Put $R=k[x]$. The relation $yr=ry+r'$ 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 $\operatorname{char}k=0$ and let $0\ne I\triangleleft A_1(k)$. Using the PBW basis $x^iy^j$, 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 139 / 1 / b / Solution

{c}
{title2=$A_n(k)$}
{wiki=Weyl_algebra}

The first Weyl algebra is $A_1(k)=k\langle x,y\rangle/(yx-xy-1)$, the algebra of polynomial-coefficient differential operators in one variable.


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