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Let $p\colon \mathbb{R}^2 \to \mathbb{R}$ be a polynomial with a non-vanishing gradient at $p^{-1}(0)$. Then, the implicit function theorem says that $S = \{(x,y) \in \mathbb{R}^2 \mid p(x,y) = 0\}$ is an embedded smooth curve. In more algebraic terms, its coordinate ring $\mathbb{R}[x,y]/(p)$ has Krull dimension $1$ and its localization at every maximal ideal is regular.

Consider now a linear differential operator $T$ in two variables (ie some finite expression involving linear combinations of $\frac{\partial^{i+j}}{\partial x^i\partial y^j}$ with coefficients in $\mathbb{R}[x,y]$), and lets say it is acting formally on the space of all power series $\mathbb{R}[\![x,y]\!]$. I am not a priori concerned with convergence, but if needed we may restrict to all analytic functions defined on some open ball centered at the origin. Is there any meaningful way of saying that the set $$S = \{f \in \mathbb{R}[\![x,y]\!] \mid Tf = 0\}$$ looks locally like a one-variable power series ring (under suitable regularity conditions on $T$)? I'm particularly interested in a more formal, algebraic approach, that is not very dependend on the base field $\mathbb{R}$.

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    $\begingroup$ It seems to be equivalent to ask whether the $\mathcal D$-module $\mathcal D/\mathcal DT$ is regular holonomic. I am not familiar with $\mathcal D$-modules thus I hope that an expert would explain this. $\endgroup$
    – Z. M
    Commented Jun 4 at 11:06
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    $\begingroup$ @Z.M restricting from power series to polynomials, to say that $\mathcal{D}/\mathcal{D}T$ is regular holonomic, it means that the graded ideal $J = \operatorname{gr} \mathcal{D}T$ of the polynomial ring $\mathbb{R}[x,y,\partial_x,\partial_y]$ is radical and the associated characteristic variety is 2-dimensional in $\mathbb{R}^4$ (ignoring any issue coming from $\mathbb{R}$ not being algebraically closed). How does this relate to the solution space $\operatorname{Hom}(\mathcal{D}/\mathcal{D}T, \mathbb{R}[x,y])$? Can one read the dimension from this vector space? $\endgroup$
    – HASouza
    Commented Jun 4 at 18:44
  • $\begingroup$ I've been thinking a bit more about this, and if $R$ denotes the coordinate ring $\mathbb{R}[x,y,\partial_x,\partial_y]/J$, then the graded object associated to the degree function on the solution space embeds in the group $\operatorname{Hom}_{\operatorname{gr} \mathcal{D}}(R, \mathbb{R}[x,y])$ of graded homomorphisms. But playing around with some examples, I don't think this says much, as for $T = \partial_i$ this graded $\operatorname{Hom}$ group is just isomorphic to $\mathbb{R}[x,y]$! $\endgroup$
    – HASouza
    Commented Jun 5 at 7:17
  • $\begingroup$ In fact, if $e(M)$ denotes the degree of the Hilbert polynomial of the solution space and $d(M)$ the dimension of the characteristic variety, the statement I'm looking after would be something like this: $e(M) + n = d(M)$. For holonomic $M$ we have $d(M) = n$ and the solution space is finite dimensional, so $e(M) = 0$. For $M$ presented by killing only one derivative $\partial_i$, the characteristic variety is a hyperplane, so $d(M) = 2n -1$, and the solution space is a space of polynomials in $n-1$-variables, so $e(M) = n-1$. $\endgroup$
    – HASouza
    Commented Jun 5 at 10:09
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    $\begingroup$ You cannot restrict to polynomials for solution space, since after restriction, you are only looking at algebraic solutions, which rarely exist. $\endgroup$
    – Z. M
    Commented Jun 5 at 11:25

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