Let $k$ be an algebraically closed field, and let $A$ be a finitely generated commutative $k$-algebra. Is the following equivalence true?
A is finite-dimensional over $k$ if and only if $A$ has finitely many maximal ideals.
The context of the question comes from considering $A$ as the coordinate (Hopf) algebra of some affine group scheme of finite type, $G$, over $k$. We say that $G$ is finite if $A$ is finite-dimensional, so I'm wondering if this is the same as saying that there are only finitely many $k$-points of $G$, that is, finitely many maximal ideals of $A$.
The forward implication is always true; see http://en.wikipedia.org/wiki/Artinian_ring#Commutative_Artinian_rings.
The backward implication is false if we do not assume $A$ is a finite-type $k$-algebra; e.g. $A$ could be an infinite degree field extension of $k$, or something like $k[[x]]$.
If we do assume $A$ is a finite-type $k$-algebra, then $Spec(A)$ is a closed subscheme of $\mathbb{A}^n_{/k}$ for some $n$. Since there are only finitely many maximal ideals, each component of this subscheme is $0$-dimensional, and so $A$ is Artinian (see the above link).