$t=1$, $J$ is prime and there is nothing to prove. Otherwise, fix $i$, write $\mathfrak{p}$ for $\mathfrak{p}_i$, and let $\mathfrak{q} = \cap_{j \neq i} \mathfrak{p}_j$. There exists some $c \in \mathfrak{q} \setminus \mathfrak{p}$. Let $a, b \in \mathfrak{p}$ be given. We have $ac, bc \in \cap_j \mathfrak{p}_j = J$, and hence $\{ac,bc\} \in J$, since $J$ is involutive by Theorem \[gabber\]. We now use the fact that the Poisson bracket is a biderivation, which gives $$\{ac,bc\} = a\{c,b\}c + a\{c,c\}b + c\{a,b\}c + c\{a,c\}b.$$ The first, second, and fourth summands on the right-hand side all have a factor of $a$ or $b$ and thus belong to $\mathfrak{p}$. Since the sum belongs to $\mathfrak{p}$, the third summand, $c^2\{a,b\}$, belongs to $\mathfrak{p}$ as well. As $\mathfrak{p}$ is prime and $c^2 \notin \mathfrak{p}$, we must have $\{a,b\} \in \mathfrak{p}$, and so $\mathfrak{p}$ is involutive.

\[zetahom\] If $M$, $J$, and $\mathfrak{p}$ are as above, we note that $J$ is homogeneous with respect to $\zeta_1, \ldots, \zeta_n$ by definition, and by [@matsumura Thm. 13.7(i)], $\mathfrak{p}$ is homogeneous with respect to the $\zeta_i$ as well.

Kernels
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In this section, we prove that the kernel $M_*$ of $\partial_n$ acting on a holonomic ${\mathcal{D}}$-module $M$ is a 