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

A ring is left Noetherian when its left ideals satisfy the ascending-chain condition, equivalently when every left ideal is finitely generated; right Noetherian is defined analogously.

Filter $S$ by word degree in $x$. The equality $R+Rx=R+xR$ lets every coefficient move past one $x$ at the cost of lower-degree terms, soFor a left ideal $I$, the leading coefficients in degree at most $n$ form an ascending chain of left ideals of $R$. Since $R$ is left Noetherian, this chain stabilizes and each term is finitely generated. Lift finitely many generators through the finitely many degrees before stabilization. Division by their leading terms reduces every element of $I$ to lower degree, and induction shows that these lifts generate $I$. Thus $S$ is left Noetherian.

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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.

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Let $e=1-XY$, the projection onto constants. Thenare matrix units: $e_{ij}{e}_{rs}={\delta }_{jr}{e}_{is}$. Consequentlyis a strictly ascending chain of left ideals; each new column is independent. Thus $T$ is not left Noetherian. The corresponding row chainshows that it is not right Noetherian.

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