${\mathcal{D}}_{n-1}$-submodule of $M$, this calculation shows that $R \cdot M_*$ is a ${\mathcal{D}}$-submodule of $M$. Furthermore, it is clear that $\ker(\partial: R \cdot M_* \rightarrow R \cdot M_*)$ coincides with $M_*$. Therefore, we may assume that $M = R \cdot M_*$. With this assumption, we conclude at once that $M = M_* + x_nM$ as ${\mathcal{D}}_{n-1}$-modules, since $M_*$ is an $R_{n-1}$-module. We claim that this sum is direct. Suppose that $m \in M_* \cap x_nM$. Since $x_nM = x_n(R \cdot M_*)$, we can write $m = x_n(\sum r_im_i)$ for some $r_i \in R$ and $m_i \in M_*$. From this, we obtain an equation $m - x_n(\sum r_im_i) = 0$ where $m$ and all the $m_i$ belong to $M_*$. By Lemma \[kerlemma\], the constant term of the left-hand side with respect to $x_n$, which is simply $m$, also vanishes. Thus $M_* \cap x_nM = 0$, and so $M = M_* \oplus x_nM$. This direct sum decomposition implies that $M_* \simeq M/x_nM$, which is a holonomic ${\mathcal{D}}_{n-1}$-module by Proposition \[bjorkxn\]. This completes the proof.

Regularity
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In this section, we define what it means for a ${\mathcal{D}}$-module $M$ to be *$x_n$-regular*. If $M$ is holonomic, this is a weak condition that is 