Dimension of $k[x_1, \ldots, x_n]/I$ I'm tasked with the following question:
Let $R = k[x_1, \ldots, x_n]/I$, where $I$ is an ideal of $R$. Show that $\dim (R) = 0$ if and only if $R$ is a $k-$vector space of finite dimension.
From Hilbert's Base Theorem, I know that $R$ is Noetherian and with $\dim(R) = 0$ I know that it is also Artinian. Is it enough to conclude that it is a $k$-vector space of finite dimension?
For the other implication I'm out of ideas. And by the way, should I suppose $k$ is algebraically closed?
 A: In fact, $R$ is Noetherian and $0$-dimensional if and only if it is Artinian.
That means in particular that the "other implication" you mention is quite straightforward, as it is enough to show that $R$ is Artinian. For that, consider a chain of ideals
$$I_0 \supseteq I_1 \supseteq I_2 \supseteq I_3 \dots$$
and note that, since each $I_j$ is a $k$-vector subspace of $R$, by counting dimensions the chain needs to stabilize.
The first implication is, however, slightly "deeper": it is, more or less, equivalent to Hilbert's Nullstellensatz, one version of which is the following:

If a $k$-algebra $A$ is finitely generated and also a field, then $A/k$ is an algebraic extension, hence $\mathrm{dim}_k A<\infty$.

This is clearly a special case of your statement (corresponding to the situation when $I=\mathfrak{m}$ is a maximal ideal). It also says that all simple $R$-modules are of finite $k$-dimension. To deduce that $R$ has finite $k$-dimension assuming it is Artinian, one can just consider the composition series of $R$,
$$R=J_0 \supsetneq J_1 \supsetneq \dots \supsetneq J_k=0.$$
Since each of the consecutive simple quotients $J_{i}/J_{i+1}$ has finite $k$-dimension, the same is true by induction for $R$.
And there is no need for $k$ to be algebraically closed in the above (the mentioned version of Nullstellensatz works for all fields).
