Weyl algebras $A_n(k)$ as tensor product of the first Weyl algebra In afew threads I've read that the Weyl algebra $A_{n+1}(k)$ is isomorphic to the $k$-tensor product of $A_n(k)$ with $A_1(k)$, why is this true?
 A: Let $A_n(k)$ (as usual!) be the $k$ algebra freely generated by letters $p_1$, $\dots$, $p_n$, $q_1$, $\dots$, $q_n$, subject to the relations $$[p_i,q_j]=\delta_{i,j},\qquad[p_i,p_j]=[q_i,q_j]=0.$$ There are algebra maps $f:A_n(k)\to A_{n+1}$ and $g:A_1(k)\to A_{n+1}$ such that $f(p_i)=p_i$, $f(q_i)=q_i$ for all $i\in\{1,\dots,n\}$ and $g(p_1)=p_{n+1}$ and $g(q_1)=q_{n+1}$ (this can be checked by using the defining relations of the domain) and, moreover, the image of $f$ commutes with that of $g$ in $A_{n+1}(k)$. It follows that there is an algebra map $f\otimes g:A_n(k)\otimes A_1(k)\to A_{n+1}$. 
On the other hand, there is an algebra map $h:A_{n+1}\to A_n(k)\otimes A_1(k)$ such that $h(p_i)=p_i\otimes 1$, $h(q_i)=q_i\otimes 1$ for $i\in\{1,\dots,n\}$ and $h(p_{n+1})=1\otimes p_1$ and $h(q_{n+1})=1\otimes p_{1}$; this can be checked again simply by verifying that this plays well with the relations defining $A_{n+1}(k)$.
Finally, one can check at once that $f\otimes g$ and $h$ are inverse isomorphisms.
