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The Weyl algebra is $$ A=k\langle x,y\rangle/\langle yx-xy-1\rangle $$ where $k$ is a field, $k\langle x,y\rangle$ is the free algebra, and $yx-xy-1$ is the defining relation. A basis for $A$ is $\{ x^i y^j|i,j\geq0\}$. What I am concerned about here is how to prove that elements of this set are linearly independent. Can we reason by contradiction that if $\sum c_{ij}x^i y^j=0$, then (after rewriting $\sum c_{ij}x^i y^j=\sum f_j(x)y^j$), $f_j(x)=0,\forall j$, thus $c_{ij}=0,\forall i,j$. This is a bit weird, I feel.

Another related question is about the $q$-Weyl algebra $$ B=k\langle x,y\rangle/\langle xx^{-1}-1,x^{-1}x-1,yy^{-1}-1,y^{-1}y-1,yx-qxy \rangle $$ where $q\in k$. How to show that that a basis for $B$ is $\{x^i y^j|i,j\in \mathbb{Z}\}$? Can someone give me some hints on these two problems? Thank you very much!

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    I edited for proper use of \langle and \rangle in TeX, i.e. where it said $$, it now says $\langle x,y \rangle$.2011-09-19
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    Can you use [Poincaré-Birkhoff-Witt theorem](http://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Birkhoff%E2%80%93Witt_theorem) to reach the conclusion of your first question? The Lie algebra would be the span of $\{1,x,y\}$, but we need a method for identifying the generator $1$ with the multiplicative identity. Moding out a suitable ideal out of the universal enveloping algebra should do it?2011-09-19
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    I tidied up the question a bit, because I found it hard to read at first. Hope you don't mind.2011-09-20
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    @GeorgeLowther:Thank you very much for the editing work!By the way,can you tell me how to edit this?I am not very familiar with using Latex here.:)2011-09-20
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    @user14242: You should see "edit" below your post, which you can click on and you can see the source.2011-09-20

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