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.