Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 139 (Noncommutative Algebra)

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

Schur lemma says that a homomorphism between simple modules is either zero or an isomorphism; consequently the endomorphism ring of a simple module is a division ring. Indeed, the kernel and image of a module homomorphism are submodules. Simplicity makes each either zero or the whole module, proving both assertions.

Solved by gpt-5.6-sol high.

After replacing $X$ by a maximal $D$-linearly independent subset with the same span, write $X=\{x_1,\dots ,x_n\}$. The standard independence lemma, proved by induction using Schur's lemma, saysOtherwise $r{x}_n$ would depend only on $(r{x}_1,\dots ,r{x}_{n-1})$, defining an $R$-map whose coordinate maps lie in ${\mathrm{End}}_R(V)$ and forcing $x_n\in {\sum }_{i<n}{x}_{i}D$. Applying the same argument with $y$ shows that ${\bigcap }_i{\mathrm{ann}}_R(x_i)y=0$ implies $y\in XD$, proving the requested claim.

The Jacobson density theorem states that if $x_1,\dots ,x_n$ are $D$-independent and $y_1,\dots ,y_n\in V$, there is $r\in R$ with $r{x}_i=y_i$ for every $i$. Induct on $n$. First match the first $n-1$ values. The independence lemma makes $I{x}_n$, for $I={\bigcap }_{i<n}{\mathrm{ann}}_R(x_i)$, a nonzero submodule and hence all of $V$; an element of $I$ supplies the final correction.

If $R$ is primitive, choose a faithful simple module $V$. When ${\dim }_{D}V=n<\infty$, density makes $R\to {\mathrm{End}}_D(V)\cong {\mathrm{Mat}}_n(D)$ surjective and faithfulness makes it injective. If ${\dim }_{D}V$ is infinite, choose an $n$-dimensional $D$-subspace $W$ and let $R_n=\{r:rW\subseteq W\}$. Density makes restriction $R_n\to {\mathrm{End}}_D(W)\cong {\mathrm{Mat}}_n(D)$ surjective.

Solved by gpt-5.6-sol high.

For a simple $R$-module $S$, matrix units show that every nonzero vector of $S^n$ generates all coordinates, so $S^n$ is a simple ${\mathrm{Mat}}_n(R)$-module. Conversely, if $V$ is simple over the matrix ring, $e_{11}V$ is a simple $R$-module andthrough the maps induced by $e_{i1}$ and $e_{1i}$. These constructions are inverse on isomorphism classes; this is the basic Morita equivalence for a matrix ring.

The Jacobson radical is the intersection of annihilators of all simple left modules. On $S^n$, a matrix annihilates every vector exactly when each entry annihilates $S$. Intersecting over all simple $S$ gives

Solved by gpt-5.6-sol high.

A module $E$ is injective if every map $A\to E$ extends across every inclusion $A\hookrightarrow B$. Baer criterion says it suffices to test inclusions of left ideals $I\hookrightarrow R$.

Necessity is immediate. Conversely, order all extensions of a given map $A\to E$ to intermediate submodules of $B$. A maximal one exists by Zorn's lemma. If its domain $C$ is not $B$, choose $b\notin C$ and let $I=\{r:rb\in C\}$. The map $I\to E$, $r\mapsto f(rb)$, extends to $R$ by the hypothesis; its value at $1$ extends $f$ to $C+Rb$, contradicting maximality. Thus $C=B$.

Solved by gpt-5.6-sol high.

If $R$ is left Noetherian and $I\to {⨁}_{\alpha }{E}_{\alpha }$ is a map from a left ideal, finitely many generators of $I$ have support in one finite set of summands. The map therefore lands in a finite direct sum of injectives and extends to $R$. Baer's criterion proves that the full direct sum is injective.

Conversely, let $I_1\subseteq I_2\subseteq \cdots$ and put $I=\bigcup I_n$. Embed each $R/I_n$ in an injective module $E_n$. The maphas finite support. If the direct sum is injective, it extends to $R$; the extension's value at $1$ has finite support, forcing $a\in I_n$ for every sufficiently large $n$ and every $a\in I$. Hence the chain stabilizes. This is the Bass-Papp theorem.

Solved by gpt-5.6-sol high.

Over a commutative PID, Baer's criterion reduces to maps $(r)\to E$. Such a map extends to $R$ exactly when every equation $re=x$ with $r\ne 0$ is solvable. Thus injective modules are exactly the divisible modules.

Let $Q$ be the fraction field. The indecomposable injectives arefor one representative $p$ of each associate class of irreducibles. The latter is the $p$-primary Prüfer module, the union of the cyclic modules generated by $p^{-n}+R$. The structure theorem for divisible modules decomposes every divisible module into copies of $Q$ and these Prüfer modules, proving that the list is complete.

Solved by gpt-5.6-sol high.

Because $N$ is essential in $M$, every associated prime of $M$ occurs in $N$. Since $JN=0$, all these primes contain $J$. For a finitely generated module over a commutative Noetherian ring,Hence $J\subseteq \sqrt{\mathrm{ann}M}$; finite generation of the ideal $J$ gives $J^{n}M=0$ for some $n$.

Now $R/J$ is essential in its injective hull. For $x\in E(R/J)$, the finitely generated module $Rx+R/J$ has essential submodule $R/J$, so the result just proved gives $J^{n}x=0$ for some $n$. The reverse inclusion is tautological, and therefore

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

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.