Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
In what papers? –  Mariano Suárez-Alvarez Apr 26 '11 at 2:56
    
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-Alvarez Apr 26 '11 at 4:10
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.