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?](https://mathoverflow.net/questions/79959/krull-dimension-transcendence-degree/79974#79974) 


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: 


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



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


