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It is a well know fact that if $k \subset R$ is an extension of rings such that $R$ is a finite dimensional vector space over $k$, then every point of $Spec(R)$ is closed (i.e., equivalently every prime ideal of $R$ is maximal), and $Spec(R)$ is finite as a set. Here $Spec(R)$ = {prime ideals of $R$}.

If on the other hand we assume that $R$ is a ring and not a field, $Spec(R)$ consists of only closed points and $Spec(R)$ is finite as a set, then can we say anything about $dim_k(R)$, i.e., whether $R$ is a finite dimensional vector space over $k$? If not, what is a counterexample?

Also, what additional hypotheses might we need to make this statement work, if it doesn't?

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Pleas remind us what $\text{Spec}(R)$ means? –  Henning Makholm Nov 5 '11 at 16:46
    
Sorry, I have edited the question with the definition of $Spec(R)$ and also what I mean when I say a point of $Spec(R)$ is closed. –  Rankeya Nov 5 '11 at 16:56
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Consider a field extension $k\subset K$. Then $Spec (K)$ consists of just one point, necessarily closed. However $dim _k(K)$ can be chosen to be at least as large as any given cardinal $\aleph $ by taking $K=k(X_a)_{a\in \aleph}$, where the $X_a$ are indeterminates.

If you want $R$ to have dimension at least $\aleph$ but not to be a field (as in your edited question), then take $R=k[X_a]_{a\in \aleph}/I=k[x_a]_{a\in \aleph}$, where $I$ is the ideal $I=<X_a.X_b > _{a,b\in \aleph}$ generated by all products of two (maybe equal) indeterminates. Again the spectrum of $R$ is a singleton set : $$Spec(R)= \lbrace \mathfrak m \rbrace, \quad \text {where} \; \:\mathfrak m= <...,x_a,...>_{a\in \aleph}$$

The good news
After these depressing examples, I'm happy to report that if $R$ is a finitely generated algebra over a field $k$, then if all the points of $Spec(R)$ are closed you can conclude that $R$ is finite-dimensional over $k$, and as a bonus that the set $Spec(R)$ is actually finite.
Sketch of proof
The ring $R$ is noetherian of dimension zero ($\iff$ all points of $Spec(R)$ are closed), hence artinian (Atiyah-Macdonald, Theorem 8.5), and so it is finite dimensional over $k$ ( Atiyah-Macdonald, Chapter 8, Exercise 3).

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