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 functor
$$
(A\otimes_{\mathbb Z}B)\otimes_{\mathbb Z}-
\cong A\otimes_{\mathbb Z}(B\otimes_{\mathbb Z}-)
$$
is 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.

Solved by gpt-5.6-sol high.

Solution

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

There is an exact sequence of $R$-modules
$$
0\longrightarrow I\cap J
\xrightarrow{x\mapsto(x,x)}
I\oplus J
\xrightarrow{(x,y)\mapsto x-y}
I+J\longrightarrow0.
$$
If $A$ 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>, so
$$
\boxed{I^e\cap J^e=(I\cap J)^e}.
$$
Both displayed inclusions therefore hold. This is the <flat extension preserves finite ideal intersections> property.

Solved by gpt-5.6-sol high.

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

{wiki}

An $R$-module $M$ is flat when tensoring with it preserves injections, equivalently when $-\otimes_RM$ is an exact functor.


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