Apologies for the late reply. Thank you so much for the reference and the explanation!

That was my fault! Very, very sorry! I initially opted to move it to the chat and then deleted the (automatic) comment because I thought that having a discussion here was better, only to change my mind again!

@LSpice I am very sorry, but could you please elaborate a bit more on that? I am having difficulty understanding exactly what you mean.

@LSpice What if I quotient it by the ideal generated by $x^p$ and $\partial^p$? My hunch is that this will become a matrix algebra. Do you think this makes sense? If so, do you see any direct way of proving it?

@abx Do you have any ideas for an elementary proof of the statement in the case of the Weyl algebra?

@LSpice Thank you! Is there any simple way to prove the statement from the paper for the Weyl algebra?

@abx If I consider $n=1$ and consider $k[x]/x^p$ and let $x$ and $\partial$ define the standard actions on $k[x]/x^p$. Then since the centre of the Weyl algebra is just $k[x^p, \partial^p]$, each element of the quotient will define an endomorphism of $k[x]/x^p$. Will/can this be an isomorphism?

Thank you very much for the quick reply. Even if that is the case, will the quotient of the Weyl algebra by its centre be isomorphic to a matrix algebra?