# 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

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
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It's enough to assume the fraction field of $R$ has finite transcendence degree over $k$, since this implies $R$ has finite Krull dimension. – François Brunault Dec 2 '12 at 21:24
Quest 2: A somewhat stronger result than $(\ast\ast)$ is given in Prop. 1.1 of the paper [Giral: Krull dimension, transcendence degree and subalgebras of finitely generated algebras. Arch. Math. 36(1981), 305-312] that can be found online here: link.springer.com/article/10.1007%2FBF01223706?LI=true#page-1 – Ralph Dec 4 '12 at 14:46

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.
For some readers (like me) let me add that the fact that $x_1,...,x_e$ can be choosen from $R/P_1$ (and hence from $R$) and not just from $Quot(R/P_1)$ is explained in mathoverflow.net/questions/75219/…. // BTW: I think P=0 isn't (*) but the tautology trans-deg Quot(R) <= trans-deg Quot(R). – Ralph Dec 4 '12 at 14:58
@Ralph : Right, the case $P=0$ isn't (*), I don't know why I wrote that. I should also have said that all this doesn't require the axiom of choice, since we are working in an extension of finite transcendence degree over $k$. – François Brunault Dec 4 '12 at 15:36