Edit: It really is impossible! Found the answer here: Transcendence degree and Krull dimension of finitely generated algebras . But I'm not deleting my question because the linked question was hard to find, and maybe my wording will help future people searching.
The question:
- Let $k$ be a field (not necessarily algebraically closed).
- Let $A$ be a $k$-algebra generated by $x_1,...,x_n$ (not necessarily an integral domain).
- Assume that $A$ has Krull dimension $d<n$.
Is it then possible that $x_1,...,x_{d+1}$ are algebraically independent over $k$?
I think it's impossible.
If $A$ is an intergral domain then I get that the field of fractions of $A$ has transcendence degree at least $d+1$ over $k$, a contradiction.
But I'm not assuming that $A$ is an integral domain.
If $k$ is algebraically closed, then the injective map $k[x_1,...,x_{d+1}]\to k[x_1,...,x_n]$ induces a dominant map from $\operatorname{maxspec}(A)$ to $\mathbb{A}_k^{d+1}$, and I think that's a contradiction again.
But what happens generally? My geometric argument probably generalizes to the case where $k$ is not algebraically closed, but I'm don't know the needed facts. Also, I'd like to have an algebraic proof (but a geometric one is fine too).
