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Let $k$ be a field, and $A$ a $k$-domain, so that the fraction field of $A$ has transcendence degree $n$ over $k$.

If $A$ is finitely-generated over $k$, then $A$ has Krull dimension $n$ (Theorem A in Eisenbud).

However, if $A$ is infinitely-generated, then it is possible for the dimension of $A$ to be less than the transcendence degree of its fraction field. Take, for example, rational functions in one variable $A=k(x)$. Dimension 0, transcendence degree 1.

Is it always true that the dimension of $A$ is less than or equal to $n$?

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up vote 36 down vote accepted

It looks to me like the answer is yes.

Fix any strictly increasing chain of primes $P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_m$ in $A$ of length $m$; we'll prove that $m \leq n$. Choose elements $x_i \in P_i \setminus P_{i-1}$ for $i = 1,\dots,m$.

Let $B \subseteq A$ be the $k$-subalgebra generated by $\{x_1,\dots,x_m\}$. The primes $Q_i = P_i \cap B$ form a strictly increasing chain of length $m$ in $B$, because $x_i \in Q_i \setminus Q_{i-1}$.

The fraction field $Q(B)$ of $B$ lives inside of $Q(A)$ and therefore has $k$-transcendence degree at most $n$. Because $B$ is finitely generated, this means that it has Krull dimension at most $n$. It now follows that $m \leq n$.

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Nice answer! Much simpler than that I wrote at the same time. – Qing Liu Nov 3 '11 at 20:38
Thank you! (I guess there was no error...) That certainly is a very annoying feature of answering. It's happened to me, too! – Manny Reyes Nov 3 '11 at 20:40
This is a real gem, Manny: crisp, elegant, implacable. – Georges Elencwajg Nov 3 '11 at 21:39

There is already a perfect, accepted answer, but I thought I add this to point out that there is a geometric reason for the dimension to possibly drop versus the transcendence degree.

For simplicity assume that the fraction field of $A$ is a finitely generated field extension of $k$ and that $A$ is integrally closed. If we are doing geometry these assumptions seem reasonable (and there is already a complete answer in the general case). I think with a little more work one can follow a similar argument without these assumptions, but this post's goal is mainly to explain the geometry behind this phenomenon, so I am not striving for completeness especially that (as I already mentioned twice) since there is already a complete solution there is no need to do that.

So, first, $A$ contains a transcendence base for its fraction field over $k$, in other words we have $$ k\subseteq k[x_1,\dots,x_n]\subseteq A \subseteq \mathrm{Frac} (A) $$ where $\mathrm{Frac} (A)$ is a finite algebraic extension of $k(x_1,\dots,x_n)$. Now let $B$ be the integral closure of $k[x_1,\dots,x_n]$ in $\mathrm{Frac} (A)$. Since $A$ is assumed to be integrally closed, $B\subseteq A$ and since $\mathrm{Frac} (A)$ is a finite algebraic extension of $k(x_1,\dots,x_n)$, $B$ is a finitely generated $k[x_1,\dots,x_n]$-module and a finitely genrated $k$-algebra and $\mathrm{Frac} (B)=\mathrm{Frac} (A)$.

The integral $k$-algebra extension $k[x_1,\dots,x_n]\subseteq B$ corresponds to a finite surjective morphism $X\to \mathbb A^n_k$ of $k$-varieties, so for $X$ we have the quoted result: $\dim X$ equals the transcendence degree of the fraction field of $A$.

Now we have that $B\subseteq A \subseteq \mathrm{Frac} (B)=\mathrm{Frac} (A)$. In geometric situations this typically happens if $A$ is a localization of $B$. In that case we get that the Krull dimension of $A$ is at most the Krull dimension of $B$ because the dimension of $A$ is just the maximum height of the points of $\mathrm{Spec}B$ contained in $\mathrm{Spec} A$.

This is really why I wrote this whole answer: the way the Krull dimension drops versus the transcendence degree is if we localize at non-closed points: In your original example $k(x)$ corresponds to the generic point of the line $\mathbb A^1$, so its geometric dimension = transcendence degree is $1$, but its algebraic dimension is $0$ because it is a field (or more generally the local ring of a non-closed point).

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@YACP: you're right. I managed to confuse myself by adding some conditions that I didn't assume. However, that was not the point of my post. I edited it accordingly. Thanks for pointing out the error. – Sándor Kovács Jan 5 '13 at 0:30

[Edit] My answer is unnecessarily complicated regarding that of Manny.

The answer is yes. First suppose $A$ is noetherian. You can find $t_1,\dots, t_n\in A$ algebraically independent over $k$. Then you get a dominant morphism $\operatorname{Spec}A\to \mathbb A^n_k$ of integral noetherian schemes which is generically finite. This then implies that $\dim A\le \dim \mathbb A^n_k$ (see this answer to related question).

If $A$ is not noetherian, write $A$ as an inductive limit of noetherian subdomains $A_i$ with $\mathrm{Frac}(A_i)=\mathrm{Frac}(A)$. Let $\mathfrak p_0\subset \cdots \subset \mathfrak p_r$ be a chain of prime ideals in $A$. Then for some $A_i$, the $\mathfrak p_j\cap A_i$ is a chain of prime ideals (take $A_i$ big enough to contain an element of $\mathfrak p_j\setminus \mathfrak p_{j-1}$ for all $j\le r$). Then by the above $r\le \dim A_i\le n$. Hence $\dim A\le n$.

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