TITLE: On a proof of the basis for the Weyl algebra
QUESTION [13 upvotes]: 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!

REPLY [5 votes]: You just have to find a representation of the algebra (as Qiaochu discussed). A simple way of doing this is to consider the Weyl algebra acting on the (commutative) polynomial ring $k[u,v]$. Regarding this as a vector space over $k$ with basis $u^iv^j$ ($i,j\ge0$), you can define $x,y$ as linear operators on $k[u,v]$,
$$
\begin{align}
&x(u^iv^j)=u^{i+1}v^j,\\
&y(u^iv^j)=iu^{i-1}v^j+u^iv^{j+1}.
\end{align}
$$
You can check that $xy-yx=1$, so $k\langle x,y\rangle$ is a representation of the Weyl algebra. Also, if $\sum_{ij}c_{ij}x^iy^j=0$, then applying this to $1\in k[u,v]$ gives $\sum_{ij}c_{ij}u^iv^j$, so $c_{ij}=0$.
Note: All I did here was to look at the multiplication rules $x\cdot x^iy^j=x^{i+1}y^j$ and $y\cdot x^iy^j=ix^{i-1}y^j+x^iy^{j+1}$, then replace these by operations on $u^iv^j$.
The same thing works for the q-Weyl algebra. Consider $V=k[u,v,u^{-1},v^{-1}]$ as a vector space over $k$ with basis $u^iv^j$ ($i,j\in\mathbb{Z}$). Define $x,y$ acting on $V$ as
$$
\begin{align}
&x(u^iv^j)=u^{i+1}v^j,\\
&y(u^iv^j)=q^iu^iv^{j+1}.
\end{align}
$$
Then, $x,y$ are invertible linear operators with $yx=qxy$, so $k\langle x,y\rangle$ is a representation of the q-Weyl algebra. As before, if $\sum_{ij}c_{ij}x^iy^j=0$ then applying this to $1\in V$ gives $\sum_{ij}c_{ij}u^iv^j=0$, and $c_{ij}=0$.