Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 1

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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