Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 139 / 4 / c

Let $D={\partial }_t$. Sincethe assignments ${\theta }_{\lambda }(y)=tD+\lambda$ and ${\theta }_{\lambda }(x)=D$ respect $[y,x]=-x$ and extend through the universal enveloping algebra.

The operator $y$ has distinct eigenvectors $t^n$ with eigenvalues $\lambda +n$, while $x(t^n)=n{t}^{n-1}$. Therefore the submodules are exactlyThe character $x\mapsto 0$, $y\mapsto \lambda$ has kernel $I_{\lambda }$, so $U(\mathfrak{g})/I_{\lambda }\cong k$ and $I_{\lambda }$ is maximal.

By part (a), the $S$-torsion in every module is a submodule. On $V=k\oplus kt$, $x(t)=1$ and $y$ has eigenvalues $\lambda ,\lambda +1$. If some $s\in S$ vanished on the $(\lambda +1)$-character, then $s(t)\in k$. Since $s(1)$ is the nonzero scalar given by its image modulo $I_{\lambda }$, subtracting a suitable constant from $t$ would produce an $S$-torsion vector $v$ with $xv=1$. Submodule closure would make $1$ torsion, contradicting $S\subseteq \mathcal{C}(I_{\lambda })$. ThusApply the same argument to each adjacent two-dimensional quotientwhere $x(t^{n+1})=(n+1)t^n\ne 0$ because $\mathrm{char}k=0$. Induction gives

Solved by gpt-5.6-sol high.