# Number of generators of ideal if quotient field has certain transcendence degree

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.

• – jeq Feb 20 '16 at 17:24

Unless I'm being stupid, it seems to me the result is false since it would imply (together with ) that all height-$s$ primes of $\mathbb{Q}[X_{1},\dotsc,X_{n}]$ can be generated by $s$ elements .
 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.
 This is true if $s = 0,1,n$ but the twisted cubic gives a counterexample for $n=4$ and $s=2$.
• Now, again, I am not sure about one detail of this argument. If $P$ is a prime ideal of $\mathbb{Z}[X_1,\ldots,X_n]$, then $P\mathbb{Q}$ is a prime ideal of $\mathbb{Q}[X_1,\ldots,X_n]$. But does the converse necessarily hold? The twisted cubic $(XZ-Y^2 , YW-Z^2, XW-YZ)$ is prime in $\mathbb{Q}[X,Y,Z,W]$, but is it obvious that it is also a prime ideal of $\mathbb{Z}[X,Y,Z,W]$. If not, it wouldn't provide a counterexample. – Lothar Sebastian Krapp Mar 1 '16 at 17:15
• If $A$ is any ring, then the ideal $\mathfrak{a} = (XZ-Y^{2},YW-Z^{2},XW-YZ)$ of $A[X,Y,Z,W]$ is the kernel of the map $A[X,Y,Z,W] \to A[S,T]$ sending $X \mapsto S^{3}$, $Y \mapsto S^{2}T$, $Z \mapsto ST^{2}$, $W \mapsto T^{3}$, so in case $A$ is an integral domain, the ideal $\mathfrak{a}$ is prime. (Also: if $S$ is a multiplicatively closed subset of $A$ and $\mathfrak{a}$ is an $S$-saturated ideal of $A$, then $\mathfrak{a}$ is prime in $A$ if and only if $\mathfrak{a}S^{-1}A$ is prime in $S^{-1}A$.) – Minseon Shin Mar 2 '16 at 0:46