Let $\mathfrak m$ be maximal in $\mathbb Q[T_1,\ldots,T_n]$ and put
$$
K=\mathbb Q[T_1,\ldots,T_n]/\mathfrak m.
$$
The <Zariski lemma> makes $K$ a finite extension of $\mathbb Q$. Base change gives
$$
\mathbb C[T_1,\ldots,T_n]/\mathfrak m\mathbb C[T_1,\ldots,T_n]
\cong\mathbb C\otimes_{\mathbb Q}K.
$$
This 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 $\mathbb C[T_1,\ldots,T_n]$ containing $\mathfrak m\mathbb C[T_1,\ldots,T_n]$. Its contraction to the rational polynomial ring contains $\mathfrak m$ and is proper, so maximality of $\mathfrak m$ forces
$$
\mathfrak n\cap\mathbb Q[T_1,\ldots,T_n]=\mathfrak m.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 101 / 3 / ii / Solution

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A flat module $M$ is faithfully flat when $N\otimes_RM=0$ implies $N=0$. Every field extension is faithfully flat over its base field.


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 Incoming links: Faithfully flat module