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

A multiplicatively closed set $S$ is a Left Ore set if for every $s\in S$ and $r\in R$ there are $s^{\prime }\in S,r^{\prime }\in R$ with $s^{\prime }r=r^{\prime }s$. This condition gives common left annihilators, so $t_S(M)$ is closed under addition and scalar multiplication in every module.

Conversely apply the assumed submodule property to $M=R/Rs$. The element $1+Rs$ is $S$-torsion, hence so is $r+Rs$. Thus some $s^{\prime }\in S$ satisfies $s^{\prime }r\in Rs$, say $s^{\prime }r=r^{\prime }s$, which is precisely the Ore condition.

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A proper two-sided ideal $P$ is prime when $AB\subseteq P$ for two-sided ideals implies $A\subseteq P$ or $B\subseteq P$. The nilpotent ideal $k{e}_{21}$ lies in every prime of $T_2(k)$, and the quotient is $k\times k$. Hence the two primes areHere $\mathcal{C}(P_a)=\{a\ne 0\}$ and $\mathcal{C}(P_d)=\{d\ne 0\}$.

Direct multiplication of triplesshows that all of $\mathcal{C}(P_a)$ satisfies the left Ore equations. For $P_d$, the largest left Ore subset isIndeed units always form an Ore set. If $d\ne 0$ but $a=0$, applying the Ore equation successively to $e_{21}$ and $e_{22}$ gives incompatible equations, so no left Ore subset can contain that element.

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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

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