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Let $K$ be a field of characteristic zero. Let $\Omega = K[x_1, \dots, x_n, dx_1, \dots, dx_n]$ be the differential ring of algebraic differential forms over $K[X_1, \dots, X_n]$.

Is there an algorithm (e.g. by Gröbner basis-like techniques) that solves the ideal membership problem for $\Omega$? That is: given a finitely generated differential ideal $I \subseteq \Omega$, is it decidable whether $f \in I$ or not for a given $f \in \Omega$?

Is calculating the differential radical of $\Omega$ or eliminating variables like in ordinary commutative algebra with Gröbner bases equally possible?

(The reason why I am interested in these things is, of course, algebraic handling of partial differential equations in exterior form.)

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  • $\begingroup$ What is $D(dx_1)$? $\endgroup$
    – Will Sawin
    Commented Dec 25, 2013 at 16:06
  • $\begingroup$ The ring $\Omega$ is super-commutative; the generators $x_i$ are of degree $0$, the generators $dx_i$ are of degree $1$. There is just one differential operator, the derivative $d$ of degree $1$, which fulfills $d \circ d = 0$, that is $d(dx_i) = 0$. $\endgroup$
    – Marc Nieper-Wißkirchen
    Commented Dec 26, 2013 at 6:48
  • $\begingroup$ Somewhat related but probably harder: mathoverflow.net/questions/47657 $\endgroup$
    – Michael Bächtold
    Commented Dec 27, 2013 at 16:16

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Yes. There is such an algorithm. This ring is a finitely generated free module over a polynomial ring, and it is sufficient to solve the submodule membership problem for these modules. But this is easy - the Grobner basis idea works perfectly. Just order the monomials times generators and find a basis for the initial submodule.

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    $\begingroup$ One might add the (maybe trivial) remark that if $I$ is differentially generated by forms $\omega_1,\ldots,\omega_k$ then it's generated as a $K[x_1,\dots,x_n]$ submodule by $\omega_i,\omega_i\wedge dx^I, d\omega_i,d\omega_i\wedge dx^I$. $\endgroup$
    – Michael Bächtold
    Commented Dec 27, 2013 at 16:13

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