Is there a connection between compactness and the Hilbert basis theorem? I was just talking with a few friends and we noticed something interesting. Namely, a straightforward corollary of the Hilbert Basis Theorem is:

If $R$ is noetherian and $V \subset R^n$ is defined as the zero set of some (possibly infinite) collection $\{f_\alpha\}$ of polynomials in $R[x_1,\ldots,x_n]$, then $V$ is the zero set of some finite set of polynomials.

This had a striking resemblance to the definition of compactness:

A topological space $X$ is compact if every cover by a collection of open sets $\{U_\alpha \} $ has a finite subcollection.

My question is: is there a deeper significance to this similarity? Is "noetherian" some sort of analog of compactness?
 A: They are related.  A topological space is called Noetherian if every open subset is compact in the induced topology. 
Let's not consider zero sets of polynomials and instead consider the Zariski topology on the prime spectrum of a ring $S$.  When $R$ is an algebraically closed field, and $S = R[T_1, ... , T_n]$, everything I'm writing corresponds directly to a statement about zero sets of polynomials.  
Let  $X$ be the topological space $\textrm{Spec}(S)$.  This is the space of prime ideals of $S$ whose closed sets are of the form $V(I) = \{ \mathfrak p \in X : I \subseteq \mathfrak p\}$ for an ideal $I$ of $S$.  Assume $S$ is a Noetherian ring (for example, $S = R[T_1, ... , T_n]$). Then one can show that $X$ is a Noetherian topological space.
Let $I$ be an ideal of $S$ generated by elements $s_1, s_2, ... \in S$.  Let $I_n = \langle s_1, ... , s_n \rangle$.  The inclusion $I_1 \subseteq I_2 \subseteq \cdots \subseteq I$ induces a descending chain of closed sets $V(I_1) \supseteq V(I_2) \supseteq \cdots \supseteq V(I)$.
Already we know that $I_n = I$ for some $n$, since $S$ is a Noetherian ring.  But let's directly use the property that $X$ is a Noetherian space to get a similar result.  The complements $X \setminus V(I_1) \subseteq X \setminus V(I_2) \subseteq \cdots$ form an open cover of $X \setminus V(I)$, and since $X \setminus V(I)$ is compact, there is a finite subcover, i.e. $X \setminus V(I_n) = X \setminus V(I)$ for some $n$.
Then $V(I_n) = V(I)$.  Actually, this doesn't tell you that $I_n = I$, or even that the zero set of $s_1, ... , s_n$ is the same as the zero set of $I$ if $S = R[T_1, ... , T_n]$ (unless $R$ is an algebraically closed field).  It does, however, tell you that $\sqrt{I_n} = \sqrt{I}$.  
