Is this height-transcendence-degree inequality true without AC ? 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 

 A: Here is a proof of (**) by induction on the height of $P$.
If $P=0$, the inequality (**) is obvious. Let $P$ be a prime ideal of $R$ of height $d \geq 1$, and consider a chain of prime ideals $0=P_0 \subset P_1 \subset \ldots \subset P_d = P$ of length $d$ in $R$. The domain $R/P_1$ has finite Krull dimension, and the prime ideal $P/P_1$ has height $d-1$ so by the induction hypothesis
\begin{equation*}
{\rm height}(P/P_1) + {\rm trdeg}_k (R/P) \leq {\rm trdeg}_k(R/P_1).
\end{equation*}
It remains to prove that ${\rm trdeg}_k(R/P_1) \leq{\rm trdeg}_k(R)-1$. Let $x_1,\ldots,x_e$ be elements of $R$ whose images in $R/P_1$ are algebraically independent over $k$. Let $y$ be any element of $P_1 \backslash \{0\}$. Consider the map $\phi : k[X_1,\ldots,X_e,Y] \to R$ sending $X_i$ to $x_i$ and $Y$ to $y$. If $Q = \sum_j Q_j(X_1,\ldots,X_e) Y^j$ lies in the kernel of $\phi$, then reducing modulo $P_1$ gives $Q_0(x_1,\ldots,x_e)=0$, so that $Q_0=0$. Since we are working in a domain and $y \neq 0$, an easy induction gives $Q=0$. Thus $x_1,\ldots,x_e,y$ are algebraically independent over $k$, which concludes the proof.
