I am trying to prove a result which is used in *A. Macintyre and A. J. Wilkie (1995), 'On the decidability of the real exponential field', in Odifreddi, P.G., Kreisel 70th Birthday Volume, CLSI*, p. 463f.:


Let $P$ be a prime ideal of $\mathbb{Z}[X\_1,\ldots,X\_n]$ with $\mathbb{Z}\cap P = \{0\}$. Suppose that $K=\mathrm{ff}(\mathbb{Z}[X\_1,\ldots,X\_n]/P)$, the field of fractions of the quotient of the ring $\mathbb{Z}[X\_1,\ldots,X\_n]$ with its ideal $P$, has transcendence degree $r$ over $\mathbb{Q}$. Then there exists $h \in \mathbb{Z}[X\_1,\ldots,X\_n]\setminus P$ such that $hP$ is generated as an ideal by $n-r$ elements of $\mathbb{Z}[X\_1,\ldots,X\_n]$.


The case $P = 0$ is easy. 
Macintyre and Wilkie say that the statement can be proved by induction on $n$. The initial step $n=1$ is simple by observing that $K \cong \mathrm{ff}(\mathbb{Q}[X]/(f))$ for some $f \in \mathbb{Z}[X]$. The expression on the right-hand-side is an algebraic extension of $\mathbb{Q}$.


In general $K \cong \mathrm{ff}(\mathbb{Q}[X\_1,\ldots,X\_n]/P\mathbb{Q})$, and $P\mathbb{Q}$ is a prime ideal of $\mathbb{Q}[X\_1,\ldots,X\_n]$. 


I do not see at all how to do the induction step. I asked this question on Stackexchange, where it wasn't answered, and I thought it would be suitable for Overflow.



Number of generators of ideal if quotient field has certain transcendence degree
Unless I'm being stupid, it seems to me the result is false since it would imply (together with [1]) that all height-$s$ primes of $\mathbb{Q}[X\_{1},\dotsc,X\_{n}]$ can be generated by $s$ elements [2].


[1] Let $A$ be a ring, $\mathfrak{a}$ an ideal of $A$, and $f$ a nonzerodivisor of $A$. Then $\mathfrak{a}$ and $f\mathfrak{a}$ have the same number of minimal generators.


[2] This is true if $s = 0,1,n$ but the twisted cubic gives a counterexample for $n=4$ and $s=2$. 


