Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 1

The Tensor-Hom adjunction is the natural isomorphismFor an R-module homomorphism $f$, it is given explicitly byConversely, an $R$-linear map $g:M\to {\mathrm{Hom}}_R(N,P)$ determines the balanced map $(m,n)\mapsto g(m)(n)$, so the universal property of the tensor product of modules givesThese formulas are inverse to each other because pure tensors generate the tensor product of modules.

Solved by gpt-5.6-sol high.

Let $u:M^{\prime }\to M$ be an R-module homomorphism. Naturality in the left argument means that precomposition by $u\otimes 1_N$ on the left corresponds under the Tensor-Hom adjunction to precomposition by $u$ on the right. For $f:M{\otimes }_{R}N\to P$,Thus the naturality square commutes pointwise on every $m^{\prime }\in M^{\prime }$ and $n\in N$.

Solved by gpt-5.6-sol high.

No. Let $Q$ be the quiver $1\to 2$ over a field $k$, and take the representation of a quiverAn endomorphism is a pair of scalar maps $(a,b)$ satisfying $b=a$, so its endomorphism ring is $k$. Every nonzero endomorphism is therefore an isomorphism, making this representation a brick module. It nevertheless has the proper nonzero subrepresentation $0\to k$, so it is not an irreducible module.

Equivalently, this is a nonsimple module over the path algebra $kQ$ whose endomorphism ring is a division ring.

Solved by gpt-5.6-sol high.

Write the nonsplit short exact sequenceFor an endomorphism $g:M\to M$, the composite $qgj:L\to N$ vanishes because ${\mathrm{Hom}}_R(L,N)=0$. Hence $g$ restricts to an endomorphism $f$ of $L$ and induces an endomorphism $h$ of $N$, giving a commutative diagram of short exact sequences.

Because $L$ and $N$ are brick modules, each of $f,h$ is either zero or an isomorphism. If both are isomorphisms, the short five lemma makes $g$ an isomorphism. If both vanish, $g$ factors successively through $N$ and through $L$, hence through a map $N\to L$; this map is zero, so $g=0$.

The mixed cases would split the sequence. If $f$ is invertible and $h=0$, then $qg=0$, so $g=j\alpha$ for some $\alpha :M\to L$; the identity $\alpha j=f$ makes $f^{-1}\alpha$ a retraction of $j$. If $f=0$ and $h$ is invertible, then $gj=0$, so $g=\beta q$; the identity $q\beta =h$ makes $\beta h^{-1}$ a section of $q$. Both contradict nonsplitting. Thus every endomorphism of $M$ is zero or invertible, and $M$ is a brick.

Solved by gpt-5.6-sol high.