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 $n$?**



Krull dimension less or equal than transcendence degree?
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 $m \leq n$.


