Bernstein–Sato polynomial

Remarks

Bernstein proved that this ideal is nonzero, so $b_f$ exists. For a nonconstant $f$, $s + 1$ divides $b_f(s)$, as one sees by setting $s = -1$.

Example: $f = x_1^2 + \cdots + x_n^2$ and $P = \Delta = \partial_1^2 + \cdots + \partial_n^2$. One has $\partial_i (f^{s+1}) = 2 (s + 1) x_i f^s$, and then $$\Delta (f^{s+1}) = 2(s+1) \left( n f^s + 2s \sum_i x_i^2 f^{s-1} \right) = 4 (s + 1) \left( s + \tfrac{n}{2} \right) f^s.$$ So $(s+1)(s + n/2)$ lies in the ideal; it is in fact $b_f(s)$.

Non-example: for $f = x_1$, the polynomial $(s + 1)(s + 2)$ lies in the ideal, with $P = (s + 2) \partial_1$, but it is not $b_f$. It is not of least degree, since $\partial_1 (x_1^{s+1}) = (s+1) x_1^s$ gives $b_f(s) = s + 1$.

The same definition works for a germ of holomorphic function, with the ring of differential operators with convergent power series coefficients; this gives the local $b$-function at a point.