We know that when $F$ is a field, the ring $F[x_1,...,x_n]$ is a Hilbert ring, because the field $F$ is a Hilbert ring. My questions:
Is any (non-trivial) affine algebra over an algebraically closed field a Hilbert ring? If this is so, is the algebraically closedness necessary?"
Grateful!
Consider any finitely generated $F$ algebra $A$. So there is a surjective $F$-algebra homomorphism
$$\varphi: F[x_1, \dots, x_k] \twoheadrightarrow A$$ for some $k \in \mathbb{N}$. This induces an isomorphism of the complete lattices of
(1) ideals of $F[x_1, \dots, x_k]$, containing $\ker \varphi$, and
(2) ideals of $A$.
Furthermore, in this correspondence, prime ideals from (1) correspond precisely to prime ideals from (2) and maximal ideals of (1) correspond precisely to maximal ideals from (2).
Since being Hilbert ring is (equivalent to) the property "every prime ideal is an intersection of a family of maximal ideals" and $F[x_1, \dots, x_k]$ is a Hilbert ring (no matter whether $F$ is alg. closed or not), it easily follows that $A$ is a Hilbert ring as well.