I don't know scheme theory, and I am doing a problem and the solution involves making conclusions based on the Zariski topology, and I want to make sure that I am "intuiting" things correctly when making my statements.
Let $k$ be an algebraically closed field, $I$ a nonzero radical ideal of $k[x_1, \dots, x_n]$, and assume $A \equiv k[x_1, \dots, x_n]/I$ has a finite number of maximal ideals. I want to show that $A$ is Artinian.
For one, $A$ is Noetherian. Since $I$ is radical, $A$ is a coordinate ring of $\mathbb{A}^n_k$, say $A = k[S]$. Hence, since $A$ has only a finite number of maximal ideals, $S$ is finite.
I know the conclusion I'm supposed to draw is that $A$ has dimension $0$, hence is Artinian, and I know this is "because if the maximal ideals of $A$ contained proper prime ideals, then $S$ has to have proper subvarieties and hence can't be discrete". Regarding scheme theory, is the point that if a closed set in the Zariski topology on $Spec$ has more than one element, then the corresponding variety in affine space has "dimension" equal to the cardinality of that closed set (in this case > $0$)? (I have studied manifolds but not varieties.)
Is this how I should be thinking about things before I learn the subject more thoroughly?
Summarizing one of the lines of argument in the comments, $A$ is Jacobson because rings of the form $k[x_1, \dots, x_n]/I$ are Jacobson. Hence, by the assumption that $A$ has only a finite number of maximal ideals, every prime is maximal (because a prime cannot be the finite intersection of more than one distinct ideal). Therefore, $\dim A = 0$. Since $A$ is Noetherian as well, $A$ is Artinian.
As a corollary, $A$ is finitely generated as a module over $k$. This follows because $A$ is a finitely generated $k$-algebra; that is, $A$ is a finite-type $k$-algebra. (See Atiyah-MacDonald, chapter 8, exercise 3.)
Note: This argument is valid for all fields $k$ and all ideals $I$.