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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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What is $D(dx_1)$? –  Will Sawin Dec 25 '13 at 16:06
    
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 $d(dx_i) = 0$. –  Marc Nieper-Wißkirchen Dec 26 '13 at 6:48
    
Somewhat related but probably harder: mathoverflow.net/questions/47657 –  Michael Bächtold Dec 27 '13 at 16:16
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up vote 4 down vote accepted

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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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 $\omega_i,\omega_i\wedge dx^I, d\omega_i,d\omega_i\wedge dx^I$. –  Michael Bächtold Dec 27 '13 at 16:13
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