Let $k$ be a field and let $A_n(k)$ be the $n^\text{th}$ Weyl algebra (i.e. the algebra generated by indeterminates $X_1,\cdots,X_n,Y_1,\cdots,Y_n$ such that $Y_iX_i-X_iY_i = 1$ and everything else commutes).
I read that $A_n(k)$ is simple and I am wondering how one can prove this result.
Looking on Google I only found this question on Stackexchange, where the simplicity was proven for the $n=1$ case by an argument using differential operators.
However, for arbitrary $k$, this argument does not work because we cannot use differentiation. Do you know a proof for arbitrary $k$ and $n$?
Thank you very much!
Step one prove the normal ordered term like this $\prod_{i}X_i^{\lambda_i}\prod_{j}Y_j^{\sigma_j}$ forms a basis of algebra $A_n$. Step two check with commutation relations for a polynomial $p:=\sum_{[\lambda],[\sigma]} c_{[\lambda], [\sigma]}\prod_{i}X_i^{\lambda_i}\prod_{j}Y_j^{\sigma_j}$ $$ [Y_i, p] = \frac{\partial}{\partial X_i} p $$ and $$ [X_i, p] = -\frac{\partial}{\partial Y_i} p $$ then we get for any none empty ideal $I\subset A_n$ it contains $1$.