Conversely, assume condition (b), and let $M,N$ be nonzero. The stated property gives
$$
M/\mathfrak mM\ne0,
\qquad
N/\mathfrak mN\ne0.
$$
These are nonzero <vector spaces> over the <residue field> $k=R/\mathfrak m$, so their tensor product over $k$ is nonzero. Associativity and base change give
$$
\begin{aligned}
(M\otimes_RN)\otimes_Rk
&\cong(M\otimes_Rk)\otimes_k(N\otimes_Rk)\\
&\cong(M/\mathfrak mM)\otimes_k(N/\mathfrak mN)\ne0.
\end{aligned}
$$
Therefore $M\otimes_RN\ne0$. This proves the reverse implication and the <local tensor nonvanishing 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 / iv / b / Solution

{title2=$\kappa(\mathfrak m)=R/\mathfrak m$}
{wiki}

The residue field at a maximal ideal $\mathfrak m$ is the quotient field $\kappa(\mathfrak m)=R/\mathfrak m$. At a prime ideal $\mathfrak p$, the residue field is the fraction field of $R/\mathfrak p$.


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