Currently I'm reading *"On the Decidability of the Real Exponential Field"* by Macintyre and Wilkie and the Proof of Theorem 1.1 (page 462-464) uses two algebraic results that involve the notion of *transcendence degree* - something I'm not very familiar with.

The first algebraic result was used in the following passage:

Then $\tilde{\alpha}$ is a non-singular zero of the system $\tilde{f}_1(\tilde{x}) = \ ... \ = \tilde{f}_n(\tilde{x}) = 0$, from which it follows (using elementary differential algebra - see Lang [1965], chapter 10, §7) that the field $\mathbb{Q}(\tilde{\alpha})$ has transcendence degree (over $\mathbb{Q}$) at most n.

Note that $\overline{\alpha} \in \mathbb{R}^n$ is a non-singular solution of $f_1(\overline{x}) =\ ...\ = f_n(\overline{x}) = 0$ with $\overline{x} := (x_1,\ ...\ ,x_n)$ and $f_1,\ ...\ ,f_n \in \mathbb{Z}[x_1,\ ...\ ,x_n,\exp(x_1),\ ...\ ,\exp(x_n)]$ interpreted as functions from $\mathbb{R}^n$ to $\mathbb{R}$. Further, $\tilde{x} := (x_1,\ ...\ ,x_{2n})$, $\tilde{\alpha} := (\alpha_1,\ ...\ ,\alpha_n, \exp(\alpha_1),\ ...\ ,\exp(\alpha_n))$ and $\tilde{f}_i$ is the function from $\mathbb{Z}[x_1,\ ...\ ,x_{2n}]$ satisfying $f_i(x_1,\ ...\ ,x_n) \equiv \tilde{f}_i(x_1,\ ...\ ,x_n,\exp(x_1),\ ...\ ,\exp(x_n))$ for all $i \in \lbrace 1,\ ...\ ,n \rbrace$.

Since I do not have access to the 1965-edition of Lang's book, I checked the 2002-edition and the closest result I could find was Proposition 5.3. on page 371. But I'm not quite sure why that would yield $\text{trdeg}(\mathbb{Q}(\alpha) \mid \mathbb{Q}) \leq n$. A short explanation or a reference where this is more immediate would be very helpful.

The second algebraic result that was used is the following: Let $m,r \geq 1$.

[...] easily proved by induction on $m \in \mathbb{N} \setminus \lbrace 0 \rbrace$, that if $Q$ is a prime ideal of $\mathbb{Z}[x_1,\ ...\ ,x_m]$ such that $Q \cap \mathbb{Z} = \emptyset$ and such that the field of frations of $\mathbb{Z}[x_1,\ ...\ ,x_m]/Q$ has transcendence degree $r$, then for some $h \in \mathbb{Z}[x_1,\ ...\ ,x_m] \setminus Q$, $hQ$ is generated by $m-r$ elements.

I do not know why this statement holds. I feel like I am missing some result on transcendence degrees here. I would be very grateful for a short explanation or a reference that yields this statement.

Algebraseems to be here: openlibrary.org/books/OL5950910M/Algebra. $\endgroup$1more comment