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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.

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    $\begingroup$ Don't you need assumptions on the $f_i$? One could take all $f_i=0$. $\endgroup$
    – François Brunault
    Commented Jan 8, 2022 at 9:20
  • $\begingroup$ At one point in the proof the $f_i$ and $\overline{\alpha}$ are introduced at the same time with the property that $\overline{\alpha}$ is a non-singular zero of the functions $(f_1,\ ...\ f_n)$. Also, those $f_i$ are introduced after applying a result that esures the existence of such $f_i$ together with an appropriate $\overline{\alpha}$; so the $f_i$ cannot be chosen freely - Sorry, I didn't clarify that. But having all $f_i = 0$ shouln't be possible by the non-singularity condition. $\endgroup$
    – Bytegear
    Commented Jan 8, 2022 at 12:08
  • $\begingroup$ Thank you very much for explaining. I guess that the first result is not just algebraic (by this I mean commutative algebra, like the second result) but also using differential properties, since the exponential function comes into play. $\endgroup$
    – François Brunault
    Commented Jan 8, 2022 at 18:18
  • $\begingroup$ Yes, in order to verify that $\tilde{\alpha}$ is a non-singular zero of $(\tilde{f}_1,\ ...\ ,\tilde{f}_n)$ I definitely needed differential properties of the exponential function. I'm struggling with the part where the author uses "elementary differential algebra". I don't understand how he got the statement on the transcendence degree; but commutative algebra alone will not yield this result. I should probably add another tag - Thank you for pointing this out. $\endgroup$
    – Bytegear
    Commented Jan 8, 2022 at 19:51
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    $\begingroup$ Lang's 1965 Algebra seems to be here: openlibrary.org/books/OL5950910M/Algebra. $\endgroup$
    – user44143
    Commented Jan 10, 2022 at 12:02

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For the second result, you can reduce to the same question with $A = \mathbb{Q}[x_1,\ldots,x_n]$. It is known that this polynomial ring is regular, which means that its localisation $A_Q$ at the prime ideal $Q$ has the property that the maximal ideal $QA_Q$ can be generated by $\mathrm{dim}(A_Q)$ elements. Here $\mathrm{dim}(A_Q)$ denotes the Krull dimension of $A_Q$, which is also the height $h(Q)$ of the prime ideal $Q$. By considering chains of prime ideals, one shows that $h(Q)+\mathrm{dim}(A/Q) = \mathrm{dim}(A)=n$. Finally, by the Noether normalisation lemma, the dimension of $A/Q$ is equal to its transcendence degree over $\mathbb{Q}$.

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  • $\begingroup$ Thank you for this helpful answer. I hope it's ok if I ask two small questions. (1) How is the reduction from $\mathbb{Z}[x_1,\ ...\ ,x_n]$ to $\mathbb{Q}[x_1,\ ...\ ,x_n]$ justified. (2) Why does $h(Q) = n-r$ yield that $gQ$ is generated by $n-r$ elements for some $g$? $\endgroup$
    – Bytegear
    Commented Jan 9, 2022 at 22:45
  • $\begingroup$ Actually I don't think their second algebraic result is true: if $hQ$ were generated by $m-r$ elements then so would be $Q$. But a prime ideal is not always generated by a family of size equal to the height. This is false even over $\mathbb{C}$: there are algebraic curves in 3-space which cannot be defined by 2 equations alone. What is true (and may suffice for their applications ?), is that $hQ$ is contained in such an ideal. $\endgroup$
    – François Brunault
    Commented Jan 10, 2022 at 20:13
  • $\begingroup$ (1) and (2) are special cases of localisation. Let $R'=S^{-1} R$ be a localisation of a ring $R$. Let $I$ be an ideal of $R$. If the ideal $I'=S^{-1} I \subset R'$ satisfies $hI' \subset \langle f_1,\ldots,f_m\rangle$ for some $h,f_1,\ldots,f_m \in R'$, then by clearing denominators everywhere we get the analogous statement for $I$. Case (1) is $S = \mathbb{Z} \backslash \{0\}$ and (2) is $S = A \backslash Q$. $\endgroup$
    – François Brunault
    Commented Jan 10, 2022 at 20:16

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