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Let $A_n$ denote the Weyl algebra - the algebra of partial differential operators in $n$ variables with polynomial coefficients.

In papers I've read the following definition:

A left ideal $I$ of $A_n$ is called zero-dimensional, if $\dim_{K(x_1,...,x_n)} A_n/I \lt \infty$.

My question: How can we view $A_n/I$ or generally $A_n$ as $K(x_1,...,x_n)$-vector space?

Arturo Magidin
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user7475
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  • In what papers? – Mariano Suárez-Álvarez Apr 26 '11 at 02:56
  • Notice that it is clear that $A_n$ is not a vector space of that field if one requires that multiplication by the $x_i$ do the obvious thing. – Mariano Suárez-Álvarez Apr 26 '11 at 04:10
  • Are you certain that you don't want the Weyl algebra to be a left module over $K[x_1,\ldots,x_n]$? – Dan Petersen Apr 26 '11 at 12:40
  • I think i understand now what was meant: Instead of – user7475 Apr 26 '11 at 19:54
  • I think i understand now what was meant: Instead of $A_n$ one can consider $A_n(x_1,...,x_n)$ which is also written as $K(x_1,...,x_n)<\partial_1,...,\partial_n>$, the ring of differential operators with coefficients rational functions. This is isomorphic to $K(x_1,...,x_n) \otimes_K A_n$. Now a left ideal I of $A_n$ (which corresponds to a ideal J = $K(x_1,...,x_n) \otimes I$ of $A_n(x_1,...,x_n)$) is called zero-dimensional, if $dim_{K(x_1,...,x_n)} A_n(x_1,...,x_n)/J$ is finite. Is it possible to formulate it this way? Thanks for your help. – user7475 Apr 26 '11 at 20:05

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