Hilbert Basis Theorem for Artinian Ring

Question

Does there exist an analog of the Hilbert Basis Theorem for Artinian Rings?

$$R ... \text{Artinian} \rightarrow R[X] ... \text{Artinian}$$

This question came up in whether the prime spectrum $\text{Spec} (A)$ of a ring $A$ is artinian as a Zariski Topological space when $A$ is artinian.

Partial Answer: (Wrong per Ross P)

For a ring $R$ (commutative with unit), Noetherian is equivalent to Artinian. This is proven in this answer. Hence, $\text{Spec}(A)$ is Noetherian and Artinian if $A$ is a field.

Answer 1

The partial answer you state is not true. The answer in that post is for modules over $R$ satisfying certain properties on some maximal ideal $m\lhd R$. Certainly for commutative rings $R$ is Artinian if and only if $R$ is Noetherian and has Krull dimension 0. Also for a field $K$ and finite type $K$-algebra $R$ it can be shown that dim$(R)$ is the transcendence degree of $R$ over $K$. (see, for instance, Atiyah-MacDonald 'Introduction to Commutative Algebra')

We can use these facts to note that a field $K$ is Artinian, but $K[x]$ has dimension 1, so cannot possibly be Artinian. A fundamental boundary to your proposed statement.

Answer 2

Suppose $R$ is not the zero ring. Let $\mathfrak{p}$ be a prime ideal of $R$. Then $\mathfrak{p}[x]$, the ideal of polynomials with coefficients in $\mathfrak{p}$, is a prime ideal of $R[x]$. But $(\mathfrak{p}[x], x)$ is a prime ideal properly containing $\mathfrak{p}[x]$ so we see that $\dim(R[x]) > \dim(R) \ge 0$. So $\dim R[x]$ is always at least $1$ and $R[x]$ is never ever Artinian (unless $R$ is the zero ring).