= Solution
Let $D=\partial_t$. Since
$$
[tD+\lambda,D]=-D,
$$
the 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 $x(t^n)=nt^{n-1}$. Therefore the submodules are exactly
$$
0,\quad k\oplus kt\oplus\cdots\oplus kt^m\ (m\geq0),\quad k[t].
$$
The character $x\mapsto0$, $y\mapsto\lambda$ 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 $S\subseteq\mathcal C(I_\lambda)$. Thus
$$
S\subseteq\mathcal C(I_\lambda)\cap\mathcal C(I_{\lambda+1}).
$$
Apply the same argument to each adjacent two-dimensional quotient
$$
(k\oplus\cdots\oplus kt^{n+1})/(k\oplus\cdots\oplus kt^{n-1})
$$
where $x(t^{n+1})=(n+1)t^n\ne0$ because $\operatorname{char}k=0$. Induction gives
$$
S\subseteq\bigcap_{n\in\mathbb N_0}\mathcal C(I_{\lambda+n}).
$$
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