# Bernstein's theorem

In the book "An introduction to the theory of local zeta functions" prof. Igusa presents Bernstein's theorem as follows:

Let $K_0$ be a field, and write $K=K_0(s)$. Let $f\in K_0[x_1,\dots,x_n]\setminus\{0\}$. We consider the $D$-module $$K[x]_f=K_0(s)[x_1,\dots,x_n,1/f]$$ where $D$ acts by $$x_i\cdot\phi(x)=x_j\phi(x)\quad \frac{\partial}{\partial x_i}\phi(x)=\frac{\partial\phi(x)}{\partial x_i}+s\phi(x)\frac{\partial f}{\partial x_i}\frac{1}{f}$$ In other words $D$ acts on $\phi$ as it would on $\phi f^s$ for $s\in\mathbb{Z}$. Now the formulation of Bernstein's theorem is:

There exists an element $P_0\in D$ such that $P_0\cdot f(x)=1$

I understand what this means, but I don't understand how the definition of the Bernstein polynomial of $f$ makes sense once we have this theorem. The Bernstein polynomial is defined to be the generator of the ideal of polynomials $b\in K_0[s]$ such that there exists $P\in D\otimes K_0[s]$ such that $$P\cdot f=b(s)$$ However, according to the formulation of Bernstein's theorem as above we would have that $b(s)=1$, which cannot be the point of this... In other words, there is something that I do not understand about either Bernstein's theorem or the definition of the Bernstein polynomial, and some clarification would be appreciated.

Turns out I overlooked something: $P_0\in D$ means that the coefficients of $P_0$ will generally be elements of $K_0(s)$, in the definition of $b_f(s)$ we consider only $P$ with coefficients in $K_0[s]$.