Author: Eiko

Tags: Differential Algebra, Cyclic Module, Cyclic Vector, Cyclic Vector Theorem, Differential Equation

Time: 2024-09-15 22:59:44 - 2024-09-15 23:00:01 (UTC)

Cyclic Vector Theorem

Cyclic Modules

Recall that in the theory of ring and modules, we have the concept of modules generated by one element $M = R m$ . Such modules are isomorphic to $R / I$ where $I$ is the kernel of the map $R \to R m = M$ , or the annihilator of $m$ . i.e.

$R m ≅ R / ann (m) .$

We say such $m$ is a cyclic vector for the cyclic module $M$ .

Cyclic Differential Modules

Let $K$ be a differential field, for a differential $K$ -module $M$ , or a $K [\partial]$ -module, we say $M$ is a differential cyclic module if there is a cyclic vector $m$ that is actually a cyclic vector over $K [\partial] / K$ , i.e. such that $M$ is generated by the $K$ -basis $m, \partial m, \dots, \partial^{n - 1} m$ as an $n$ -dimensional $K$ -vector space. In this case the matrix of $\partial$ under this basis is

$(\begin{matrix} - a_{0} \\ 1 & - a_{1} \\ 1 & \dots \\ 1 & - a_{n - 1} \end{matrix}), \partial^{n} m + a_{n - 1} \partial^{n - 1} m + \dots + a_{0} m = 0.$

i.e. let $e_{i} = \partial^{i} m$ , we have

$\partial e_{i} = e_{i + 1}, i = 0, \dots, n - 2; \partial e_{n - 1} = - \sum a_{i} e_{i} .$

If we start with an $n$ -dimensional $K [\partial]$ -module $M$ , i.e. $\dim_{K} M = n$ , if $M$ is cyclic as a $K [\partial]$ -module, then by the fact that ideals in $K [\partial]$ are principal (warning: $K [\partial]$ is not commutative), we can show actually $M ≅ K [\partial] / (P)$ for some

$P = \partial^{n} + a_{n - 1} \partial^{n - 1} + \dots + a_{1} \partial + a_{0} .$

Then take the preimage $m$ of $1$ of that isomorphism, we see that $m, \partial m, \dots \partial^{n - 1} m$ actually generates $M$ as $K$ -vector space.

That is to say, a cyclic differential module over differential field $K$ is just a finite dimensional cyclic $K [\partial]$ -module.

Cyclic Vector Theorem

Let $K$ be a differential field of characteristic $0$ and with non-zero derivation $\partial$ . Then any finite dimensional (over $K$ ) differential module $M$ is cyclic, i.e. $M$ has a cyclic vector.

Proof

The idea of the proof is similar to the primitive element theorem, which considers a generic linear combination $a + t b$ .

Normalize

For a differential module $M$ , $m$ is the cyclic vector in $(M, \partial)$ as $K [\partial]$ -module iff $m$ is a cyclic vector in $(M, u \partial)$ as a $K [u \partial]$ -module, because $(u \partial)^{n} = u (\partial^{n} + (\dots) \partial^{n - 1} + \dots)$ which is monic. If $\partial^{i} m$ generates $M$ , so is $(u \partial)^{i} m$ and vice versa.

This means by assuming $\partial x = y \neq 0$ , and replacing the derivation with $\frac{x}{y} \partial$ , we can assume there exists $x$ such that $\partial x = x$ .

Finding the maximal cyclic submodule

Assume $\dim M = n$ and let $r$ be the dimension of a maximal cyclic submodule. If $r = n$ there is nothing to prove. Assume for now $r < n$ , so any $r + 1$ elements of the form

$a, \partial a, \dots, \partial^{r} a$

must be linearly dependent, the wedge form $a \land \partial a \land \dots \land \partial^{r} a = 0$ vanishes. Take a one-parameter family of elements $a + t b$ where $t \in Q$ , into the wedge form

$(a + t b) \land \partial (a + t b) \land \dots \land \partial^{r} (a + t b) = 0,$

by the characteristic $0$ assumption the field is infinite so each coefficients of $t^{i}$ vanishes as well, taking the coefficient of $t$ we have

$\sum_{i = 0}^{r} a \land \partial a \land \dots \land \partial^{i - 1} a \land \partial^{i} b \land \partial^{i + 1} a \land \dots \land \partial^{r} a = 0.$

Now let $b = x^{s} c$ , compute and plug in

$\partial^{i} b = \sum_{j = 0}^{i} (\binom{i}{j}) s^{j} x^{s} \partial^{i - j} c,$

to get a polynomial in $s$ which vanishes on integers. It has to be zero by the assumption that $K$ has characteristic $0$ . Now consider the $s^{r}$ coefficient

$a \land \partial a \land \dots \land \partial^{r - 1} a \land (s^{r} x^{s} c) = 0 \Rightarrow a \land \partial a \land \dots \land \partial^{r - 1} a \land c = 0.$

Since $c$ is arbitrary, we have in fact for any $a$ ,

$a \land \partial a \land \dots \land \partial^{r - 1} a = 0.$

This means the maximal cyclic submodule have dimension $< r$ , which is a contradiction. So we can only have $r = n$ and $M$ is cyclic.

Dense Generation

The above proof actually tells us that a generic linear combination $a + t b$ is a cyclic vector, so in some sense the vectors which does not generate the module form a subvariety of the projective space of $M$ .