Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 (Commutative Algebra)

For $R$-modules $M,N$, the tensor product of modules is an $R$-module $M{\otimes }_{R}N$ together with the balanced mapsuch that every balanced map $b:M\times N\to P$ factors through one unique $R$-linear map $\tilde{b}$:Equivalently,naturally in $P$. This is the universal property of the tensor product of modules.

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Yes. The integers form a principal ideal domain, and over a principal ideal domain a module is flat exactly when it is torsion-free. Thus the torsion-free modules $A$ and $B$ are flat modules. The functoris a composite of two exact tensor functors, so $A{\otimes }_{\mathbb{Z}}B$ is flat. Applying the converse direction of the same characterization shows that it is torsion-free. This is the torsion-free module over a principal ideal domain is flat criterion.

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No. Take $R=\mathbb{Z}$, $M=\mathbb{Q}$, andEvery $N_i$ is a torsion group, so localization at the nonzero integers givesand hence ${\prod }_i(\mathbb{Q}{\otimes }_{\mathbb{Z}}{N}_i)=0$.

By contrast, the element $x=(1\mathrm{mod}i{)}_{i\ge 2}$ of ${\prod }_i{N}_i$ has infinite order: no positive integer is divisible by every $i$. The canonical descriptionshows that $1\otimes x$ is nonzero, because an element dies in this localization only if one nonzero integer annihilates it. Thus the left module is nonzero while the right module is zero, giving the tensor product and infinite direct product counterexample.

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Assume tensor products of two nonzero modules never vanish. If $\mathfrak{m}\ne \mathfrak{n}$ were distinct maximal ideals, then $\mathfrak{m}+\mathfrak{n}=R$ andalthough both residue fields are nonzero. Hence $R$ has one maximal ideal $\mathfrak{m}$ and is a local ring.

For every module $M$,If $\mathfrak{m}M=M$, this tensor product vanishes. Since $R/\mathfrak{m}\ne 0$, the assumed property forces $M=0$. Thus condition (a) implies condition (b).

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Conversely, assume condition (b), and let $M,N$ be nonzero. The stated property givesThese are nonzero vector spaces over the residue field $k=R/\mathfrak{m}$, so their tensor product over $k$ is nonzero. Associativity and base change giveTherefore $M{\otimes }_{R}N\ne 0$. This proves the reverse implication and the local tensor nonvanishing criterion.

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There is an exact sequence of $R$-modulesIf $A$ is a flat module over $R$, tensoring this sequence with $A$ preserves its left exactness. The image of each tensor product inside $A$ is the corresponding extension of an ideal, soBoth displayed inclusions therefore hold. This is the flat extension preserves finite ideal intersections property.

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The inclusionalways holds: each generator coming from $I\cap J$ lies in both extended ideals.

The reverse inclusion can fail. Letfor a field $k$. Under $A\cong k[x]$,whereas $I\cap J=(xy)$ in $R$, soThus statement (2) is true in general and statement (1) is false in general.

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A multiplicative subset $S\subseteq R$ contains $1$ and satisfies $s,t\in S\Rightarrow st\in S$. The localization of a ring consists of fractions $r/s$ modulo the relationIts structure map $\iota :R\to S^{-1}R$ makes every $s\in S$ invertible. The universal property of localization says that if $f:R\to T$ is any ring homomorphism for which every $f(s)$ is a unit, there is one unique homomorphism $\tilde{f}:S^{-1}R\to T$ satisfying $\tilde{f}\iota =f$, namely

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The local criterion states that an $R$-module $M$ is flat if and only if $M_{\mathfrak{p}}$ is flat over $R_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$. It is enough equivalently to test maximal ideals.

If $M$ is flat, localization of an exact sequence and the natural isomorphismshow immediately that every $M_{\mathfrak{p}}$ is flat.

Conversely, let $N^{\prime }\to N$ be injective and let $K$ be the kernel ofAfter localization at any prime $\mathfrak{p}$, flatness of $M_{\mathfrak{p}}$ gives $K_{\mathfrak{p}}=0$. A module whose localization at every maximal ideal is zero must itself be zero: if $0\ne x\in K$, its annihilator is contained in a maximal ideal $\mathfrak{m}$, and then $x/1\ne 0$ in $K_{\mathfrak{m}}$. Hence $K=0$, tensoring by $M$ preserves every injection, and $M$ is flat. This proves that flatness is local.

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Put $S=A\setminus \mathfrak{p}$ and definewhere $A$ acts on $B$ through the given inclusion. By the prime ideal correspondence for localization, primes of $B_{\mathfrak{p}}$ correspond to primes $\mathfrak{q}$ of $B$ satisfying $\mathfrak{q}\cap S=\varnothing$, equivalently $\mathfrak{q}\cap A\subseteq \mathfrak{p}$.

The localized extension $A_{\mathfrak{p}}\subseteq B_{\mathfrak{p}}$ remains integral. If $\mathfrak{q}\cap A=\mathfrak{p}$, thenis an integral domain integral over the field $A_{\mathfrak{p}}/\mathfrak{p}{A}_{\mathfrak{p}}$. An integral domain integral over a field is a field, so $S^{-1}\mathfrak{q}$ is maximal.

Conversely, if $\mathfrak{n}$ is maximal in $B_{\mathfrak{p}}$, the contraction of a maximal ideal under an integral extension is maximal in the local ring $A_{\mathfrak{p}}$, hence equals $\mathfrak{p}{A}_{\mathfrak{p}}$. Contracting once more to $A$ gives $\mathfrak{p}$. Extension and contraction are inverse under localization, proving the required fiber primes of an integral extension bijection.

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TakeThis extension is not integral. Since every nonzero integer is already invertible in $\mathbb{Q}$,has the maximal ideal $(0)$. On the other hand, the only prime ideal of $B$ is $(0)$, whose contraction to $\mathbb{Z}$ is $(0)$ rather than $(2)$. The set of primes of $B$ lying over $\mathfrak{p}$ is therefore empty while $\mathrm{MaxSpec}{B}_{\mathfrak{p}}$ is not.

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Let $\mathfrak{m}$ be maximal in $R/I$ and let $\mathfrak{n}\subseteq R$ be its contraction. The ideal $\mathfrak{n}$ is maximal and contains $I$. It is disjoint from $S=1+I$: if $1+i\in \mathfrak{n}$ with $i\in I\subseteq \mathfrak{n}$, then $1\in \mathfrak{n}$. The prime ideal correspondence for localization therefore defines the proper ideal $S^{-1}\mathfrak{n}$, andis a field. Thus $f(\mathfrak{m})=S^{-1}\mathfrak{n}$ is maximal.

Conversely, let $\mathfrak{q}$ be the contraction to $R$ of a maximal ideal of $S^{-1}R$. Then $\mathfrak{q}$ is maximal among ideals disjoint from $S$. Since $\mathfrak{q}+I$ is disjoint from $S$—otherwise an equation $q+i=1+j$ would put $1+(j-i)\in \mathfrak{q}\cap S$—maximality gives $I\subseteq \mathfrak{q}$. If a proper ideal strictly contained $\mathfrak{q}$ in a maximal ideal $\mathfrak{n}$ of $R$, then $\mathfrak{n}$ would also contain $I$ and hence remain disjoint from $S$, a contradiction. Thus $\mathfrak{q}$ is maximal and contains $I$.

The two standard extension-contraction bijections, first for $R\to R/I$ and then for $R\to S^{-1}R$, now give inverse maps

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For an ideal $I\subseteq k[T_1,\dots ,T_n]$, defineFor $X\subseteq k^n$, defineand recall that $\sqrt{I}=\{f:f^r\in I\text{ for some }r\ge 1\}$ is the radical of an ideal.

For an algebraically closed field $k$, the Weak Hilbert Nullstellensatz says that every maximal ideal of $k[T_1,\dots ,T_n]$ isfor a unique $a\in k^n$, equivalently every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says

To prove the weak form, let $\mathfrak{m}$ be maximal. The residue fieldis a field finitely generated as a $k$-algebra. By the Zariski lemma, $K/k$ is finite algebraic; algebraic closedness gives $K=k$. If $a_i$ is the image of $T_i$, the quotient map is evaluation at $a=(a_1,\dots ,a_n)$ and its kernel is $(T_1-a_1,\dots ,T_n-a_n)$. This proves the assertion.

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Let $\mathfrak{m}$ be maximal in $\mathbb{Q}[T_1,\dots ,T_n]$ and putThe Zariski lemma makes $K$ a finite extension of $\mathbb{Q}$. Base change givesThis ring is nonzero because the field extension makes $\mathbb{C}$ a faithfully flat module over $\mathbb{Q}$. Choose a maximal ideal $\mathfrak{n}$ of the quotient, or equivalently a maximal ideal of $\mathbb{C}[T_1,\dots ,T_n]$ containing $\mathfrak{m}\mathbb{C}[T_1,\dots ,T_n]$. Its contraction to the rational polynomial ring contains $\mathfrak{m}$ and is proper, so maximality of $\mathfrak{m}$ forces

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LetCertainly $\mathfrak{a}\subseteq (X)$, so $\sqrt{\mathfrak{a}}\subseteq (X)$. Conversely, cubing the second generator shows that $\mathfrak{a}$ contains $X^6(Y-3{)}^3$, while multiplying the first generator by $X^5$ shows that it contains $X^6{Y}^3$. The ideals generated by $Y^3$ and $(Y-3{)}^3$ are comaximal ideals, so a polynomial combination of the two polynomials is $1$. Multiplication by $X^6$ gives $X^6\in \mathfrak{a}$, hence $X\in \sqrt{\mathfrak{a}}$. Thereforegenerated by the single polynomial $X$.

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Write the finitely generated algebra asBy the Weak Hilbert Nullstellensatz, $\mathbb{C}$-algebra homomorphisms $A\to \mathbb{C}$ correspond exactly to the points of the affine algebraic set $V(I)$.

If $V(I)$ is finite, its cardinality is finite. If it is infinite, the complex affine algebraic set cardinality dichotomy giveswhich is uncountable. This also covers the zero algebra, whose homomorphism set is empty. Therefore

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For a prime ideal $\mathfrak{p}$, its height is the supremum of lengths $r$ of strict chainsThe height of an ideal $I$, without assuming $I$ prime, isThe Krull dimension of a ring $R$ is

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For the maximal ideal $\mathfrak{m}$ of $(A,\mathfrak{m})$, the associated graded ring isIf $x+{\mathfrak{m}}^{r+1}$ and $y+{\mathfrak{m}}^{s+1}$ are homogeneous classes, their product isIt is a graded algebra over the residue field $k=A/\mathfrak{m}$.

The Hilbert series isThe Hilbert-Serre theorem makes this rational. The number $d(G_{\mathfrak{m}}(A))$ is the order of its pole at $t=1$, as recorded by the pole dimension of an associated graded ring.

The Dimension theorem for Noetherian local rings states

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Put $\overline{A}=A/(x)$ and $\overline{\mathfrak{m}}=\mathfrak{m}/(x)$. Applying the dimension theorem to $A$ and $\overline{A}$ givesSince $x$ is a non-zero-divisor, it belongs to no minimal prime of the Noetherian ring $A$. Any chain of primes in $A/(x)$ lifts to a chainof primes of $A$ containing $x$. A minimal prime $\mathfrak{q}\subseteq {\mathfrak{p}}_0$ cannot contain $x$, so the inclusion is strict. Prepending $\mathfrak{q}$ gives a chain of length $r+1$ in $A$. Thus the dimension drop by a non-zero-divisor givesCombining these equalities proves

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The polynomial ring $\mathbb{C}[X,Y]$ is a two-dimensional unique factorization domain. Let $\mathfrak{p}$ be a prime ideal containing $(f,g)$. Since $f\ne 0$, the prime $\mathfrak{p}$ is nonzero. If it were not maximal, it would have height one, so the height-one prime in a unique factorization domain would give $\mathfrak{p}=(h)$ for an irreducible polynomial $h$. Then $h$ would divide both $f$ and $g$, contrary to the hypothesis. Every prime ofis therefore maximal.

The Hilbert basis theorem makes $R$ Noetherian, and it has Krull dimension zero by the preceding paragraph. The Noetherian dimension-zero criterion for an Artinian ring now shows that

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A discrete valuation on a field $K$ is a surjective group homomorphismsatisfyingwhenever $x+y\ne 0$. Its discrete valuation ring is

One standard characterization defines a Dedekind domain as a Noetherian integrally closed domain of Krull dimension one. Equivalently, all its localizations at nonzero prime ideals are discrete valuation rings.

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Choose $0\ne a\in \mathfrak{m}$. Since the one-dimensional local domain has no nonzero prime ideal other than $\mathfrak{m}$, one has $\sqrt{(a)}=\mathfrak{m}$. Finite generation of $\mathfrak{m}$ therefore gives some $n\ge 1$ with ${\mathfrak{m}}^n\subseteq (a)$. Choose $n$ minimal andIn the fraction field of $A$, put $y=x/a$. Then $y\notin A$ but

If $y\mathfrak{m}\subseteq \mathfrak{m}$, multiplication by $y$ would preserve the nonzero finitely generated faithful $A$-module $\mathfrak{m}$. The determinant trick would make $y$ integral over $A$, contradicting that $A$ is integrally closed and $y\notin A$. Hence some $t\in \mathfrak{m}$ satisfies $yt\notin \mathfrak{m}$. Since $yt\in A$ and $A$ is local, $yt$ is a unit.

For any $z\in \mathfrak{m}$, one has $yz\in A$, and thereforeThus $\mathfrak{m}\subseteq (t)$, while the reverse inclusion follows from $t\in \mathfrak{m}$. ConsequentlyThis proves the principal maximal ideal in a one-dimensional normal local domain result.

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Let $J$ be an ideal of $S^{-1}A$ and contract it to an ideal $I$ of $A$. Every element $a/s\in J$ has $a/1=(s/1)(a/s)\in J$, so $a\in I$ andBecause $A$ is a Noetherian ring, write $I=(a_1,\dots ,a_r)$. Thenso every ideal of $S^{-1}A$ is finitely generated. Hence every localization of a Noetherian ring is Noetherian, proving the Localization of a Noetherian ring theorem.

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Because $0\notin S$, the localization $S^{-1}A$ remains an integral domain. Part (a) makes it Noetherian. Integral closedness is preserved by localization: if $x$ in the common fraction field is integral over $S^{-1}A$, clearing the finitely many denominators in a monic equation shows that $sx$ is integral over $A$ for some $s\in S$, whence $sx\in A$ and $x\in S^{-1}A$.

The prime ideal correspondence for localization shows that every chain of primes in $S^{-1}A$ comes from a chain in $A$, soIf its dimension is one, it is a Noetherian integrally closed domain of dimension one and hence a Dedekind domain. If its dimension is zero, its zero ideal is maximal, so the domain is a field. This proves the Localization of a Dedekind domain alternative.

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