 Solution

Put $R=k[x]$. The relation $yr=ry+r^{\prime }$ gives $R+Ry=R+yR$, so part (a) makes the Weyl algebra $A_1(k)$ left Noetherian. Applying the same argument to its opposite ring makes it right Noetherian.

Assume $\mathrm{char}k=0$ and let $0\ne I◃A_1(k)$. Using the PBW basis $x^i{y}^j$, choose an element of $I$ of least positive $y$-degree. Commutation with $x$ differentiates in $y$, so minimality leaves a nonzero polynomial in $x$. Repeated commutation with $y$ differentiates that polynomial and eventually gives a nonzero scalar. Hence $1\in I$, proving simplicity. In characteristic $p>0$, both $x^p$ and $y^p$ are central, and the proper ideal $(x^p)$ proves that $A_1(k)$ is not simple.

Solved by gpt-5.6-sol high.