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I am trying to prove the following statement by induction on $n$:

Let $P$ be a prime ideal of $\mathbb{Z}[X_1,\ldots,X_n]$ with $\mathbb{Z}\cap P = \{0\}$. Suppose that $K=\mathrm{Frac}(\mathbb{Z}[X_1,\ldots,X_n]/P)$, the field of fractions of the quotient of the ring $\mathbb{Z}[X_1,\ldots,X_n]/P$, has transcendence degree $r$ over $\mathbb{Q}$, i.e. $K$ is isomorphic to $M(Y_1,\ldots,Y_r)$ for some algebraic extension $M$ of $\mathbb{Q}$. Then there exists $h \in Z[X_1,\ldots,X_n]\setminus P$ such that $hP$ is generated as an ideal by $n-r$ elements of $\mathbb Z[X_1,\ldots,X_n]$.

I managed the initial step: For $n=1$ one can quickly show that $\mathrm{ff}(\mathbb{Z}[X]/P)$ is isomorphic to $\mathrm{ff}(\mathbb{Q}[X]/P\mathbb{Q})$ via a natural homomorphism. Since $\mathbb{Q}[X]$ is a principal ideal domain, $P\mathbb{Q} = (f)$ for some polynomial $f$, which we can assume to be in $\mathbb{Z}[X]$, as we can clear fractions by multiplying with a suitable integer. Then $P = (f)$ as an ideal in $\mathbb{Z}[X]$, so $hP = (hf)$, i.e. we have one generator. Moreover, $f$ is irreducible, as $P$ is prime, so $\mathrm{ff}(\mathbb{Z}[X]/P)$ is isomorphic to an algebraic extension of $\mathbb{Q}$. Hence, $r=0$, and we are done ($n-r = 1-0 =1$).

For the induction step my idea was to somehow use the fact that $\mathbb{Z}[X_1,\ldots,X_{n}]$ is obtained from $\mathbb{Z}[X_1,\ldots,X_{n-1}]$ by adjoining one more variable. Hence, we could consider $P' = P \cap \mathbb{Z}[X_1,\ldots,X_{n-1}]$, use the fact that this is a prime ideal of $\mathbb{Z}[X_1,\ldots,X_{n-1}]$, apply the induction hypothesis to get $h'P' = (p_1,\ldots,p_{n-r-1})$, and somehow find $h$ such that $hP = (p_1,\ldots,p_{n-r-1},p_{n-r})$.

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This question was answered on MathOverflow https://mathoverflow.net/questions/231693/number-of-generators-of-ideal-if-quotient-field-has-certain-transcendence-degree

It turns out that it is false, and the actual result I was looking for is less general.

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