have differential domains, differential fields, etc.
A differential module over a differential ring \((R, \partial)\) is an \(R\)-module \(M\) equipped with an additive map \(D: M \to M\) such that \(D(rm) = \partial(r) m + r D(m)\) for all \(r \in R\) and \(m \in M\).
Modules isomorphic to \((R,d)^{\oplus n}\) are called trivial. Successive extension of trivial modules are called unipotent.
Elements of \(R_C = R^\partial = \ker(\partial)\) are called constants, elements of \(M^D=\ker(D)\) are called horizontal. In Kedlaya’s book they use \(H^0(M):=\ker(D)\) and \(H^1(M):=M/D(M)\).
Remark. The ring \(D_X\) in \(D\)-modules are non-commutative because it fuses all derivations and functions \(\mathcal{O}_X\). By comparison, the differential ring we are talking about here only relate to functions \(\mathcal{O}_X\) which does not include derivations, and only care about one specific derivation. Think of it as the regular function ring equipped with a given vector field. We expect it to be much easier to work with.
Lemma. \(K\) be a differential field with constants \(K_C\). Then for any differential \(K\)-module \(M\), the canonical map
\[M^D\otimes_{K^\partial} K\xhookrightarrow{} M\]
is injective, and \(\dim_{K^\partial} M^D \le \dim_K M\). We are extending the flat sections as a ‘basis’ of \(M\), but unfortunately we cannot guarantee we have enough flat sections to generate \(M\), in practice they can be less than \(\dim_K M\).
Proof.
Find a set \(\{m_i\}\) of linear independence sections of \(M^D\) over \(K^\partial\) and let \(c_1m_1+\dots+c_nm_n=0\) be the shortest linear dependence relation of \(\{m_i\}\) in \(K\), where \(c_i\in K\).
Then \(n\ge 2\), assume \(c_1\neq 0\) and divide by \(c_1\) we can assume \(c_1=1\).
\[ m_1 + c_2 m_2 + \dots + c_n m_n = 0 \]
Differentiation gives \(0+c_2'm_2+\dots+c_n'm_n=0\) where \(c_i'=\partial(c_i)\). Since we cannot get a shorter linear dependence relation, all of \(c_i\) have to be constants, contradicting our assumption that \(m_i\) are linearly independent over \(K^\partial\).
Let \(M\) be a finite free \(R\) differential module, if \(M\cong \bigoplus_{i=1}^n R e_i\) we can compute for \(v = \sum_i r_i e_i\)
\[ D(v) = \sum_i \partial(r_i) e_i + \sum_i r_i D(e_i)\]
where \(D(e_i) = \sum \Lambda_{ji} e_j\). So in the coordinate form \(D = \partial I + \Lambda\) and the information of \(D\) is encoded in a matrix \(\Lambda \in M_n(R)\).
Remark This matrix is only taking derivative to one direction since the differential module only remembers one derivation. In the context of \(D\)-module or vector spaces with connections, it will produce a matrix of one forms, waiting to be evaluated at all directions.
Warning We say a free differential \(R\)-module is just a free \(R\)-module and at the same time a differential module, not meant to be a direct sum of \((R,d)\), the latter is called trivial instead.
With a (commutative) differential ring \(R\), let us denote by \(\mathbf{DM}(R)\) the category of differential \(R\)-modules (not D-modules, they are \({}_{D_X}\mathbf{Mod}\)). Note that there is no left or right differential modules, unlike the case of \(D_X\)-modules.
We have the following useful functors that are familiar in the theory of modules:
Direct Sum
\[\oplus_R : \mathbf{DM}(R) \times \mathbf{DM}(R) \to \mathbf{DM}(R).\]
\[D(m_1\oplus m_2) = D(m_1)\oplus D(m_2)\]
Forgetful functor
\[U: \mathbf{DM}(R) \to {}_{R}\mathbf{Mod}.\]
Tensor Product
\[\otimes_R : \mathbf{DM}(R) \times \mathbf{DM}(R) \to \mathbf{DM}(R).\]
Note that these modules are bi-\(R\)-modules so tensoring make sense, and the result of tensoring two differential modules is a differential module, whose structure is given by \(D(m\otimes n) = D(m)\otimes n + m\otimes D(n)\).
Hom functor
\[\mathrm{Hom}_R(-,-): \mathbf{DM}(R)^\text{op} \times \mathbf{DM}(R) \to \mathbf{DM}(R).\]
\[ D(f)(m) = D(f(m)) - f(D(m)) \]
Exterior Power
\[\wedge^k_R : \mathbf{DM}(R) \to \mathbf{DM}(R),\quad M \mapsto \wedge^k_R M.\]
\[D(m_1\wedge\dots\wedge m_k) = \sum_{i=1}^k m_1\wedge\dots\wedge D(m_i)\wedge\dots\wedge m_k\]
Symmetric Power
\[\mathrm{Sym}^k_R : \mathbf{DM}(R) \to \mathbf{DM}(R),\quad M \mapsto \mathrm{Sym}^k_R M.\]
\[D(m_1 \dots m_k) = \sum_{i=1}^k m_1\dots D(m_i)\dots m_k\]
We have said that differential modules are weaker version of \(D_X\)-modules in which you only take one direction of differentiation. But this points that it can be seen as a module over some weaker \(D\)-ring.
Differential modules can be seen as modules over a non-commutative ring. For example a differential module \((M,D)\) over \((k[t],\partial)\) can be seen as a left module over \(k[t][\partial]\). The non-commutativity of derivation and function multiplication is encoded in the ring structure as follows
\[ [\partial, t] = \partial t - t\partial = \partial(t) = 1. \]
This includes examples like \(K=k(t)\) where \(R=k(t)[\partial]\) is a Euclidean ring.
Theorem If \((K,\partial)\) is a differential field, then \(K[\partial]\) is left-Euclidean and right-Euclidean (this can be obtained from the left-Euclidean structure by taking adjoint).
This implies that \(R=K[\partial]\) be a left and right principal ideal ring, any (left/right) ideal can be written as \(Rf\) or \(fR\) for some \(f\in R\).