# Trying to use the Zariski topology in a problem without knowing scheme theory.

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?

• (Coming from someone who has yet to see the geometry of algebraic geometry) Sounds like you're getting it to me! Only the latter question I am questioning. I do not think the dimension is equal to the cardinality. I think the dimension is kind of like the height of a prime ideal. Say the prime ideal is a subvariety of dimension $n$. Then I expect it to have subvarieties of dimension $n-1$, $n-2$,..., $0$. If each of these correspond to prime ideals, then I expect the height of my original prime ideal to be $n$. Or something... (Like a cone contains lines, a line contains points...) – Eoin Jun 30 '15 at 22:25
• So I do not think it has to do with the cardinality of the closure of the prime ideal but, the length of the longest chain among the chains in the closure of the prime ideal. (Simply because, certainly a cone has more than one line, certainly a line has more than one point, etc.) – Eoin Jun 30 '15 at 22:27
• From the commutative algebra point of view this property is easily proven: A is a Jacobson ring, so every prime ideal is an intersection of maximal ideals; since there are only finitely many maximal ideals it follows that every prime ideal (contains) is maximal. (No need to assume k algebraically closed and I a radical ideal.) – user26857 Jun 30 '15 at 22:51
• It's unclear to me what you're asking. Points in $S$ are closed. It follows that $S$ is discrete and hence zero-dimensional. We conclude that $A$ is zero-dimensional as desired. What's crucial here is recognizing the correspondence between irreducible subsets and prime ideals. In particular, there is a correspondence between points and maximal ideals. This follows from the Nullstellensatz. – Ayman Hourieh Jun 30 '15 at 23:14
• @pseudoname123456 $A$ is a finitely generated algebra over a field, so it is Jacobson. For what it's worth, my argument doesn't require $I$ to be radical either. – Ayman Hourieh Jul 1 '15 at 13:12

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 $I$.