Author: Eiko

Tags: differential algebra, p-adic differential equations

Time: 2024-11-27 22:09:03 - 2024-11-28 13:16:47 (UTC)

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.

Differential Modules And Rings

• 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 $D(rm)=\partial (r)m+rD(m)$ 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.

Flat sections work as basis

Lemma. $K$ be a differential field with constants $K_C$. Then for any differential $K$-module $M$, the canonical map

$$M^D{\otimes }_{K^{\partial }}K\stackrel{}{\hookrightarrow }M$$

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 $c_1{m}_1+\cdots +c_n{m}_n=0$ be the shortest linear dependence relation of $\{m_i\}$ in $K$, where $c_i\in K$.

• Then $n\ge 2$, assume $c_1\ne 0$ and divide by $c_1$ we can assume $c_1=1$.

  $$m_1+c_2{m}_2+\cdots +c_n{m}_n=0$$

• Differentiation gives $0+c_2^{\prime }{m}_2+\cdots +c_n^{\prime }{m}_n=0$ where $c_i^{\prime }=\partial (c_i)$. 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 $K^{\partial }$.

Matrix Of Action

Let $M$ be a finite free $R$ differential module, if $M\cong \underset{i=1}{\overset{n}{⨁}}R{e}_i$ 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 $D(e_i)=\sum {\mathrm{\Lambda }}_{ji}{e}_j$. So in the coordinate form $D=\partial I+\mathrm{\Lambda }$ and the information of $D$ is encoded in a matrix $\mathrm{\Lambda }\in M_n(R)$.

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.

Functors

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

  $${\wedge }_R^k:\mathbf{DM}(R)\to \mathbf{DM}(R),M\mapsto {\wedge }_R^{k}M.$$

  $$D(m_1\wedge \cdots \wedge m_k)=\sum _{i=1}^k{m}_1\wedge \cdots \wedge D(m_i)\wedge \cdots \wedge m_k$$

• Symmetric Power

  $${\mathrm{Sym}}_R^k:\mathbf{DM}(R)\to \mathbf{DM}(R),M\mapsto {\mathrm{Sym}}_R^{k}M.$$

  $$D(m_1\dots m_k)=\sum _{i=1}^k{m}_1\dots D(m_i)\dots m_k$$

Differential Operators / Polynomials

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

$$[\partial ,t]=\partial t-t\partial =\partial (t)=1.$$

$K[\partial ]$ is Euclidean

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).

• For $a,b\in K[\partial ]$ where $b\ne 0$ we can always find $q,r\in K[\partial ]$ such that $a=bq+r$ and $\mathrm{deg}(r)<\mathrm{deg}(b)$.

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 $f\in R$.