While reading about the Hilbert's Nullstellensatz, the following question naturally posed itself:
Question. Let $k$ denote a field, not necessarily algebraically closed.
For which ideals $I$ is the quotient ring $k[x_1,\ldots,x_n]/I$ a finite dimensional vector space over $k$?
One way to describe such ideals is that they are exactly the ideals $I$ such that $\sqrt{I}$ is a finite intersection of maximal ideals. Indeed, suppose $\sqrt{I}$ is a finite intersection of maximal ideals, say $M_1,\dots,M_k$. Let $A=k[x_1,\dots,x_n]$, let $B=A/I$, and let $C=A/\sqrt{I}$. By the Chinese remainder theorem, $C$ is isomorphic to $\prod_i A/M_i$. Each $A/M_i$ is a field and is finite-dimensional over $k$ (by Zariski's lemma), so $C$ is finite-dimensional over $k$. Moreover, $C$ is the quotient of $B$ by its nilradical $N$. For some $d$, $N^d=0$, and $N^r/N^{r+1}$ is a finitely generated $C$-module for each $r$ and hence a finite dimensional $k$-vector space. It follows that $B$ is a finite-dimensional $k$-vector space (since by reverse induction on $r$ starting from $r=d$, $N^r$ is finite-dimensional for all $r$, with the final case being $r=0$).
Conversely, suppose $I$ is such that $A/I$ is finite-dimensional. Then $C=A/\sqrt{I}$ is finite-dimensional as well, and in particular is artinian. So $C$ is a reduced artinian ring, and is therefore isomorphic to a finite product of fields $\prod L_i$. The kernels of the quotient maps $A\to L_i$ are then maximal ideals $M_i$ of $A$, and $\sqrt{I}$ is their intersection (the kernel of the map $A\to \prod L_i$).
The following are equivalent:
i) $k[x_1,\ldots,x_n]/I$ is a finite dimensional vector space over $k$
ii) For all $i=1\cdots n$ we have $I\cap k[x_i]\neq 0$
iii) For every field extension $k\subset k'$ the algebraic subset $V_I(k')=\{a\in k'^n\vert \forall P\in I, P(a)=0\}\subset k'^n$ is finite.
iv) For $\Omega$ an algebraic closure of $k$, the subset $V_I(\Omega)\subset \Omega^n$ is finite
Beware that it is not sufficient to suppose that $V_I(k)$ is finite to deduce i):
For example if $k=\mathbb R, n\gt 1$ and $I=\langle x^2_1+\cdots+x_n^2 \rangle $, the subset $V_I(\mathbb R)=\{(0,\cdots,0)\}$ has just one element but the $\mathbb R$-algebra $\mathbb R[x_1,\ldots,x_n]/\langle x^2_1+\cdots+x_n^2 \rangle $ is infinite dimensional over $\mathbb R$.
Edit: proof of equivalence
$\:i)\Rightarrow ii):$
Since the canonical linear map $\; k[x_i]/I\cap k[x_i]\hookrightarrow k[x_1,\ldots,x_n]/I$ is injective, we see that $\; k[x_i]/I\cap k[x_i]$ is finite dimensional, which implies $I\cap k[x_i]\neq 0$
$\:iv)\Rightarrow ii):$
Since the i-th projection $F_i:=pr_i(V_I(\Omega))\subset \Omega$ is a finite set, there is a polynomial $0\neq P_i(x_i)\in k[x_i]$ which vanishes on that set. [For example take the product of the minimal polynomials over $k$ of the elements of $F_i$]
A fortiori the polynomial $P_i(x_i)$ seen as a polynomial in $k[x_1,\ldots,x_n]$ vanishes on $V_I(\Omega)$.
But then the Nullstellensatz tells us that for some power $r$ we have $(P_i(x_i))^r\in I$ and we have found our polynomial $0\neq(P_i(x_i))^r\in I\cap k[x_i]$, as required by ii).
$ii)\Rightarrow i):$
Since $k[x_i]$ is a principal ring, $I\cap k[x_i]=\langle f(x_i)\rangle$ for a unique monic polynomial of degree $d_i$.
But then it is clear that every monomial $x_1^{s_1}\cdots x_n^{s_n}$ is congruent modulo $I$ to a product $Q_1(x_1)\cdots Q_n(x_n)$ where $Q_i(x_i)$ is a polynomial in $x_i$ with $\operatorname {degree}Q_i(x_i) \lt d_i$.
Hence the obvious $k$-linear map $\bigotimes_{i=1}^n k[x_i]/\langle f_i(x_i)\rangle \to k[x_1,\ldots,x_n]/I$ is surjective
and $k[x_1,\ldots,x_n]/I$ is finite-dimensional over $k$, with dimension bound $\operatorname {dim}_k k[x_1,\ldots,x_n]/I \leq d_1\cdots d_n$.
$\:ii)\Rightarrow iii):$
Let $0\neq f(x_i)\in I\cap k[x_i]$ and let $Z_i\subset k'$ be the finite set of zeros of $f(x_i)$ in $k'$.
Then $V_I(k')\subset Z_1\times \cdots Z_n$ and thus $V_I(k')$ is also finite.
$\:iii)\Rightarrow iv):$
Trivial.
