Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 (Commutative Algebra)

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.

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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 $n\in N$.

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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.

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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.

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A minimal primary decomposition is an expressionin which every $Q_i$ is a primary ideal, the prime ideals $\sqrt{Q_i}$ are pairwise distinct, and the decomposition is irredundant: deleting any $Q_i$ changes the intersection.

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The Second uniqueness theorem for primary decomposition says that in a minimal primary decomposition of an ideal in a Noetherian ring, every primary component belonging to an isolated prime is unique. Here an isolated prime is a minimal member of the set $\{\sqrt{Q_i}\}$.

Let $\mathfrak{p}=\sqrt{Q_i}$ be isolated and apply localization at a prime ideal. If $j\ne i$, minimality of $\mathfrak{p}$ gives $\sqrt{Q_j}⊈\mathfrak{p}$, so some element of $Q_j$ becomes a unit in $R_{\mathfrak{p}}$. ConsequentlyBecause $Q_i$ is $\mathfrak{p}$-primary, multiplication by any $s\notin \mathfrak{p}$ cannot carry an element outside $Q_i$ into $Q_i$. ThereforeThe right side depends only on $I$ and $\mathfrak{p}$, proving uniqueness.

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No. Take the Noetherian ring $R=k[x]$, the finitely generated $R$-moduleand $N=0$. Its annihilator of a module iswhich is a prime ideal and hence a primary ideal. But with $r=x$ and $m=(0,\overline{1})$ we have $m\ne 0$ and $rm=0$, while $x^{k}M\ne 0$ for every $k$ because the free summand survives. Thus $N$ is not a primary submodule.

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Pass to $A=R/I$. It is enough to prove that the zero ideal of $A[t]$ is primary. The zero ideal of $A$ is primary, so every zero divisor of $A$ is nilpotent element.

Suppose $fg=0$ in $A[t]$ with $g\ne 0$. By McCoy theorem, some nonzero $a\in A$ satisfies $af=0$. Hence every coefficient of $f$ is a zero divisor and therefore nilpotent. There are only finitely many coefficients, so the ideal they generate is nilpotent; consequently some power of $f$ is zero. This proves that $(0)$ is primary in $A[t]$, and the coefficientwise quotient of a polynomial ringshows that $I^e$ is primary.

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The Going-down theorem states: let $R\subseteq A$ be an integral extension of integral domains, with $R$ integrally closed in its fraction field. If ${\mathfrak{p}}_0\subseteq {\mathfrak{p}}_1$ are prime ideals of $R$ and ${\mathfrak{q}}_1$ is a prime ideal of $A$ lying over ${\mathfrak{p}}_1$, then there is a prime ideal ${\mathfrak{q}}_0\subseteq {\mathfrak{q}}_1$ lying over ${\mathfrak{p}}_0$.

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Put $K=\mathrm{Frac}R$, letbe the minimal polynomial of an algebraic element $y$ over $K$, and let $B$ be the integral closure of $R$ in a finite normal extension containing all roots of $f$. Since $y$ is integral over $R$ and $R$ is integrally closed domain, every $a_i$ belongs to $R$.

Write $y={\sum }_j{p}_j{z}_j$ with $p_j\in \mathfrak{p}$ and $z_j\in A$. Every $K$-embedding into the normal extension fixes the $p_j$ and sends each $z_j$ to an element integral over $R$. Thus every conjugate of $y$ lies in the extended ideal $\mathfrak{p}B$. Each nonleading coefficient of $f$ is, up to sign, an elementary symmetric polynomial in those conjugates, so it lies in $\mathfrak{p}B\cap R$.

For an integral extension, extension followed by contraction preserves a prime ideal:Indeed, the determinant trick gives $r^m\in \mathfrak{p}$ for $r\in \mathfrak{p}B\cap R$, and primality then gives $r\in \mathfrak{p}$. Hence $a_i\in \mathfrak{p}$ for every $i<d$.

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As an $A^{\prime }$-algebra, $A{\otimes }_R{A}^{\prime }$ is generated by the elements $a\otimes 1$. If $a$ obeys a monic relationover $R$, then $a\otimes 1$ obeys the same monic relation after applying the structure map $R\to A^{\prime }$. Thus every generator is an integral element. The subalgebra generated by finitely many integral elements is finite as a module, and therefore integral; each tensor involves only finitely many generators. Hence $A{\otimes }_R{A}^{\prime }$ is integral over $A^{\prime }$.

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If the coefficients of $p$ are integral over $R$, they generate a finite $R$-algebra $B$. Then $B[t_1,\dots ,t_n]$ is a finite $R[t_1,\dots ,t_n]$-module, so every one of its elements, including $p$, is integral.

Conversely, use the fact that the integral closure of a graded ring is graded. Give $A[t_1,\dots ,t_n]$ its ${\mathbb{N}}^n$-grading and regard $R[t_1,\dots ,t_n]$ as a graded subring. If $p$ is integral, each homogeneous component $a_{\alpha }{t}^{\alpha }$ is integral. Applying the evaluation homomorphism $t_1=\cdots =t_n=1$ shows that every coefficient $a_{\alpha }$ is integral over $R$.

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The inverse limit is the submodule of the direct product consisting of compatible families:Its projection to $N_i$ sends $(x_j)$ to $x_i$; these projections satisfy the universal property of an inverse limit.

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Two decreasing filtrations of a module $(M_n)$ and $(M_n^{\prime })$ are equivalent when each contains a fixed shift of the other: there are $a,b\ge 0$ such thatfor every $n\ge 0$.

For an $I$-filtration, $I{M}_n\subseteq M_{n+1}$, so $I^n{M}_0\subseteq M_n$. If it is stable from $s$ onward, thenIt is therefore equivalent to the I-adic filtration. Any two stable $I$-filtrations are consequently equivalent to each other.

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No. Give $R=k[x]$ the x-adic filtration $R_n=(x^n)$, take $M=R$, and let $N=(x)$. At $n=1$ the intersection filtration giveswhereas the induced filtration of $N$ givesThus intersection with a submodule need not equal the induced filtration.

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Yes. Scalar multiplication in the quotient module gives directlyHence the displayed filtration is precisely the induced filtration.

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Work modulo $I$. Putand let $A$ be the image of $I^{\prime }/I$ in $C$. The separating condition $I^{\prime }=I^{\prime }\cap (I+J)$ modulo $I$ says that $I^{\prime }/I\to C$ is injective, so we may regard $A$ as a submodule of $C$.

Since $r\in \sqrt{(I:I^{\prime })}$, some power $r^q$ annihilates $A$. Apply the Artin-Rees lemma to $A\subseteq C$ and the principal ideal $(r)$. There is $s$ such that for every $m\ge s$,For $m\ge s+q$, the right side is zero. Pulling the equality $A\cap r^{m}C=0$ back to $R$ givesas required.

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For an $\mathfrak{m}$-primary ideal $I$ in a Noetherian local ring, the functionagrees for all sufficiently large $n$ with a polynomial in $n$. This is the Hilbert-Samuel polynomial, also called here the characteristic polynomial of $I$. Using $R/I^{n+1}$ instead merely shifts its variable.

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Suppose $I$ has $s$ generators. Its associated graded ringis generated in degree one by their initial forms, so there is a graded surjectionBecause $I$ is $\mathfrak{m}$-primary, $R/I$ has finite length of a module. The degree-$n$ piece on the left has lengthso $\mathrm{length}(I^n/I^{n+1})$ grows with degree at most $s-1$. Summing these lengths shows that $\mathrm{length}(R/I^n)$ has polynomial degree at most $s$.

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The weighted Hilbert series of $K[x,y,t_1,\dots ,t_n]$ isThe homogeneous polynomial $x^k-y^k$ has degree $k$ and is a non-zero-divisor, so quotienting by it multiplies the series by $1-z^k$. Therefore the requested Poincare series of a graded module is

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Let $P=p_1\cdots p_n$ and $A=\mathbb{Z}/P\mathbb{Z}=R_0$. By the Chinese remainder theorem, $A\cong {\prod }_i{\mathbb{F}}_{p_i}$, so ${\mathrm{length}}_A(A)=n$.

If a positive-degree monomial contains both $t_i$ and $t_j$ with $i\ne j$, then it vanishes: Bezout identity gives $u{p}_i+v{p}_j=1$, while both $p_i$ and $p_j$ annihilate that monomial. Thus the degree-$d$ component for $d\ge 1$ isand every summand has length one. Hence every $R_d$ has length $n$, including $d=0$, and

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The denominator has degree one, independently of the number $n$ of variables. This reflects the fact that all mixed monomials vanish and each component of the ring supports only one polynomial direction; equivalently, the Krull dimension of this graded ring is one.

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