# Gaps in Dimension Polynomials

There are several notions of rank/dimension defined on differential fields. However, we do not have a reasonable way to estimate these typically ordinal valued invariants. Especially, we do now know a lower bound for the Lascar Rank, an invariant coming from model theory. It turned out that the question of finding a lower bound is related to the following question.

Let $K\subseteq K\langle \eta\rangle$ be a partial differential field extension of characteristic zero. Suppose that the Kolchin polynomial $\omega_{\eta/K}(t)$ is of degree $n>0$. Is it true that for any $k < n$ there is a $\nu$ in $K\langle \eta\rangle$ such that the degree of $\omega_{\nu/K}(t)$ is $k$? (Here $\eta$ and $\nu$ are finite tuples).

Now, as differential algebra is not a very popular subject and most people do not know what a Kolchin polynomial is, I will also ask a very similar question in commutative algebra.

Let $S$ be a graded commutative algebra over $K[X_1,\ldots,X_d]$ where $K$ is a field of characteristic zero and let $H_S(t)$ be its Hilbert polynomial. Suppose that $\deg (H_S(t))=n>0$. Is it true that for any $k< n$ there is a graded subalgebra $T$ of $S$ such that $\deg(H_T(t))=k$?

-
Why there is a (nontrivial) graded subalgebra at all?! –  Wadim Zudilin Jun 28 '10 at 11:54

"Especially, we do now know a lower bound for the Lascar Rank, an invariant coming from model theory."

Do you actually mean to write "not" in place of "now"? But, this confused me since, in the case of one derivation, $\omega c$ where $c$ is the leading coefficient of the Kolchin polynomial should be a lower bound on Lascar rank.

-
@James: Yes, you are right. I meant not. There is an easy lower bound in the case of ordinary differential fields as you mentioned but the case of several derivations is still open. There are a few things we know though. For instance it is possible to fix the Lascar rank at $\omega$ and increase the leading coefficient of the Kolchin polynomial arbitrarily. Just look at the generic type of the equation $\delta_1(y)=\delta_2^n(y)$. Actually we do not even know if there is a type $p$ of infinite transcendence degree with ${\rm RU}(p)=1$. Even this seems like a nontrivial problem. In the case of difference-differential fields one can construct a type $p$ of infinite transcendence degree with ${\rm SU}(p)=1$ but I couldn't manage to adapt the argument to the partial differential case.