# Reference request for results that involve the transcendence degree

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.

• Don't you need assumptions on the $f_i$? One could take all $f_i=0$. Jan 8, 2022 at 9:20
• 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. Jan 8, 2022 at 12:08
• 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. Jan 8, 2022 at 18:18
• 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. Jan 8, 2022 at 19:51
• Lang's 1965 Algebra seems to be here: openlibrary.org/books/OL5950910M/Algebra. Jan 10, 2022 at 12:02

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 $$\mathbb{Q}$$.
• 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 $g$? Jan 9, 2022 at 22:45
• 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. Jan 10, 2022 at 20:13
• (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 $S = A \backslash Q$. Jan 10, 2022 at 20:16