# Equivalence of definitions of dimension in integral domains.

If $R$ is a finitely generated algebra over a field $k$ that is an integral domain, it is known that the Krull dimension of $R$ is equal to the transcendence degree of the field of fractions of $R$ over $k$. See for example this question.

Now, let $R$ be a finitely generated algebra over a domain $A$ that is also an integral domain, and such that the homomorphism from $A$ to $R$ is injective. Let $F, K$ be the quotient fields of $R$ and $A$ respectively. I want to determine whether Krull dimension of $R$ coincides with trascendence degree of $F$ over $K$.

I think there must be cases in which they don't coincide, because I have always seen this result with $A$ being a field, but I am not able to find an example. I have thought in $R$ as the polynomial ring with $n$ indeterminates over $A$.

Passing to the quotient field loses any information about the dimension of $A$, but of course the dimension of $A$ plays crucial role for the dimension of $R$. So they will not coincide in most cases.
For example $\mathbb Z[X]$ is two dimensional. Passing to the quotient fields gives you $\mathbb Q \hookrightarrow \mathbb Q(x)$ with transcendence degree $1$.