# 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

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.