Let $R$ be a $k$-algebra ($k$ a field) and a domain of finite Krull dimension. In

$\quad$ Krull dimension less or equal than transcendence degree?

it is shown that
$$\text{Krull-dim}(R) \le \text{trans.deg}_k Quot(R).\tag{*}$$

From the paper [1] I learned the inequality
$$\text{height}(P) + \text{trans.deg}_k Quot(R/P) \le \text{trans.deg}_k Quot(R)\tag{**}$$
for all prime ideals $P \subseteq R$. Obviously $(\ast\ast)$ strengthens $(\ast)$.

The paper gives as reference for $(\ast\ast)$ a combination of two results: First, in [2, Chap. IV, §3, Cor. 1] the inequality is proved for valuation rings. The general case then follows by an embedding theorem for domains into valutation rings [3, Chap. I, (11.9)].

However, the proof of the embedding theorem uses Zorn's Lemma. Since $(\ast)$ can be proved without the Axiom of Choice, I wonder:

**Question 1:** Is $(\ast\ast)$ also true without assuming the Axiom of Choice ?

**Question 2:** Are there alternative references for $(\ast\ast)$ than whose given by Wadsworth ?

References:

- Wadsworth: The Krull dimensions of tensor products of commutative algebras over a field. J. London Math. Soc. (2), 19(1979), 391-401
- Zariski, Samuel: Commutative Algebra II
- Nagata: Local Rings